An atom's wavefunctions (orbitals) are orthogonal. This follows from the fact that the Hamiltonian is Hermitian, i.e eigenfunctions (wavefunctions) of quantum mechanical operators are orthogonal.  If they are on different atoms then they are not orthogonal and the overlap clearly depends on how far apart the atoms are.

Mathematically orthogonality means that (in one dimension) two wavefunctions $ \psi_i$ and $\psi_j$ satisfy the condition
$$\int_{-\infty}^{\infty} \psi_i\psi_jdx =\delta_{ij}$$

where $\delta_{ij} = 1 \text{ if } i=j \text { else } 0$ and is the Kroncker delta. (In this case if $i=j$ then the wavefunctions are also normalised as the product is $1$, this joint condition, orthogonal & normalised  is sometimes called 'orthonormal')

In the case of atomic orbitals there are three quantum number, the principle one n and angular ones l and m. The orthogonality condition is then
$$ \int_0^{\infty}r^2\int_0^{\pi}\sin(\theta) \int_0^{2\pi}\psi^*_{n'l'm'}(r,\theta,\phi)\psi_{nlm}(r,\theta,\phi)~dr ~d\theta ~d\phi=\delta_{nn'}\delta_{ll'}\delta_{mm'}$$

In your examples $n=2$ and $n'=3$ so the orbitals are orthogonal on these grounds alone.

As a check the 2s orbital has the form
$$ \psi_{2s} = A(2-\sigma)\exp(-\sigma/2) $$
where A contains only constants and $\sigma=Zr/a_0$ where Z is the atomic number, $a_0$ the Bohr radius and r the radial distance. One 3p orbital ($l=1, m=0$) has the form
$$\psi_{3p} = A'\sigma(6-\sigma)\exp(-\sigma/3)\cos(\theta)$$

Integrating the product over the angles $\theta$ which is  the integral $\int_0^{\pi}\cos(\theta)d\theta $. We can see that this integral is zero from a plot of cosine.

Edits:

As the question is now amended, the answer above holds for a single atom, but when there are two atoms then the overlap depends on their separation as the wavefunctions are no longer orthogonal because they have different coordinate origins, one on one atom the other on the other.

The overlap integral is written as
$$S= \int_{-\infty}^{\infty} \psi_i\psi_j~d\tau$$

where $\tau$ is the set of coordinates. Even in the simplest case of $\ce{H2^+}$ the coordinates needed are prolate spheroidal ones so to give an example it is simplest to use a one dimensional example since the principle is the same but technically more complex in 3D.

Starting with H atom 1s orbitals
$$ \psi_{1s} = a_0^{-1/2}\exp(-x/a_0)$$

if the atoms a and b are separated by 2R then the wavefunctions are

$$ \psi_a = a_0^{-1/2}\exp(-|x-R|/a_0) \\
\psi_b = a_0^{-1/2}\exp(-|x+R|/a_0)$$

and the overlap is the integral
$$ S=\int_{-\infty}^{\infty} \psi_a\psi_b dx =a_o^{-1}\int_{-\infty}^{\infty} \exp(-|x-R|/a_0)\exp(-|x+R|/a_0) dx$$

which evaluates to $$S=\frac{(a_0+2R)\exp(-2R/a_0)}{\pi a_0^3}$$

The wavefunctions with $a_0=1$, (and $R=1$ for the first two plots), $\psi_a$ and $ \psi_b$ vs. x  overlap $\psi_a.\psi_b$ vs. x and overlap integral S vs. the atom separation R are shown in the figures.


Why does the water in a Powerade bottle tastes a little like Powerade after many refills?

Typically water bottles used in sports or biking are manufactured from high-density polyethylene (HDPE), which is essentially like a longer-chain version of paraffin wax. In short, it's basically a solid nonpolar hydrocarbon (with the exception of the Zeigler-Natta catalysts, which are only present in trace amounts).

Meanwhile, a quick stroll through Fenaroli's Handbook of Flavor Ingredients shows that many of the flavor compounds typically found in foods are either somewhat nonpolar, have nonpolar sections, or have (mostly unsubstituted) aromatic rings present. For example, ethyl valerate, one of the primary constituents of apple aroma, has a butyl chain, is fairly nonpolar, and is nearly insoluble in water.

However, it may be soluble in HDPE, and so it may infuse the bottle with the flavor agents so that the taste can remain for a long time.

Of course, doing a double-blind test may be your best bet for checking to see if this is a real phenomenon, or your imagination. But I've personally noticed this as well, so it could very well be real.


1) Is this true?

It's more or less true in the cases were hybridization methods \ techniques work.  For example, when the hybridization of water is discussed it is often said that the central oxygen is more or less $\ce{sp^3}$ hybridized and therefor the two lone pairs on oxygen are in (roughly) $\ce{sp^3}$ orbitals.  Current thinking based on photoelectron spectral data suggests that, instead of this traditional view, one lone pair appears to be in an orbital that is predominantly $\ce{P}$ and the other in an orbital that is predominantly $\ce{S}$.  So hybridization models don't always give the correct answer.  I'm not sure what the hybridization of an atom with a lone pair that's attached to a double bond really is.  But it is probably OK to say what the teacher said in an introductory class in order to keep things simple for a while.

  2) Does this rule imply that both the oxygens in the below diagram are
  sp2 hybridized?

Assuming that the traditional hybridization approach is applicable here (e.g. is the ester oxygen behaving like water or is it "traditionally" $\ce{sp^3}$ hybridized?) then yes, both oxygens are roughly $\ce{sp^2}$ hybridized.  Remember, hybridization is not a step function.  There are hybridization indices between 2 and 3.

  3) Even if we didn't know that the rule existed, we should still know
  that both oxygens are sp2 hybridized because of the following
  resonance possibility, correct?

Some data on esters suggests that the $\ce{C-O-C}$ angle is around 115 degrees.  Somewhat between 109 degrees ($\ce{sp^3}$) and 120 degrees ($\ce{sp^2}$).  And maybe an intermediate hybridization is what we should expect based on the resonance structures for an ester.


What does "gamma" in gamma-oryzanol mean?

No designations such as alpha-oryzanol, beta-oryzanol, etc. appear to be in use, and according to this WebMD entry, as well as this PubChem document, another name for gamma-oryzanol is simply "oryzanol".

There seems to be some ambiguity as to just what oryzanol, or gamma-oryzanol refers to. According to Chemical Book:

  Gamma Oryzanol is a mixture of 6 substances derived from rice bran oil, including sterols and ferulic acid, called Cycloartanyl Ferulate or Triterpene alcohol ferulate.

But then other sources, including this US National Institutes of Health publication, consider gamma-oryzanol composed of four components:

  There are four main components of γ-oryzanols, namely, cycloartenyl ferulate, 24-methylene cycloartenyl ferulate, campesteryl ferulate, and sitosteryl ferulate [original publication:  Lerma-García M. J., Herrero-Martínez J. M., Simó-Alfonso E. F., Mendonça C. R. B., Ramis-Ramos G. Composition, industrial processing and applications of rice bran γ-oryzanol. Food Chemistry. 2009;115(2):389–404. doi: 10.1016/j.foodchem.2009.01.063

Further confusing the issue is that the PubChem article referenced above gives a single structure for gamma oryzanol (the same as that given in the OP), but does not include any physical properties that one would expect for a single pure compound (i.e. no melting point, solubility, vapor pressure data) although it does give the molecular weight. Again, PubChem lists "oryzanol" as a synonym and does not have any other oryzanol-named compound in it's database.

Summary TL/DR:
The gamma in "gamma-oryzanol" doesn't refer to a specific structure or isomer and there is no corresponding compounds named alpha-oryzanol, beta-oryzanol, etc. Rather, the term gamma-oryzanol is also simply called "oryzanol" and it refers to a combination of several different moieties as described above.


In laboratory, nobody cares. It's too cheap to care about.

In industry, other catalysts are usually used (heterogeneous ones).


Can air inside a high temperature (1300C) Kiln cause an explosion?

I think in general the physics would make it unlikely that the air pressure would build up faster than it leaks out, attempting to seal a furnace is unlikely to happen by accident.

A bomb calorimeter or pressure vessel made to specifically seal the inside from outside does not have a loose fitting lid over a cast refractory edge, it is made to much finer tolerances.

Chemical reactions are not likely unless you have a significant mount of vaporisable fuel in the charge and this is contained until at some point it lifts the lid and lets vapour out and air in with glowing elements to ignite the mixture, a flame may result, explosion still unlikely as volume small and not contained any longer.

In practice most kilns have a hole for temperature sensor that is not airtight and would let pressure equalise as they are heated.

Unless a specific atmosphere (inert, reducing, oxidising, carburising) is required in a kiln there is no real need to make them air tight and they rarely are, even then it is often achieved within a protective wrapping of the load and supporting chemistry inside the wrapping like charcoal, cyanides etc).


Why can't I balance this reaction algebraically?

Here's how to solve the system in python:

>>> import numpy as np
>>> stoich_mat = np.array([[-1, 0, 1, 0, 0],  # copper balance
                           [0, -1, 0, 0, 2],  # hydrogen
                           [0, -1, 2, 1, 0],  # nitrogen
                           [0, -3, 6, 1, 1],  # oxygen
                           [1, 0, 0, 0, 0]])  # impose constraint that a == val
>>> rhs = np.array([0, 0, 0, 0, 1])  # last number is assumed val of a
>>> np.linalg.solve(stoich_mat, rhs)
array([ 1.        ,  2.66666667,  1.        ,  0.66666667,  1.33333333])

Thus this solution says:

$$\ce{Cu   +   2.67 HNO3   ->  Cu(NO3)2    +   0.67 NO   +    1.33 H2O}$$

Multiplying through by 3:

>>> rhs_times_3 = np.array([0, 0, 0, 0, 3])
>>> np.linalg.solve(stoich_mat, rhs_times_3)
array([ 3.,  8.,  3.,  2.,  4.])

$$\ce{3 Cu   +   8 HNO3   ->  3 Cu(NO3)2    +   2 NO   +    4 H2O}$$

If your question is how computers (or people) can solve systems of linear equations, you can read up on Gaussian elimination and related techniques.


Vodka is a solution of water and ethanol.

Water boils at 100 °C
Ethanol boils at 78.37 °C

A mixture of both liquids will start boiling close to the boiling point of ethanol (otherwise distillation wouldn't work).

The lowest boiling temperature is a mixture of:

95.63 mass-% ethanol
4.37 mass-% water

which boils at 78.2 °C.

Read more about this effect at: Azeotrope There is also a picture of a phase diagram there.


What is the chemistry behind heat produced when blood is bleached?

This reaction produces heat for the same reason as burning methane in oxygen. There's a lot of chemical energy stored in the reactants, and comparatively little in the products.
The hypochlorite is oxidising the organic content of the blood - not just breaking proteins but then breaking down the amino acids too, as well as anything else that happens to be in the sample.


How to derive the pressure dependency for the Gibbs free energy?

This question has a lot of parts! I hope I manage to answer most but probably edits will be required.

$$\mathrm{d}G=\left(\frac {\partial G}{\partial p}\right)_T+\left(\frac {\partial G}{\partial T}\right)_p$$

What does this mean physically?
This is simply an expression that shows the contributions to the change in the Gibb's free energy, which varies with both pressure and temperature (i.e., $G$ is a function of $p$ and $T$). However, it is possible to separate the exact differential of $\mathrm{d}G$ into two terms using the mathematical expression of partial derivatives, first term is the variation of $G$ with pressure at constant temperature, and the second term is the variation of $G$ with temperature at constant pressure.

What does it mean for $\Delta G$ to be independent of the path? (same question: what does it mean for $\Delta G$ to be a state function?)
You can calculate a change in $G$, or equivalently, $\Delta G$, for any reaction. So for example for a reaction like a combustion process for glucose

$$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 CO2~(g) + 6 H2O~(g),}$$

you can calculate $\Delta G$ for this reaction at any particular temperature/pressure that you would like. And in fact, if you don't do the combustion directly but somehow do it via completely dissociated gaseous atoms

$$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 C~(g) + 18 O~(g) + 12 H~(g) -> 6 CO2~(g) + 6 H2O~(g)}$$

You will still end up with the same $\Delta G$ for the overall reaction. Obviously if you calculate the $\Delta G$ for each individual step/arrow in the gaseous atoms case it will be very different, but they will combine to give the same overall $\Delta G$ as the direct combustion. In other words, $\Delta G$ won't be able to tell you whether you have done the combustion directly or done it laboriously via gaseous atoms, it just tells you what is the difference between the two states, first state being glucose + oxygen, second state being carbon dioxide + water. $\Delta G$ for glucose that you have eaten and combined with oxygen you breathe in and breathe out carbon dioxide + water would still the same $\Delta G$ as for combustion, even though now it is all done through cells, Kreb's cycle and a lot of biochemistry.

How would something be dependent on the path?
As you can see from my example above, the amount of effort you need to make $$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 CO2~(g) + 6 H2O~(g)}$$ happen can differ a lot, and for each of the cases, what you "get back" from doing that reaction is different, such as heat released when burning glucose vs keeping your cellular biochemistry going with respiring glucose. In physics/physical chemistry, this is formally "work done". Work is dependent on path.

I am a little unsure whether you went from

$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T \tag1$$

to

$$\left(\frac {\partial G}{\partial p}\right)_T=V \tag2$$

in the right way, so I am going to go through it again.

At constant $T$, $\mathrm{d}T=0$. So $(1)$ becomes

$$ \mathrm{d}G=V\mathrm{d}p \tag3$$

which after rearranging is equivalent to $(2)$.

The integration then should go from $(3)$ to give

$$ \int_{p_1}^{p_2}\mathrm{d}G=\int_{p_1}^{p_2}V\mathrm{d}p \tag4$$

The reason why what you wrote look a little weird

  $$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=\int_{p_1}^{p_2}V\mathrm{d}p$$

is that you usually don't just cancel partial differentials like you did. Whereas $(3)$ is an exact differential, which you integrate over in $(4)$, and that is OK. Unfortunately I am not sure whether what you wrote is wrong, it is just something I would not write myself.

The rest of the derivation is fine.

Final question. I believe you are fine with mathematically going from $\int_{p_1}^{p_2}\mathrm{d}G$ to $G(p_2)-G(p_1)$, however, if you think about this result physically, $\Delta G$ does not depend on any other $p_3, p_4, p_5...p_n$ that the system might pass through on the way from $p_1$ to $p_2$. If $G$ is not a state function, then you would expect to have to consider at the very least all these different points of $p$ that you might have gone through, and different ways of doing the reaction passing through different $p$ would change $G$ even if the initial and final $p$ are the same. In that case, an integral like what we have written here would be a very poor way of finding $\Delta G$ because we are missing the contribution from all the intervening $p_n$, and we would have to use some different mathematical tools. In fact, I believe we would have failed to derive the master equation $(1)$ much earlier on, so the rest of the derivation would simply not follow.


Molecular orbitals used with CCSD(T) geometries

Pulling together some of my comments into an answer:

The principal question at issue is "what do you mean by orbitals?", which is partially related to "what do you want to look at them for?" You can obtain an corresponding 1-particle density matrix from all of these methods, which can be diagonalized to yield natural orbitals (which will have non-integer occupation numbers. There are also certainly reference orbitals used to compute these methods (usually Hartree-Fock, but there are methods like orbital optimized-MP2 (OO-MP2) and Brueckner doubles (an orbital optimized CC variant)). Those cited statements are, to some extent, meaningless without the answers to those questions.

If you run a single-point, you have generated an energy, not orbitals. All orbitals are an implementational tool to obtain this energy and, in that sense, no orbitals "exist", in that they are not observable and can be replaced with an alternative method to obtain the same energy.

For your purpose, since you are just exporting to DMC, yeah wouldn't expect the results to vary too much by orbital, the DMC procedure just needs a mostly reasonable starting point. CC natural orbitals would probably produce a more compact wavefunction and therefore maybe run faster than DFT orbitals, but they are enough more expensive to compute that it's probably not worth it. Probably the orbitals you got from the CC calculation you did are just the HF reference orbitals. It's also worth pointing out that natural orbitals are probably not so different from the DFT orbitals unless there is a lot of correlation (in which case, DFT method will probably produce poor orbitals). For systems where the reference state is qualitatively wrong (e.g. diradical system that might suffer from a lot of spin-contamination in HF), OO-MP2 would improve that and those orbitals would probably be substantially better and that would be a case where the cost would probably be justifiable.


Suppose the carbon was $sp^3$ hybridised. The carbon would have 4 hybridized $sp^3$ orbitals which would point towards the corners of a tetrahedron; they are $109.5^\circ$ degrees apart from each other.

You have to form two bonds between the same two carbon atoms. Suppose you use one $sp^3$ hybridised orbital to form one $\sigma$-bond. How are you going to form another bond? The other bonds of the carbon are $109.5^\circ$ apart. This is not feasible.

The other option for the carbon is to assume $sp^2$ hybridization and have three hybridized orbitals stay $120^\circ$ apart. The one remaining $p$-orbital can be in any direction ($x$-axis or $y$-axis) except along the bonding axis ($z$-axis). In this way, you can form a double bond using one $sp^2$ hybridized orbital and one $p$-orbital.


Why are the radial wavefunction and radial distribution function different?

The radial wave function $R(r)$ is simply the value of the wave function at some radius $r$, and its square is the probability of the finding an electron in some infinitesimal volume element around a point at distance $r$ from the nucleus.

But, the infinitesimal volume of space at radius $r$ is $4\pi r^{2} dr$ (it's a spherical shell with thickness $dr$ at radius $r$). That means that probability of finding an electron at radius $r$ is proportional to $R^{2}(r)4\pi r^{2}$. But the behavior of this function is such that the probability of finding the electron at radius 0 is also 0.


For every reaction with an incomplete dissociation process in aqueous solution, you have to consider water's autoprotolysis.
\begin{aligned}\ce{
(1)&&HA + H2O &<=> H3+O + A^{-}\\
(2)&&2H2O &<=> H3+O + {}^{-}OH \\\hline
(3)&&HA + 3H2O &<=> 2H3+O + A^{-} + {}^{-}OH
}\end{aligned}

The degree of dissociation is usually defined as
\begin{aligned}
&&\alpha &= \frac{c(\ce{A^{-}})}{c_0}&\text{with}&&c_0&=c(\ce{A^{-}})+c(\ce{HA})\\
\therefore&&\alpha &= \frac{c(\ce{A^{-}})}{c(\ce{A^{-}})+c(\ce{HA})}&\text{with}&&
c(\ce{A^{-}})&=K_a\frac{c(\ce{HA})}{c(\ce{H3+O})}\\
(4)\therefore&&\alpha &= \frac{K_a}{K_a + c(\ce{H3+O})}
\end{aligned}

At infinite solution, the concentration of hydronium ions will be governed by the autoprotolysis $(2)$ and can therefore be rewritten as
\begin{aligned}
&&K_w&=c(\ce{H3+O})c(\ce{{}^{-}OH})\\
(5)\therefore&&c(\ce{H3+O})&=\sqrt{K_w}
\end{aligned}

At $25~^\circ{}C$ you will find your given equation:
\begin{aligned}
\alpha[\%] &= \frac{K_a}{K_a + 1\cdot10^{-7}}\cdot100\%
\end{aligned}


Shortcut method to find atomic term symbols ignoring the spin orbit coupling

After searching on the internet, and going through textbooks, I understand there is no "shortcut-method" to get the atomic term symbols in atoms with a large number of electrons.


Mass of iron from the reduction of iron(III) oxide

After finding the moles of $\ce{Fe2O3}$ we see that after balancing the equation, $2$ moles of $\ce{Fe}$ are formed for every mole of $\ce{Fe2O3}$.

But that would be considering that the equation moves completely to the RHS, which is assumed if nothing is explicitly mentioned but not true.


Melting points and boiling points of primary alcohols do not follow the same trend

Melting points are a bit trickier to compare than boiling points, especially when you're looking at the lightest examples of a group of molecules. This graph for the unbranched alkanes exemplifies nicely how the trends differ in smoothness.

As you increase the weight of molecules, the observed trend depends on how you increase the weight. For example, if you start with methanol and increase the weight by adding more $\ce{-HCOH -}$ units (methanol, ethylene glycol, glycerol, and so on) you would find a very clear increase in both melting and boiling points. In some sense this is a more legitimate trend to analyse, because the relative amounts of different types of intermolecular interactions stays approximately constant (in particular, all molecules have one hydroxyl per carbon atom, and hence they all have about the same amount of hydrogen bonding per atom in the molecule). In your sequence of alcohols, the strong hydrogen bonding allowed by the hydroxyl group becomes "diluted" as the molecule gains an ever larger alkyl chain, which can only sustain much weaker van der Waals interactions. In the limit of alcohols with very large alkyl chains, the alcohols' melting and boiling points do not differ tremendously from their parent saturated hydrocarbons. Compare the melting and boiling points of 1-hexadecanol (49°C and 344°C) and hexadecane (18°C and 287°C).

Also, the temperature at which a substance freezes is dependent not only on the strength of intermolecular interactions, but how the solid packs. Even substances with strong interactions may only freeze at very low temperatures if they pack very poorly when solidified. Extreme examples to which this effect contributes are ionic liquids. They tend to have a very low vapour pressure at reasonable temperatures (as you would expect of many salts), but some melt even below 0°C. The freezing of such bulky entities would entail lots of ordering and therefore a very large decrease in entropy, so it is disfavoured (or equivalently, their melting is highly entropically favoured, so the free energy of the liquid phase becomes lower than the free energy of the solid phase even at relatively low temperatures). The exact geometry of the molecules in the solid is also important, as some favourable interactions may be suppressed in the solid state due to steric hindrance which may be less present in the more freely-moving molecules of a liquid. Furthermore, molecules with a high amount of symmetry tend to have high melting points, because freezing is entropically favoured due to molecules more readily fitting in the solid lattice.

Yet another effect to consider is how large a surface area each molecule has available for intermolecular interactions. This is often brought up to explain differences between melting and boiling points of branched and unbranched organic compounds; when comparing structural isomers, branched compounds have a lower available area for interaction than unbranched compounds, so the latter tend to have higher melting and boiling points (branching may sometimes create a combination of opposing packing and surface area effects, though).

So why does your sequence of alcohols behave the way it does? The fact that the trend in boiling point is monotonic while the melting point is not suggests that even though there is a relative shift in importance of types of intermolecular interaction as the alkyl chain increase (from hydrogen bonding to van der Waals interactions), it is probably not the source of the melting point drop. Hence it likely has to do with geometric factors in the solid. The alcohol's alkyl chain probably disrupts the hydrogen bonding network of the solid as it grows from methyl to propyl in a very severe fashion, by making the solid pack less well or by partially hindering the number or strength of hydrogen bonds in the geometry of the solid. Perhaps looking at the crystalline structures of the solid alcohols may provide further insight.

Edit: I wrote a wrong answer initially. I've updated it, but now I realize I'm still not quite sure it's correct. I originally meant to write just a comment, and as I kept writing more I forgot that my very first thought was actually incorrect, so the whole text wasn't built as well as it should be. Anyone is free to pick apart or repurpose what I wrote!

Edit 2: User Uncle Al posted an interesting list of compounds and their melting/boiling points, showing the importance of molecular symmetry and solid packing.


What is the difference in dehydration of cis- and trans-1,2-cyclohexanedicarboxylic acid in anhydride formation?

The mechanism of formation of anhydride from a dicarboxylic acid is as follows: [1]

So as you can see, it involves a nucleophilic attack at the $\ce{C=O}$ of one of the carboxylic acids by the other one, following which internal rearrangement occurs and a water molecule is kicked out.
Stereo chemically, we are interested in finding out in which configuration can the lone pairs of oxygen from one of the $\ce{-COOH}$ attack the $\ce{π^* C=O}$ type antibonding orbital of the other $\ce{-COOH}$ group.
To achieve this, the nucleophile( oxygen in this case) has to align properly with the $\ce{π^*}$ orbital at the Bürgi–Dunitz angle. This configuration is necessary whenever a nucleophile has to successfully add to a carbonyl group. The interacting diagram will look like this for formaldehyde: [2]

(Green ball= nucleophile| Blue and Red balls= carbonyl)
We can also roughly estimate that the situation will be similar for the $\ce{C=O}$ part of the carboxylic acid group.
To see which of the two reactants in question will form the required anhydride, one can simply look at the 3-D conformers of both in the appropriate orientation, and compare to see which of the cases resembles the standard Burgi-Dunitz interaction more closely as illustrated above
Here is the cis configuration: [3]

And this is the trans configuration: [4]

1= oxygen| 2= carbonyl of acid group
As you can see, the oxygen seems much more aligned with the carbonyl in the trans configuration as compared to the cis. So, 2 might form the anhydride more easily than 1 (that is, with a lower activation energy barrier due to reduced torsional strain)
But again, it's not a strictly inaccesible barrier, as a few bond rotations in the cis configuration can also lead to a somewhat favorable configuration for the reaction. So it may ultimately lead to two different kinds of anydrides by each of the isomers like you said, 2 can lead to formation of 4, while 1 can lead to formation of 3


The "counter argument" you link to cites E = mc^2 and the higher energy content of hot water.

First off, note the "c^2" in that formula. That's the speed of light, squared. It is a huge number, meaning that you need huge differences in energy to get tiny changes of mass.

The other way around, you only need little mass to generate lots of energy, something nuclear physicists have been exploiting for quite some time. ;-)

So, while that effect is "true" to some extend, it is tiny. But yes, one molecule of H2O has more mass when hot than the same molecule when cold.

But when we talk about "weight", we usually mean "density", i.e. weight per volume. And here, a different, much more pronounced effect comes into play: Most substances expand when heated, reducing their density. Each molecule might be a little bit heavier, but the molecules are much farther apart, more than offsetting the E = mc^2 effect.

Cooling down increases the density -- the same amount of matter takes up less volume. An effect that home preservers have exploited for generations when canning things: You put hot produce into a container, and close it. The contents cool down and contract, producing underpressure in the container, sealing it. That's the "pop" you hear when opening a glass of marmelade the first time.

So, per molecule hot H2O is (a bit) heavier, but per volume hot H2O is (much) less dense.

At least down to about 4 degrees celsius, where a peculiar quirk to the H2O molecule makes it become less dense again with further cooling. That's why a pond starts freezing at the surface, not the bottom, where the 4 degree water is up to the point where the whole pond is frozen solid. An effect that fish have been exploiting for quite some time now. ;-)


Bromonium ion or Mesomeric effect ( intermediate during addition to alkenes)

You are not going to see the formation of a primary carbocation—the activation energy is too high. Even though it is true the following are resonance structures of one another:

the geometry in which the primary carbocation is formed does not allow for the simultaneous formation of the bromonium ion:

The bromonium would only be able to form after a rotation of the $\ce{C-C}$ bond, so we need to consider the primary carbocation sans-bromonium as the real product of the protonation of the alkene. This is a very high energy product, so it will be slow to form at low temperatures and is out-competed by the other reaction. Since positive charge on carbon-1 can immediately be delocalized to the bromine, this product of protonation is more stable (meaning it forms faster).


How to extract elemental bismuth from pepto bismol?

I'm not sure you did do the experiment wrong. I performed this experiment and it closely matched this video which had a light yellow liquid and foam.

Here are my results

I filtered it a couple more times to remove more foam. Then added the aluminum foil.

This is what I got at the end before I filtered the liquid out at the end.

In the end I was able to extract 6.2 grams of bismuth powder from 67 tablets of pepto (a bit less than the theoretical ~10 grams I was hoping to get).


Doesn't the isothermal irreversible expansion/compression of an ideal gas law violate the ideal gas law?

The ideal gas law describes the behavior of an ideal gas exclusively for thermodynamic equilibrium states.  During an irreversible expansion or compression, the gas passes through non-thermodynamic-equilibrium states, except at the very beginning and at the very end.  In these irreversible non-equilibrium states, the ideal gas law does not correctly describe the gas behavior.  For rapid expansions and compressions, the force that the gas exerts on the piston depends not only on the gas volume, but also on the rate of change of gas volume.  So yes, in irreversible/expansion of an ideal gas, the ideal gas law is violated.

That said, if we can control the force between the gas and the piston by some external means (say by using an automatic control system in conjunction with a flush mounted pressure transducer in the inner piston face), we can calculate the amount of work that the gas does on the piston (without trying to resort to the ideal gas law) even for an irreversible expansion/compression.  This then would enable us to use the first law of thermodynamics to calculate the final state of the gas (at least for an adiabatic process).


Is it possible to prevent agglomeration and sedimentation of microparticles just by increasing the viscosity of the medium?

If you can make this, it will be stable. At equal volume of the phases, this will already be solid-like (0.5 is above percolation threshold, which is in the range of 0.25-0.3, depending on the geometry of your particles.).

See https://en.wikipedia.org/wiki/Percolation_threshold#Thresholds_for_3D_continuum_models

The absolute particle size is irrelevant, only the size distribution matters. When you reach the percolation threshold, the only thing that matters about the continous phase is that it's incompressible, which is the case for practically all condensed matter. If it's viscosity is very low (and the solid particles large, > ~100 µm), it could flow out of the compound, but surface tension would usually preclude that.

Also a very lowly viscous continuous phase could mean that the compound yields already to gravity. https://en.wikipedia.org/wiki/Yield_(engineering)

However it is not so simple to make such a suspension. If you try to mix powder and polymer, your stirrer (of whatever kind) will stop or break long before you get a homogeneous mixture, because your sample becomes solid, and there is no way of melting it or anything. A helper solvent can lower the solid fraction, below the percolation threshold, but is hard to get rid of later.

So you'll have to find out where the percolation threshold actually is, and just use a bit more. It varies a bit with temperature (your polymer phase will have a larger expansion coefficient, not sure if that is enough to make the difference). Anyway the yield stress is still tolerable just above the threshold, and processing is possible.


How come I get different Gibbs free energy for the same reduced equation?

The change in Gibbs free energy is written per mole. For reactions, consider it "per mole of reaction." Since the stoichiometry of the second reaction is exactly double the first, "one mole of the first reaction" is half "one mole of the second reaction." Therefore, the value of the change in free energy is doubled as expected.

For an arbitrary reaction, there is no standard.

For the standard free energy of formation of a substance, the canonical reaction should generate one mole of the substance from its constituent elements (the first reaction here).


What happens when an alkylborane is treated with acetic acid?

Acetic acid effects protonolysis of the borane:

If AcOD is used then RD (instead of RH) is formed.


Why can two carbon atoms not form more than triple bond with each other?

Unfortunately, an answer to this requires a quick dip into orbital theory. You may have heard of s, p and d orbitals; carbon as a main group element of group 14 has the electronic configuration of $[\ce{He}]\,\mathrm{2s^2\,2p^2}$, meaning two of its electrons are in an s orbital and two in p orbitals in its atomic ground state. The process by which these orbitals form bonds is simply overlap: If you have two orbitals that can be moved together and these two orbitals overlap, you generate a bonding and an antibonding orbital from these two. (This is a mathematical operation known as linear combination; plugging two orbitals into a linear combination means that exactly two orbitals are mathematically required to come out of the linear combination.)

The bonding (or antibonding) orbitals that are generated by this process are labelled σ or π orbitals. Note that these are the Greek letters corresponding to s and p: a σ orbital is basically an s orbital streched in one direction and a π orbital is a p orbital streched in the same manner.

Due to a number of constraints that you will learn about soon enough™, a carbon–carbon triple bond $\ce{C#C}$ will consist of one σ bond and two π bonds as shown in the image below.

Figure 1: $\ce{H-C#C-H}$ showing the σ and π orbitals required to make the respective bonds. Image taken from jahschem.wikispaces.org.

See how two of the three possible p orbitals on each carbon (figure 1’s left half) overlap to form two π bonds; the third possible p orbital takes part in both the σ bond to the other carbon and that to the hydrogen — it has been linear combined with the s orbital to form an sp hybrid orbital outside the scope of this answer.

The main point of interest here is that it is, for reasons of symmetry, not possible for more than two p orbitals of a single atom to form π bonds to a single other atom at any one time. And carbon does not have any additional accessable orbitals that it could use for further bonds. Thus, it is not possible for carbon to form quadruple bonds.

The next orbital available would be a d orbital; the transition metals are those in which the d orbitals are valence orbitals and take part in bonding. Once you have a better understanding of orbitals, it may be worthwhile to look up quadruple bonds of transition metals such as in $\ce{[(H2O)Cr({\unicode{x3bc}-}OAc)4Cr(H2O)]}$ or dirhenium octachloride. The latter is particularly interesting because the fourth — δ, in analogy to d — bond can be cleaved by irradiation causing a structural change that clearly shows there must have been a quadruple bond previously. To the best of my knowledge, quadruple bonds have not been shown to exist between main group elements though.

And finally, $\ce{C2}$ does exist; however, explaining it requires orbital theories on a much higher level than in the scope of this answer.


Naming of a Hydrocarbon with a Benzene Functional Group

First, by your logic, it would be 3-phenylbut-1-ene. Note that the 1 is right before the ene, indicating that is where the double bond is. It is a perfectly understandable name, one that many chemists would use for this compound.

The other way to name it is to consider it a substituted benzene. Then you get (but-3-en-2-yl)benzene. This is the preferred IUPAC name for the compound, because benzene is the base molecule by IUPAC priority rules (rule P-52.2.8 of the 2013 edition of the Blue Book: “Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name.”).


Why oxygen leaks from tires faster than nitrogen?

The rate of oxygen, $\ce{O2}$, permeation through tire rubbers is about 3.4 times faster than nitrogen.  The reference for this is "Permeability Properties of Plastics and Elastomers" by L. W. McKeen.

The reason is that $\ce{O2}$ is a smaller molecule than $\ce{N2}$.  The size of a molecule is not determined by only the bond lengths between atoms but the size of the electron cloud around the atoms.  The kinetic diameter of $\ce{O2}$ is 0.346 nanometers and $\ce{N2}$ is 0.364 nanometers.  So, $\ce{N2}$ is larger than $\ce{O2}$ even though the $\ce{N-N}$ bond length is shorter than the $\ce{O-O}$ bond length.


Why does a neutral atom of beryllium not have any electrons in a p orbital?

We need to first consider n, the quantum number for the shell, and then l (~the classical angular momentum of the electron), which tells us the shape of the electron density. The first shell (n=1) contains two s electrons with paired spins. That shell is closed, and you begin over again with n=2: the possibilities for n=2 are l=0 (s electron shape, spherical) and l=1 (p electron shape, lobed). Then, since a lobed shape can point in different directions, when l=1, ml can equal +1, 0 or -1 , defining px, py, pz electron shapes, each pointing in one of three orthogonal directions.

So the filling process puts the first two (lowest energy) electrons in the 1s level (when full, it is 1s2); then the next level is 2s (filling to 2s2), then 2p. Since we have run out of electrons, the 2p levels do not get filled, so we don't have to consider 2px, 2py or 2pz designations.

Now, for boron, there will be a p electron. Whether you call it px, py or pz will depend on how you define the directions in a molecule. In an individual free boron atom, 2px, 2py and 2pz all have the same energy and are degenerate.

This is all theoretical. If you ever heated up beryllium so that there were free atoms, they would be so energetic that quantum levels up to 10 or 100 would be occupied, and there would be beryllium ions floating around too!


How to predict the basicity of organic amines?

The actual number of resonance structures is inconsequential: what matters is whether they are significant contributors to the resonance hybrid. This issue has been discussed at length, particularly in the case of carboxylic acids vs phenols: see e.g. Why is acetic acid more acidic than phenol?

Along this line of thought, the resonance structures that involve cationic carbon with six valence electrons are not really worth considering. Octet-complete resonance structures are always much more major contributors than octet-deficient ones.

If we only count octet-complete resonance structures of the conjugate acids, then guanidine (A) wins hands down with three; the two amidines (B) and (C) both have two; and dimethylamine (D) only has one.

Between (B) and (C) the choice is tougher. You might be inclined to think that this is where the minor contributors come in: surely benzamidine (C) is more basic than acetamidine (B), since its conjugate base has some of these minor resonance contributors. However, experiment shows that the order of acidity is the other way round: benzamidine (pKa 11.6 in water at 20 °C) is actually a weaker base than acetamidine (pKa 12.52),[1] making for an overall basicity order of (A) > (B) > (C) > (D).

The reason for this is not a resonance effect, but rather an inductive effect. It is true that the benzamidinium ion enjoys marginally more stabilisation by resonance. However, in general, sp2 carbons (and aryl groups) are more electron-withdrawing than sp3 carbons (and alkyl groups). So, the positive charge on nitrogen is also destabilised by the inductively electron-withdrawing phenyl group.

A rare case of inductive effects outweighing resonance effects, but there you are.

Reference

Albert, A.; Goldacre, R.; Phillips, J. The strength of heterocyclic bases. J. Chem. Soc. 1948, 2240–2249. DOI: 10.1039/JR9480002240.


Effect of concentration on molar conductance

The molar conductivity of a strong electrolyte $(\Lambda_m)$ is given as the ratio of measured conductivity $(\kappa)$ to the molar concentration $(c)$:
$$\Lambda_m=\frac{\kappa}{c}$$

However, as you note, the value of $\Lambda_m$ is not independent of concentration. The reason for this is that $\kappa$ does not scale linearly with concentration.

$$\Lambda_m = \Lambda_m^\circ -K\sqrt{c}$$

In this equation, $\Lambda_m^\circ$ is the limiting or intrinsic conductivity of the electrolyte and $K$ is an empirical constant. The interpretation of $\Lambda_m^\circ$ is the molar conductivity at infinite dilution as $c\rightarrow0$.

$$\lim_{c\rightarrow 0}{\Lambda_m}=\Lambda_m^\circ$$

The infinite dilution scenario is different from the scenario when there were no ions in solution to begin with, just as mathematically $\lim_{x \rightarrow 0}{f(x)}$ does not always equal $f(0)$ (often because $f(0)$ is undefined).

In the limiting case of $c\rightarrow 0$, the ion concentration never reaches zero, but it does keep getting close. You can never dilute a solution containing a solute to zero concentration (i.e. pure solvent).

If you graph $\Lambda_m$ as a function of $\sqrt{c}$, you should get a straight line with a non-zero intercept $(\Lambda_m^\circ)$. This value is an empirical parameter used to model data. It can also tell you something about the intrinsic conductivity of ions (see this Wikipedia article). However, as you say, at exactly $c=0$, he conductivity should be negligible. At any other value of $c$, however the conductivity should follow from the equation given. Thus, the conductivity is not necessary linear in the infinitesimal range around $c=0$:

$$\Lambda_m=\begin{cases}
10^{-8}\approx0, &c=0\\
\Lambda_m^\circ - K\sqrt{c}, &c>0 \end{cases}$$


What would happen if we had a different kind of metal in our bloodstream?

To start, transition metal complexes with large porphyrin-like molecules are so electronically complex that standard chemical intuition doesn't apply and you have to either do the experiment or run a (nearly impossible) calculation to determine what effect a non-physiological metal would have on the protein activities. In the case of hemoglobin, your best bet would be to think about metals which have two accessible oxidation states because the interconversion of Fe(II) and Fe(III) is critical to the function of hemoglobin.

To give a broader answer, let's think about the general question of substituting non-native metals into a protein. This is, in fact, an very interesting and open question. On smaller metalloproteins there have been studies probing whether nature 'picked' the best metal for a task or just a convenient/abundant one [1,2]. The focus of those studies is on catalysis, and there are cases where many non-physiological metals can function in the protein and even catalyze the reaction as good or even better than the wild type.

[1] https://pubs.acs.org/doi/10.1021/ar500227u

[2] https://pubs.acs.org/doi/abs/10.1021/acs.jpcb.6b00997


Understanding Joule-Thomson expansion apparatus

The purpose of the porous plug is to cause a viscous frictional pressure drop, resulting from the gas flow through the tiny interstitial pores of the porous medium.  This results in a dissipation of mechanical energy (pressure and kinetic energy) to internal energy, which translates into a "viscous heating" to essentially offset the adiabatic expansion cooling of the gas.  The net result is no change in temperature for an ideal gas, and small change in temperature for a real gas.


The aldehyde is both deshielded by reduction in electron density about the carbon it is bonded to through electron withdrawal and by the magnetic anisotropy of the double bond. Although a benzylic hydrogen is slightly more deshielded than the hydrogen attached to a two electron pi bond like a carbonyl or an alkene, it does not also have electron density withdrawn from its adjacent carbon. Generally electron withdrawal through resonance is more important than inductive electron withdrawal, so the difference in chemical shift between an aldehyde and an alkene will be greater than the difference in chemical shift between the proton proton geminal to a halogen in  a halogen-substituted alkene and the proton in its unsubstituted analogue. All that is to say the additive effects of resonance electron withdrawal and magnetic anisotropy make an aldehyde proton more shifted than a benzylic proton.


Number of possible microscopic states and their probabilities

Some notes I had easily at hand may help. You have already calculated the probability of obtaining the chance with a number $k$ of type of ball (or molecule) out of a total on $n$ is
$$\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n} \tag{25c}$$
This distribution is a maximum when $k=n/2$. This can be seen with a straightforward argument. The factorial terms are symmetric, $k!(n-k)!=(n-k)!k!$ and always positive.  When $k=0$ or $k=n$ the probability is very small tending to zero; i.e $1/2^n$ is small so there must be a maximum somewhere in the range $0\to n$. The symmetric nature ensures that this will be at $k = n/2$. This can also be determined by differentiation. It is not possible to differentiate a factorial, as $n$ is discrete, but replacing $n!$ with the Sterling approximation $(n!=n^ne^n)$ and setting the derivative in $n$ to zero allows the maximum to be found.
The second feature of this probability distribution is that it becomes extremely narrow as $n$ increases. This is shown in the figure which shows the normalised probability $p/p_{max}$ vs $k/n$ which is a fraction between $0$ to $1$. At large $n$ the distribution becomes so narrow, or peaked at $n/2$, that almost all the probability is described by that at $n/2$, i.e. this value is so great compared to others that $p_{max}$ is in effect greater than all others combined.

The normalised probability function  $\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n}$ vs $k/n$ for different $n$. The the data only exists for integer values as shown as circles for $n=20$, at larger $n$ a line is drawn as the points become too congested. The curve for $n=10000$ is in red. At large $n$ the distribution becomes very narrow and the maximum value at $n/2$ is greater than all others combined and the distribution can be approximated by its maximum value.
You next need to choose $k/n$ corresponding to 2% either side of $k/n=0.5$ and calculate $p/p_{max}$.


The color of the precipitate is strongly reminiscent of copper(II) hydroxide. I hypothesize:

Chloride corrosion of copper from exposed brass (ref 1 || ref 2)

a. Per ref 2 above, sulfate may also participate
Alkaline precipitation of $\ce{Cu(OH)2}$ in the neutrally-buffered soap
Insufficient $\ce{Na_4EDTA}$ to complex the amount of $\ce{Cu}$ corroded from the brass

It's also possible the blue material is $\ce{Ni(II)}$ complexed in such a way that its color is changed from the typical green-blue of $\ce{Ni(OH)2}$.


That ain't MnSOD, it's arginase.

B = boron

The molecule in red is 2(S)-amino-6-boronohexanoic acid.

See figure 2 of Arginase−boronic acid complex highlights a physiological role in erectile function Nature Structural Biology 6, 1043 - 1047 (1999)


Stereochemical outcomes in opening of vinyl epoxides

While @orthocresol was busy finding the Carreira paper, I was busy constructing my own thoughts on the stereochemical consequences of the reactions of 1 (iii) and 5 (ii). Yes, it is an issue of conformation. It is far easier to rationalize the results after the fact. To predict the most reactive conformation of epoxide 1 the Newman projection 2, viewed along the red bond of 1, places the two carbon chains anti to one another. Intermediate 3 undergoes straightforward SN2 displacement at the allylic site.
In the case of epoxide 5 (ii), the two alkyl chains maintain the anti positions in conformation 6 but now the carbon-bound chlorine acts as a neighboring group in an intramolecular SN2 displacement to form intermediate 7. This species suffers a second SN2 displacement to afford 8 with retention of stereochemistry at the reacting site. Compound (i) follows the same pathway as 1 (iii).    The role of water is to cleave the silylether.


Relative energies between reactants and products

That's interesting; I just did this experiment (I'm actually writing my lab report up now) and for me the trans-bis(glycinato)copper(II) monohydrate was the thermodynamically favoured product and the cis-bis(glycinato)copper(II) monohydrate the kinetically favoured. As far as the reactant energy levels are concerned, I attached a photo of a rough graph my lab instructor gave us that illustrates the energy levels for the two pathways to the final thermodynamically favoured product. I hope that helps!!


Out there in the real world, university students and school pupils alike favour strict rules that are true as often as possible (or at least have clear, easily remembered exceptions), while their professors and teachers on the other hand try to tell them that these rules are merely to be considered ‘guidelines’ rather than actual rules (and attempt to welcome them aboard the Black Pearl).

And from this point on is where the question turns philosophical very quickly. One side will say

  Ah, so the rule is still valid in this surrounding; I just need to remember to do A and B when applying it!

while the other side will reply

  Ah, but this shows us that the so-called ‘rule’ can no longer be strictly applied because you need to consider data C and D that point towards the real truth.

The thing is, both sides are somehow correct while parts of their descriptions remain wrong. Maybe we can think of these as two mesomeric descriptions of the same principle?

As we all (should) know, all our concepts of describing bonding and electron location are mere approximations of the actual underlying truth. Hence, all the rules that were ever laid out — like a bond consists of two electrons between two atoms, oxidation states or the octet rule — are again just mere approximations. They cannot be correct nor can they be properly ‘valid’. Thus, what isn't properly true can't be fulfilled or invalidated.

I think that is the most theoretical answer to the question.

From a practical point of view, it mostly depends on how strict you postulated the octet rule, and how quickly you would consider it violated.

Traditionally, the octet rule wasn't valued highly for elements that weren't carbon, nitrogen, oxygen or fluorine. As such, anything that violated the octet rule in its strictest sense was considered a violation (and proof for its invalidity).

Nowadays, my observations imply that the octet rule is gaining support again, and some of the simpler cases that were hither considered violations are now something more like special cases or maybe just octet rule with additional mesomery. In the eyes of those supporting this side, these compounds comply with the octet rule.

And of course, there is the large grey space between the black and the white.

I for my part tend to favour rules like the octet rule very much — although I tend to use a not-as-standard defintion for it:

  s- or p-block elements will always favour surroundings where the number of valence electrons is 8 or less. For the two elements of the first period, this reduces itself to 2 or less.
  Mesomeric structures that allow an all-octet are to be favoured over those that do not.

Under this definition, which relies heavily on mesomery for cases such as $\ce{IF7}$ or $\ce{PCl5}$ while it allows for a single, unambiguous description of $\ce{SO4^{2-}}$, these cases comply to the octet rule and actually further prove its validity. Others will surely disagree.


Second order reaction half life equation?

What you did isn't wrong, but instead of:

$$v=-\frac12\frac{d[A]}{dt}=k[A]^2$$

some write:

$$v=-\frac{d[A]}{dt}=k[A]^2$$

which is also true.

Twice a constant is still as constant (a different constant of course).

See http://www.chem.arizona.edu/~salzmanr/480a/480ants/int2ord/int2ord.html


How long does a wine sauce need to simmer for to reduce the alcohol to less than one percent

The original source of the data is Alcohol retention in food preparation Journal of the American Dietetic Association vol. 92, April 1992, pages 486-488.

After 2.5 hours of simmering at 85 degrees C, pot roast Milano still retained 4-6% of the original alcohol content.


Why does hydrolysis of ADP to AMP yields the same amount of energy as ATP to ADP, if ADP is more stable?

ATP has two high energy bonds which are phosphoanhydride bonds. Both bonds release the same amount of energy. Therefore, when ATP converts to ADP, breaks one phosphoanhydride bond, and releases -30.5 kJ/mol. When the 2nd bond is broken in ADP releases -30.5 kJ/mol because it is the same phosphoanhydride bond. However, AMP has only a phosphate ester bond which releases -61 kJ/mol.
The "less stability" of ATP doesn't affect energy yield.


The “X” factor simply indicates that the solution is in a concentrated form that must
usually be diluted to a “1X” concentration for use. For example, a 5X concentrated solution must
be diluted 5-fold, while a 100X concentrated solution must be diluted 100-fold. The dilutions
are usually done using water.

See more  here under part 3:

http://csmbio.csm.jmu.edu/biology/courses/bio480_580/mblab/SolutionPrepReview.html


Discrepancy when calulating mol weights with ChemSketch and Python RDKit

In first place, you must clarify with the users of the database the purpose of the entries about molecular weight because average molecular weight and exact molecular weight differ in their meaning and subsequent use.  It may be that the documentation of the database provides this information.  Later, once this target is set, use either one of the program to compute the entries for all compounds of this database to ensure consistency in the data.  RDKit's current release 2021_03_1 is just some weeks old and probably the easier route to cover eventually all entries in your database.

Molecules consist of atoms (like oxygen, nitrogen, etc).  For some of them, say nitrogen, despite they are all nitrogen, there might be more than one form and these differ slightly in their mass per individual atom (isotopes).

The mass of these isotopes is published.  Equally, biannually, there are updated public records about the probability that an atom of e.g., nitrogen either is $\ce{^{14}N}$ (about 99.6%) or $\ce{^{15}N}$ (about 0.4%).

These probabilities are equally valid for molecules, if you have 1000 molecules each containing one atom of nitrogen, than 4 of them should contain the heavier isotope $\ce{^{15}N}$.  Similar there are such probabilities about other atoms, like carbon, oxygen, etc., which are all independent of each other.  Among carbon atoms, for example, 98.9% are $\ce{^{12}C}$, and 1.1% are $\ce{^{13}C}$, respectively.  (Except in radiochemistry / nuclear chemistry.)

Exact molecular weight is used, e.g. in databases eventually used for e.g., mass spectroscopy.  An analysis of your sample will recognize these different probabilities and may yield different, separate signals about the molecules with $\ce{^{14}N}$ from the ones about $\ce{^{15}N}$.  Thus exact molecular weight is about molecules of one isotopic composition.
Say our sample were $\ce{(CN)2}$ of two carbon atoms and two nitrogen atoms per molecule, than there would be four mass signals for all combinations of the two isotopes of carbon and nitrogen possible.  Their intensity  correlate with these probabilities.  Exact molecular weight is a monoisotopic mass, typically to consider only the mass of the molecules consisting with the most frequently observed isotopes per element, i.e., only about $\ce{^{12}C}$ and $\ce{^{14}N}$.  Depending on the intended analysis, it may be worth to retain six (or even more) decimals.

Average molecular weight now accounts not only for the molecules consisting of the most frequently seen isotopes per element, but for any combination of the elements with each other.  These weights are are averaged by the probability that these combinations occur, e.g that there is a molecule $\ce{(^{13}C^{14}N)2}$ consisting only of the heavier carbon isotope and the lighter nitrogen isotope among all $\ce{(CN)2}$ molecules of your sample.  Average molecular weights are typically used in to set up a reaction in the lab, to know how many grams of compound A shall react with how many grams of compound B.  Depending on the scale of the reaction and the needs of your clients, to retain two decimals of this value may suffice.

The pace these IUPAC publishes the data about these isotopes is independent from the one the programs are updated by the developers, and later by the users. This contributes why an old program still may include outdated values.


What is the oxidation state of sulphur in sulphur dioxide?

Your teacher used an interesting example that doesn't obey the rules cleanly. Sulfur can have many oxidation states in this method. In general, larger elements can have oxidation states different than suggested by the group they are in.

If you have learned the electronegativity rules yet, here is a good place to use. If you haven't, higher electronegativity basically means that the element has stronger driving force to reaching an octet; you can look up an electronegativity table if you like. The more electronegative element takes the electrons first. So, in your example, $\ce{SO2}$ has $\ce{2O^2-}$, because of the group oxygen is in. Then, because the $\ce{SO2}$ is neutral, there must be a $\ce{S^4+}$.

But, in a compound such as $\ce{Na2S}$, you can see that S is closer to reaching an octet than Na. So, this becomes $\ce{S^2-}$ while each sodium become $\ce{Na+}$.

Try finding the oxidation states (numbers) for $\ce{PF5}, \ce{WO3}, \ce{SOCl2}$ for understanding. Solutions in the spoiler below.

   $\ce{PF5~is~P^5+,5F-}$
 $\ce{WO3~is~W^6+,3O^2-}$
 $\ce{SOCl2~is~ S^4+, O^2-, 2Cl^-}$


Hybridization of nitrogen in trisilylamine, (SiH3)3N?

Ordinarily and according to Bent’s rule, we would expect nitrogen’s lone pair to be in an $\mathrm s$ orbital and nitrogen using its three $\mathrm p$ orbitals to form three bonds to the three silicon atoms. This configuration would allow for the greatest stabilisation.
However, due to nitrogen’s small size, this perfect world already falls apart for ammonia ($\ce{NH3}$), where nitrogen is bound to three otherwise tiny hydrogen atoms. Because an electronically perfect angle of $90^\circ$ would generate much too much steric strain between the hydrogen atoms, $\mathrm s$ contribution is mixed into the bonding $\mathrm p$ orbitals to a certain extent; for ammonia, this extent happens to be almost perfect $\mathrm{sp^3}$ — results for other amines will vary. This electronic situation is not ideal, however it is clearly better than having $\mathrm{sp^2}$ hybridisation and the lone pair in a $\mathrm p$ type orbital. An $\mathrm{sp^2}$ hybridisation of nitrogen in ammonia can be reached, but only as the transition state of nitrogen inversion.
Carrying on to the compound $\ce{N(SiH3)3}$, we would be inclined to again assume a hybridisation of $\mathrm{sp^3}$ in line with the previous paragraph. However, Beagley and Conrad performed electron diffraction studies on $\ce{N(SiH3)3}$ and found the molecule to be practically planar within experimental error.[1,2] A planar molecule without doubt means that nitrogen is $\mathrm{sp^2}$-configured in $\ce{N(SiH3)3}$.
The question remains why. There must be some kind of stabilising interaction of nitrogen’s remaining $\mathrm p$ orbital with something else to keep that molecule planar. Beagley and Conrad suggest — in line with what was thought at the time — that this be due to π bonds with silicon’s remote $\mathrm d$ orbitals.[1]. Numerous evidence, much of which is collected on this site, speaks the opposite (namely that $\mathrm d$ orbitals do not play any role in the bonding situation of main group metals).
Instead, I think we are dealing with something you may call ‘inverse hyperconjugation’. Remember that $\chi(\ce{Si}) = 1.9$ which is less than hydrogen, meaning that the $\ce{Si-H}$ bonds are polarised towards hydrogen. This in turn means that $\sigma^*_{\ce{Si-H}}$ is a silicon-centred orbital with its primary lobe pointing towards nitrogen. Therefore, nitrogen’s $\mathrm p$ orbital can favourably interact with the antibonding $\sigma_{\ce{Si-H}}^*$ orbital, increasing the $\ce{Si-N}$ bond order and decreasing the $\ce{Si-H}$ bond order. The effects are the same as with hyperconjugation stabilisation of secondary or tertiary carbocations but the electronic demand is reversed. We could attempt to draw the following resonance structures in Lewis formalism to explain this:
$$\ce{H-SiH2-N(SiH3)2 <-> \overset{-}{H}\bond{...}SiH2=\overset{+}{N}(SiH3)2}\tag{1}$$
In this Lewis formalism, the double bond would be generated from a $\mathrm p$ orbital on both silicon and nitrogen.

Note and reference:
[1]: B. Beagley, A. R. Conrad, Trans. Faraday Soc. 1970, 66, 2740–2744. DOI: 10.1039/TF9706602740.
[2]: Actually, $\angle(\ce{Si-N-Si}) \approx 119.5^\circ < 120^\circ$. The authors state:[1]

The apparant slight deviation from planarity is associated with a shrinkage effect11 on the $\ce{Si\dots Si}$ distance of about $\require{mediawiki-texvc}\pu{0.007 \AA}$ (see [table]). Spectroscopic results12 are entirely in agreement that the molecule is planar.
11 A. Allmenningen, O. Bastiansen and T. Munthe-Kaas, Acta Chem. Scand., 1956, 10, 261. [sic!]
12 E. A. V. Ebsworth, J. R. Hall, M. J. Mackillop, D. C. McKean, N. Sheppard and L. A. Woodward, Spectrochim Acta, 1958, 13, 202. [sic!]


Advanced Level Chemistry Book Recommendation

Here are a few recommendations for A-level chemistry, made by fellow students such as yourself. TSR is a good student resource for A-level prep since there is are masses of students talking about the same topics.

(Zumdahl is a good general chemistry textbook as well, with quite the number of practice problems. However it is a full-fledged textbook you probably don't need a few days before an exam.)


Ogawa et al. [1] were first to report a crystal structure (CSD-YAFNOI) of a compound with quaternary sulfur, bis(2,2′-biphenylylene)sulfurane:

Figure 1. Molecular structure of bis(2,2'-biphenylene)sulfurane (CSD-YAFNOI). Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$.

  Compound 1 was synthesized as follows (Scheme 1). Dibenzothiophene 5-oxide (200 mg, 1.0 mmol) in anhydrous tetrahydrofuran (THF, 10 ml) was treated with trimethylsilyl trifluoromethanesulfonate (0.25 ml, 1.3 mmol) under an $\ce{N2}$ atmosphere at −78 °C. After stirring at 0 °C for 30 min, the
  mixture was cooled to −78 °C and was treated with $\pu{1.0 mol dm-3}$ 2,2'-dilithiobiphenyl (1.0 ml, 1.0 mmol) in diethyl ether solution. The whole mixture was stirred at −78 °C for 1 h and at 0 °C for 30 min under an $\ce{N2}$ atmosphere. After evaporation of the solvent, the residue was washed with anhydrous diethyl ether (10 ml) and was extracted with anhydrous benzene (10 ml) under an $\ce{N2}$ atmosphere. The solvent was removed under reduced pressure, and the crude product was recrystallized from anhydrous THF at −20 °C to give 1 as orange rods in 96% yield.

  Scheme 1 Reagents: i, trimethylsilyl trifluoromethansulfonate in THF;
  ii, 2,2'-dilithiobiphenyl in diethyl ether-THF

Further work by Sato et al. [2] resulted in a synthesis and crystal structure (CSD-NEDCEE) of bis(2,2′-biphenylylene)sulfuranyl bis(tetrafluoroborate).

Structurally, it's a similar compound with a greater, nearly 90° (in contrast to 60° twist angle in neutral bis(2,2′-biphenylylene)sulfurane), twist angle between 2,2′-biphenylylene ligands, however water molecules and $\ce{[BF4]}$-counterions appear heavily disordered:

Figure 2. Fragment of the molecular structure of bis(2,2′-biphenylylene)sulfuranyl bis(tetrafluoroborate) (CSD-NEDCEE) showing the bis(2,2′-biphenylylene)sulfuranyl cation. Oxygen atoms from water molecules as well as tetrafluoroborate anions are omitted for clarity. Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$.

  Recently, we have succeeded in the first isolation and structural determination of bis(2,2′-biphenylylene)sulfurane [10-S-4(C4)] (1) as a stable sulfurane(IV) having only carbon ligands.[…] We considered that this sulfurane would be a suitable precursor to provide the desired dication. Therefore, we tried the reaction of bis(2,2′-biphenylylene)sulfurane (1) with xenon difluoride ($\ce{XeF2}$) in the presence of $\ce{BF3 * OEt2}$ and indeed obtained the bis(2,2′-biphenylylene)sulfurane dication, [8-S4(C4)]²⁺ (2) as an amazingly stable bis(tetrafluoroborate) salt.[…] Here, we communicate the first isolation and structural determination of bis(2,2′-biphenylylene)sulfurane dication (2) having only carbon ligands. […]

  The sulfurane 1 was reacted with 1 mol equiv of xenon difluoride in the presence of $\ce{BF3 * OEt2}$ in dry $\ce{CH3CN}$ at −40 °C (Scheme 1). After the solvent was removed at room temperature, the residue was washed with $\ce{CHCl3}$ at room temperature, and bis(2,2′-biphenylylene)sulfurane bis(tetrafluoroborate) (2) was isolated as a stable moisture-insensitive yellow powder in 62% yield.

  Scheme 1

Subsequently, hexacoordinated derivatives – bis(2,2′-biphenylylene)dimethyl- and diphenylpersulfuranes – were synthesized and their molecular structures were elucidated [3].

References

Ogawa, S.; Matsunaga, Y.; Sato, S.; Iida, I.; Furukawa, N. First Preparation of a Sulfurane with Four Carbon–Sulfur Bonds: Synthesis and Molecular Structure of Bis(2,2′-Biphenylylene)Sulfurane. J. Chem. Soc., Chem. Commun. 1992, 0 (16), 1141–1142. https://doi.org/10.1039/C39920001141.
Sato, S.; Ameta, H.; Horn, E.; Takahashi, O.; Furukawa, N. First Isolation and Molecular Structure of Bis(2,2′-Biphenylylene)Sulfuranyl Bis(Tetrafluoroborate) [8−S−4(C4)]²⁺. J. Am. Chem. Soc. 1997, 119 (50), 12374–12375. https://doi.org/10.1021/ja971336k.
Sato, S.; Matsunaga, K.; Horn, E.; Furukawa, N.; Nabeshima, T. Isolation and Molecular Structure of the Organo-Persulfuranes [12−S−6(C6)]. J. Am. Chem. Soc. 2006, 128 (21), 6778–6779. https://doi.org/10.1021/ja060497y.


"Berthollide" seems to be a word for "non-stoichiometric compound" that has fallen out of use.

You have to look at the structure, as a rule things happen for a reason.  The formula can be derived from the structure.

Consider the series of alums, which is  KAl(SO4)2·12H2O, or more generic:  X+Y3+(SO4)2·12H2O.  The generic structure can accomodate a monovalent cation, provided it's sufficiently large (lithium is out of the question, sodium is an edge case), and a tervalent cation (Al, Cr, Fe, pretty much anything), and the proportion can be freely chosen within the constraints of the structure.


Does rearrangement take place in the reaction between ethenylcyclohexane and iodine chloride?

Yes, I believe the rearrangement can happen. However, the product from the rearrangement is probably not the major one.
The closest reaction to what you have drawn that I could find on SciFinder is this:

The paper is mentioned below in the reference. In the paper, the researchers performed similar reactions of alkenes with $\ce{I-Cl}$ and in cases where there is a possibility of a rearrangement to a tertiary carbocation, the by-product similar to $\text{(C)}$ is generated (for example in case of the cyclohexane ring being replaced with iso-propyl, the percentages are $\text{(A)=55%; (B)=2%; (C)=38%}$).
For the reactions drawn above, the researchers did not mention the A:B ratio, however, in most cases A is the major product, which is expected (Markovnikov's rule). Nevertheless, the product formed from the rearrangement is not the major one in any of the reactions.
The mechanism is not mentioned in the paper, but my guess is that it will not be much different from the usual carbocation rearrangement mechanism. Only in this case, the hydride shift will break the iodonium ring, while creating a tertiary carbocation. The Cl can then attack the carbocation and form the product.
This is my proposed mechanism:

To reiterate, this is likely not the major pathway. The usual addition to the double bond will be the major pathway and form the major product.
Reference:
T. Horibe, Y. Tsuji, and K. Ishihara, "Thiourea–I2 as Lewis Base–Lewis Acid Cooperative Catalysts for Iodochlorination of Alkene with In Situ-Generated I–Cl", ACS Catalysis, 2018, 8(7), 6362–6366


Why does the ionization enthalpy of elements across a period not follow a regular pattern while the atomic size always decreases?

Put very roughly, we can boil this down to why aluminum and sulfur are depressed with respect to the average of the surrounding elements.
Aluminum is $\ce{[Ne]{3}s^2{3}p^1}$.  Its first ionization removes the $3p$ electron while leaving the more stable $3s$ orbital intact, whereas magnesium has to break its more stable $3s$ orbital.  So by comparison, the first ionization energy of aluminum drops off.
This does not mean aluminum will form ionic compounds more readily than magnesium.  Despite aluminum's subshell advantage, both elements have first ionization energies that are too high to be offset by the electron affinity of a nonmetallic element and electrostatic attraction between the resulting ions.  There is the possibility of retaining metal-metal bonding between +1 ions, which has become fairly well-known with magnesium, but ordinarily a reasonably low second or third ionization energy is needed to get a favorable energy balance with these elements.  So to form ionic compounds aluminum ultimately has to give up its $3s$ electrons, which are more tightly bound than magnesium's, and thereby the subshell difference is obscured.  Magnesium ultimately wins when it comes to ionic compound formation.
Turning now to sulfur, that element has the ground state configuration $\ce{[Ne]{3}s^2{3}p^4}$.  In this case loss of an electron gives a half-filled $3p$ subshell with all the electrons unpaired and having identical spins.  This stabilizes the half-subshell configuration through a quantum-mechanical exchange interaction between the electrons with like spin and total angular momentum.  Sulfur forms this favored configuration when it ionized once, while phosphorus has to break it, and so the first ionization energy of sulfur (but again, not second or higher ionization energies) is depressed as we saw with aluminum.  Note that the effect here is less than that involving full or empty subshells that we saw with aluminum versus magnesium.


Let's use the n-propyl carbocation as an example.  It has two hydrogens alpha to the cationic carbon and two resonance structures involving hyperconjugation of those alpha hydrogens can be drawn.


Will entropy increase in adiabatic expansion?

The answer to your question is yes and no. You are correct in supposing that as an ideal gas expands the entropy will increase as it has more space to occupy and so the number of ways the molecules can be placed in the total space has increased. (Try to not think of entropy in terms of randomness but in the number of ways molecules can be positioned in space, or occupy different energy levels). However, accompanying expansion is cooling and if the adiabatic process is carried out reversibly the decrease in entropy due to cooling exactly matches that increase due to expansion, and so the entropy change is zero.
If the expansion is irreversible and adiabatic, the gas does less work, looses less internal energy and so there is entropy increase as the cooling is now not as large as when expansion is adiabatic and reversible, and so the two process, entropy increase due to expansion and decrease due cooling no longer match.


This answer is meant to address your second question about computerized representations of molecule structure since comments have already addressed your first question.

To truly represent the structure, mol and sdf formats are common: https://en.wikipedia.org/wiki/Chemical_table_file

Common string representations include SMILES and InChI. An issue with smiles is that there are multiple valid ways to represent a given molecule, which means a simple string match is not a good test for sameness. Canonical SMILES and InChI attempt to address this issue via additional rules. InChI additionally provides a standard way to convert the sometimes very long InChI strings into 27 character hashed version called InChIKey, which are better for searching.

Example Acetoin
SMILES string: CC(C(=O)C)O
InChI string: InChI=1S/C4H8O2/c1-3(5)4(2)6/h3,5H,1-2H3
InChIKey string: ROWKJAVDOGWPAT-UHFFFAOYSA-N


Based on Molecular Orbital Theory , how is it decided that the double bond in O2 molecule has one sigma bond and one pi bond?

$\ce{O2}$ MO diagram
From the MO diagram, we can see that it has 2 electrons in a $\sigma$ orbital, 4 in $\pi$ orbitals, and 2 in $\pi$* orbitals. Taking this together gives a bond order 2, with 1 coming from the sigma electrons and another 1 being the net value for the $\pi$ and $\pi$* electrons.


Speed distribution of lighter vs heavier gases

To my understanding, the speed distribution is wider for lighter molecules. The Wikipedia page for the Maxwell Boltzmann distribution has two images at the top that convey this very well.

In the first plot, $a=\sqrt{k_bT/M}$ where $M$ is the molar mass and $k_b$ Boltzmann's constant. For a Maxwell-Boltzmann distribution, the variance is given by $$\sigma^2= \frac{a^2\cdot(3\pi-8)}{\pi}=\frac{k_b\cdot T\cdot(3\pi-8)}{M\cdot\pi}$$ So, this shows that the speed distribution will be wider for higher temperatures and smaller molar mass.


Like aniline, phenol too reacts to a very less extent during Friedel-Crafts reaction.

The reason being that the oxygen atom of phenol has lone pair of electrons which coordinate with Lewis acid.

In fact most substituents with lone pair would give poor yield.

The two pathways involved in the reaction with phenol reduce the overall yeild:

Phenols are examples of bidentate nucleophiles, meaning that they can react at two positions:

on the aromatic ring giving an aryl ketone via C-acylation, a Friedel-Crafts reaction

or,

on the phenolic oxygen giving an ester via O-acylation, an esterification


Can a half-reaction be oxidation if it has the higher reduction potential?

Redox reactions are reversible. Despite the forward reaction is not spontaneous (as indicated by the unfavourable potential), it can still proceed.


This is true that all the aromatic protons come near the region having $\delta$: $\pu{7\!-\! 8 ppm}$. But you can always easily find examples where all the aromatic protons doesn't have the same chemical shift (which definitely include your examples). In fact the cases where all the aromatic protons have exactly similar chemical shift is hardly found.

The main idea is the difference between the terms Chemical Equivalence & Magnetic Equivalence, which is often misinterpreted. In the first example you have given, all the protons don't have the same chemical shifts and this is seen from the multiplet pattern as well. What you can obviously predict is that the two ortho protons are chemically equivalent, and the two meta protons are also chemically equivalent, and the para proton is a separate different type of proton. For observing these differences, what you need to do is to record the NMR at high frequency (e.g., $500$ or $\pu{600 MHz}$). There you can see the three different peaks clearly. Now in this case only, the two ortho protons are also magnetically equivalent as they couple with all the other protons equally, i.e. the $^3\!J_{H-H}$ and $^4\!J_{H-H}$  values are same for the two ortho protons, but in general this may not be the case at all (which is simply to consider another different para substitution at the ring, then the two ortho protons will become magnetically non-equivalent and therefore will have different chemical shift which is observable). Similar is the case with meta potons i.e. they are also magnetically equivalent. So those two protons will have same chemical shift. But ortho, meta, and para protons will definitely have different chemical shifts in your first example. If you zoom in or record the spectra in a high frequency machine, you should see a doublet of doublet peak (due to $^3\!J_{H-H}$ and $^4\!J_{H-H}$) for both ortho and meta protons, and a triplet of triplet for the para proton.

In the second example also, the two protons ortho to $\ce{-N(CH3)2}$ are chemically as well as magnetically equivalent as they both couple with the remaining proton in the same way. So, applying the similar logic you should see a doublet peak (due to $^4\!J_{H-H}$ only) for those two protons and a triplet peak for the remaining one and these two peaks should be at two different chemical shifts.

To encourage you more here are some spectra from the book Introduction to Spectroscopy which emphasize on aromatic regions and consider different spin-spin splitting and show how complicate a structure can be:

The above spectra depicts the differences in the spectra between ortho, meta, and para substitution in the benzene ring. The below two are some typical examples of complicated spectra due to spin-spin splitting as well as magnetic non-equivalence.

From these examples you can see that the aromatic protons can definitely come at distinct regions and what you need is to just record them in high field NMR machine and zoom into that region.


Is the given statement regarding Radial Probability Distribution Function correct or not?

Conceptually you are right as the commenters have mentioned, but since we are on a thread about nitpicking, we might as well go the extra distance.
Technically, $r^2R^2$ itself is not a probability but a probability density. In order to get the actual probability, you need to integrate it over a region. The probability of finding an electron between $r = r_1$ and $r = r_2$ would therefore be
$$P = \int_{r_1}^{r_2}r^2R^2 \, \mathrm dr \tag{1}$$
For an infinitesimal region, it suffices to multiply by $\mathrm dr$, such that the infinitesimal probability of finding an electron in the region $(r, r + \mathrm dr)$ is
$$\mathrm dP = r^2R^2 \,\mathrm dr \tag{2}$$
Notice how if you integrate $(2)$ over a finite region you get $(1)$.
Also, some books give this extra factor of $4\pi$, which I never really understood. You have two choices: either the probability density function is $r^2R^2$ (which works for all orbitals), or $4\pi r^2\psi^2$ (which works only for spherically symmetric orbitals, i.e. s-orbitals). I've written about this in more detail at  What is the exact definition of the radial distribution function?, so won't repeat myself here.


Counterintuitive Reaction of Aluminium On Steel Balls

This is actually related to the thermite reaction, where the rust on the balls reacts with the pure aluminum in the foil. Here is a pdf detailing the experiment.

The activation energy for the reaction is reached through the friction and pressure, and the rust on the balls is actually very important, since it is the exchange of oxygen between the iron oxide and aluminum that releases the heat as iron and aluminum oxide form.


Cyanide complexes in Molecular Orbital Theory

In this answer of Martin’s, you can find a molecular orbital diagram of $\ce{CO}$. The corresponding diagram for cyanide, $\ce{CN-}$ is essentially identical, there will only be different orbital energies and very slightly different extends of the lobes.

When forming a coordinate bond to a metal centre, cyanide will primarily attack with its highest occupied molecular orbital, the HOMO, since it is a nucleophile (attacking a positively polarised metal centre). You can also see the HOMO depicted as the bottom image of the following set as taken from Wikipedia:

Note the large lobe extending outwards on carbon’s side. An attack with this lobe is greatly preferred, hence why isocyanides are comparatively rare.


That is the optical density (absorbance). It's just a range of between 0.6 and 0.7. Without the context, it's hard to say why they don't have a precise value, but in something like a protocol, it's common to see that kind of thing when there is some natural variation or the exact number isn't important.


The first step is understanding what you have.  The seven crystal systems are derived from decreasing symmetry (lifting degeneracies) of the cubic system by altering lengths and angles.  Example: Cube volume is $\ce{a^3}$.  Parallelepiped volume with all symmetries explicit is

http://en.wikipedia.org/wiki/Parallelepiped

http://www.rockhounds.com/rockshop/xtal/index.shtml
Good reading about crystal systems and their expression.
http://en.wikipedia.org/wiki/Space_group
As usual, nomenclature is a mess.
http://en.wikipedia.org/wiki/Hermann%E2%80%93Mauguin_notation
That's a winner.

Crystallographers often translate the unit cell within the lattice to make its contents  "nice." The rhombohedral system is particularly nasty for choosing the unit cell itself.  65 Sohnke space groups of 230 overall contain chiral formula units.  Racemic screw axes make that interesting, ${2_1}$, ${4_2}$, ${6_3}$.  Within the 65, 11 enantiomorphic space groups - themselves chiral without regard to unit cell contents - allow the same substance to crystallize in two mirror-image space groups.


What's the effect of temperature on buffering capacity?

You question is too broad to answer since there is no "universal answer." In chemistry there is a "tends too" answer and then a list of exceptions.

By and large the root of the question centers around the change in the equilibrium constant as a function of temperature. The equilibrium constants do depend on temperature, but how much "tends" to depend on the particular equilibrium under consideration. However the Arrhenius equation can offer some insight.


Mineral spirits goes by many names, including "white spirit" and "paint thinner", and is basically just a hydrocarbon solvent mixture. According to this Wikipedia article:

  White spirit is a mixture of aliphatic and alicyclic C7 to C12 hydrocarbons with a maximum content of 25% of C7 to C12 aromatic hydrocarbons. A typical composition for mineral spirits is > 65% C10 or higher hydrocarbons,[6] aliphatic solvent hexane, and a maximum benzene content of 0.1% by volume, a kauri-butanol value of 29, an initial boiling point of 145 °C (293 °F) to 174 °C (345 °F), and a density of 0.79 g/ml.

There are also three different types based on whether or not it's undergone processing to remove sulfur, solvent extraction or hydrogenation. Then each type has a separate grade: low flash, regular and high flash grades. So with three different types each having three different grades, there are nine different sorts of mixtures that "officially" fall under the term white spirits, or mineral spirits.


How to determine enzyme inhibitor efficacy?

If the inhibitor is a competitive inhibitor, to quantify inhibition the $K_m$  of the inhibitor ($K_I$) is determined.

$K_I$ is compared to the $K_m$ of the substrate.

Then, the degree of inhibition is a function the inhibitor and substrate concentrations.

$V_{max}$ is unaffected by a competitive inhibitor.

In the presence of an inhibitor the kinetics are modified such that the enzyme has an effect $K_m$ as:

$K_m(1 + [I]/K_I)$


Obtaining number of functional groups in a compound from SMILES, SDF or MOL2 format

There are numerous programs, but it's pretty easy to use the obgrep tool in Open Babel to do this using SMARTS for the various functional groups:

obgrep -c '[CX3]' file.sdf # count the number of methyl groups
obgrep -c '[#6][OX2H]' file.sdf # count the number of hydroxyl groups

etc. Examples of functional groups using SMARTS can be found on the Daylight SMARTS examples page.

I've also written scripts in Python for these purposes - depends on how you want to run the SMARTS searches.


It's hard to tell what is the answer you were supposed to provide, but yes, generally speaking, you are correct.
According to CRC handbook of chemistry and physics [1, p. 4-14] (emphasis mine):

Francium, the heaviest known member of the alkali metal series, occurs as a result of an alpha disintegration of actinium. […]
Because all known isotopes of francium are highly unstable, knowledge of the chemical properties of this element comes from radiochemical techniques. No weighable quantity
of the element has been prepared or isolated.

As the result, the majority of physical properties of francium, including density, were obtained by extrapolation and not by direct measurements.
According to Bonchev and Kamenska [2], $ρ_\mathrm{calcd.}(\ce{Fr}) \approx \pu{2.9 g cm-3}$, which is in a good agreement with the trend for the measured densities across the group.
For comparison, the following data (NTP conditions) is collected from the table PHYSICAL CONSTANTS OF INORGANIC COMPOUNDS [1, p. 4-43]:
$$
\begin{array}{cl}
\hline
\text{Element} & ρ/\pu{g cm-3} \\
\hline
\ce{Li} & 0.534 \\
\ce{Na} & 0.97 \\
\ce{K} & 0.89 \\
\ce{Rb} & 1.53 \\
\ce{Cs} & 1.873 \\
\hline
\end{array}
$$
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; CRC Press, 2017; Vol. 97. ISBN 978-1-4987-5429-3.
Bonchev, D.; Kamenska, V. Predicting the Properties of the 113-120 Transactinide Elements. J. Phys. Chem. 1981, 85 (9), 1177–1186. https://doi.org/10.1021/j150609a021.


Why can acidification of solution lead to selective precipitation of ions such as Fe2+ and Zn2+?

FeS has 1500 times the solubility of ZnS.  Given the slow addition of a small molar fraction of acid that solubilizes both Zn(II) and Fe(II), that fraction will retain at equilibrium.  It gets complicated if soluble polysufides exist. Iron gives pyrite and marcasite ($\ce{FeS2}$) via mackinawite ($\ce{FeS}$) and greigite ($\ce{Fe3S4}$), but they are insoluble.


REDOX spontaneity using electrode potentials

The reason is simply semantics. Iodi(n)e which is the neutral valence state of the element will not donate an electron to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$.   Iodi(d)e which is the anionic form of iodi(n)e has a negative charge and can donate an electron to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$.


Is a 2-D periodic structure isomorphic with the surface of a torus, a sphere, neither or both?

Yes, a 2D-periodic space can be mapped to a torus, but that's more a question for the math.SE…

Regarding your bonus question, why would there be? What would you do with it? Molecular structures are intrinsically 3D, so I don't see what you would do in a 2D (periodic or not) space? Even when we talk about planar or pseudo-2D structures (buckyball, nanotube, etc.) they are 3D objects with 3D electronic densities and wavefunctions.

Edit: 3D structures that are periodic in two dimensions and finite in the other one can be studied by many computational chemistry codes. They are often referred to as slab calculations or surface calculations. The most common issue is that of Coulombic interaction (or Poisson equation solver), which typically requires special treatment in the 2D case.


Yes this reaction will definitely work. Use a nice leaving group like iodide and the reaction should go easily. Although the nitrogen lone pair is delocalised into the benzene ring it is still quite nucleophilic. Additionally, the delocalisation is restored when the nitrogen loses a proton to get the substituted amine. As such there probably won't be a big problem with forming a quaternary ammonium salt (which is what usually happens if you try to alkylate amines) because doesn't have any opportunity to maintain delocalisation in the product and so will be disfavoured compared to the tertiary amine.


It appears it can be decomposed in the presence of Hydrogen Peroxide, with decomposition being noticeable even at 3% (extremely slowly), but working well around 12%. I didn't do anything beyond basic tests, but a white salt (likely Urea?) will precipitate out, embedding in the pores, if any, of the plastic; this can interfere with this clean up method. It also produces Aldehydes, seemingly mostly Formaldehyde from the odor, but with reactions based on the environment. Some Acetaldehyde was produced in my case, but the rate was slow enough that only small short lived flames were produced.

This was able to be sped up with heating or HCl. I would shy away from adding HCl, as it produced a strong Chlorine smell, without any obvious yellow gas.


Why does the rate of reaction increase less than proportionally to the increase in mass of catalyst?

There is a lot going on here. Part of the problem is poor experimental design. I'll just throw this out without evaluating all the energies. (Too lazy...)

(1) The potassium iodide has to be dissolved into the solution since you added it as a solid. This means there is an energy change and this could be endothermic or exothermic. Adding more salt increases the effect obviously.

(2) It takes some time to dissolve the potassium iodide. So before all the potassium iodide is dissolve some is already reacting with the hydrogen peroxide.

Greater surface area (smaller particles of potassium iodide) would allow the crystals to dissolve faster.

(3) The reactions are:
$\ce{H2O2(aq) + I-(aq) = H2O(l) + OI-(aq)}$
$\ce{H2O2(aq) + OI-(aq)= H2O(l) + O2(g) + I-(aq)}$

Overall the reaction is:
$\ce{2H2O2(aq) =  2H2O(l) + O2(g) + \text{heat}}$

But the heat is the combination of the two reactions.

If $\ce{[H2O2](aq) >> [I-](aq)}$ then both reactions will happen.

If $\ce{[H2O2](aq) << [I-](aq)}$ then only the first reaction will happen. The iodide isn't acting as a catalyst but as a reactant.

3.3 % Hydrogen peroxide is

$0.033 \cdot 15 \approx 0.5$ g $\ce{H2O2}$

Moles $\ce{H2O2}$ = $\dfrac{0.5}{34} = 0.015 \text{moles} = 15 \text{millimoles (mM)}$

0.2 g KI = $\dfrac{0.2}{166}$ = 1.2 mM

1.0 g KI = 6.0 mM

So you're adding so much KI that some might get left as $\ce{OI-}$.

(4) so to do the experiment better:

cut the KI concentration
dissolve the KI in 5 ml of water & equilibrate  temp to that of 3.3% H2O2
mix solutions with stirring


When is it true that more nodes equals higher energy?

General case

There is indeed a mathematical theorem that deals with the number of nodes an eigenfunction corresponding to a certain eigenvalue can possess.
It was laid down by Courant$^{[1, 2]}$ and it states the following:

  Given the self-adjoint second order (partial) differential equation

  \begin{equation} \left(\hat{L} + \lambda \rho(\mathbf{x}) \right)
 u(\mathbf{x}) = 0 \end{equation}

  (where $\hat{L} = L(\mathbf{\Delta}, \mathbf{x})$ is a linear,
  hermitian differential operator, $\rho(\mathbf{x})$ is positive and
  bounded, and $\lambda$ is the eigenvalue) for a domain $G$ with
  homogeneous boundary conditions, that is $u(\mathbf{x}) = 0$ on the
  boundary of the region $G$; if its eigenfunctions are  ordered
  according to increasing eigenvalues, then the nodes of the $n^{\text{th}}$
  eigenfunction divide the domain into no more than $n$ subdomains. The
  nodal set of $u(\mathbf{x})$ is defined as the set of points
  $\mathbf{x}$ such that $u(\mathbf{x}) = 0$. No assumptions are made
  about the number of independent variables.

The proof is rather involved and so I won't show it here. But if you want you can look it up in [1] or here.

So, Courant's nodal line theorem tells us, that if we order the possible energy eigenvalues of the time-independent Schroedinger equation as $\lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$, then (depending on precisely how you set up the numbering) the  $n^{\text{th}}$ eigenfunction, $\Psi_{n}$ (the one with energy eigenvalue $\lambda_n$) has at most $n$ nodes (including the trivial one at the boundary $\mathbf{x} \to \infty$).
Unfortunately, this gives you only an upper bound for the number of nodes a wave function with a certain energy eigenvalue may possess. So, all we know is that the ground state wave function $\Psi_{1}$ cannot have any nodes within the region $G$ (in total it has one node, namely the one at $\mathbf{x} \to \infty$). Wave functions for higher $n$ may possess up to $n-1$ nodes within $G$ but may as well have less. Thus, we cannot in general say that if a wave function has more nodes than another one it will automatically correspond to a state with higher energy.

Special case: Schroedinger equation in one dimension

There is however a special case:
For the Sturm-Liouville eigenvalue problem (and thus for ordinary second order differential equations with homogeneous boundary conditions) we can strengthen Courant's nodal line theorem such that if we order the possible eigenvalues as $\lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$, then the  $n^{\text{th}}$ eigenfunction (the one with energy eigenvalue $\lambda_n$) has exactly $n$ nodes (including the trivial one at the boundary $\mathbf{x} \to \infty$).

This is useful since the one-dimensional time-independent Schrödinger equation is a special case of a Sturm-Liouville equation. So, in the case of the inhomogeneous radial Schrödinger equation with a local potential and node-less inhomogeneity such as the radial Schrödinger equation for the hydrogenic atom

\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0
\end{equation}

it is generally true that a wavefunction with more (radial) nodes must always correspond to a state of higher energy than a wavefunction with less radial nodes. Also, it is clear that the wavefunctions of the one-dimensional particle-in-a-box must follow this rule. But for the three-dimensional particle-in-a-box this is not true anymore, since in that case the Schroedinger equation of the system is not an ordinary second-order differential equation but a partial-differential equation for which only the general version of Courant's nodal line theorem holds.

Some concluding remarks

For real-world system like molecules or crystals the Schroedinger equation is a partial differential equation for which the special case outlined above doesn't apply so that only Courant's nodal line theorem in its general form holds which doesn't give a strict justification for the statement that more nodes mean higher energy. Yet it is very often observed that the number of nodes indeed increases with increasing energy.
The reason for this can be motivated in the following way: The kinetic energy $E_{\mathrm{kin}}$ of a state is proportional to $\int \Psi \Delta \Psi \, d^{3} r$. Via Gauss's theorem it can be shown that $\int \Psi \Delta \Psi \, d^{3} r \propto \int |\nabla \Psi |^{2} \, d^{3} r$ and so $E_{\mathrm{kin}} \propto \int | \nabla \Psi |^{2} d^{3} r$. Now, nodes force a wavefunction to change it's sign. This often means that the value of $\Psi$ has to increase/decrease rather rapidly thus leading to areas with high absolute values of the gradient and thus to high kinetic energy. Since the potential energies shouldn't differ too much between the different states the higher kinetic energy usually also entails a higher total energy. As an example consider the bonding and antibonding wavefunctions of a homonuclear diatomic molecule whose atoms are placed at the positions $r_{\mathrm{A}}$ and $r_{\mathrm{B}}$.

The bonding wavefunction has no nodes. Its value between the atoms doesn't have to undergo a rapid change and thus the slope is rather low. The antibonding wavefunction has one node between the atoms. Its value between the atoms must change rapidly from its positive to its negative maximum thus entailing a very high slope. The slopes of the tail regions are comparable for the bonding and antibonding wavefunctions since it can smoothly fall of to zero at infinity and is not required to go from a maximum value to zero within a very confined region of space - thus even if one wavefunction has to start of at a higher maximum value the gradient will not be much higher. It follows that the antibonding wavefunction has a higher kinetic energy than the bonding wavefunction.

References

[1] R. Courant, D. Hilbert, Methods of Mathematical Physics, Vol. 1, Interscience, New York, 1953, p. 451-455.

[2] R. Courant, "Ein allgemeiner Satz zur Theorie der Eigenfunktionen Selbstadjungierter Differentialausdrücke", Nachr. v. d. Ges. d. Wiss. zu Göttingen 1923, p. 81.


These are basic terms of radiochemistry:

Isotopes are nuclides with a common number of protons $Z$.
Isobars are nuclides with a common number of nucleons $A$.
Isotones are nuclides with a common number of neutrons $N$.
Isodiaphers are nuclides with a common neutron excess $N-Z=A-2Z$.

Thus, in the chart of the nuclides, isodiaphers are found in diagonals from bottom left to top right. For example, the mother and daughter nuclides of alpha decay are isodiaphers.