Why does Avogadro's Law hold true for ideal gases?

Avogadro's Law actually states that the volumes of ideal gases at a fixed temperature and pressure are proportional to the number of molecules that are present. This is actually a sub-case of the more general ideal gas law, which states $PV = nRT$ (or $PV = nkT$, if you're working on a per-particle basis) .

The ideal gas law actually falls out of a basic statistical mechanics analysis of the behavior of a collection of molecules which interact solely by elastic collision. (Here's a 25 minute video of one such derivation.)

Basically, the more particles you have in a given region with a certain fixed speed (i.e. a fixed temperature), the more frequently they collide - not only with each other, but with the walls of the container. This means that putting more particles in the (same size) container (at the same temperature) increases the outward pressure on the container walls. In fact, the rate of collisions (and thus the pressure) is directly proportional to the number of particles (doubling the number of particles at a fixed temperature doubles the rate of collisions).

But we want to hold the pressure constant! If we can't change the temperature, one way to reduce the pressure is to increase the volume (e.g. let the container expand like a balloon.) As the volume grows, each particle has to travel farther before it can collide with the walls of the container, meaning that the rate of collisions and thus the pressure on the walls gets less and less.

It just so happens that this relationship between pressure and volume is inversely proportional (that is, Boyle's Law). So to halve the pressure, we need to double the volume. So if we double the number of particles, to keep the same pressure we need to also double the volume. Or more generally, the volume is increased in direct proportion to the number of particles - that is, each mole of gas takes up a certain volume (at a fixed temperature and pressure). If we try and put it into a different volume, the pressure (or the temperature) will change.

Okay, but why doesn't a fixed number of different particles take up different volumes? For example, why doesn't doubling the mass of the particles make it take up half the space (or double, or a quarter ...)? - Heavier particles might hit the walls of the container harder, but at any given temperature heavier particles are moving slower then lighter ones, so the collision rate is less. When you run the numbers and calculate the velocity of the particles, the rates of collision, and then how hard they hit the container walls, the mass cancels out. At a fixed temperature, the contribution to pressure from a single particle colliding with the walls of the container is independent of its mass.

You can do the same sort of analysis with other properties. They are either 1) are explicitly discounted by the assumption of "ideal gas" (e.g. no particle interactions) 2) don't factor into the calculation at all (e.g the absorption spectra of the atoms) or 3) precisely balance out - leaving you with $PV = nRT$ being the relation for an ideal gas.


How Would A Vast Mixture (Like an Ocean) of Ammonia-Water Behave in Terms of Evaporation and Freezing?

Freezing

  I've heard that the freezing and boiling points of the mixed liquid would exist on some continuum between the freezing and boiling points of its constituent liquids (ammonia and water), varying depending on the exact mix ratio and of course pressure. Is that correct?

For freezing points, that is incorrect. Pure water freezes at 273 K and pure NaCl (table salt) "freezes" at 1074 K. However, if I dissolve NaCl in water, the freezing point of water will be lowered. In you water/ammonia scenario, water will freeze as pure water, but the freezing point will be lower. Ice formation would increase the concentration of ammonia in solution.

Evaporation

Water and ammonia mix in the liquid state, and everything mixes in the gas state. Evaporation (and condensation) depend on the ratio of components in the liquid state, the partial pressure of components in the gas state (or ratio of components and total pressure), and the temperature. At equilibrium at earth-like temperatures, there would be more ammonia than water in the atmosphere (the boiling point of ammonia is much lower than that of water, so the vapor pressure of ammonia is higher than that of water). A sudden drop in temperature would make it rain ammonia, mixed with a bit of water. So condensation and precipitation would increase the concentration of ammonia in the liquid state, and a sunny day would lower it.

  If the later behavior is correct I could imagine an ocean of water-ammonia varying substantially in its water/ammonia ratio from pole to equator thanks to the greater heat near the equator and the ammonia molecules evaporating.

You are making a lot of assumptions here: that the exoplanet rotates fairly quickly around its own axis, that its axis is roughly perpendicular to the star, and that the star is close enough to heat up the planet.

  What I'd like to most wrap my head around is whether the water-ammonia mix would behave as a single uniform liquid, freezing and evaporating at a rate determined by the average of the mix's freezing/boiling point, or whether the ammonia molecules would start to evaporate before the water molecules, and the water molecules start to form ice before the ammonia molecules.

No, they would do different things. Pure water at terrestial atmospheric pressure freezes at 0 $^\circ$C, while ammonia already boils at -33 $^\circ$C. Mixing them will lower the freezing point of water and increase the boiling point of ammonia, so there is no temperature where you would separate them completed by freezing all water and boiling off all ammonia. On the other hand, they don't freeze or boil in lock step. If you lower the temperature, water will form ice and the ammonia concentration in the liquid will increase. If you increase the temperature, more ammonia than water will turn into gas (evaporating or boiling), so the ammonia concentration in the liquid will decrease.


I'm a layman, but I'd like to know to to use GAMESS for some specific tasks

For GAMESS, the command is not $\$STATTP$, but it is $\$STATPT$. Also, please change the commas to dot, otherwise it considers it to be a part of the number.

Please refer to the GAMESS manual for better understanding. The link is: www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf


Short Answer: Yes, based on the electrochemical/reactivity series, it is possible to percipitate gold from solution using metals higher on the series like copper or zinc. However, it isn't done often because gold is highly unreactive, and when gold salts are formed, they are formed deliberately under atypical conditions like very high temperatures or acidities.
It would be possible to displace gold from it's ionic salt (gold (iii) chloride) using zinc or copper. This is because, as you mentioned copper is higher on the activity series than gold.
The likely reason to why companies don't use single-displacement reactions to retreive gold from solution is be cause gold salts rarely exist. Gold is extremely unreactive (see here) Gold (III) chloride is normally synthesized by reacting gold metal with chlorine gas at 180 degrees Celsius. This temperature is higher than anywhere you'd find on Earth's surface, and chlorine gas is too reactive to exist in that state in the atmosphere. These conditions are typically found only in the lab and not out in nature. Another example of a man-made gold salt is aqua regia (a 3:1 ratio of hydrochloric acid and nitric acid) which can dissolve gold. To recover the gold, ions are displaced using ferrous sulfphate reducing agent (source).
If you are interested in gold recovery, here is an excellent article on a variety of methods.


Contribution of vibrational degrees of freedom

You have to be careful to make a distinction between the coordinates needed to describe a molecule, and the degrees of freedom which the molecule can actually access. For this purpose, one might choose to define the number of degrees of freedom as the minimum number of spatial coordinates needed to fully describe the positions of atoms relative to a fixed reference frame. This will be $3N$ total coordinates which corroborates the other answer you link. We then recognize that motion for non-linear (linear) systems, motion in six (five) of these coordinates correspond to translation and rotation. Therefore, the other $3N-6$ ($3N-5$) coordinates describe the vibrations of the system.

Now, if one is concerned with the total energy of a system, then we should keep track of the Hamiltonian. In a classical context, the Hamiltonian is a function of all coordinates which are capable of holding energy in one form or another, $H(\mathbf{q},\mathbf{p})$. For molecules this means we must keep track of $3N$ spatial coordinates, $\mathbf{q}$, and $3N$ momentum coordinates, $\mathbf{p}$.

Now, if we want to know about the average kinetic energy of atom 1 in the $x$ direction, we perform an ensemble average

$$
\langle T_{x,1}\rangle=\frac{1}{2m}\frac{\int p_{x,1}^2e^{-\beta H(\mathbf{q},\mathbf{p})}d\mathbf{q}d\mathbf{p}}{\int e^{-\beta H(\mathbf{q},\mathbf{p})}d\mathbf{q}d\mathbf{p}}
$$

It is a nice exercise to show that if all the degrees of freedom in the Hamiltonian are quadratic, then this separates into lots of integrals over gaussians and the top and bottom will cancel in all coordinates except for the $x$ direction motion of atom 1. This integral will give you the average energy as $1/2kT$.

But, since the integral over the quadratic momentum and the quadratic position as in the harmonic potential are exactly the same, we have just shown that every type of energy which depends quadratically on a coordinate has an average energy of $1/2kT$.

Therefore, if we have a particle in 1-D subject to a harmonic potential, the Hamiltonian is,
$$
H=\frac{p^2}{2m}+\frac{1}{2}kx^2
$$

which means the average energy is $kT$ since there are two coordinates which the Hamiltonian depends on quadratically.

So, hopefully now it is clear why we double to count of vibrations and arrive at the answer,
$$
\gamma=1+\frac{2}{13}=1.154
$$

I know this was very long-winded and overly detailed, but I hope this gives you a little better picture into where the equipartition theorem comes from and how this convenient counting of coordinates we do is really a coincidence of the fact a harmonic potential and kinetic energy are each quadratic in their respective coordinates.

As an aside, the distinction I have made between coordinates and degrees of freedom is completely arbitrary. Indeed, people often refer to both a harmonic potential and the kinetic energy as quadratic degrees of freedom. I prefer this language, but then one must be careful to clarify between degrees of freedom of position and degrees of freedom of momentum because this is more coordinates than are needed to describe the system completely, so it seems there are more degrees of freedom than there really are.


No, deuterium is completely stable. I found the answer at Hyperphysics, and it has to do with the mass energies of the products and reactants of this hypothetical reaction.

The decay of deuterium would be
$$\ce{D -> P + N + e + \bar{\nu}}_e$$
where $\ce{D}$ is deuterium, $\ce{P}$ is a proton, $\ce{N}$ is a neutron, $\ce{e}$ is an electron and $\ce{\bar{\nu}}_e$ is an electron antineutrino. The combined mass energies total to $1877.05\,\mathrm{MeV}$ (mega electron volts). But the mass of $\ce{D}$ is $1875.6\,\mathrm{MeV}$, so the decay cannot happen because conservation of mass/energy would be broken. There's no way to get around this, not even imbuing the initial deuterium atom with extra energy of some other sort. Deuterium will never decay unless, for some reason, the proton turns out to be unstable.

If, for some reason, deuterium did decay, it could go to $\ce{^2He}$ … which would then decay back to deuterium.


As commented by SteffX, less the $\mathrm{p}K_\mathrm{a}$, more stronger the acid. More stronger acid means that the conjugate base or the anion formed after donating the proton is more stable compared to the carbanion of an acid of a  higher $\mathrm{p}K_\mathrm{a}$ value. So, if the anion formed is a carbanion, then lower the $\mathrm{p}K_\mathrm{a}$, the stable the carbanion is.


Reason for prescence of oil bubbles while frying

I am extremely skeptical that a typical cooking oil and water form a significant azeotrope. Such a claim demands an authoritative reference. Most known azeotropes have boiling points only fractions of a degree below the lower boiler, making little difference in the boiling point. Likewise, most known azeotropes contain over 99 wt% of the lower boiler again making most azeotropes insignificant.

It is simply my opinion, but I believe the oil transported from the bubbling liquid is in the form of an aerosol or perhaps even larger droplets.

Those issues aside I agree the bubbling is boiling of lower bp chemicals, almost certainly mostly water. As cooking occurs, decomposition into lower bp compounds is possible (and if significant amounts form, they may form azeotropes with the oil (but see comments above)).

I am also not aware of the solubility of the various atmospheric gasses (N2, O2, CO2) in oil. Gasses are in general less soluble as temperature increases (eg carbonated beverages), so they also would boil out if present in significant amounts (I'd guess they aren't very soluble at 1 atm pressure).

I should also add that while most azeotropes aren't "significant", most liquids do form them.


You might experiment with fresh-ground apples, which contain polyphenol oxidase and catechol oxidase, enzymes that turn brown in air (a tiny amount of copper or ferrous salt is also present, acting as a catalyst).

For that matter, dilute ferrous salts are often almost colorless (pale green) but darken to brown on exposure to air. (Digging in a podzol, soil of a pine forest, shows that color change.) So, you could extract a bit of liquid from greenish podzool and load your pen or marker with that.


Substances comprised of large molecules are liquids at room temperature?

You may give porphyrins a try.  These dyes are of interest for molecular electronics, and to render them less crystalline / more glassy, alkyl chains are attached to the central motif.  One recent example are the room temperature liquid porphyrins developed in the group around professor Gryko:

(source, Chem Asian J. 2010, 5, 904-909; doi: 10.1002/asia.200900693.)
Besides standard spectroscopic characterisation, the publication contains the results of DSC:

so compound 4d ($\ce{C176H294N4O12}$) may be a good candidate for an organic molecule, without charges and metal coordination, and suitable for ellipsometry / VASE measurments on surfaces.


Calculating the molar mass of a solute from freezing point

The freezing point depression is based on the molal concentration (moles of solute per kg of solvent).
$$\Delta T_f = -k_f \cdot m$$
You know the freezing point depression of the solution and the cryoscopic constant, so you can calculate the molality:
$$m = -\dfrac{\Delta T_f}{k_f}=-\dfrac{\pu{-1.23 ^\circ C}}{1.86 \frac{\textrm{kg}\cdot ^\circ \text{C}}{\textrm{mol}}}=\pu{0.66 molal}$$
Since molality is the number of moles of solute divided by the mass of the solvent in kg, you can calculate the moles of the solute:
$$n_\textrm{solute}= \pu{0.66 m} \cdot \pu{ 0.105 kg}= \ ...$$
Divide your mass of solute by the moles of solute and you will get the molar mass (and not 2.989 g).


EDIT:

I read the paper linked in your comment, and see that my initial guess was incorrect (see below) - for this system, it is as you say - for temperatures above the bulk pure polymer glass transition temperature, you would not expect to see shrinkage stress due to evaporation.

END EDIT

I don't have access to that paper - so this is just an educated guess, but usually physicists use the glass transition temperature to characterize polymer behavior as a function of temperature because things roughly (very roughly) scale with $T_g$. In other words, if I know how the behavior of one linear homopolymer depends on temperature, and I know a certain response curve scales with $T_g$, then I can scale the temperature curve of another linear homopolymer using the $T_g$ and sometimes get accurate results.

They could be linking it to shrinkage stress because $T_g$ also characterizes the dynamic behavior of polymers (how they "move" in response to thermal fluctuations and external stress). This is essentially the principle of time-temperature superpostion - or the idea that as temperature increases, polymers move the same way, only faster. So heating up is like fast-forwarding time, and cooling down is like slowing time down.

The glass transition temperature is a discontinuity in this behavior - at a certain low temperature, the polymers freeze in place (on large length scales). This phenomenon is not well understood, but we have found empirically that the temperature at which it occurs can be used to scale response curves in certain situations, and since it is fairly easy to measure, it gets used a lot in these types of models.

So in summary, and again this is an educated guess, I would imagine that the glass transition temperature would be used in this model regardless of the temperature you were actually operating at - it is probably a fundamental parameter of the curve that is used to put the "reality" (the type of polymer) into the model.


The authors use a 3D printer to generate the labware, and to place the reagents in its inlets.

So we're not printing standard ink on paper but using a the print head to exactly place very amounts of monomers at certain positions. These monomers cure (polymerize) upon heat or irradiation to a solid resin. By this, small reactors of arbitrary design (think in 3-neck reactors) with a volume of ~ 10 mL may be printed layer by layer.

Apparently, it is even possible to create reactors with windows to allow for later reaction monitoring or print and assemble electrodes for spectroelectrochemistry.

  How can one "print reagents"? What is meant by this?

This is all about precise positioning of the printhead! Thus same printhead used for the ink can be used to fill the inlets (think in dropping funnels of normal labware) of the reactor with componends whose reaction you want to monitor. Instead of dropping, material transfer is achieved by applying a vacuum to the reactor through a self-healing hole.

  Most synthetic reactions require hours of refluxing and whatnot [...]

Well, a lot of reactions do proceed quite well at room temperature, think in

Diels-Alder reactions of furanes and maleimides
electrochemically triggered reactions
photoinduced electron transfer reactions

just to mention a few.

  [...] not to mention isolation/purification afterwards. How can all of these techniques be performed with a printer?

Oh, you got that one slightly wrong. The "printer" is just used to build the reaction sphere (the flask, if you like it) and to place the reagents in the right inlet of your vessel. The rest is "just chemistry" ;)

I hope i made myself clear enough - at least after the n-th edit ;)

I think the concept is pretty cool.


Why is sodium bisulfite needed in syn-dihydroxylation?

You need to reduce remaining oxidizer (OsO4) and maintain pH of the solution. NaHSO3 acts both as the reducing agent and a buffer.


Why do people still use the mole (unit) in chemistry?

Simply speaking, because it's an appropriate unit to use.

Let's imagine I wanted to measure the length of a rope. What would be an appropriate length to use? Inches? Centimeters? Feet, maybe? It would really be awkward to express it as 0.000189393 miles, or as 304,800,000 nanometers.

(Note: if you can't see why these units are awkward, take any page discussing things like this (e.g. biomass of certain species) and change all the units so that they're nonsense like this. Then put it away for a week and try to read it later.)

Now let's say I've changed my mind and I'd like to measure all the rope created in the world in a year. Now an appropriate unit is almost certainly miles or kilometers, and not inches or centimeters.

Let's consider something else: so far, I've been using length to measure ropes. Would it make sense to measure their combined mass instead? Maybe not for small amounts, since I think I would throttle you if you told me to cut off half a pound of rope, but for global-scale things, tons or metric tons may make sense.

On the other hand, using measurements like the average density of the rope or the combined diameters of all the ropes really wouldn't be much use at all.

What we've seen here is that when we're measuring things, there are measures that make sense (for rope, length, maybe weight) and some measures that really don't (color, average diameter, etc.). In those measures, there are units that are convenient (inches, feet), and some that aren't (nanometers).

This is exactly the problem with chemical units, but much more magnified. When I measure the energy released on hydrolysis of a sample, I'm not measuring the energy of one bond, or two bonds, or a thousand bonds, or even a million. I'm measuring a collection of so insanely many molecules that there's really no number in common language for it.

What is an appropriate way to measure things in these molecules? If I'm only interested in how much I have and not what's in it, per gram or per volume is often a good way to do it. Since when I'm measuring the temperature change of something, I don't particularly care what it's made of, measures of this (specific heat, for instance) are done in units of something per gram.

On the other hand, if I do care what my sample is made of, then I need a better way to measure it. For example, sodium chloride and calcium chloride look similar, but will have very different energies since there's three ions in a $\ce{CaCl2}$ unit and only two in an $\ce{NaCl}$ unit.

Since you can't have half a molecule, the simplest way to do it is to count the darn things. Unfortunately, measuring things per atom is a really awkward way to do things because of the problems I described above. The energy it takes to melt ice is 0.000000000000000000001029 J/molecule.

What we need then, is some count of molecules that's convenient. We could go by 1000 molecules, or a million molecules, but it's a pain in the butt to convert between that and macroscopic units (how much does 1 million calcium atoms weigh?) Now it just so happens that $6.022 \times 10^{23}$ atoms of carbon-12 weigh 12 grams, and a single carbon-12 weighs 12 amu. A single molecule of water weighs roughly 18 amu, $6.022 \times 10^{23}$ molecules of water weigh 18 grams.

Let's take a look at what using $6.022 \times 10^{23}$ molecules offers us over other units:

It is appropriate. $6.022 \times 10^{23}$ molecules will typically
fall in the gram to kilogram range of substance, which can easily be
measured out on a balance. It's also within the range of what you'd
usually use in a lab.
It allows for easy conversion. You can easily tell when you have
$6.022 \times 10^{23}$ molecules because your atomic mass is equal to
your macroscopic mass.
It makes sense. Intuitively, we'd like to measure things per atom or
per molecule, but doing so leads to some ugly units. Instead, if we
measure per $6.022 \times 10^{23}$ molecules, we still are doing a
counting-based measurement, but the numbers themselves are a lot
nicer.

You probably know this already, but we call $6.022 \times 10^{23}$ molecules a mole. These are the advantages of using the mole as a unit, and not for backwards-compatibility. I am of the opinion that if we shed all our units today and started from scratch, we would start off with grams and liters, but we would probably start using the measure of a mole again within a year. It's just a very powerful, useful way to measure things.

P.S. This is much longer than I was initially intending and was not written while I was in the most awake state of mind. Please let me know in the comments if you think this answer is unclear, or stupid, or just plain wrong.


What is the most compact way to store Oxygen?

Solid ozone has a density of 1.728 g/cm3 according to The Density of Solid Ozone J. Am. Chem. Soc., 1959, 81, pp 805–806.

Best I can think of at atmospheric pressure, assuming your goal is to store and recover as much pure oxygen as possible in a given volume.

If your goal is just to pack as many oxygen atoms in a given volume, BeO, which has a density of 3.01 g/cm3 is better.

At high pressure there is metallic oxygen.


How to visualise the total electron density from a Gaussian 16 calculation?

When GaussView gives you the error

  Surface build failed! Processing mesh, No data to plot. IsoValue may be out of range

it means that the data in your .cube file doesn't have values of density at that isovalue. So when it tries to plot a surface, it cannot plot anything.

The IsoValue (default is 0.00040, I believe) means that you construct a surface at every place in the density where the density equals the IsoValue ('iso' = 'same'). So for the default, it opens up your .cube file and looks for the cube coordinates that have density at (or near) 0.00040, and plots a surface that connects these points (this is your 'mesh').

Most likely, your density values are too small large to plot an isosurface at this value. Try making the default density isovalue in GaussView to be something smaller larger and go from there. (Practically, I'd keep shifting an order of magnitude until a surface builds, e.g. try a value of 0.004 next).

Even better, you should look into the .cube file itself to see what your volumetric data looks like. The first few lines define the grid spacing and dimensions of the cube, but then you will see a bunch of floating point values. Here's an example .cube I took from this site (also a good reference):

CPMD CUBE FILE.
 OUTER LOOP: X, MIDDLE LOOP: Y, INNER LOOP: Z
    3    0.000000    0.000000    0.000000
   40    0.283459    0.000000    0.000000
   40    0.000000    0.283459    0.000000
   40    0.000000    0.000000    0.283459
    8    0.000000    5.570575    5.669178    5.593517
    1    0.000000    5.562867    5.669178    7.428055
    1    0.000000    7.340606    5.669178    5.111259
 -0.25568E-04  0.59213E-05  0.81068E-05  0.10868E-04  0.11313E-04  0.35999E-05
      :             :             :           :            :            :
      :             :             :           :            :            :
      :             :             :           :            :            :
        In this case there will be 40 x 40 x 40 floating point values
      :             :             :           :            :            :
      :             :             :           :            :            :
      :             :             :           :            :            :

Look at the value of your floating point numbers at the end. As a starting point, pick a number for your isovalue that is about an order of magnitude less than your largest values, and adjust from there.

If your values are all at or near zero (so, practically speaking, on the order of 1E-8 or lower) then you need to revisit how you are generating the .cube file, since your cube is not capturing any density---then it's a cubegen issue not an isosurface issue.


Why does capillary action occur better in narrow tubes than in wide tubes?

You have to look which forces act where. You have a force that tries pull the liquid up the capillary, which stems from the gain in energy due to adhesion. At the same time, however, gravity acts against this force. In equilibrium, both forces have the same value:

$$F_\text{g} = F_{\text{surf}}$$
$$ mg=\rho Vg= \rho R^2\pi h g=2\pi R\sigma_s \cos(\varphi)$$

Which leads to the formula presented by Babounet in another answer:

$$ h = \frac{2\sigma_s\cos(\varphi)}{R\rho g}$$

So, to boil it down, if you lower the radius $R$, you lower the force due to gravity $F_g$, but the force due to surface energy $F_\text{surf}$ does not fall as fast, which leads to a higher height $h$.

(source, $W=F_g$ here, naming isn't perfect, $F_{\text{surf}}$ is not drawn into)


Effect of chloromethyl substituent vs methyl substituent on yields of nitration

The following is the abstract of K.M. Zhang "Selective Nitration Of Benzyl Chloride And Its Application In The Synthesis Of Nitrobenzaldehyde"(link). The full pdf is behind paywall.

  The selective nitration of benzyl chloride and its application in the synthesis of 4-nitrobenzaldehyde were studied. Nitration of benzyl chloride with $\ce{HNO3}$ in the presence of $\ce{Nb2O5}$(6% weight percent) at 0-5℃ for 1.5 hours gave mononitro products in 94% yield with 77-80% para selectivty. $\ce{Nb2O5}$ was recovered by filtration and reused in the nitration. The yield of mononitro products and para selectivity were above 90% and 76% respectively for 5 times repeated use of $\ce{Nb2O5}$. Nitration of benzyl chloride with urea nitrate(UN) in $\ce{H2SO4}$ was also studied. The nitration was carried out at 25℃ for 24 hours and gave mononitro products in 98% yield with 90% para selectivity. Base catalyzed hydrolysis of 4-nitrobenzyl chloride at 75-80℃ for 22 hours gave 4-nitrobenzyl alcohol in 77% yield with 98% purity. Oxidation of 4-nitrobenzyl alcohol in dilute nitric acid at 55℃ for 3 hours gave 86% 4-nitrobenzaldehyde, compared with industrial oxidation of 4-nitrobenzyl alcohol,whichwas carried out at 70℃for 4 hours and gave 70% 4-nitrobenzaldehyde. 4-nitrobenzaldehyde was also obtained in 86% yield with 92% purity via ionic liquid immobilized $\ce{TEMPO/CuCl}$ catalyzed oxidation of 4-nitrobenzyl alcohol with molecular oxygen in ionic liquid $\ce{[bmim]PF6}$. The reaction condition is mild, the selectivity is high, and no 4-nitrobenzoic acid is detected in any case.


Question about the role of d-orbitals in polarisation

Well, please be more specific about the molecule(s) in question for that quotation.  d-orbitals are certainly involved in some bonding, especially for the transition metals, but you are right that the participation of d-orbitals in the bonding of main group covalent molecules and polyatomic ions has been overstated in the past and is in fact fairly minimal.

As you mentioned, there is very little, if any, d-orbital participation in the bonding of sulfate.  As for d-orbitals being involved in polarization but not bonding, that literally does not make sense for something like sulfate.  Sulfur's d-orbitals are empty, so it is impossible for them to polarize without bonding.  Only when there are occupied d-orbitals (as in transition metals) can non-bonding d-orbitals affect things like the shape and polarity of the compound (as in crystal field theory/LFT).

So, long story short, I think that comment you read is simply wrong, and your original understanding is correct.  See the below references if you want to read more about it.

References

Reed, A. E.; Schleyer, P. v. R. Chemical Bonding in Hypervalent Molecules. The Dominance of Ionic Bonding and Negative Hyperconjugation over d-Orbital Participation. J. Am. Chem. Soc. 1990, 112 (4), 1434–1445. https://doi.org/10.1021/ja00160a022.
Cunningham, T. P.; Cooper, D. L.; Gerratt, J.; Karadakov, P. B.; Raimondi, M. Chemical Bonding in Oxofluorides of Hypercoordinate Sulfur. Faraday Trans. 1997, 93 (13), 2247–2254. https://doi.org/10.1039/a700708f.


The dilithium dimer exists- but it isn’t particularly common, and we usually only see it in the gas phase.

Essentially, the 2s electrons interact and form a bonding orbital. (s-p mixing is particularly pronounced here, but I presume you don’t know MO Theory.)

Due to the diffuse nature of lithium’s atomic orbitals, the bonding is pretty weak (Wikipedia gives a binding energy of $102\text{ kJ/mol}$, about one fourth that for a hydrogen molecule); and the inter nuclear distance between the 2 lithium atoms is actually quite large.


Explanation for increasing wave number trends for hydrocarbons

As orthocresol suggested in his comment, you are correct. First of all, methane does not contain a $\ce{C-C}$ bond, so that answer could not explain the wave number form methane. In addition, you can think of bonds as tiny harmonic oscillators. Specifically, they behave like a mass on a spring, absorbing frequencies that match the compound's frequency of oscillation. Remember that for a mass on a spring, $\omega = \sqrt{\frac{k}{m}}$, where k is the spring's stiffness (related to the strength of the bond) and m is its mass. This is why stronger bonds (such as a $\ce{C=C}$ bond over a $\ce{C-C}$ bond) have a higher frequency, and therefore wavenumber (these two quantities are directly proportional). Likewise, the $sp$ orbitals in acetylene are stiffer than the $sp^2$ orbitals in ethylene, which are in turn stiffer than the $sp^3$ orbitals in methane. This explains why methane has a lower wave number than ethylene, which in turn has a lower wave number than acetylene.


What is the safe level of energy output for hydrogen combustion at home?

First things first: hydrogen gas and oxygen gas do not spontaneously react but need activation. Activation can be provided by a flame, an electric spark or similar.

That said, here is the important part. The reaction will not turn into a fireball but even small volumes will explode with pretty loud bangs. It’s not the loudest bang known to chemistry instructors but it’s certainly something to be aware of – you don’t want your neighbours ringing the police because they think a shooting has been going on.

If you’re going to try this, I recommend starting with the smallest volumes possible, not bigger than a shot glass. My chemistry teacher in high school used a glass cylinder approximately the size of a small (200 ml) water glass. That gave an even stronger bang but the cylinder itself remained unharmed, probably because there was a large opening on one side.

However, the experiment is unlikely to show you much because the water vapour is obviously gaseous and will dissipate rather than being visible anywhere. It would almost be more instructive to identify the two gases as hydrogen (burns when brought to a flame) and oxygen (ignites glimming wood) by the standard procedures used in high school (at least in mine).

Tl;dr: I suggest you don’t try igniting a 2:1 mixture unless you’ve seen it performed recently and know what you’re doing.


Reaction of Silicon Dioxide with Hydrofluoric Acid

It is surprising that Silicon halides behave differently with water according to the choice of the halogen fluor or chlorine. The following reactions are known to happen : $$\ce{SiO_2 + 4 HF -> SiF_4 + 2 H_2O} ......... (1)$$
$$\ce{SiCl_4 + 2 H_2O -> SiO_2 + 4 HCl}........ (2)$$ These equations are exactly the opposite from one another. Why does reaction ($1$) not proceed in the other way like ($2$) ? I don't know, but I would like present a personal interpretation. Maybe it is due to the fact that reaction ($1$) proceeds in two steps, with the intermediate formation of $\ce{H_2SiF_6}$. Maybe ($1$) proceeds like that $$\ce{SiO_2 + 6 HF -> H_2SiF_6 + 2 H_2O}$$ $$\ce{H_2SiF_6 <=> SiF_4(g) + 2 HF } $$$\ce{H_2SiF_6}$ is well known to be formed if $\ce{HF}$ is in excess. And of course HCl cannot form $\ce{H_2SiCl_6}$, because the chlorine atom is too big to act as a ligand this octahedral structure. $\ce{H_2SiCl_6}$ does not exist.
This is a personal idea. I would be pleased to hear about any comment.
Sorry for presenting these lines as an Answer. Alas the text is much too long for a comment...


New York Times Crossword: Is "electrically balanced" in chemistry "apolar" (given), "neutral," or something else entirely?

Apolar is certainly a valid alternative to non-polar.
Either is natural in English as it is for, e.g. "asymptomatic", or its use in chemistry for non-polar has been introduced by non native English speakers quickly translating in their works basically the same word, "apolaire" in French and "apolare" in Italian, for instance.
Let me skip this discussion as matter for another SE site.
However, a quick Google chemical oriented search showed me that apolar is less diffused than what I expected, and seems to be more recurrent in biochemistry and astrochemistry, or as suggested in a comment, computational chemistry.
For the point more relevant to Chemistry SE,
"1) Is "apolar" an accepted or commonly-used word in chemistry as 2) a synonym for "electrically neutral," and if so, in what context has this appeared?"
For 1) see above. For 2) the answer is NOT. It just means, closer to the general definition that OP has found and reported, that a chemical entity has not (electrical) poles, i.e. no excess or partial electrical charges are present over its part.
While speaking of bonds, or simple biatomic molecules, apolar and non polar means that negligible or no electrical dipole moment is present. Still, an ensemble of polar bonds can nevertheless lead, as per vectors summation, to an apolar, or non-polar if you prefere, molecule.
While apolar chemical entities are indeed neutral, the same can be true for very polar uncharged molecules.
Another use of the terms is to classify solvents. In this case partial or overall electrical dipole moments are just one of the many factors influencing the interaction between molecules of different solvents or between them and those of a solute. The interested reader can refer to the section "Other polarity scales" of this Wikipedia article https://en.m.wikipedia.org/wiki/Solvent for more details.
(In the context of the puzzle, the given word definition is somewhat acceptable, because an apolar molecule won't present much of charge whatever finely examined, unless one discerns electrons and nuclei. And the fact that - as for the title rather than the OP text - they gave a somehow vague "electrically balanced" description, makes it quite accurate for apolar, in a crossword.)


Can somebody please explain this statement (about the greenhouse effect)?

"Methane has a large effect (24 times as strong as carbon dioxide per unit mole) for a brief period (having an estimated lifetime of 8.9±0.6 years in the atmosphere),[12] whereas carbon dioxide has a small effect for a long period (over 100 years). Because of this difference in effect and time period, the global warming potential of methane over a 20-year time period is 72."[ Wikipedia, quoting Jain, Briegleb et al ].

Simplified, that means:

$\ce{CH4}$ has a shorter half-life in the atmosphere than $\ce{CO2}$, reducing its effect as a green-house gas over time, but
$\ce{CH4}$ is so much more potent in retaining heat than $\ce{CO2}$ that this more than compensates for its short lifetime, making $\ce{CH4}$ a stronger green-house gas over the long term.


Three different types of dissociation can occur here, the two mentioned by the OP:

$$\ce{K[Ag(CN)2] <=> K+ + Ag+ +2CN-}$$
$$\ce{K[Ag(CN)2] <=> K+ + Ag(CN)2^-}$$

And one where only one cyanide ligand dissociates from the silver:

$$\ce{K[Ag(CN)2] <=> K+ + AgCN + CN-}$$

However, as Klaus points out in his comment dissociation $\ne$ ionization. Ionization is a chemical reaction that generates ions where there were none before. Consider the ionization reaction of ammonia $\ce{NH3}$ in water:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

I'm not sure if this type of reaction is possible, although the cyanide anion is the conjugate base of a weak acid $(\ce{HCN}, \ \text{pK}_a =9.21)$. Thus, any liberated cyanide ion could react with water (assuming this is an aqueous solution!), though I would hardly call this an ionization (as there is no net increase in ions).

$$\ce{CN- + H2O <=> HCN + OH-}$$


Nicolau already outlined it quite clearly in his answer.

  Notice that after mixing the solutions, the number of moles of species containing $\ce{A}$ ($\ce{HA}$ or $\ce{A−}$) is simply $n_A = C_A V_A$

so you have $n_{\ce{HA}} + n_{\ce{A-}} = C_A V_A$

  We will assume that volumes are additive (total volume $V_T = V_A + V_B$)

and given that $[\ce{HA}] = n_{\ce{HA}}/V_T$ and $[\ce{A-}] = n_{\ce{A-}}/V_T$ it should be fairly clear that

$$[\ce{HA}] + [\ce{A-}] = \frac{C_A V_A}{V_A + V_B}$$

as desired.


is bleach residue (after evaporation) safe?

Bleach, especially employed in laundry applications, may have, in addition to the active ingredient sodium hypochlorite (NaOCl), an equal concentration of NaCl (as it is prepared from the action of Cl2 on NaOH) and at times, an alkaline, like lye (NaOH) or Washing Soda (sodium carbonate, Na2CO3). The latter can cause skin rashes and irritation (and will be present on drying).

The rationale for the latter is to keep the pH elevated in aqueous sodium hypochlorite as upon exposure to CO2, there is the possible introduction of hypochlorous acid. While HOCl is more powerful (and reportedly nearly a hundred times more potent disinfectant than the associated hypochlorite) is far less stable (a short shelf-life). Further, the decomposition product is O2 and also HCl (and not NaCl).

The created hydrochloric acid can further react with the sodium hypochlorite or hypochlorous acid as follows:

$\ce{HCl + NaOCl -> NaCl + HOCl}$

$\ce{HCl + HOCl <=> Cl2 + H2O}$

leading to a chain decomposition of the bleach, hence the advantage of adding NaOH or Na2CO3. To quote the cited source:

  The pH has a significant effect on the stability of sodium hypochlorite solutions. Below pH 11 the decomposition of sodium hypochlorite is significant due to the shift in the equilibrium in favor of the more reactive hypochlorous acid. A pH between 12 and 13 gives the most stable solution. This equates to 0.4 to 4.0 grams per liter (gpl) excess NaOH. Greater concentrations will not improve the stability. Excessively high improve the stability. Excessively high alkalinity will damage textiles and retard the bleaching and disinfecting actions of the hypochlorite.

I would suggest that you can even dilute your current NaOCl dosing and add some lemon juice, just prior to use, to create (per the source reference noted above) a much more powerful mix (and a more potent bleaching agent, so be careful where you select to employ it) albeit, with a much shorter shelf-life.

To rapidly remove bleach smell (likely Cl2O from the dehydration of HOCl), one can apply a spray of dilute H2O2, which rapidly converts NaOCl to NaCl and oxygen. Caution: this further promotes bleaching action via intermediate radical formation (see 'Hydroxyl radical from the reaction between hypochlorite and hydrogen peroxide'), so again be careful where you apply it.


Equilibrium constant in terms of partial pressures

However, surely $P_j$ alone is dimensionless as it's a partial pressure defined by dividing a pressure by a pressure - leaving no units. How does that work? Or am I wrong by assuming that a partial pressure has no units?

You are wrong to assume that partial pressure has no units.  Partial pressure has units of pressure, such as mmHg, atm, bar, psi, etc.

For ideal gas mixtures, $y_j=\frac{P_j}{P_{tot}}$ or equivalently $y_j P_{tot}=P_j$, i.e. the partial pressure $P_j$ of a gas is equal to its mole fraction $y_j$ times the total pressure $P_{tot}$.

The expression you gave for $K_p$ is dimensionless because $P^\Theta$ has the same dimension as $P_j$.

  Do I assume that $\frac{P_C}{P^Θ}=1$....

That part sounds good.

  leaving me with $K_p=\frac{P_{CO}}{(P_{O_2})^{1/2}}\frac{1}{P^\Theta}$

Not sure about that one.  How is $K_p$ dimensionless in what you wrote?


Functional variation in derivation of the Hartree-Fock equations

Is it not just because of the fact that $[aa|bb]$ is in chemists notation, meaning that $[aa|bb] = [a(1)a(1)|b(2)b(2)]$ and so you can just switch them around?


Atmospheric Humidity Effects on Non-water Evaporation

Yes, because though the liquid phase initially contains no water, water molecules may enter the liquid phase from the gas phase.  The more hygroscopic the liquid is, the more the evaporation rate will be affected by humidity.  If the liquid is very hygroscopic and has a low vapor pressure, the initial evaporation rate may even be negative in humid conditions, as water enters the liquid phase at a rate greater than liquid enters the gas phase.


Choosing an electrolyte for a galvanic cell

The point of an electrolyte is to allow ion flow between the anode and the cathode in order to balance the charges on either side so the reaction can continue. What you are looking for in a good electrolyte is that it won't react with either of your metals (i.e. won't form a precipitate), and that the ions will readily move (that's what makes soluble ionic compounds a great choice).

With $\ce{ZnSO4}$, for example, you wouldn't want to have $\ce{Pb^2+}$ ions anywhere in your cell, since $\ce{PbSO4}$ would precipitate and your concentrations of your ions would change, therefore changing your overall cell potential.


Why is the outermost part of the Bunsen burner flame not the hottest region?

When you have the intake valve (at the bottom of the bunsen burner) opened, the gas can mix with incoming oxygen. By the time is gets to the top of the tube, you've got a nice continuously-flowing source of well-mixed gas and oxygen. That allows combustion to occur very quickly because the methane and oxygen molecules don't have to travel very far to react, increasing the reaction rate, and therefore the rate at which heat is released.

In contrast, when the intake valve is closed (or when it's open and you're considering the outer non-luminous flame), the oxygen for the combustion reaction comes from the surrounding air, which is not thoroughly mixed with the gas at the time of ignition. That results in a slower reaction, and therefore a slower rate of heat being released.

Periodic videos just did a nice thermal imaging of a bunsen burner with this explanation: https://www.youtube.com/watch?v=in9BGz958jg


Effect of pH on volatility (odor) of organic acid

Acetic acid is a weak acid (pKa 4.8). Even if you neutralize it, some acetate ions will react with water to form acetic acid and hydroxide ion (hydrolysis). Since acetic acid has a quite strong smell, we are able to detect its presence in neutral acetate solutions. If you will increase pH to a high value with a strong base, the concentration of acetic acid will drop (Le Châtelier's principle).


Reactivity of Benzaldehyde between aldehydes

For a carbonyl group to show reactivity toward nucleophilic reagents there should be positive charge on the carbon of the carbonyl group. The existence of positive charge on the carbon allows the reagent to donate electrons much easily.

Alternatively, A compound with more stabilized positive charge will show less reactivity towards nucleophilic reagents. The positive charge could either be stabilized by resonance or electron donating groups. In case of benzaldehyde, the positive charge is highly stabilized (due to resonance) than in other simpler aliphatic aldehydes.


Why should animal fats not be used for cooking?

For one thing, saturated fats tend to have higher melting points, such that at room or body temperatures, they are often solid (though exceptions obviously exist). Therefore, unsaturated fats and oils will be easier to use since they can be poured.

There is also a longstanding, though possibly unjustified, link between saturated fats and oils, and cardiovascular disease. Avoiding saturated fats is supposedly healthier.
Because livestock are more expensive and slower to grow than plants, vegetable or plant oils are easier and faster and cheaper to obtain.
Finally, animal fats often absorb flavour from the animal meat, giving them an unwelcome taste in many contexts. By contrast, vegetable and plant oils either don't absorb anything or the result is attractive pr characteristic in most cooking contexts. So vegetable or plant oils are preferred for versatility and for flavour.


How is the volume kept constant in an NVE MD simulation?

The answer to this is linked to how you define the “volume” of your system, and the boundary conditions. The simplest case is simply to consider an isolated systems, like a molecule in gas phase. This corresponds to $N$ atoms (or molecules or particles) with no restrictions of their coordinates and no boundary conditions. This is essentially an infinite box. Its volume naturally does not and cannot vary.

Note that the volume of the system is not defined as “the volume actually occupied by the atoms”, such as for example the Connolly volume of your system. Instead, the volume of the system is the boundary of space where the atoms can go.

If you take a finite volume for your simulation, for example of cube of size $L$ and volume $V=L^3$. You have to put periodic boundary conditions, for example reflexive (your box is surrounded by a wall, and particles bounce on it) or periodic (look up for periodic boundary conditions in a textbook). Then from the initial positions and velocities of the atoms, Newton's equation allows you to propagate those quantities and determine their evolution in time. But nothing in Newton's law affects the system size $L$ (and thus $V=L^3$) or the boundaries. They will stay fixed and the volume will be constant.

Also, in all cases, because of conservation of matter, $N$ is constant. And if the forces acting on the system are conservative (as are all interatomic forces), then the total energy in the system $E$ is conserved.* Thus, the system samples the $(N, V, E)$ thermodynamic ensemble.

* the momentum of the system is also conserved… and it is not as trivial as it may sound :)


Salt solubility and solvent dissociation

Common ion effect is the key.
A solution of $\ce{NH4OH}$ and $\ce{NH4Cl}$ is basically(no pun intended) a basic buffer solution, which has an almost  constant $\mathrm{pH}$ and thus a fixed concentration of hydroxide ions.
The solubility product of $\ce{Al(OH)3}$ as you say is also constant. So as to maintain the constant nature of both $[\ce{OH-}]$ and $K_{\textrm{sp}}$, there will be a fixed equilibrium concentration of $\ce{Al^3+}$ for a given $\mathrm{pH}$ above which it will start precipitating as $\ce{Al(OH)3}$. By increasing the hydroxide concentration we are forcing the $\ce{Al^3+}$ cation to precipitate earlier than it would have done in a normal, neutral aqueous solution.
This process of forcefully changing the $\mathrm{pH}$ of a solution to alter the solubility of different compounds is very common. In this case it was used to reduce the solubility and precipitate a substance, in other case it might be used to increase the solubility.
For example, during the same qualitative analysis schedule you mentioned, $\ce{HCl}$ is added alongside $\ce{H2S}$ during the detection of group 2 cations(like $\ce{Cu^2+, Hg^2+}$) to prevent precipitation of group 4 cations(like $\ce{Zn^2+, Mn^2+}$) in the form of sulphides. As the acidic conditions increase the solubility of sulphides, added with the fact that group 4 sulphides are not that insoluble prevents them from precipitating with the group 2 cation sulphides.


Why does pH decrease when diluting a base?

You are overcomplicating it. The concept of pH and pOH stems from the concentration of $\ce{OH-}$ or $\ce{H+}$ per solvent unit $\ce{H2O}$, like so $\ce{\frac{[OH-]}{[H2O]}}$ or $\ce{\frac{[H+]}{[H2O]}}$. There is also a logarithm involved, but lets keep it simpler than that. You are adding far more of the denominator than the numerator, hence the fraction is becoming smaller and the pH is going towards pH of pure water.


What conditions can cause a material to exist as a gas?

Generally, when a substance is at > 0 K (0 Kelvin or -273°C) the atoms/molecules making up that substance possess kinetic energy, which means that each atom/molecule are [almost] constantly moving.  If that atom/molecule is in a vacuum, it would move in [almost] a straight line.  However, in the ordinary matter that we encounter everyday, atoms/molecules exist in macroscopic phases which are composed of a humongous number of particles.  Which means that each particle experiences frequent collisions with neighbouring particles, making the trajectory of each particle to appear zig-zag like.
The higher the temperature in the vicinity of an atom/molecule, the higher the kinetic energy of the particle, and the faster that particle's motion becomes.
Also, depending on the properties of the atoms/molecules in the substance, each atom/molecule experiences electrostatic forces (either attractive or repulsive) exerted by neighboring atoms/molecules.
At the interface between a liquid and a gas (e.g., water in a tumbler sitting on a table in a room at sea level), the surface of the liquid is constantly being bombarded by gaseous particles that have much higher kinetic energies than the molecules of the water in the tumbler.  This bombardment imparts enough kinetic energy to some of the water molecules at the surface.  Some of the bombarded molecules gain enough kinetic energy to enable them to leave the water surface to become gaseous water molecules.  Some of the bombarding particles (oxygen molecules, nitrogen molecules, water vapour molecules, etc.) from the air lose enough kinetic energy so that they become stuck into the liquid surface.  At equilibrium (maximum humidity saturation of the air in the room), the number of water molecules leaving the liquid surface is equal to the number of molecules entering the liquid surface.

It is the net result of factors 2, 3 and 4 that determines the phase of a substance.  For example, substances like water, which are composed mainly of polar molecules (molecules possessing an asymmetrical distribution of charge around it, making it behave like a bipolar/tripolar/multipolar magnet), have relatively high melting and boiling points.  Below the melting point, the molecules exist in a [3D] crystalline lattice.  Each molecule are in constant motion but are at the same time held together by electrostatic forces (hydrogen bonds).  Since in a solid, the kinetic energy or velocity of each molecule is not high/fast enough, those particles cannot overcome/escape from the electrostatic attraction exerted by their immediate neighbours.  Therefore, the molecules stay stuck in the crystal lattice, exhibiting their kinetic energy as vibrational motion.

When the melting point is reached, the kinetic energy of the molecules starts to overcome the electrostatic attraction from neighbouring particles, allowing them to move away from their neighbours.  This is the liquid state, in which the kinetic energy of the particles are enough to prevent the particles from being stuck in one place, allowing them to move about, making the phase fluid; however, the kinetic energies are not enough to allow the particles to move farther away from each other.  In the liquid state, the distances between neighbouring particles are almost the same as that in a solid.

For a liquid to become a gas, two factors must be overcome: (a) the intermoleculer forces in the liquid, and (b) the pressure above the liquid surface.  At the boiling point, the kinetic energies of the particles become high enough to overcome (a) and (b) to enable the particles to move farther away from each other, and we get a gaseous phase.

Factor (a) is overcome, for example, by heating the liquid in the container.

You can appreciate the effect of factor (b) when you go mountain-climbing (5000 feet or higher).  Take a thermometer with you to the mountain top.  You will observe that water at high altitudes boil at temperatures lower than 212°F (100°C).

Another way to experience the effect of factor (b) is through the following experiment:  Get a 50-mL (1 to 2-oz) syringe, remove the needle, draw water to about 25% capacity of the syringe barrel, plug the inlet with your index finger, then with your other hand, pull the plunger until it is almost at its maximum position.  You should see the water in the syringe boil without the application of heat.

This seems to be a simple question but a full answer can be appreciated only with some knowledge of statistical thermodynamics.

BTW, the pressure of a gas is the macroscopic result of a humongous number of particles impinging upon a surface (the diaphragm of a barometer, the surface of a manometer liquid, the walls of a container, etc.).


The cold salt water is going to take heat out of the ice cube just as you say, (and so the salt water will be heated a little to an extent depending on their relative amounts  etc.) The ice surface in contact with the salt water will dissolve as the freezing point of the salt water is lower than that of the ice. Eventually all the ice  will dissolve, but this will have taken some small amount of energy to do as explained next. (I assume little ice and lots of solution but whether all the ice will in fact melt will depend on the exact composition and could be found from a temperature vs. mole fraction plot for a solid liquid equilibrium at constant pressure. There could be solid salt-ice & ice, solution & ice, solution & solid salt-ice or just solution.)

Dissolving a solid in a liquid, say salt in water, can cause a drop in temperature (provided the container is insulated) because it takes  energy to dissolve compounds and this can only come from the liquid. (The ions or molecules in the solid must be attracted to one another, by ionic or van der Waals forces, and so to be separated takes energy). The measure is called the heat of fusion (or melting), or enthalpy of fusion which is a positive quantity, for common salt it is $\approx 30 \pu{~kJ mol^{-1}}$ and of ice $\approx 6 \pu{~kJ mol^{-1}}$.


Is there a commercial liquid with refractive index greater than n=2?

Definition of refractive index:

$n=\frac {c}{v}$

If n>2 then this means that the velocity of light in the material is less than half than in vacuum so there must be a strong interaction with the material.

 One of the highest refractory index in liquids is 1.79(20°C) possessed by a solution of sulfur in methylene iodide($CH_2I_2$). Liquids are less dense than solids and this explains why it's difficult to have a refractory index greater than 2.
I found a nature article that says:

  Selenium monobromide, Se2Br2, has a higher
  refractive index than that of any pure liquid hitherto
  recorded. Prepared by direct combination, the value
  for $\nu$ is 1.96±0.01 rising to 2.02 on exposure to the
  atmosphere, owing to decomposition of the bromide,
  with separation and reabsorption of selenium.

https://www.nature.com/articles/133066b0

EDIT: From a more recent paper(2016), selenium monobromide is confirmed as one of the liquid with the highest refraction index(>2). However, diiodomethane is more easily accessible commercially.

  A promising candidate solvent is the organic liquid diiodomethane (CH2I2), which is one of the liquids with the highest
  known refractive index values (n = 1.74). While other high
  refractive index liquids exist (phenyldi-iodoarsine (C6H5AsI2)
  with n = 1.85 and selenium monobromide (Se2Br2) with n=
  2.1), diiodomethane has the key advantage of being commercially available. In addition, diiodomethane is an excellent
  solvent, and many liquid formulations using salts dissolved in
  diiodomethane are reported to increase the refractive index
  and are even available commercially.

https://www.osapublishing.org/ao/abstract.cfm?uri=ao-55-12-3165


Does 4-methylbenzaldehyde give haloform and aldol reactions?

First, Cannizzaro reaction is not given by ketones (your statement says ketones give cannizzaro reaction).

Counter to your analysis, the following reaction will not occur since $\ce{H}$ bonded to methyl group is not sufficiently acidic for hydroxide ion to attack. Refer to $\mathrm{p}K_\mathrm{a}$ values given by Mathew Mahindaratne.

Under the given conditions  P on ozonnolysis gives formaldehyde and 4-methylbenzaldehyde. Both are nonenolizable aldehydes capable of undergoing Cannizzaro reaction. Both compounds do not undergo haloform reaction.

Scheme 1

Under the given conditions R gives acetophenone and formaldehyde. Acetophenone gives haloform test and it also undergoes aldol condensation, but not cannizaro reaction.

Scheme 2

Choice B also meets the conditions given.
Thus choices A & B are correct.


What is the logic behind complex splitting?

Spin coupled nuclei are affected by the spin states of their coupled partners. That is to say, they can detect whether their spin coupled partner is spin up or spin down (for spin 1/2 nuclei), and the energy level transitions are different, depending on whether that other spin is up or down. If a nucleus is coupled to more than one spin, it is affected by the spin state of each of those nuclei. 'Complex' coupling or splitting is a qualitative term used to describe first order couplings that involve more than two coupled nuclei. They can easily be described using standard splitting diagrams.

Consider the example below for the spin system AMX

In the absence of any spin coupling, spin A will appear as a singlet - it has only one energy level transtion available. If it is coupled to spin M, the energy levels of spin A will depend on whether spin M is up or down, and so there are two non-equivalent energy level transitons possible for spin, different in energy determined by the magnitude of the coupling constant. Now, if Spin A is coupled to M AND X, then each of the combinations of the AM transitions will also depend on the spin state of X, and so each of the AM transitions will also have two possible transtions - a total of 4 lines, or what we would call a doublet of doublets.

A useful reference for interpreting complicated, but first order, splitting patterns is by Hoye:

http://pubs.acs.org/doi/abs/10.1021/jo001139v


The temperature rise after neutralisation

One calorie is the amount of energy (heat) needed to raise 1 gram of water 1 degree Celsius at a pressure of one atmosphere. (It is almost constant to 3 significant figures, but there is some temperature dependence...) Also assume that 1 cm^3 of water is 1 gram which is more than good enough for the calculations needed for the problem.

So for the reaction in the problem given, 50 mL of water by 6 degrees is 300 calories.

For the other reactions I'll give calories and temperature rise for product solution.

(a) 600 calories, 12 degrees C

(b) 600 calories, 6 degrees C

(c) 300 calories, 3 degrees C

(d) 300 calories, 1.5 degrees C


Concentration of gas above aqueous ammonia

If the container is open, as shown below, then you have a total mess.

Obviously eventually all the ammonia and water will evaporate into the room eventually. There is absolutely no way to tell how much ammonia gas would be with the volume of the container at any given time, and the ammonia concentration within the container wouldn't ever be homogeneous.

If the container is closed, then you have a somewhat more tractable problem.

However there are two equilibriums to consider. First the gas phase must be saturated with water vapor, and second the ammonia must be distributed between the gas and liquid phases (if any liquid phase remains).

Just out of curiosity, 3x3x1 inchs has a volume of 9 cubic inches, or about 147 mL. So if the air in the lab was at $\pu{20 ^\circ C}$ and completely dry then about $\pu{2.5 \mu L}$ of water would be needed to saturate the volume to 100% humidity.

  Calculation:
  water vapor pressure at $\pu{20 ^\circ C} = \pu{17.5 torr}$
  atmospheric vapor pressure = 760 torr
  molar volume at $\pu{20 ^\circ C}$ and 760 torr = 24.1 liters
  Moles water = $\dfrac{17.5}{760}\cdot \dfrac{0.147}{24.1} = \pu{1.4\cdot10^{-4} moles}$
  water is $\pu{18 g/mole}$ and $\pu{1 g/ml} = \pu{0.001 g/\mu L}$
  $ \pu{\mu L water} = \dfrac{\pu{1.4\cdot10^{-4} moles}\cdot \pu{18 g/mole}}{\pu{0.001 g/\mu L}} = \pu{2.5 \mu L} $

NOTE: If the container is sealed there is still the problem of estimating how long it takes the system to reach equilibrium.


There is no sensible single answer to this: it depends on the chemical (and you really should read the safety notes before using anything or you will cause severe trouble.)

Consider these scenarios.

You are working with a solution of common salt and you spill it on your lab coat. No problem at all; you could eat your dinner off the lab coat as it is the same slat you might sprinkle on food. But you spill a solution containing some mercury salts on your coat, get it cleaned and decontaminated immediately as it will eventually poison you even if you don't use it to wrap your dinner.

Or you spill some ether on your lab coat. Don't stand near naked flames until it has evaporated, but once it has gone, it has gone with no lingering effects. Spill some nitric acid/sulphuric acid mixtures on it and even if you wash it off before wearing it again you might still have the problem that it has created some nitrocellulose, which can spontaneously combust, taking you with it. Equally badly, you spill some tellurium salts on your coat. They might not kill you but you will be shunned for weeks as you will smell strongly of something like garlic, but more offensive.

Bottom line: read the safety instructions and act accordingly: there is no general rule.


In theory and especially synthetically there is no real reason why one should restrict oneself to polysaccharides where all monomers are in cyclic forms. A linear monosaccharide could, for example, act as a bifurcating branching site with two different sugars attached to the aldehyde carbon to build up the required O,O-acetal. Another variant may be to have a methyl group block the second acetal oxygen. In any case, you want to avoid the aldehyde becoming a hemiacetal because these are too reactive and would degrade quickly.

$$\begin{align}\ce{\underset{aldehyde}{R-CH=O}} && \ce{\underset{hemiacetal}{R-CH(OH)(OR)}} && \ce{\underset{acetal}{R-CH(OR)(OR)}}\end{align}$$

You may want to look up acetal and acetalisation in the general organic chemistry textbook of your choice to learn more about why hemiacetals are much less stable than acetals.

In practice, however, linearised monomers in polysaccharides are not interesting. To the best of my knowledge, they do not occur in nature due to the labile hemiacetal intermediate that would be required — but to a much greater extent due to the fact that nature optimised the connection of cyclic monosaccharides into polysaccharides and is fully content with what it has. There is no need to evolve a synthase that is able to shield the highly reactive intermediates from the surroundings and thus it didn’t happen. While the formation of oligosaccharides can also be a spontaneous process given a catalyst, linear monomers in oligosaccharides will not form that way.

For most natural product or bioorganic chemists, it is by far more interesting to synthesise species that are known in nature or logical extensions thereof. While a linearised monosaccharide might qualify as an extension, it is not that logic an extension and thus of much lesser value.


To decide if a molecule is aromatic or not you must know some rules-

The Molecule Must Be Cyclic and the system should be conjugated.
The molecule must follow Huckle's rule i.e. it should have (4n+2) Pi Electrons participating in resonance.

3.The molecule must be planar.

For details see-https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_15%3A_Benzene_and_Aromaticity/15.03_Aromaticity_and_the_Huckel_4n___2_Rule

Now if you try to draw the resonance structure for your 1st molecule then you will see that 4 electrons are participating in resonance at a time. So the system is not fully conjugated. It will be a cross conjugated system.So both the structures you have mentioned are non aromatic.

Stability is a result of aromaticity. You can not comment that if a compound is stable then it is due to aromaticity.

Here is an image showing resonance in para xylylene.

It can be seen from the resonating structures that some of the resonance structures are not aromatic making the molecule non-aromatic in total.

Any recommendations on making the image better are welcomed.


Is Hoffman elimination anti or syn elimination?

There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another. For example, in some elimination reactions, the E1 and E2 pathways can operate in competition with one another. Activation energy is associated with each of these 3 reaction pathways. Whichever pathway has the lowest activation energy will be the major pathway followed. By changing solvent, reaction temperature, the relative strength of the nucleophile, the relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other.

(Taken from Ron's answer, in order to start an explanation)(1)

From Wikipedia:

  The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favour an E1cB mechanism.

So on further thinking about the Hofmann elimination, you notice the following

$\ce{NR_3^+}$ is a poor leaving group (Smith, March. Advanced Organic Chemistry 6th ed. (501-502))
Due to this, the rate of cleavage of the $\ce{C-H}$ bond cleavage exceeds the rate of the $\ce{C-N}$ bond cleavage
In fact, the three well known poor leaving groups namely $\ce{-F}$, $\ce{-SMe^{2+}$, $\ce{NR^{3+}$ all have the same observation

What's the conclusion?

Of course, these undergo $E1cb$ when heated to a suitable temperature

There are sources which infer the same -

Wikipedia
Organic mechanisms by Xiaoping Sun (3)

  Some examples of compounds that contain poor leaving groups and can undergo the E1cB mechanism are alcohols and fluoroalkanes.

  The $\ce{-NMe3}$ group is electronegative, it can activate beta hydrogen to make it acidic, vulnerable to attack by a base. This gives rise to carbanion  character (e1cb) for the transition state.

Hence, $E_1cb$ is the correct answer


Carbon 11 radiolabeled inhaled Beta-2 agonist or corticosteroids, is it possible?

Positron emission tomography (PET) has been taken using $\ce{^{11}C}$- and $\ce{^{18}F}$-labelled reagents, although they are not $\beta_2$-agonists (Ref.1). The abstract states that:

Positron emission tomography (PET) provides three-dimensional images of the distributions of radionuclides that have been inhaled or injected into the lungs. By using radionuclides with short half-lives, the radiation exposure of the subject can be kept small. By following the evolution of the distributions of radionuclides in gases or compounds that participate in lung function, information about such diverse lung functions as regional ventilation, perfusion, shunt, gas fraction, capillary permeability, inflammation, and gene expression can be inferred. Thus PET has the potential to provide information about the links between cellular function and whole lung function in vivo. In this paper, recent advancements in PET methodology and techniques and information about lung function that have been obtained with these techniques are reviewed.

It is clear that unto 2007, there are no evidence for the use of radiolabeled $\beta_2$-agonist or corticosteroids for PET in literature. However, there are other radiolabeled reagents have been use for the imaging. For instance, the mixed-venous content is obtained by imaging the right heart during the infusion, and the alveolar concentration is obtained by imaging the lung, after the blood volume is subtracted from the image using a separate $\ce{^{11}C}$-labeled carbon monoxide ($\ce{^{11}CO}$) inhalation.
References:

R. Scott Harris, Daniel P. Schuster, "Visualizing lung function with positron emission tomography," J. Appl. Physiol. 2007, 102(1), 448-458 (DOI: https://doi.org/10.1152/japplphysiol.00763.2006)(PDF).
C. G. Rhodes, S. O. Valind, L. H. Brudin, P. E. Wollmer, T. Jones, P. D. Buckingham, J. M. Hughes, "Quantification of regional V/Q ratios in humans by use of PET. II. Procedure and normal values," J. Appl. Physiol. 1989, 66(4), 1905–1913 (DOI: https://doi.org/10.1152/jappl.1989.66.4.1905).


Why "initial" corrosion studies only, rather than longer time-spans?

Here's an attempt at an answer -

Initial corrosion may be taken to mean when the electrochemical reactions occurring are the main driving force for the material to corrode, meaning changing chemical and/or physical properties. As corrosion continues, physical factors (stresses, strains, grain defects, crystal defects, etc.) can contribute and dominate the reason why a material might fail. Studying the initial mechanism can lead to solutions that inhibit or prevent corrosion since it is currently not likely you can salvage a heavily corroded piece of metal and return it to its original service.

As for the duration variance of the studies, it could be that mechanisms change with time or that some systems are slow to corrode but still need characterization to allow for the material in question to be more accurately used. This, however, is more my opinion than actual experience.


How does the derivative of concentration make sense?

There are two ways to look at this. In the first, we note that all thermodynamic and kinetic analyses rely on the assumption of very large populations of molecules. The curve of concentration vs time is made of discrete points, but the large population ensures that the change in concentration resulting from addition or subtraction of one molecule is so small that it can be treated as infinitesimal. Thus, we can create a smooth curve by interpolating between points, and the result will be continuous and differentiable. From a practical standpoint, any experiment to determine the concentration will result in discrete points separated by much much more than the change of one molecule, so we are implicitly assuming that the behavior between points can be treated as a well-behaved continuous curve.

The second, and perhaps more mathematically rigorous, view is that we never differentiate the actual concentration v time "curve". Instead, we are claiming that the change in concentration over time can be represented by a function which is continuous and differentiable. The only difference between this and the first view is semantic. Does $[A](t)$ refer to the actual concentration of A or to a function that is a proxy for the concentration? It doesn't matter.

Either way, the key point is that traditional kinetic and thermodynamic analyses assume a population large enough that the discrete behavior can reasonably be treated as equivalent to a continuous function.

[On a side note, there are those who would extend your argument to the $dt$ part of the derivative, as there is no proof that time is infinitely divisible, but that's a discussion for another time and place]


Benzene from cyclopropane-1,2,3-triyltrimethanol in acidic medium

I have not written the steps for carbocation formation, simple dehydration.


Terminological questions in chemistry usually are better answered using IUPAC recommended definitions and the IUPAC Gold Book defines orbital (atomic or molecular) as

  Wavefunction depending explicitly on the spatial coordinates of only
  one electron.

But to be honest, I don't like this particular definition. The problem is that a one-electron wave function is, strictly speaking, a function of not only the spatial coordinates of an electron ($\vec{r}$) but also the spin one ($m_s$). In quantum chemistry we distinguish the so-called spin orbitals, which are one-electron wave functions, from the so-called spatial orbitals, which can be though of wave functions that describes a hypothetical spin-less electron.

The connection between spin and spatial orbitals is not so trivial. In the most general case a spin orbital $\psi$ can be represented as follows,
$$
  \psi(\vec{x}) = \phi_{\alpha}(\vec{r}) \alpha(m_s) + \phi_{\beta}(\vec{r}) \beta(m_s) \, ,
$$
where $\vec{x} = \{ \vec{r}, m_s \}$ is the joint spin-spatial coordinate of an electron, and $\phi_{\alpha}$ and $\phi_{\beta}$ are two different spatial orbitals. What we have here is essentially the linear combination of two possible spin states of an electron ("spin-up" and "spin-down") each with its own spatial probability.

And since OP mentioned in comments the fully relativistic Dirac's treatment of a single electron, it is a good idea to note an alternative way of expressing the idea of a spin orbital: in a form of the  so-called two-component wave function, or Pauli spinor, that represents the state of a single electron in a non-relativistic theory,
$$
    \left| \psi \right>
    \longleftrightarrow
    \begin{pmatrix}
        \phi_{\alpha}(\vec{r})  \\
        \phi_{\beta}(\vec{r})
    \end{pmatrix} \, .
$$
This is to be contrasted with a four-component Dirac spinor that describes a state of an individual electron in a relativistic theory,
$$
    \left| \psi \right>
    \longleftrightarrow
    \begin{pmatrix}
        \phi_{\alpha}(\vec{r})  \\
        \phi_{\beta}(\vec{r})   \\
        \chi_{\alpha}(\vec{r})  \\
        \chi_{\beta}(\vec{r})
    \end{pmatrix} \, ,
$$
in which the last two components describe a "spin-up" and a "spin-down" states of the associated positron.

Now, back to chemistry, we note that in general chemistry we always think of each and every electron as being in a pure spin sate: either "spin-up" or "spin-down", rather than in a superposition of these two. Besides, we assume that spin orbitals come in pairs: for each and every spatial orbital $\phi(\vec{r})$ we have two spin orbitals given as follows,
$$
  \psi(\vec{x}) = \phi(\vec{r}) \alpha(m_s), \quad \psi(\vec{x}) = \phi(\vec{r}) \beta(m_s) \, .
$$


The violation of a certain rule or principle in the following electronic configuration

If we talk only about electronic configurations and rules violation then you are correct. The 2nd one (excited state) is referred for hybridization that is sp2. An excited state  configuration is meant only when the particular atom gets ready or excited for forming a bond. Like boron does it with 3 fluorine atoms to form $\ce{BF3}$. So, if we talk about electronic config and rule violations we must go for ground states exceptional cases are $\ce{Cu}$ , $\ce{Cr}$ and $\ce{Mo}$.


The problem is where the energy for heating comes from not the gas or the mechanism of heating

Rocket engines often use burning fuels to produce thrust because chemicals are a very good store of energy and burning often produces hot gaseous products.

Yes, you can produce hot gases by other means (there are electrically driven plasma rockets and you could, maybe, heat gases with microwaves or similar radiation). But the problem you have is where to get the energy for that process? A self-contained rocket has to have an energy source and that source has to be compact and a slight as possible (a long electrical cord to an electric socket isn't practical in orbit or in space). Just about the best way to store energy is various combinations of reactive chemicals (eg oxygen and kerosine) and the easiest way to turn that into thrust is burning. The only other approach know to work well is to put a nuclear reactor on the rocket as an energy source but that has obvious disadvantages for other reasons.

Rocket scientists have spent a lot of heroic effort to optimise mixtures of (often dangerous) chemicals to achieve better thrust. For a good summary read John D Clark's Ignition where he traces the history of trying to make use of extreme oxidising agents like chlorine trifluoride to get better performance.


Effects of Ionizing a Radioactive Isotope

There is no effect. Imagine you are demolishing a building with high explosives; does it matter whether the small window on the third floor was open or closed prior to the explosion? No, it doesn't. The building will go crashing down all the same.

I believe this is a correct metaphor of the relation between chemistry and radioactivity. Of course the product will be ionized. Quite probably, it will be multiply ionized via Auger events, and also will certainly ionize quite a few other atoms around. Whether it was ionized before is about as unimportant as that small window on the third floor.


The cubic primitive, cubic body-centered and the cubic face-centered are just the three possible Bravais lattices of the cubic crystal system.

In total, we know seven crystal systems, namely (in order of increasing symmetry) triclinic, monoclinic, orthorhombic, tetragonal, trigonal, hexagonal and cubic.

Among these, the triclinic crystal system has the lowest symmetry ($a \ne b \ne c;\ \alpha, \beta, \gamma \ne 90°$), the cubic crystal system has the highest symmetry ($a = b = c;\ \alpha = \beta = \gamma = 90°$).

A careful look at illustrations of the hexagonal closest packing ($7+3+7$ atoms in $ABA$ layers) in a hexagonal prism and a bit of thinking gives the number of atoms:

$3$ atoms in the $B$ layer are not shared by any other prism. They fully count
as $\mathbf{3}$.
$2$ atoms in the centre of the top and bottom $A$ layer are each shared by any other prism (above and below). Each of them counts $\frac{1}{2}$. So there's another $2 \times\frac{1}{2}= \mathbf{1}$ atom.
Top and bottom $A$ layers each have $6$ atoms on corner positions. Each atoms is shared by three neighbors in the the same layer and with another hexagonal prism on top or below. So they count as $2\times6\times\frac{1}{3}\times\frac{1}{2}=\mathbf{2}$ atoms.

The total count thus is $\mathbf{6}$.

These diagrams of the hexagonal closest packing, do not represent the the unit cell as the most simple unit from which the crystal is build by symmetry operations.

Cutting the hexagonal prism (along the axis through the atoms centering the top and bottom faces) into three pieces gives identical rhombohedra. These can be transferred into each other by rotations along the axis described above.


Are there any observable effects of casein binding to capsaicin?

Capsaicin doesn't dissociate in water and can't act as a base (like well-known bases such as Sodium Hydroxide or Ammonia) and won't react with the casein in milk. So there won't be any curdling/precipitation of casein protein. Casein will precipitate through pH adjustment if you lower its pH to the isoelectric point (from 6.6 to 4.6).

Notes:

Casein isn't a base, neither an acid, it has an amphoteric nature.
Bases generally don't react with each other.
Pure capsaicin is a powder, not an oil.


What is defined by part by weight in the definition of equivalent weight?

The term "parts by weight" is how you would specify in a recipe the relative proportions of the ingredients. So if the recipe called for one part by weight of flour and two parts by weight of sugar, you could use 1 ounce of flour and two ounces of sugar (to make a small batch) or 1/2 pound of flour and 1 pound of sugar to make a large batch.

So the "recipe" for water calls for exactly 1.008 parts by weight of hydrogen and 8.000 parts by weight of oxygen. If you use too much of either ingredient, there will be an excess of it after the reaction is run.

"Parts by volume" would apply if you were combining liquids and measuring them by volume rather than by weight.


Solvent for recrystallization of benzoic acid?

It can be difficult to guess at what a given compound will crystallise out from, and often there is a lot of trial and error involved.

Generally, the most 'common' type of recrystallisation involves finding a solvent in which the compound of interest is sparingly soluble at room temperature, but highly soluble at elevated temperature such that slow cooling of a saturated solution affords recrystallised material.

If you look up the entry for benzoic acid in Purification of Laboratory Chemicals, Armagero and Perrin, several solvents are suggested:

  Benzoic acid has been crystallised from boiling water (charcoal), aqueous acetic acid, glacial acetic acid, benzene, aqueous EtOH, petroleum ether (b 60-80C), and from EtOH solution by adding water.


The step that was missing was basically just human error in the creating of the graph (wrong column for x) on top of a 10Dq/B to Dq/B conversion. The Wikipedia graph was correct and the raw data was correct, the creator of the graph on the right (me) made a mistake. My bad.


Emission Spectrum of Hydrogen and Helium?

We have solved the Schrödinger equation for hydrogen-like atoms such as $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, and $\ce{B^4+}$. The energy levels are neatly arranged:

$$E_n = - \frac {E_0} {n^2}$$

where:

$E_n$ is the energy of the $n$-th energy level
$n$ is a positive integer ($1$, $2$, $\dots$)
$E_0 = 13.6~\mathrm{eV}$

This is possible because we only need to consider two particles: the nucleus which has a positive charge and the electron which has a negative charge. We only need to consider the interaction between those two particles. Also, the nucleus is not moving, which simplifies some calculations

However, for higher atoms such as helium, where there are more than one electron, the Schrödinger equation contains two terms for the two nucleus-electron attractions and a term for the electron-electron repulsion. The term for the electron-electron repulsion makes things difficult because both electrons can be moved. That term makes the equation impossible to solve analytically.

For hydrogen, the energy of an energy level is determined solely by its principal quantum number. That means, the $2s$ orbital has the same energy as the $2p$ orbital (ignoring the virtual particle effect).

This is not true in higher atoms such as helium. The energy levels of helium can be seen here:

(The energy levels of hydrogen is on the right of the diagram for comparison.)

Here, the azimuthal quantum number has an effect on the energy of the energy levels, which makes for more possible differences between energy levels, which makes for more lines as well as different on the emission spectrum.


Why only molecules with three or more atoms undergo unimolecular reaction?

I'm guessing this quote is given in the context of classical transition rate theory. Although the statement may seem very strong at first, it is generally true but has some important caveats.
Energy Landscapes
Lets consider a diatomic molecule.  It's geometry can entirely be described by one degree of freedom - the bond length.  As we move along that degree of freedom, stretching or compressing the bond we can develop a surface of the energy of the bond at different distances.  A common model used to approximate this surface is the Morse Potential.

These surfaces are often call "energy landscapes".  Minima in the energy landscape are stable structures - our starting materials and different products.  If you push them a bit they move back down the surface to where they started.
And maxima are our transition states between minima, push a little and they'll move down one side of the barrier or the other and into a minima.
Also, a system can easily explore the parts of the energy landscape with less energy than the system currently has - the "classically allowed region" - like a ball rolling up and down the side sides of a valley.
Over time a system tends to move along the energy landscape to positions of lower energy if allowed time to equilibrate and release energy to the outside - like a ball rolling to rest at the bottom of the valley, losing energy to friction.
So normally, below the dissociation energy, the system will roll around in the bottom of the energy landscape, vibrating the bond.
In classical transition state theory, a reaction happens when the atoms move over the energy landscape  along a particular degree of freedom or group of degrees of freedom.
In order for a diatomic molecule to dissociate, it needs to have enough energy moving along the bond vibration degree of freedom to explore parts of the energy landscape at large $r$.  Which means it has to have a lot of energy in the vibrational mode as mentioned in your quote, an amount about equal to the dissociation energy.  If it mode has that much kinetic energy though then, as mentioned above the molecule will fall apart incredibly quickly.
Changing Energy Landscapes
However, rather that adding thermal energy we can change the shape of the energy landscape.  Many different energy landscapes exist at different energies.  the Difference between different landscapes is the arrangement of the electrons in the molecule.
By shining high energy light at a molecule, you can excite the electrons in it into a different configuration of orbitals, leading to a different shape .  This is what happens in photodissociation.  You excite the molecule into a different electronic configuration, and onto different potential energy surface.  On the new surface the molecule is no longer sitting in a minimum, but near a maximum and the system releases energy by moving down the surface - this time towards extending the bond length and dissociating, rather than towards a stable minimum.


ELSD is an acronym for Evaporative Light Scattering Detector. It is used as a quasi-universal detector for HPLC.
The first step is the nebulization of the column effluents with a dry gas as you have mentioned. If the column effluents contain any non-volatile analyte, then those will form minute particles. The mintue particles become suspended in the gas stream and form an aerosol. The gas stream is made to pass through a focused light beam (usually a diode laser) which is perpendicular to the flow direction.
If aersol is present, the light will be scattered, and can be detected/analysed by a detector photodiode which is placed perpendicular to both the light beam and flow direction.
There is a nice picture that I got from this site:

Reference- Analytical Chemistry, G. D. Christian, P. K. Dasgupta, K. A. Schug, 7th ed., Wiley, 2013


Can anyone help me with this question on vibrational modes of free base porphyrin (C20H14N4)?

If you rotate the molecule by $45^\circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$.

Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. You can ignore the position of the double bonds because it is an aromatic system.

Whenever an atoms lies on a mirror plane or a pure rotation axis, its position will be unaffected by the symmetry operation.


Reduction and oxidation in organic chemistry

In organic chemistry reduction is usually the gain of hydrogen or the loss of oxygen and oxidation is usually the gain of oxygen or the loss of hydrogen.

Reduction of alkenes (gain of hydrogen):

Oxidation of secondary alcohols (loss of hydrogen):

The strength of an oxidising or reducing agent can be evaluated by measuring its reduction potential. Good oxidising agents have highly positive $E^\circ$ values, meaning they are 'more keen' to gain electrons, whereas good reducing agents have highly negative $E^\circ$ values. For example:

$$
\begin{align}
\ce{MnO4- + 8 H+ + 5 e- &<=> Mn^2+ + 4 H2O} &\quad E^\circ &= \pu{+1.51 V} \\ \ce{Cr2O7^2- + 14 H+ + 6 e- &<=> 2Cr^3+ + 7 H2O} &\quad E^\circ &= \pu{+1.33 V}
\end{align}
$$

Since the redox potential for the $\ce{MnO4-}$ system is more positive, $\ce{MnO4-}$ is a stronger oxidising agent than $\ce{Cr2O7^2-}$ under acidic conditions.

The reduction potential is affected by the $\mathrm{pH}$ (along with many other factors — temperature, concentration of reactants — see the Wikipedia article for more discussion). Under neutral or alkaline conditions, $\ce{MnO4-}$ undergoes a different reaction with a less positive reduction potential, meaning it is a weaker oxidising agent under alkaline conditions that under acidic conditions.

$$\ce{MnO4- + 2 H2O + 3 e- <=> MnO2 + 4 OH-} \quad E^\circ = \pu{+0.59 V}$$

Reactive metals, such as zinc, tend to be good reducing agents. Zinc amalgam and hydrochloric acid are used in the Clemmensen reduction, to reduce ketones to alkanes.

$$\ce{Zn^2+ + 2e- <=> Zn(Hg)} \quad E^\circ = \pu{−0.7628 V}$$

Many commonly used reducing agents in organic chemistry, such as $\ce{LiAlH4}$ and $\ce{NaBH4}$, serve as sources of the hydride ion $\ce{H-}$. As an example, in the reduction of ketones using $\ce{NaBH4},$ hydrogen is added across the $\ce{C=O}$ double bond though a tetrahedral intermediate.


Explaining redox reactions to a 6 year old?

I might suggest you take a look at What's Chemistry All About by Frith, Gillespie and Larkum.  (ISBN: 1409547078)  It is a very approachable book that in my mind pays close attention to the science jargon introduced to its audience.  Redox reactions are described in the section "Let's make things happen" under the types of reactions.  A bit of history about where the name oxidation comes from and the safety symbol that you will see on an oxidizing agent are both included in the brief discussion.

The example reactions given are combustion and firecrackers, which I think are appropriate and accessible to a child.

As for experiments to do, I'll suggest another book, the Illustrated Guide to Home Chemistry Experiments by Thompson. (ISBN: 0596514921) There is a chapter on electrochemistry with two experiments that might be of interest to a young learner: the electrolysis of water to make oxygen and hydrogen gas, and the oxidation of iron.  Of course, there's also the "make your battery" type experiments which are always fun.

As for simulations, I would suggest heading on over to the PHet website and have a stroll through all the activities there.  I don't recall off hand what they have for electrochemistry, but there's plenty to keep a budding scientist interested.