How to calculate the vapour density of a mixture?

Was it safe to assume that the amount of substance of mixture equals that of the amount of substance of both the gases?

Amount of substance is additive, just like mass. You have to be careful with volume, though. So yes, it is safe to make that assumption.

  I have to figure out the vapour density of the mixture

I'm not sure you are starting with a safe question, i.e. if the vapour density quantity was intended to describe mixtures. Wikipedia has "vapour density = molar mass of gas / molar mass of H2", and I'm not sure whether a mixture has a defined molar mass. The same source states that vapour density is sometimes based on the density of air rather than dihydrogen gas (to assess whether certain gases collect on the floor or ceiling of a room when released).


How to reduce propyne selectively for the trans product

As you point out, a dissolving metal reduction using $ \ce{Na}$ and $ \ce{NH3}$ in the presence of a proton source such as $ \ce{tBuOH}$ is able to reduce a disubstituted alkyne to the corresponding trans alkene.

The mechanism is understood to go via the following stepwise mechanism:

Both protons on the product  come from the $ \ce{tBuOH}$, rather than the $ \ce{NH3}$ as suggested in the comments, so in theory, using $ \ce{tBuOD}$ would give you the desired product.

One issue with this however, is the ease of exchange between $ \ce{H}$ and $ \ce{D}$– any residual water present would be able to convert $ \ce{tBuOD}$ to $ \ce{tBuOH}$, which would then give you the protonated rather than deuterated product.  Experimentally, you'd probably end up getting a mixture of products with varying degrees of deuteration (mono and di).


The best chromatographic technique to analyse a solution of benzene, toluene, and ethylbenzene in n-hexane

The choice of analytical technique is based on three simple factors: what is available, and how low concentration do you want to go, and how much time do you have? Do you have a gas chromatograph with flame ionization detector or a UV high performance liquid chromatograph or both? This type of mixture is readily amenable to both GC and reversed phase HPLC. The only problem is hexane in reversed phase HPLC, but if less than 1 uL is injected it should be fine.


For what reasons, is isopropyl alcohol widely used as a cleaning solvent?

From this Wikipedia article

  Isopropyl alcohol dissolves a wide range of non-polar compounds. It also evaporates quickly, leaves nearly zero oil traces, compared to ethanol, and is relatively non-toxic, compared to alternative solvents. Thus, it is used widely as a solvent and as a cleaning fluid, especially for dissolving oils.

 You should try Googling first :-)

Also, the antiseptic properties of isopropyl alcohol are an added benefit (over the other lower alcohols).

As you pointed out, methanol is notoriously toxic, so using that as an industrial cleaning agent is a no-no.

Ethanol for industrial purposes is often available as denatured alcohol; which has added impurities to render it toxic and/or alter its taste and smell.
These impurities may interfere with the equipment (you mention the electronics and aeronautics fields) you clean it with.

However commonly available isopropanol isn't denatured (then again, there's no real need to denature it). Since it is lacking in added impurities, it presents itself as a better choice over ethanol for cleaning such equipment.


Why is a squared cosine window function commonly used for 2D spectra?

The difference in window function is a result of the parameter being optimized for the respective experiment.  Exponential multiplication of the Free-Induction Decay (FID) weights the beginning of the FID where the amplitude is at it's greatest.  Weighting the beginning of the FID results increases the signal-to-noise ratio, since the noise should have the least impact with the strong signal.

For 2D NMR, we are usually more interested in resolution to aid in differentiating the various peaks and cross-peaks.  We can always take more scans to try and increase the signal strength, so resolution enhancement becomes the key objective.  A pure sine function offers increased resolution at the loss of signal enhancement.  The shifted sine function (or pure cosine function with SSB=2), provides a good mixture of the two objectives.  The enhancement doesn't weight the early parts of the FID as heavily as exponential weighting, but more than a sine function.

The use of SSB=2 for a cosine function is used most often in 2D experiments where we need a usually more significant amount of signal enhancement, such as TOCSY, NOESY etc.  sine enhancement is used most often for COSY, to reduce filtering out peaks from overlap.

--UPDATE--

To answer some further questions posted in the comments.  Using a gaussian weighting for 2D-NMR data is not as preferable as cosine or sine weighting.  The gaussian weighting has a much more narrow distribution, and rapidly approaches zero.  This can prematurely truncate the FID (by essentially weighting the later portion as zero), which introduces artificats into the resulting spectrum known as "truncation wiggles".

Squaring the cosine (which remember is just the sine function shifted) is done both to slow the increase and decrease of the weighting function over the course of the FID.  At the beginning of the FID, this allows for a greater amount of sensitivity to the early parts, while sacrificing a small amount of resolution.  Additionally, the slow decay also reduces the truncation artifacts discussed in the first question of the update.


What is the difference between cross-linked bond and secondary bond in polymers?

This reference indicates that primary bonds in a polymer are the intramolecular bonds (the covalent bonds) that hold the polymer together. Cross-linking different chains of polymer together does occur through a covalent bond, such as a disulfide bond.

The secondary bonds that help to give the polymer its physical properties are intermolecular forces, such as hydrogen bonding for polymers that contain hydroxyl or primary or secondary amine groups. These secondary bonds can be disrupted by changing solvent conditions or heating, for example, which wouldn't disrupt the primary bonds.

So, I suppose that you could say, loosely, that there are "two kinds of bonds between polymers in solid material" if you mean between the polymer chains before they have an opportunity to cross-link (by a change in pH, for example). After they cross-link, the physical properties of the bulk sample of the polymer will still be affected by secondary bonds, the intermolecular interactions, between the cross-linked chains.

The primary bonds will be stronger and can only be affected by chemical changes, the secondary bonds (intermolecular forces) can be affected by changing the physical conditions.


Difference between actual position of electron and Radial Distribution Probability

Ok, let's start out by looking at the hydrogenic atom. The wave function $\Psi(r, \theta, \varphi)$ can be split into a radial part $R_{n,\ell}(r)$ and an angular part $Y_{\ell,m} (\theta, \varphi )$, so that $\Psi(r, \theta, \varphi) = R_{n,\ell}(r) Y_{\ell,m} (\theta, \varphi )$, where $n$, $\ell$ and $m$ are the Principal, Azimuthal and Magnetic quantum number, respectively, and the functions $R_{n,\ell}(r)$ are the solutions of the radial Schroedinger equation

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

with the nuclear charge $Z$ and the mass of an electron $m_{\mathrm{e}}$.

The probability of finding the electron in the hydrogen-like atom, with the
distance $r$ from the nucleus between $r$ and $r + \mathrm{d}r$, with angle $\theta$ between $\theta$ and $\theta + \mathrm{d}\theta$, and with the angle $\varphi$ between $\varphi$ and $\varphi + \mathrm{d}\varphi$ is

\begin{align}
  | \Psi(r, \theta, \varphi) | \, \mathrm{d} \tau = \left[R_{n,\ell}(r) \right]^{2} \left|Y_{\ell,m} (\theta, \varphi ) \right|^{2} r^{2} \sin \theta \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\varphi \ .
\end{align}

To find the probability $D_{n, \ell}(r) \, \mathrm{d}r$ that the electron is between $r$ and $r + \mathrm{d}r$ regardless of the direction, you integrate over the angles $\theta$ and $\varphi$ to obtain

\begin{align}
  D_{n, \ell}(r) \, \mathrm{d}r &= r^{2} \left[R_{n,\ell}(r) \right]^{2} \, \mathrm{d}r \underbrace{\int_{0}^{\pi} \! \! \int_{0}^{2\pi} \left|Y_{\ell,m} (\theta, \varphi ) \right|^{2} \sin \theta  \, \mathrm{d}\theta \, \mathrm{d}\varphi}_{= \, 1} \\
  &= r^{2} \left[R_{n,\ell}(r) \right]^{2} \, \mathrm{d}r \ .
\end{align}

Since the spherical harmonics are normalized, the value of the double integral
is unity.

The radial distribution function $D_{n, \ell}(r)$ is the probability density for the
electron being in a spherical shell with inner radius r and outer radius $r + \mathrm{d}r$.
For the $2\ce{s}$ and $2\ce{p}$ states, these functions are

\begin{align}
  D_{2, 0}(r)  &= \frac{1}{8} \left( \frac{Z}{a_{0}} \right)^{3} r^{2} \left( 2 - \frac{Z r}{a_{0}} \right)^{2} \mathrm{e}^{-Zr/a_{0}} \\
  D_{2, 1}(r)  &= \frac{1}{24} \left( \frac{Z}{a_{0}} \right)^{5} r^{4}  \mathrm{e}^{-Zr/a_{0}} \ ,
\end{align}

where $a_{0}$ is the Bohr radius.
The graph of those two function can be seen here:

The most probable value $r_{\mathrm{mp}}$  of $r$ is found by setting the derivative of $D_{n, \ell}(r)$ with respect to $r$ equal to zero. From the graph above it can be seen that $r_{\mathrm{mp}}(2\ce{p})$ is smaller than $r_{\mathrm{mp}}(2\ce{s})$ (and if you calculate it you can check that for yourself).

But $r_{\mathrm{mp}}$ is not the quantity you are actually interested in. Because although $r_{\mathrm{mp}}$ gives you the most probable distance of the electron from the nucleus you want to know the average distance $\langle r \rangle$ between the electron and the nucleus as that is the distance you will get when you measure $r$ repeatedly and average over your results.
The radial distribution functions may be used to calculate the expectation values
of functions of the radial variable r. You can get them via

\begin{align}
  \langle r \rangle_{n, \ell}  &= \int_{0}^{\infty} r D_{n, \ell}(r) \, \mathrm{d} r \ .
\end{align}

If you evaluate this integral for the $2\ce{s}$ and $2\ce{p}$ orbitals you get

\begin{align}
  \langle r \rangle_{2\ce{s}} = \langle r \rangle_{2, 0}  &= \frac{6 a_{0}}{ Z } \\
  \langle r \rangle_{2\ce{p}} = \langle r \rangle_{2, 1}  &= \frac{5 a_{0}}{ Z } \ .
\end{align}

You find that the $2\ce{s}$ electrons are on average further away from the nucleus than the $2\ce{p}$ electrons.

So, for the hydrogenic atom your claim that $2\ce{p}$ electrons are further away from the nucleus than the $2\ce{s}$ electrons is wrong. But in real atoms you have more than one electron and those electrons will influence each other. One important effect here is, that the inner electrons (or electron density) screens nuclear charge from the outer electrons. Thus the outer electrons feel a lower effective nuclear charge $Z^{\mathrm{eff}}$ than the inner electrons. The extend of this screening is distance dependent, so $Z^{\mathrm{eff}}
= Z^{\mathrm{eff}} (r)$.
Now, if you look back at the formula for $\langle r \rangle_{n, \ell}$ things will certainly change since $D_{n, \ell}(r)$ will have a different form and the additional $r$-dependence of $Z^{\mathrm{eff}}$ will make things more difficult. Assuming that the general form of $D_{n, \ell}(r)$ will be similar in the real atom, it can be expected that the parts of $D_{n, \ell}(r)$ for small $r$ now have a higher weight in the integral of $\langle r \rangle_{n, \ell}$ than before since these parts of the electron density suffer less from the screening effect. Since $2\ce{s}$ electrons have a small portion of the electron density that lies very close to the nucleus ("penetrates deep into the core region") and $2\ce{p}$ electrons don't this will lead to the observation that $2\ce{p}$ electrons are on average further away from the nucleus than the $2\ce{s}$ electrons.

A more involved explanation of the effects at work in multi-electron atoms can be found at this answer of mine. The important bit starts at the paragraph "Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?"


Which easy-to-acquire clear liquid would be best for testing whether or not a twenty-sided die is balanced?

You don't want a denser liquid. You want a liquid with just the right density so the dice would barely float, almost fully submerged. To this end, you mix the solutions of different concentration and so fine-tune the density, as is beautifully described in the accepted answer to your rpg.SE question.

A denser liquid is worse for your purpose. Ultimately, with a very dense liquid (like mercury) the dice would just sit on one face, much like they do on a tabletop.

In the relatively rare case your dice might be made of different kind of plastic and wouldn't float at all. That's when you genuinely do need a denser solution. Chemistry knows a lot of salts; which of these are available to you is another question. Saturated solution of $\ce{MgSO4}$ has the density of 1.3 or so; $\ce{CaCl2}$ will get you to ~1.4, and other salts farther yet, but they are also more difficult to come by.

As for the reference books with some density values, you have to read those really, really careful to see what they actually refer to, and then read again, just in case. 1.028 looks like the density of seawater; it is salty all right, but that's pretty far from saturation. The solution you use is a lot more dense. 1.44 for hydrogen peroxide looks like the density of the pure compound, which you surely can't acquire (which in turn is probably good for your survival); what you had is probably the 30% solution which is a lot less dense.

As for the density at which your d20 will float... well, how are we supposed to guess that? For all we know, your die may be carved out of wood; also, chances are it is cast in bronze. The resulting figures may vary pretty widely. Upd. So it is plastic, after all. Well, that still leaves a pretty wide room for wiggling. You can't tell the density until you measure it. Chemistry is an experimental science, they say.

So it goes.


Equilibrium between R and S configuration?

Because even though neither the R nor the S isomer is thermodynamically favored (in an achiral environment), the free energy barrier (or activation energy) between those two states is huge. This is the most common case, when R/S interconversion requires breaking and reforming a chemical bond. Then, even though the thermodynamic equilibrium would be to have a racemic mixture, kinetics prevent the solution from reaching this equilibrium state.


How can a conductivity meter be used to determine the equivalence point?

I’m assuming you’re talking about acid/base titrations.

In these, you typically start with an acid (maybe strong) and slowly add a strong base until you reach the equivalence point (classical, using indicator) or are sufficiently past it (for plotting, using a conductivity meter). Of course, switching acid and base is possible but for the ease of discussion I will consider only acids to be titrated with bases.

Say for example your mixture included hydrochloric acid and acetic acid. Hydrochloric acid (strong) can be considered fully dissociated, acetic acid (weak) mainly undissociated in the initial solution. Thus, your solution contains significant concentrations of:
$$\ce{H3O+\qquad Cl- \qquad CH3COOH}\tag{1}$$
Now you start adding a base such as sodium hydroxide. The most significant macroscopic observation is that the concentration of hydronium ions $\ce{H3O+}$ decreases while additional sodium ions are added. At half-way towards having titrated away all hydrochloric acid, you would have the following species in solution:
$$\ce{\underset{less}{H3O+}\qquad Cl- \qquad Na+ \qquad CH3COOH}\tag{2}$$
At the equivalence point of hydrochloric acid, we can consider the significant portion of hydronium to be consumed; remember that only a small percentage of acetic acid is dissociated, meaning most of it is not. Thus, we have the species:
$$\ce{Cl- \qquad Na+ \qquad CH3COOH}\tag{3}$$
Henceforth, any additional hydroxide added will, in a macroscopic and sighlty simplified view, only serve to deprotonate the acetic acid to acetate. Thus, half-way towards the equivalence point of acetic acid, we can observe the following species in significant concentrations:
$$\ce{Cl- \qquad Na+ \qquad \underset{less}{CH3COOH} \qquad CH3COO-}\tag{4}$$
Finally, we will reach the second equivalence point at which all acetic acid has been consumed to acetate, meaning the solution features:
$$\ce{Cl- \qquad Na+ \qquad CH3COO-}\tag{5}$$
Hitherto, we have not observed any hydroxide because it was immediately quenched. Now, however, all species that would be able to quench hydroxide have been consumed. Thus, any additional sodium hydroxide will remain sodium hydroxide, giving us the following species:
$$\ce{Cl- \qquad Na+ \qquad OH- \qquad CH3COO-}\tag{6}$$

Now that the types of species in each solution has been established, which influence the conductivity and how? Water by itself does not conduct electricity but water with dissolved ions does. Therefore, we would immediately expect the solution’s conductivity to rise from point (3) to point (5) since a non-ion is converted to an ion (and sodium ions are constantly being added). This is indeed observed: a slow but steady rise in conductivity from (3) to (5).

The key, however, is to realise what happens from (1) to (3) (which also influences our understanding of (6)). Here, it may seem that hydronium ions are constantly being replaced by sodium ions, meaning that the total ion count should remain constant. Indeed, it approximately does. But hydronium (and hydroxide) is a much stronger conductor than any other ion.

Typical ions would conduct electricity by being drawn towards the anode or cathode depending on their charge. This is also, in general, what happens to hydronium (or hydroxide) ions. However, because they are defined merely by an excess (or missing) proton added to (or substracted from) a water molecule, they have the ability to move much more rapidly following the mechanism below:

$$\begin{multline}\ce{cathode^-\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-OH2+} \\\ce{-> cathode-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}OH2}\end{multline}\tag{7}$$

Because this proton-exchange mechanism (the Grotthuss mechanism) allows both hydronium and hydroxide travel speeds much more rapidly than diffusion control, these two ions will induce a much higher conductivity if present. Therefore, our overall graph will feature:

a high-conductivity starting poing
a strong, linear decrease in conductivity while hydrochloric acid is consumed
a global minimum
a slow, linear rise (as discussed before)
a turning point
a strong, linear increase in conductivity while hydroxide is added.

To find points 3 and 5, the best method is to determine sufficiently many data points of the linear sequences 2, 4 and 6 and extend the straight lines. The crossing points represent the equivalence points 3 and 5.

What I have outlined here with two acids also, naturally, works for a single acid; depending on the nature of the acid you will either not see 1/2 or not see 4.


How to ionize a liquid and apply a force using electrohydrodynamics

Though it is possible to ionize the liquid, it is doubtful you want to do that: the effect of ionization would likely be equivalent to exposing the photopolymer resin to bright light.

A System for Cooling Electronic Elements with an EHD Coolant Flow from iopscience.iop.org describes setting up and testing a pump using isopropanol as the working fluid.

If you do decide to try electrohydrodynamic pumping, either of an atomized spray or EHD pumping of dielectric liquids, take precautions against ignition of the liquid and test whether ionization ruins the photoresist.


How to explain the concept of mol and reactions involving mol quantities

The fundamental misunderstandings here are related to two concepts:

Conservation (atoms are conserved, molecules are not)
A mole is just a number (like the chemist's dozen)

I would start with conservation, then move on to moles.

Explaining why atoms are conserved, but not moles

For second-graders, hands-on activities are probably best, so molecular models or something similar would be a good approach. I like your example using two molecules - if you can get some models, you could try having groups of students start with two (or more - 4+ would make it easier to see) molecules of $\ce{SO3}$. Have them count the number of molecules and the number of each atom. Write the results down where everyone can see them. Then, have them do the "reaction" - let them take apart the molecules and put them back together to make$\ce{O2}$ and $\ce{SO2}$. Again, have them count the molecules and the atoms. Show them that the number of atoms stays the same (for each element and in total), but the number of molecules can change.

Once your students are comfortable with this idea, introduce the concept of moles as a way of counting very large numbers. This is the more difficult part, and in my experience is the "make or break" concept for most chemistry students.

A mole is just a very large number

Goal: The number of atoms in macroscopic samples is huge

You can motivate the discussion by calculating the number of molecules/atoms in very small samples. In my classes (college level) I use the number of molecules in one mL of hydrogen gas at STP, and show that if you counted one molecule per second, it would take longer than the age of the universe (a lot longer) to count them all. Sometimes I use the number of silicon atoms in a bag of sand and show them that it is larger (much larger) than the number of stars in the universe. Other examples would work as well - the idea is to get them thinking about the largest quantity that they can possibly imagine and show them that the number of atoms in even a small amount of matter is much larger than that.

Goal: A mole is just a number

Now you want to link this to moles. I start with the concept of a dozen. I ask things like "how many donuts are in a dozen?" "Ok, how many cookies?" "How many cats?" "How many people" until someone points out that it's always 12. Then I ask them why we say a dozen instead of 12. If they don't figure it out quickly on their own, I tell them that for bakers and their customers, it is a convenient way to count stuff. Then I tell them chemists also have a convenient way to count atoms and molecules, which is good because there are a lot of them! The number they use is called a mole, and similarly to how a dozen = 12 of anything, a mole = 6.022 x 10^23 of anything.

Goal: A mole is a HUGE number

Depending on how good the students are with scientific notation, you might want to actually write out that number of zeroes - it's an impressive demonstration! When there is time, I also like to demonstrate the scale by comparing it to different things as the number of zeroes grows. For example, I say (or have them say) "thousand," "million," "billion," "trillion," for every three zeroes. At around 11 zeroes I start comparing it to unimaginably large things. As examples:

The number of stars in the Milky Way galaxy is about 100 billion ($1 \times 10^{11}$).
The number of sheets of paper that would fit in a stack between Earth and the moon is about 4 trillion ($4 \times 10^{12}$).
The number that would fit between Earth and Mars (at their average distance) is about $2 \times 10^{15}$.
Age of the universe in seconds is $4 \times 10^{17}$.
There are $10^{22}$ stars in the universe.

I think that this is effective because it connects the concept of moles being huge and the number of atoms being huge to a bunch of astonishing facts - people tend to remember weird little surprising facts, and because of the way memories form, the moles stuff gets stored along with it, leading to an artificially created "intuitive" understanding of the size scale of moles and their relationship to counting atoms.

Now you need to tie it all together.

Goal: Moles are used to count atoms and molecules

I connect the concept of a mole being a very large "convenience" number used for counting to the stoichiometry and conservation concepts with a simple exercise. I write out a chemical equation, which in your case would be:

$$
\ce{2SO3 <=> 2SO2 + O2}
$$

Underneath, I write the number of molecules of each, starting with the smallest. So, I would write...

$$
\begin{array}{|c|c|c|c|c|}\hline
\ce{2SO3} & \ce{<=>} & \ce{2SO2} & \ce{O2} \\ \hline
2 \space \rm{molecules} &  &  2 \space \rm{molecules} & 1 \space \rm{molecules} \\ \hline
\end{array}
$$

... and say "If we have two molecules of $\ce{SO3}$, we will end up with two molecules of $\ce{SO2}$ and one of $\ce{O2}$." Then I ask them: "What if I start with 10 molecules $\ce{SO3}$?" and let them figure it out, writing on the board as I go. Next, I ask

"What if I start with 12?"
"What about a dozen?"
"How about 100?"
"A billion?"
"$6.022 \times 10^{23}$?"
"A mole?" - at this point, most of them will have the idea
"10 moles?"

At the end, the board looks like this:

$$
\begin{array}{|c|c|c|c|c|}\hline
\ce{2SO3} & \ce{<=>} & \ce{2SO2} & \ce{O2} \\ \hline
2 \space \rm{molecules} &  &  2 \space \rm{molecules} & 1 \space \rm{molecules} \\ \hline
10\space \rm{molecules} &  &  10\space \rm{molecules} & 5 \space \rm{molecules} \\ \hline
12\space \rm{molecules} &  &  12\space \rm{molecules} & 6 \space \rm{molecules} \\ \hline
1\space \rm{dozen} &  &  1\space \rm{dozen} & 0.5 \space \rm{dozen} \\ \hline
100 \space \rm{molecules} &  &  100 \space \rm{molecules} & 50 \space \rm{molecules} \\ \hline
1 \times 10^9 \space \rm{molecules} &  &  1 \times 10^9\space \rm{molecules} & 0.5 \times 10^9 \space \rm{molecules} \\ \hline
6.022 \times 10^{23} \space \rm{molecules} &  &  6.022 \times 10^{23}\space \rm{molecules} & 3.011 \times 10^{23} \space \rm{molecules} \\ \hline
1 \space \rm{mole} &  &  1 \space \rm{mole} & 0.5 \space \rm{mole} \\ \hline
10 \space \rm{mole} &  &  10 \space \rm{mole} & 5 \space \rm{mole} \\ \hline
\end{array}
$$

At this point most everyone will "get it" - and the advantage of this method is that they sort of figure it out for themselves along the way, which means they will remember it more clearly. If anyone still struggles with the conservation part, you can have them calculate the number of each atom and each molecule for all or some of the examples above, in the same manner as at the beginning of the lesson.


Is there an online database of acid and basic dissociation constants?

There are a number of resources available that provide dissociation constants for acids and bases.  The following are examples:

UC Davis Chemwiki pages E1: Acid Dissociation Constants at 25°C and E2. Base Dissociation Constants at 25°C provide many examples.  These are based on information taken from CRC Handbook of Chemistry and Physics, 85th Edition.

The WikiBook Chemical Principles/Solution Equilibria: Acids and Bases provides a reasonable database of dissociation constants of acids and bases (Tables 5.3 and 5.4), as well as some further information about how the values are calculated.

The California State University webpage p-functions for common acids and bases, also at 25C, of a series of acids and bases.


Alternative PCB etching solutions - Is KMnO4 or NaClO suitable oxidizer?

If you don't want to use the classical solid etchants, such as

Iron(III)chloride ($\ce{FeCl3}$) or
Ammonium persulfate ($\ce{(NH4)2S2O8}$) or
Sodium persulfate ($\ce{Na2S2O8}$)

there's the mixture of hydrogen peroxide ($\ce{H2O2}$) and hydrochloric acid ($\ce{HCl}$) in various ratios.

Typical recipes use

water
30% $\ce{H2O2}$
37% $\ce{HCl}$

in a volume ratio of 1:1:1.

Hydrochloric acid can be considered a "household chemical", it was/is available at DIY stores as muriatic acid and typically used to clean brickwork.

I would advise against mixing concentrated (glacial) acetic acid and hydrogen peroxide solutions of higher concentrations. You might end up forming peracetic acid, which is unhealthy and tends to decompose spontaneously (explode) at higher concentrations.

There's another method worth to be mentioned, although the etchant is usually not a household item.

Cupric chloride $\ce{CuCl2}$ in aqueous solution can be used at room temperature to remove metallic copper from the PCB too:

$$\ce{CuCl2 + Cu -> CuCl}$$

Thinking in terms of transition metal waste management, this seems to be pretty nice, since the resulting copper(I) solution can be regenerated (oxidized) to copper(II) (= the etchant) using hydrochlorid acid and hydrogen peroxide. This conference proceeding provides further information.

As far as the permanganate is concerned, it seems that it has been used to desmear PCBs in industry, but it has been superseeded by plasma treatment to avoid damage to the boards.


Why does hydrogen have a lower ionization energy than fluorine?

Fluorine is bigger, but there are $9$ protons in the nucleus, compared to $1$ in case of hydrogen. So, the outermost electron in fluorine experiences more attraction than what hydrogen's single electron experiences, in spite of larger size of fluorine.

However, the outermost electron in fluorine doesn't experience attraction from all the $9$ protons (nuclear charge). Due to improper shielding, the effective nuclear charge experienced by one $2p$ electron of fluorine becomes -
$\mathrm{ [9 - {(6 \times 0.35) + (2 \times0.85)}]= 5.2}$.

In comparison, the single electron in hydrogen experiences effective nuclear charge of $1.0$

Note: For hydrogen, there is no other electron to shield nuclear charge. So, effective nuclear charge = actual nuclear charge.

Source: To know how to calculate effective nuclear charge visit this Wikihow link.


Comparing the dipole moment of water and ethanol

It's probably due to the slight difference in electronegativity between H (2.2) and C (2.55)

Here's some data on gas-phase dipole moments from the CRC Handbook:

Water - 1.85 D
Methanol - 1.7 D
Ethanol - 1.66 D
n-Propanol - 1.55 D
n-Butanol - 1.66 D

Using some incorrect data, I pointed to a possible odd-even effect. Using the correct gas-phase dipole moments, we see that these are all pretty similar (i.e., +/- ~0.2D).

Certainly water has the highest dipole moment, but H is slightly more electropositive than C, suggesting that as a likely cause.


Can silica gel beads cause silicosis if crushed?

No, it won't. Silica beads dust may cause irritation to the skin and eyes. Unlike silica crystalline,  synthetic amorphous silica gel is  indurated, and so does not cause silicosis.


Conditions for spontaneity of redox reaction among redox couples with equal standard reduction potentials

Yes, it is possible to construct two half cells with the same standard reduction potentials. For example, we can half cells made of copper dipping in copper sulfate solution. The caveat is that if both the cells are under standard conditions, the resulting electrode potential would be zero.
The trick is to use different concentration of copper ions in each half cell. With the help of Nernst equation you can determine which side will be the anode and which side will be the cathode. Read more about "concentration cells" as they are called.


To start your book didn't use significant figures properly. There is no way that the weight can have 3 significant figures in the mass and then conjure 6 significant figures in the molecular weight. To make matters worse the pressure is only given to 1 significant figure. YUCK!

Use PV = nRT to determine the moles of gas present.

$\mathrm{n =}\dfrac{\mathrm{PV}}{\mathrm{RT}} = \dfrac{0.1 \text{ bar} \times 0.03405 \text{ L}}{\text{0.08314 L bar K}^{−1} \text{mol}^{−1} \times 819\text{ K}} = 5.000 \times 10^{-5}\text{ mol}$

$\text{MW} = \dfrac{\text{m}}{\text{mol}} = \dfrac{ 0.0625\text{ g}}{5.000 \times 10^{-5}\text{ mol}} = 1,249.8\text{ g/mol}$

This is of course wrong. It would seem that the pressure should have been 1.00 bar. That would have given an answer of 125 g/mol. Using 31.0 g/mol as the atomic weight of phosphorous would reveal that the phosphorous vapor was $\ce{P_4}$.


How to use Google’s Alpha-fold to predict the structure of a two-protein complex?

Are you asking how to take code developed to find the MFE (minimum free energy) structure of a single protein, and modify it to instead determine the structure of a two-protein complex?
Or are you asking if you can use this code, as is, to fold a concatenated combination of the two proteins, and in turn use this result as a surrogate for the predicted structure of the two-protein complex?
If you're asking the former, I would recommend contacting one of the code developers to ask them for guidance.  It's possible that they have already developed a prototype option to address this.
If you're asking the latter, that approach is problematic, for two reasons:

Folding the concatenated protein would constrain the last amino acid in the first protein to be next to the first amino acid in the second protein.  That's not a constraint that would make sense a priori for a two-protein complex.

You have to ask yourself if, in forming the two-protein complex, you envision both proteins unfolding and then refolding to do a global energy minimization— since that's what folding the concatenated protein would give you (with the additional constraint mentioned in point #1, above).  If, on the other hand, you would like the proteins to each fold into their respective MFE structures, and then (while essentially retaining those structures) find the MFE configuration for the two-protein complex, then folding the concatenated protein clearly would not give you that.


Does the amount of solute affect rate of solvation?

If the answers are to be exclusive, then obvious choice is agitating the mixture.

If the answers are not to be exclusive, then agitating the mixture + increased surface by adding more solute increases $\frac{\mathrm{d}c}{\mathrm{d}t}$

Adding more solute does not increase rate of solvation,  but as the absolute surface increases, it increases $\frac{\mathrm{d}c}{\mathrm{d}t}$.

Note also without affecting solubility from your comment, so increasing temperature is out.


Can an element have a third energy shell without completely filling the second energy shell?

Yes, it is possible. When an element is excited, some electrons from lower shell jumps to higher shells. For eg, in-ground state suppose an element has electronic configuration
$\ce{(1s)^2 (2s)^2 (2p)^5}$ . If we energize it, then some of the electrons jump up for a short period of time(till the energy is available) . The config becomes $\ce{(1s)^2 (2s)^1(2p)^6 }$.


What is the difference between an assay and an extemporaneous preparation?

An assay is an investigative (analytic) procedure for qualitatively
  assessing or quantitatively measuring the presence, amount, or
  functional activity of a target entity (the analyte).

E.g percentage purity of a given substance, viscosity, bulk density, total organic content etc.

The analyte can be a drug, a biochemical substance, or a cell in an organism or organic sample.

Extemporaneous preparation refers to the preparation (mixing, assembling, packaging and labelling) of a medicinal product by a pharmacist (or licensed practitioner) according to prescription requirements for a specific patient . Extemporaneous preparations  are made in  relatively small quantities  for short-term supply and use.

For example: Magesium trisilicate mixture can be extemporaneously prepared from the following:

Magnesium trisilicate
Light magnesium carbonate
Sodium Carbonate
Concentrated Peppermint emulsion
Double Strength Chloroform water

(Magnesium trisilicate mixture is an oral suspension containing 5% w/v of these excepients in suitable vehicle.)

Guidelines of extemporaneous preparations can be found in the Pharmacopoeia.

Where you want to determine the content of say one of the excepients , is assaying. For example you may wish to experimentally find the actual content of magnesium in the mixture, you may perform titration with a suitable standard solution, then you can actually be able to calculate its percentage weight in volume


According to Chemical Demonstrations: A Handbook for Teachers of Chemistry (Vol 3, p226), a similar effect can be obtained for copper using copper(II) acetate, ammonia, and hydrazine.

  The copper metal forms when hydrazine reduced and ammonia complex of
  copper(II) in basic solution. Hydrazine slowly reduces the complex to
  copper metal. When the surface of the test tube is very clean and wet,
  the copper metal will adhere to it, forming a copper-colored mirror.

  The overall reaction that occurs in this demonstration can be
  represented by the equation $\ce{2[Cu(NH3)4]2+_{(aq)} + N2H4_{(aq)} +
> 4OH-_{(aq)} -> 2Cu_{(s)} + N2_{(g)} + 4H2O_{(l)} + 8NH3_{(aq)}}$

Whether or not this can be described as "about the same effort" depends entirely upon the access you have to hydrazine, and to appropriate handling/safety equipment for working with the reagents. While hydrazine in solution is -- if I recall correctly, but I am not an MSDS -- considerably more stable than pure hydrazine, it's still toxic and corrosive (among other things).


Is strength of an acid proportional to concentration or dilution?

This is an interesting question. It is sort of like being asked "What is your size?". If you are buying shoes then size means one thing, yet it is something different if you are buying pants.

So there are two different "strengths."

Strength can mean how much acid is present. So per unit volume of solution, a solution with 50% concentration has half the strength of the 100% concentration.

Strength can also mean how acidic, in terms of pH, would a specific solution be if the acid concentration were kept constant. In this sense hydrochloric acid is a stronger acid that acetic acid.

So the gist of your problem is that you're trying to draw a comparison between two different uses of the word strength when such a comparison is pointless.


The meaning of the word "equivalent" is relative to something. Fortunately, as a chemical term, that is still true. An equivalent is more completely called "a reaction equivalent" and is relative to the reaction. There are a few standard reactions (which are typically industry or discipline based) and it might be used usefully in limited (well constrained) situations. The problem is, especially when dealing with reactive gasses, that the words "uncontrolled" and "detonation" might also occur when explaining that you chose the wrong standard - the correct standard reaction is NOT always obvious or determinable in context. So, my best advice is that when someone uses "equivalent" ask what reaction, specifically, they are referring to. And when using the word yourself, always note the context. Examples: NCO equivalents, H(0) equivalents, H(+) equivalents may all refer to the same compound but would have different values.


Giving a cursory Google e-book search gave me results particularly from old textbooks of 19th century:

Phosphate of soda (Sodium phosphate, $\ce{Na3PO4}$) - was sold under the name of tasteless salt
Sulfate of lime (calcium sulfate, $\ce{CaSO4}$) - it was named agustine because this salt was considered to have no taste. It was considered to mix with acids to form tasteless salts
Phosphates of barytes (barium phosphate, $\ce{Ba3(PO4)2}$) - considered tasteless
Phosphate of strontium (strontium phoshate, $\ce{Sr3(PO4)2}$) - considered tasteless
Superphosphate of lime (calcium superphosphate, $\ce{Ca(H2PO4)2}$) - at high temperature, it melts to a semi-transparent glass which is tasteless.
Sulfate of strontium (strontium sulfate, $\ce{SrSO4}$) - considered tasteless

@Ivan said in a comment(deleted) that calcium sulfate, barium phosphate, strontium phosphate and strontium sulfate are insoluble in water, so they cannot serve the purpose of breaking the emulsion. Finally as OP clarifies, calcium superphosphate is corrosive and hence not safe to eat, it all comes down to sodium phosphate which is both tasteless and edible.
References(in order of mentioned points)

London Medical Gazette: Or, Journal of Practical Medicine, Volume 18, 1836
The Critical Review: Or, Annals of Literature, Volume 33
Tobias Smollett, 1801
Pantologia: A New Cabinet Cyclopaedia, Comprehending a Complete Series of Essays, Treatises, and Systems, Alphabetically Arranged, J. Walker, 1819
The Cyclopædia, Or, Universal Dictionary of Arts, Sciences, and Literature, Volume 39, Longman, Hurst, Rees, Orme & Brown, 1819


In general, you should never assume a need for hybridisation unless you have a bonding situation (such as a tetrahedron or related tetravalent structures) which cannot be explained without hybridisation.
As soon as you have a lone pair on any atom, it is more favourable to have this lone pair in an unhybridised s orbital rather than in a hybridised $\mathrm{sp}^n$ orbital. This can be seen for example in phosphane, hydrogen sulphide, etc. in which practically unhybridised central atoms are possible and thus the bond angles show pure p orbitals taking part in the $\ce{A-H}$ bonds.
If you consider very small central atoms (e.g. nitrogen in ammonia) or very large ligands (e.g. chlorine in $\ce{PCl3}$), then having a perfectly unhybridised central atom — meaning a bonding angle of $90^\circ$ — would introduce too much steric strain between the ligands. Thus, the overall system is more favourable if the central atom’s bonding orbitals are partially hybridised, i.e. if the bond angles are enlargened slightly to reduce steric stress. The final bond angle and thus the final central atom’s configuration is a carefully balanced compromise between a hybridisation with a p contribution as high as possible and the ligands separated maximally.

By the way, chlorine is practically never hybridised and it is most certainly unhybridised in $\ce{ClF3}$. Rather, this molecule should be understood as having:

a single $\ce{Cl-F}$ σ bond using a p orbital both of chlorine and of fluorine

a four-electron-three-centre bond using a p orbital of each of the three atoms that form the linear $\ce{F\bond{...}Cl\bond{...}F}$ fragment (i.e. a p orbital of chlorine). This bond can be understood with the following resonance structures:
$$\ce{F-\overset{+}{Cl}\bond{...}\overset{-}{F} <-> \overset{-}{F}\bond{...}\overset{+}{Cl}-F}$$

one chlorine lone pair in a p orbital perpendicular to all $\ce{C-F}$ bonds

one chlorine lone pair in an s-type orbital.


What happens to the NADH produced by lactate oxidation?

What you have read about the direct energy costs/benefits of glycolysis and gluconeogenesis  is not wrong. Glycolysis produces 2 ATP and gluconeogenesis consumes 6 ATP.

The net reaction for coversion of glucose $\ce{C6H12O6}$ into pyruvate $\ce{C3H3O3-}$ by glycolysis is (note that I am keeping track of the charges and the magnesium counterion/cofactor on ADP/ATP):

$$\ce{C6H12O6 + 2NAD+ + 2[MgADP]- + 2HPO4^{2-} ->}$$
$$\ce{ 2C3H3O3- + 2NADH + 2[MgATP]^{2-} + 2H+ + 2H2O}$$

For conversion of pyruvate back to glucose by gluconeogenesis (note that gluconeogenesis usually uses 4ATP and 2GTP - for simiplicity let's consider that a GTP has about the energy equivalence to an ATP):

$$\ce{2C3H3O- + 2NADH + 4[MgATP]^{2-} + 2[MgGTP]^{2-} + 6 H2O ->}$$
$$\ce{C6H12O6 + 2NAD+ + 4[MgADP]- + 2[MgGDP]- + 6HPO4^{2-} +2H+} $$

If you compare the two processes directly, in absence of the source/fate of any of the components, you see that gluconeogenesis consumes 6 ATP/GTP and 2 NADH while glycolysis produces 2ATP and 2NADH. The NADH consumption and production is balanced out.

  Why would we compare these two directly and not worry about coupled processes?

Not all organisms undergo oxidative phosphorylation. Most do, but not all.
More importantly, in the context of your question, NADH can do other things. Not all NADH goes to oxidative phosphorylation.
Most importantly, the direct comparison is carbon balanced with the same organic species: glucose and pyruvate. If we try to include all possible variations and coupled processes we find ourselves eventually trying to do a general chemistry level stoichiometric and thermochemical analysis of an entire life-form. I'd rather not. There are lots of feed stocks into gluconeogenesis besides lactate and pyruvate. An amino acid or lipid that can be catabolized into acetyl-CoA can be fed into gluconeogenesis.

  As for lactate...

True, pyruvate can be generated from lactate $\ce{C3H5O3-}$, which also produces NADH:

$$\ce{C3H5O3- + NAD+ -> C3H3O3- + NADH + H+}$$

If we start with lactate, we have no net generation of NADH in gluconeogenesis. However, we are also no longer comparing the interconversion of the same carbon species. Lactate is produced from pyruvate (consuming NADH), so we are back to square one. Including lactate in the comparison instead of pyruvate leads to no net NADH production/consumption.


X-Ray Diffraction: Phase Identification Strategy

I don't have much information on your particular system: are they nanocrystals?  Are you expecting just one or two phases?  Dozens of phases? Do you have a lot of amorphous material present?

I have one publication you can look at which addresses a rather complicated sample with amorphous, nanocrystalline, and crystalline objects.  It is (hopefully) overkill for your sample, but it will give you ideas on how to partition the problem (and some of the references may help too):

Gainsforth, Z., Brenker, F. E., Simionovici, A. S., Schmitz, S., Burghammer, M., Butterworth, A. L., et al. (2013). Stardust Interstellar Preliminary Examination VIII: Identification of crystalline material in two interstellar candidates. Meteoritics and Planetary Science, 49(9), 1645–1665. doi:10.1111/maps.12148

You may also find it helpful to play with simple fitting softwares such as CrystalMaker/CrystalDiffract where you can try out different mixtures and fits.  Sometimes the peak-finding softwares are a little simplistic with a large number of false positives.  If you try manually combining the possible phases, you can probably rule many of them out and simplify your list.

A few tips:

1) Look for sets of peaks with FWHM that are similar.  This works with nanocrystalline materials.  If you have a nanocrystal with a larger crystal, then the larger crystal will have sharp peaks and the nanocrystal will have broad peaks.  Then you can select which peaks apply to Phase A and Phase B and simplify your search space drastically.

2) If Peak A is present in your experimental data, and you get a "hit" for phase X, but Phase X also predicts Peak B but you don't see it in your experimental data, then you can rule Phase X out.  You may need to understand the characteristics of the phase in a little detail because you can have situations where one atom substitutes for another and you get a change in peak intensities or positions.

3) Do you know the chemical composition of your sample?  If one of your "found" phases is a Yttrium based superconductor, and you're pretty sure you don't have Yttrium, then you can drop that phase.

4) Try looking at it with an electron microscope (SEM or TEM).  This will provide basic constraints as to whether you have many phases, or just a couple, and how big the crystals are.  This is a lot of work, but maybe less work than spending the next month chasing false positives.  TEM can also give you diffraction on any phase, no matter how small, and will be pretty unambiguous.

5) You can also provide constraints using techniques such as XANES, NMR, FTIR, etc., depending on what kind of sample you have.  Which of these are useful is also dependent on the resources you have present at your university/company/facility.

Hope that helps!


Effect of water viscosity (increased due to NaCl concentration) on dissolution rate of CO2

By dissolving NaCl you increase the ionic strength of the solution, i.e. there is more electrostatic attraction between the layers of water increasing the viscosity. Carbon dioxide is non-polar, so by increasing the polarity of the solvent, the carbon dioxide tend to form less interactions with the solvent.


Why do the protons in pentacarbonylmethylmanganese show up at –0.21 ppm?

Unfortunately, you don't give an indication of what you expect the chemical shift for the -Me to be, other than higher than is observed. However, you are correct in your assumption; the π back donation from the Mn centre to CO ligands does decrease the shielding, and therefore does increase the chemical shift, but this is a good case of how chemical shifts are all relative.

The dominating factor in shielding in d-orbital transition metal complexes comes from the paramagnetic ring currents within the incomplete valence d-shell. The circulating electron in the d-orbital produces a much greater angular momentum than for a p-orbital complexes (typically organic molecules, say). The influence of the d-orbital ring current induces a very large diatropic shift for ligands close to the metal centre. These can by very large for hydrides; up to -50ppm in some cases, and still significant for methyl ligands.

So, your starting point for the chemical shift of the -Me ligand is something that has significant shielding, and would be largely negative. What you would expect, based on your argument, and indeed what is observed by experiment, is that as you increase the number of π accepting ligands around the metal centre, you get an increase in the chemical shift of your -Me ligand. For example, as you go from M(PR3)5(CO)1 to M(PR3)1(CO)5, there would be a general trend to deshielding and an increase in the observed chemical shift.

Relative changes in chemical shift in transition metal/ligand chemistry are a great tool for determining stereochemistry and relative positioning of other ligands; a trans substitution will have a greater influence on the shielding than cis, for instance, however the overall dominating factor in the local shielding of your observed -Me still remains the diatropic shift from the metal centre.


Suppose a heated 5.00 g sample of sodium bicarbonate lost a mass of 1.00 g, what percent of impurity is in the sample?

First there is a big gotcha. We'll let $x$ be the grams of sodium bicarbonate and $y$ be the grams of the impurity. Since we were given no information about the impurity we must assume that the impurity neither loses mass nor gains mass when heated with the sodium carbonate. So: $$x + y = 5.00$$

Well the balanced equation is really:

$\ce{2NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)}$

so 2 moles of $\ce{NaHCO3}$ yields 1 mole of $\ce{Na2CO3}$. Thus the mass conversion factor is: $$\dfrac{0.5*\text{MW(}\ce{Na2CO3}\text{)}}{\text{MW(}\ce{NaHCO3}\text{)}} = \dfrac{0.5*106.0}{84.0}= 0.6310$$

So we now also have: $$0.6310x + y = 4.00$$

Subtracting the second equation from the first we get: $$0.3690x=1.000$$ so $$x=2.71g$$ and $$\text{Purity} = \dfrac{2.71}{5} = 54.2\%$$ $$\text{Impurity} = 100\% - 54.2\% = 45.8\%$$


Calculate the weight percent of limestone in a rock sample

The question is a bit misleading ;)

Your calculation of the moles of $\ce{CaCO3}$ would be correct if the rock sample would be pure 100% calcium carbonate. But then, all the titration would be a waste of time: There's 100 wt-% $\ce{CaCO3}$ in pure $\ce{CaCO3}$.

So, there's apparently something else in the rock sample (that doesn't disturb the titration).

However, what you know for sure is:

the stoichiometry of the reaction between $\ce{CaCO3}$ and $\ce{HCl}$. With other words: How many moles of $\ce{HCl}$ are needed to react one mole of $\ce{CaCO3}$
concentration and volume of the $\ce{HCl}$ used in excess to neutralize all the $\ce{CaCO3}$
concentration and volume of the $\ce{NaOH}$ to neutralize all the $\ce{HCl}$ that is left after reacting all $\ce{CaCO3}$

HINT:

Did you consider to solve the back titration experiment by starting with the second neutralisation step?

   If 11.56mL of a 1.010 M $\ce{NaOH}$ was used to neutralize the remaining $\ce{HCl}$, how much was left of it? Compare that with the initial amount...


How to calculate ion concentrations for a precipitation from the solubility product constant?

The concentration (the activity, to be precise) of solids is assumed 1. [Au+][I-]/[AuI]=[Au+][I-] = Ks. Here Ks is the solubility constant. If water originally didn't have any I- or Au+, then the amounts of [Au+]=[I-]=x. Then you have x*x=Ks and x= square root of Ks. The amount of AuI doesn't seem to play any role because they didn't specify the amount of water it is dissolved in.


My chemistry teacher Mr Ong Chiau Jin answers:

  Why are all the orbitals that have the same principal number in
  Hydrogen degenerate?

  In hydrogen, all orbitals with the same principal quantum number 'n'
  (1,2,3...) are degenerate, regardless of the orbital angular momentum
  quantum number 'l' (0,1...n-1 or s,p,d,). However, in atoms with more
  than one electron, orbitals with different values of l for a given
  value for n are not degenerate. Why is this? Surely the radial
  distribution functions are similar for hydrogen (in that there's still
  penetration of orbitals and so on). Or is it that orbitals with
  different values of l are degenerate for a value of n greater than
  would be occupied in that particular atom's ground state?

  The answer is right there in your question. It's only the interaction
  of multiple electrons in an atom like He, Li, Be, etc. that makes the
  different angular momenta wave functions differ in energy. Consider
  this. For the one electron system, why should a p or d orbital differ
  in energy from an s? What makes them differ? In the multi-electron
  case, the p orbitals have different spatial extent, different angular
  components, so the electron density caused by an electron in that
  orbital will interact differently with the other electrons. In other
  words, you need to have more than one electron for the "shape" of the
  p and d and f orbitals to matter to the other electrons. In the H
  atom, there's only one electron, so there's no electron-electron
  repulsion to differentiate the s, p, and d orbitals.

This may not be the perfect answer but it does capture the essence of the perfect answer, which is the idea of multi-electron interactions. Hope this helps.


Can these changes be reversed by changing temperature?

Both processes involved the denaturation of proteins and complex reactions so is impossible reverse them only with a change of temperature.

Boiling an Egg
Although in most case denaturation is not reversible there are some cases where it is.
However in the case of Egg white denaturation is irreversible (see here) and so the whole process is by the way irreversible.
Cooking batter
The same for the batter even if is unclear which flours are used, and the composition of the batter  you can be sure that after cooking batter on a hot griddle surely some irreversible reactions occur: denaturation, Maillard reaction, Amadori rearrangment.


Yes

As the reaction is written above, you may incorrectly conclude the bromine is in a liquid state throughout and hence is omitted from the equilbrium constants.

However as the question reminds us bromine is highly volatile and significant amounts of gas will accumulate above the liquid, hence most of the reactions will be between gas molecules of bromine and chlorine rather than gas and liquid.

The vapour pressure tells us the partial pressure of these gas.  Liquid bromine will evaporate until the sealed vessel contains bromine vapour with a partial pressure of 0.1 atm.

This means it is actually the equilibrium constant for the all three gases:
$$K_p=\frac{[\ce{BrCl}]^2}{[\ce{Br2}][\ce{Cl2}]}$$
where $[\ce{A}]$ is the partial pressure of $\ce{A}$.  As the vapour pressure of $\ce{Br2}$ is known to be 0.1 atm, that is the partial pressure of bromine you use in your equilibrium constant.

Hence the minimum amount of bromine to add to establish any equilibrium with the liquid is the point where partial pressure of bromine is 0.1 atm and all the bromine is in the gas phase.  Any more bromine added will condense out back into liquid.


Why even the most volatile solvents leave a residue after evaporating?

If the surface material is clean and neither soluble nor reactant in any way with the solvent, then it is probably a solvent purity issue. If you have a good micro-balance, you could pour a known volume of the solvent into a pre-weighed light-weight, i.e. aluminum foil, container, dry it, and re-weigh. From that you can calculate the non-volatile solids content of your solvent and compare it to the specifications.


How much heat is required to raise the temperature of 230 g CH3OH(l) from 22.0 to 30.0 ∘C?

The first thing to be sure of is that you're doing all the conversions correctly. Molar mass is the number of grams in a mol. It's found by adding the molar mass of all the elements in a compound times the number in the compound. For example the molar mass of $CH_4$ would be $$\textrm{Molar Mass of }C + 4\times\textrm{ Molar Mass of }H$$
Once you've found the correct molar mass it should have units of mass per mol - most commonly $\textrm{grams mol}^{-1}$. Thus if you have the mass of a substance and the mass required to have a mol, you can divide mass by molar mass to find moles. Then you simply:
$$Q_{heat} = \Delta T\times \textrm{ number of moles } \times C_p$$
Where $C_p$ is heat capacity. To find the heat of vaporisation again you just multiply the per mol value by the number of moles:
$$Q_{vap} = \textrm{ number of moles } \times \Delta_{vap}H^\circ$$
Where $\Delta_{vap}H^\circ$ is the heat of vaporisation. Be sure you got all these steps right and be aware of the difference between kJ and J! ($1\textrm{ kJ} = 1000\textrm{ J}$) Using this method I got a very different answer to yours. Feel free to post your own working for us to look at!

Note: Sometimes enthalpy of vaporisation values are actually the energy to vaporise from 298K (24.5C) but this can't be your problem because, as Jason Patterson said, even with just the enthalpy of vaporisation, no heating, your value is too low.


Perhaps an example of what you mean can be seen with the noble gases. As you go from helium to neon to argon, etc. you are adding inner electron shells and making the electron cloud around each atom more polarizable. That in turn increases the van der Waals attraction between atoms, resulting in a dramatic increase in boiling points down the series.

You might be able to make an atom more polarizable (and increase van der Waals forces) by electronically exciting it to a higher energy level. Can anyone else elaborate on that possibility?


Electron affinity of nitrogen and exchange energy

$ \rm N^{-}$ has the outer electron structure of $ \rm 2s^2 2p^4$. This means we have coulombic repulsion in one of the p orbitals which means adding a single electron will be endothermic. This accounts for the decrease in ionization energy between N and O.

When you get to $\rm N^{3-} $ we have 6K units of exchange energy which is enough to offset the coulombic repulsion.

You see this effect when considering the electron configurations of the transition elements.


Is selenium disulfide polar or non-polar?

I understand your assumption of selenium disulfide being a simple molecular like that. It is similar to the mistake of silicon dioxide, which at first glance we might assign it to be a linear molecule like carbon dioxide. However the crystal structure of silicon dioxide is far more complex.
In selenium disulfide, the selenium and sulfur atoms form a 8-membered ring, similar to the 8-member rings you would find in certain allotropes of either elemental selenium or elemental sulfur. The ratio of $\ce{Se:S}$ atoms is not exactly $\mathrm{1:2}$ as the chemical formula implies (that would give $\mathrm{\frac{8}{3}}$ atoms of $\ce{Se}$ and $\mathrm{\frac{16}{3}}$ atoms of $\ce{S}$ per ring). Indeed the ratio is closer to $\mathrm{3:5}$ ($\mathrm{3:6}$ is $\mathrm{1:2}$ anyways). A better representation of the compound is $\ce{Se_{n}S_{8-n}}$.


The reversible work is $W=\int{PdV}$ over the path of the process.  This integral is not zero for the cyclic process (involving V changing) even through $\Delta V$ from beginning to end of the process is zero.  So, whoever told you that $W = 0$ when $\Delta V=0$ is wrong.


The key difference between graphite and carbon fibers is the degree (or range) of ordering.  In graphite there are large, extensive planes of carbon stacked one atop another resulting in long-range order. In the fibers, there are small ribbons of graphite-like (it is not all carbon, due to the way the fibers are prepared there is still some nitrogen present in the molecular array) material oriented along the fiber axis and there is only short range order.  As a consequence graphite with its extensive 3 dimensional planar array of ordered carbon atoms is crystalline.  Whereas the fibers, being ordered over a much smaller range,  are amorphous.  The amorphous nature of the fiber allows the smaller ribbons to bend and fold, and mechanically interlock with one another.  This mechanical interlocking is what serves to increase the fiber's strength.  See here for a very nice and concise description of how the fibers are prepared.  The pictures in the link make clear how the short range ordering in the fibers results in a highly folded and interlocked structure.  Further heating of the fiber will convert it to graphite.


How to calculate the average quadratic velocity of carbon dioxide from given data?

Use the Ideal Gas Law: $PV=nRT \to T = \frac{PV}{nR}$.  From the mass you can find the number of moles, $n$. $P$ and $V$ are given, and $R$ is the universal gas constant.

Even better you could rearrange the equation to $RT = \frac{PV}{n}$, substitue the value for $RT$ and do less arithmatic.


The microstructure is retained upon cooling. After bainite forms, you can even quench it and it will remain bainite. The only way to get martensite is to cool it so fast that bainite has no time to form. The only way to get bainite is to cool it so fast that bainite can form, but not so fast you get martensite, and not so slow that you get pearlite. After the microstructures have formed, they remain, unless you reheat it.


Solubility and Evaporation - Would the concentration still be 1 ppm?

Correct, when the water evaporates you'd have more salt than is soluble in water. The excess would then fall out of solution. Adding the same amount of water that evaporated would leave you with the same kind of solution you started out with, so the concentration of dissolved salt would be the same as when you started (although you might need to stir some to get the salt back in solution).


Chemists have tables and tables and tables that correlate functional groups to numbers.

For example:
http://www2.ups.edu/faculty/hanson/Spectroscopy/IR/IRfrequencies.html

If we see a number on the graph, a certain type of functional group may be present.  The sharp one at 3056 tells me aromatic hydrogen.  The 2919 is the benzylic hyrodgens.  Since I'm a synthesis guy, knowing the exact types of vibrations occurring there was never important to me.

Be warned, chasing every peak is normally quite frustrating.  Vastly more important is normally the overall look for comparison with a known compound, which is sort of a molecular finger print.


Plants are able to store energy as carbohydrates, i.e. they can make more carbohydrates than they need for their metabolism  when there is no sunlight, and store it for later use. So the amount of energy in your equations is not equal.

Some plants are offering this stored energy as sugars as nectar or fruits in exchange for pollination and spreading their seeds or use it as energy storage for times when they can't make enough carbohydrates for quick use, e.g. in winter or as storage for high demand, e.g. quick growth to outsmart competitors for space.
In addition, quite much of this fixed CO2 goes into structures like stems, roots and leaves and thus is sequestered for some time from the atmosphere.


Resonance structures are a useful way of visualizing and representing the bonding in a molecule - they don't actually exist, but they are a useful tool to see what the bonding and structure of a compound is.

Bonding in a compound can sometimes be nicely represented using two-center, two-electron bonds - for example methane. The structure and reactivity of methane can be rationalised by representing it as a carbon with four hydrogens. Each "bond" can be represented by having two electrons, one from carbon, and one from hydrogen, in a bonding pair. You might draw this as a combination between an sp3 hybrid and a 1s orbital. In reality these hybrid orbitals don't exist either - they are merely a tool to simplify the bonding to localised bonding which is easier to understand. In reality there a multiple interactions between the carbons 2s and 2p orbitals, with the 1s orbital of the hydrogen. However, it is easier, and an accurate enough model to model methane with four localised bonds.

Resonanace is useful where this idea of two-center, two electron bonds breaks down - for example we know from experiment that the structure of benzene is a regular hexagon, with identical carbon-carbon bonds. However the idea of two-center two-electron bonds breaks down here - we can't assign a structure using the ideas we know - carbon makes four bonds, the shape of benzene is a regular hexagon with identical bonds. In this case we can represent benzene as an in-between structure between two structures that can be represented as nice two-center, two-electron bonds. When we draw benzene as a resonance structure of two rings with alternating double bonds, what we are trying to represent is that the real structure of benzene is somewhere between these two structures - it is neither of the structure, and it doesn't resonate - ie. flick really fast between the two structures - instead it is a mixture of the two structures. In benzene the structure we are trying to represent can be understood by considering each carbon to be sp2 hybridized - again these hybrids aren't real, but a tool to understand bonding more easily. Then we have the 6 p-orbitals in the plane of the molecule. In reality these combine two give 6 molecular orbitals. These give a final structure where the bonding is identical between all six carbon atoms, a kind of one and a half bond - but the shape and idea of this delocalised bonding is very important to the reactivity of benzene and other aromatics - which is where this idea of resonance starts to break down in some ways, but the structures can still be useful, but with a proper understanding of the bonding behind the structure.


Probability of the single molecule reaction to happen

When two species $A$ and $B$ react, $A + B \to product$, the rate of reaction is $k\mathrm{[A][B]}$ where $k$ is the rate constant and $\mathrm{[A]}$ and $k\mathrm{[B]}$ the concentrations at any time during the reaction, which clearly change with time. To work out what happens in each small time interval means calculating the change in the number of reactant molecules in a given volume. Doing this means calculating two things. The first is the time at which any of possibly several reactions will occur, and the second is to decide which reaction this will be.
In the reaction $A + B \to product$, there is only one type of reaction but there are very many schemes where there are two or more reactions; for example, the photo-dissociation of halocarbons with UV light and subsequent reactions to destroy ozone.

$$\mathrm{CFCl_3} \overset{h\nu}\to \mathrm{CHCl_2  +Cl}\qquad \mathrm{Cl+O_3} \to \mathrm{ClO+O} \qquad \mathrm{ClO+O_2} \to \mathrm{Cl+O_2} $$

Given such a reaction scheme, the first step is to decide what reaction will happen next given the current state of the reaction at time $t$. At the level of considering individual molecules, once the reaction has started, it does not follow sequentially as shown in the scheme. This only happens when a whole ensemble of molecules, typically as large as Avogadro's number is examined, and then normal kinetic equations apply because of the averaging over many events that occurs. In any given time interval when considering molecules one at a time, it is not possible to predict exactly which reaction will happen next. It is quite possible, for instance, that several $\mathrm{Cl+O_3}$ reactions may occur before either of the other two. The best that can be done is to guess the outcome, and repeat the calculation many times and this approach is now described.

Suppose that the reactions are labelled with a number $R$ that represents reaction 1 or 2 or 3 etc. The chance of a reaction of type $R$ occurring will be defined as

$$prob(T + dT, R) \equiv a_RdT$$

Specifically, this is the probability that a reaction of type $R$ occurs in time interval $T$ to $T + dT$ in the reaction volume, given the current state of all reactions at an earlier time $t$. The values of $a_R$ can be easily worked out, and this is shown later. What is now needed is the probability $P_0(T)$ that no reaction of any type occurs up to time $T$ past the present time $t$, multiplied by the probability that only a reaction of type $R$ occurs in the small time interval $T \to T + dT$, which is defined as $a_RdT$. The probability of reaction type $R$ happening is therefore

$$ P(R,T)= P_0(T)a_RdT$$

and shown schematically in the figure.

Calculation of reaction probability.

To calculate the probability of no reaction occurring, consider any small interval $d\tau$ after some arbitrary time $\tau$, making an interval $\tau \to \tau + d\tau$. The probability of no reaction is then

$$P_0(\tau+d\tau) =P_0(\tau)\left(1-\sum_{R=1}^n a_Rd\tau  \right)$$

where the summation is over all $n$ reaction types $R$. The summation is made because this gives the chance that any one of the $R$ reactions will occur, and therefore one minus this number is the chance that no reaction occurs. Rearranged, this equation gives

$$ \frac{P_0(\tau+d\tau)-P_0(\tau)}{d\tau}=-\sum_{R=1}^n a_R$$

and in the limit of a small time interval $d\tau$, it becomes the true derivative $\displaystyle \frac{dP_0}{d\tau}=-\sum_{R=1}^na_R$.

For clarity only, a notational change is now made so that

$$a_0=\sum_{R=1}^n a_R \tag{8}$$

and therefore $\displaystyle \frac{dP_0}{d\tau}=-a_0$.The chance of no reaction up to time $T$ is found by integrating, and is $\displaystyle P_0(T)=e^{-a_0T}$.

The final result for the probability of reaction R occurring at time $T$ is the chance of no reaction up to $T $ multiplied by the chance of reacting at $T$, which is $a_R$, giving

$$P(T,R) = a_Re^{-a_0T} \tag{9}$$

an expression that shows that these times are Poisson distributed. Equation 9 is a joint probability density function, as it is the chance that reaction occurs in the time interval and that it will be reaction $R$.

Two terms now remain undefined. The first is the individual $a_R$ and their sum $a_0$, and the second determines which reaction happens next. Calculating $a_R$ is very easy, as each $a_R$ is the product of the number of molecules present in a reaction of type $R$ multiplied with the rate constant per unit volume.

If there is just a single reaction $\mathrm{A} \overset{k_1}\to \mathrm{B}$ then $a_1=k_1N_A$ where $N_A$ is the number of $A$ molecules present at time $T$. If the reaction is $\mathrm{A+B} \overset{k_2} \to \mathrm{C}$ then $a_2=k_2N_aN_B$ and if both reactions are present in the overall reaction scheme which would then be

$$\mathrm{A} \overset{k_1}\to \mathrm{B} \qquad \mathrm{A+B} \overset{k_2} \to \mathrm{C} \qquad a_0=\sum a_R=a_1+a_2 \tag{10}$$

In the special case where a reaction combines two identical molecules $\mathrm{A+A} \overset{k_2} \to $ the number of indistinguishable combinations, i.e the number of distinct A-A pairs of the two species must be calculated and this makes $\displaystyle a = k_2\frac{N_A(N_A - 1)}{2!}$. Normally, in dealing with chemical kinetics, $N_A$ is so vast that $\displaystyle a = k_2\frac{N_A^2}{2}$, but in the Monte Carlo simulations, $N_A$ is small because individual molecules are dealt with, and the exact formula must be used. Thus, macroscopic, (kinetic), chemical equations are only valid when deviations from the Poisson distribution are negligible.

An aside

The term $\displaystyle a = k_2\frac{N_A(N_A - 1)}{2!}$ implies that these numbers are averages in the reaction volume at time $t$ and should be written as the average of the product, i.e. as $\displaystyle \frac{\langle N_a(t)(N_A(t)-1)\rangle }{2!}$. The question then arises as how to deal with this. Two approximations can be made; the first is to assume that $N_A$ is large compared to 1 and so $\langle N_A(t)(N_A(t)-1)\rangle \rightarrow \langle N_A^2(t)\rangle $, which seems to be a reasonably sensible approximation, the next step is more problematic and is $\langle N_A^2(t)\rangle \sim \langle N_A(t)\rangle^2$. This is a drastic step and assumes that the variance in the number of molecules is zero, i.e. that $N_A(t)=\langle N_A(t)\rangle$ which means that $N_A$ is a deterministic process, so that the random nature of the numbers of $N_A$ entering into the reaction volume is always the same, which is not what was assumed at the outset. (To see that the variance is zero use the definition $\sigma^2_{N_A} = \langle N_A^2 \rangle -\langle N_A \rangle^2$ where $\sigma$ is the standard deviation)

In any normal kinetic experiment the number of molecules is vast, $\langle N_A(t)\rangle \sim 10^{18}$ would not be untypical, so that the fluctuations in the  number of molecules is tiny and the approximation made above is valid.

$$ \frac{\sigma_{N_A}}{\langle N_A \rangle} \sim \frac{\sqrt{\langle N_A \rangle}}{\langle N_A \rangle} \sim \frac{1}{\sqrt{\langle N_A \rangle}}\sim 10^{-9}$$

Even if there are as few as 1000 molecules the relative fluctuation  is not too drastic but in experiments using small numbers of molecules, as in the case of many 'single molecule' experiments then clearly the rate equation approach needs to be modified.

The final part of the calculation determines which particular reaction step is going to occur and at what time this is done by choosing pairs of numbers that have the distribution $P(T, R)$ equation 9. Gillespie (1997, 2007) argues that $P(T, R)$ is equivalent to two distributions, $P_1$ and $P_2$;

$$P(T,R)= P_1\cdot P_2 = \left(a_0e^{-a_0 T} \right)\left( \frac{a_R}{a_0}\right)$$

and which can be determined by choosing two uniformly distributed random numbers, $r$, each in the range 0 to 1. The equation

$$T=-\frac{1}{a_0}\ln(r) \tag{11}$$

generates a random number $T$ according to the probability density function $\displaystyle P_1=a_0e^{-a_0 T}$.

Equation 11 produces times that are Poisson distributed, and the time $T$ is added to the current time of the reaction. The second distribution, $P_2$, is calculated by choosing a uniform random number according to the probability density function $a_R/a_0$. For example, the chance that reaction 1, (equation 10), occurs is $a_1/(a_1 + a_2)$ and for reaction 2 this chance is $1 - a_1/(a_1 + a_2) = a_2/(a_1 + a_2)$. Which reaction occurs is determined by choosing a random number between 0 and 1 and seeing which of the two regions it falls into. If it falls between 0 and $a_1/(a_1 + a_2)$, reaction 1 occurs; otherwise, reaction 2 occurs. If reaction 1 occurs, $\mathrm{A} \to \mathrm{B}$, then 1 is subtracted from the original number of $A$ molecules in each step of the reaction and 1 is added to the number of $B$ molecules.

If reaction 2 happens, $\mathrm{A+B} \overset{k_2} \to \mathrm{C}$ then 1 molecule is removed from each of the numbers for $A$ and $B$. How many molecules are added and subtracted, of course, depends on the stoicheiometry of each particular reaction.

In the case of a termolecular reaction $\mathrm{ A + B + C} \rightarrow $ when the probability of three molecules simultaneously colliding is calculated the probability the chance that this occurs in  an infinitesimal time interval $d t$ is proportional to $dt^2$ so becomes vanishingly small and thus does not occur with any physical reality. It is often convenient, however, to consider a termolecular scheme as a simplified approximation to a scheme such as $N_A+N_B \leftrightharpoons X^*; \; X^* + N_C \rightarrow$

The outline plan for the Gillespie algorithm is

(1) Define the initial amounts of each species.

  Define the arrays of data points to hold results.   Set the maximum time. Set the time to zero; t = 0.

(2) Start a loop around the calculation until the time is up or no molecules are left.

  Calculate $a_1,\;a_2,\;a_3$ etc. and $a_0 =a_1 +a_2 +a_3 +\cdots$ Calculate time $\displaystyle T = -\frac{1}{a_0}\ln(r)$
  Increment time $t = t + T$
  Calculate which reaction occurs: if a random number falls between 0 and $a_1/a_0$, reaction 1 occurs, and so forth for other reactions.
  Add or subtract numbers of molecules from the totals, depending on which reaction has occurred and its stoicheiometry.
  Store the results.

(3) Continue the loop until finished.

Refs

'Markov Processes. An Introduction for Physical Scientists' D. T. Gillespie, publ Academic Press 1992.

D. Gillespie  J. Phys. Chem. 81, p2340, 1977 eqn 29

The python code for a Lotka-Volterra predator-prey reaction is given below. I'm assuming that matplotlib and numpy, as np.blabla() etc are used.

def lotka():                # Lotka reaction Gillespie method
                            # x0 + Y = 2Y  (k1) ,  Y +  Z = 2Z (k2) , Z = Q(k3)

    #k1 =10.0   k2 = 0.01   k3 = 10.0
    #y0 =2000.0  z0 =2000.0
    ycount = np.zeros(bins,dtype=int)
    zcount = np.zeros(bins,dtype=int)
    dtime =  np.zeros(bins,dtype=float)
    for i in range(bins):
        dtime[i] = maxt*(i+1/2)/bins
        pass
    ny = y0
    nz = z0
    ycount[0] = y0
    zcount[0] = z0
    t = 0.0
    indx = 0
    while indx < bins :
        r1 = k1*ny
        r2 = k2*nz*ny
        r3 = k3*nz
        a0 = r1 + r2 + r3+1e-20
        t0 = -np.log( np.random.rand() )/a0
        t = t + t0
        indx = int(np.trunc( t*bins/maxt ))
        if indx < bins :
            r = np.random.rand()
            if  r > ( r1 + r2 )/a0  :
                nz = nz - 1
                zcount[indx] = nz          # react 3
            elif r > r1/a0 :
                nz = nz + 1
                ny = ny - 1
                zcount[indx] = nz
                ycount[indx] = ny          # react  2
            else:
                ny = ny + 1
                ycount[indx] = ny          # react 1
                pass
            pass
        pass         # events

    return zcount,ycount,dtime      # end lotka

    bins = 1000
    maxt = 200.0
    k1 = 0.5
    k2 = 0.001
    k3 = 0.5
    y0 = 400.0
    z0 = 100.0

    zcount,ycount,dtime = lotka()

    fig1= plt.figure(figsize=(10, 6))
    plt.rcParams.update({'font.size' : 16})  # set font size for plots

    plt.plot(dtime,zcount,color='red')
    plt.plot(dtime,ycount,color='blue')#'ro',markersize=2)

    plt.xlabel('time')
    plt.ylabel('population')
    plt.xlim([0,maxt])

    plt.tight_layout()

    plt.show()


Accelerated Oxidation Of Iron When Coating Breaks

The main factor for this phenomena is the the values of the standard electrode potentials for of the metal ions.

The UC Davis ChemWiki page Corrosion, uses this scenario as an example:

  the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for $\ce{Sn^2+}$ (E° = −0.14 V) and $\ce{Fe^2+}$ (E° = −0.45 V) show that $\ce{Fe}$ is more easily oxidized than $\ce{Sn}$. As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen

Tin is therefore more of a noble metal than iron - a galvanic corrosion chart is shown below, showing that tin is more 'noble' than iron (Image source):

The significance of tin being more noble than iron is that, according to the UC Davis ChemWiki page Electrochemical Corrosion, it can act as a depolariser, removing more electrons than the water/air alone - hence, accelerate the corrosion process.

Instead of the copper used in the example, it would be tin in your example.

A somewhat simplied illustration of the process is shown in the diagram below:


Only valence shell electrons participate in bonding then how is Cr in H2CrO4 forming 6 covalent bonds although it has only 1 valence electron?

The electronic configuration of $\ce{Cr}$ is: $[\ce{Ar}]4s^2 3d^4$. Because of very little energy difference between 4s- and 3d- orbitals, it can lose 6 of it's valence electrons and thus form 6 covalent bonds.


Which platonic solids have been built as carbon skeletons?

The previous answer and comments have already listed out the known platonic hydrocarbons, however, since the OP had expressed an interest in the synthesis of these platonic hydrocarbons in one of his comments, I have provided some commentary to that effect along with links to the original papers that first reported these compounds. Also, of interest would be this paper
Tetrahedrane
Technically speaking, Tetrahedrane ($\ce{C4H4}$) still remains a hypothetical compound, as an unsubstituted version hasn't been synthesised yet.

Two stable derivatives, whose synthesis has been achieved include tetra-tert-butyltetrahedrane (prepared in 1978 by Günther Maier) and tetra(trimethylsilyl)tetrahedrane (reported in 2002).
Tetra-tert-butyltetrahedrane
The original paper discussing the synthesis can be found here. The synthesis has been summarised below. The authors rationalise the stability of this compound as follows:

If  all  the  corners of the tetrahedrane unit are occupied by bulky groups then on lengthening  one bond of the  ring  skeleton , two  tert-butyl groups  tending  to increase  their  mutual  separation  will  be forced  back  together  by  the  other ones  ("corset  effect"); a tetrahedral  structure permits  the  four  substituents  to adopt a regular spherical distribution and hence a maximum mutual separation.

In simpler terms, trying to break a bond forces the bulky substituents together increasing strain.

Tetrakis(trimethylsilyl)tetrahedrane

Again, the synthetic route is given below. The original paper is also by Günther Maier.
Reaction with methyllithium has yielded the stable tetrahedranyllithium derivative. These coupling reactions with this lithium compound have allowed access to more derivatives, for example the dimer in the last step of the reaction scheme given above.
This compound is far more stable than the tert-butyl analogue. On account of the silicon–carbon bond being longer than a carbon–carbon bond, the corset effect is reduced. $\sigma$-donor character of the trimethylsilyl explains the increased stabilisation of the tetrahedrane.
The tetrahedrane skeleton is made up of banana bonds, and hence the carbon atoms are high in s-orbital character
Cubane

Cubane was first synthetically realised by Eaton and Cole in 1964. The synthetic scheme from the original paper is given below:

Note: In the original synthesis, the starting material for this, 2-bromocyclopentadienone, was obtained from 2-cyclopentenone via allylic bromination with N-bromosuccinimide in carbon tetrachloride followed by addition of molecular bromine to the alkene gives a 2,3,4-tribromocyclopentanone. After treatment of this compound with diethylamine in diethyl ether, which causes elimination of two equivalents of hydrogen bromide, we get the desired diene product.
Again, the cubane is kinetically stable because of lack of readily accessible decomposition paths.
Dodecahedrane

Dodecahedrane was synthesised by Leo Paquette in 1982, and later by Prinzbach (via pagodane) in 1987.
The pagodane synthetic route has been given below.

Paquette's 1982 organic synthesis takes about 29 steps, and starts from cyclopentadiene, and for the sake of brevity, I have not presented it here.


How to dissolve the build-up from a paraffin based oil?

Naphtha, paint thinner or mineral spirits are common names for the type of solvent suggested by @NotNicolau, and should be effective. Caution: all are flammable and could be ignited by a spark from machining, from a motor or from a switch. These solvents are considered only mildly toxic, though.

Chlorinated solvents such as trichloroethylene are less flammable and even more effective at removing this "varnish" (probably not paraffin, which would be greasy but not tacky) but they are more toxic and the fumes are more annoying.

Ventilate the area well both during cleaning and afterwards until all solvent has evaporated and blown away.


Electrochemistry - Understanding Diffusion Potential

Let's discuss the problem in a qualitative way. In such a cell, the metal will get dissolved in the right-hand side of the cell, where the concentration is low, so that $[M^{z+}]$ increases in this compartment, which becomes the anode. In the left-hand side, the ions $[M^{z+}]$ are discharged  and deposited as a metal layer on the electrode, which is the cathode.

The only problem is what is going on in the solutions. If the anion $A^-$ has a high mobility, it will flow from left to right through the membrane to compensate the chang of concentration of the metal ion. And after some time, the two solutions will have the same intermediate concentrations, and the cell will not deliver any current.

But if the ion $A^-$ has a low mobility, the cation $[M^{z+}]$ has a higher mobility and will easily cross the membrane from right to left. It will take a much longer time for the concentrations to be equalized in the two compartments. The reason is that a much smaller amount of ions $A^-$ will cross the membrane from left to right. But at the end, the two solutions will have the same concentrations.


Can decarboxylation be done under carbon dioxide?

Because nitrogen is the standard inert gas in lab practice. If you are "rich" You can use argon, but carbon dioxide is seldom used, because it won't work on alkaline solutions.

  or will the presence of carbon dioxide slow down the reaction

First, what you presumably mean is influencing the equilibrium unfavourably, don't you? You should learn to discern kinetics and equilibrium.

Answer: not really. Whether you have a carbon dioxide atmosphere or not will shift the decomposition temperature not more than some degrees.


The reverse reaction of neutralization is called hydrolysis ( X salt + H2O = acid + base )


I think vapor have a temperature near to the condensation point so it can become liquid easily but gas have much higher temperature than it's condensation point hence we need a process to make it liquid for example if steam is at around 100°c then it can be termed as water vapor and if you consider nitrogen it liquifies at -198°c or near to that hence it is referred as gas.


How can I estimate reaction rates in water with computational chemistry?

Finding solvent  effect in computational chemistry is still a crucial part and subject to deep research effort. Now there are two types of models that are being used in computational chemistry.
One is explicit solvation model: Here you will use explicit solvent molecule with your reactant or products and then using first principles you will find the equilibrium configuration but the problem is, it is very expensive and the configuration space is so much high that it is practically impossible. For that you have to use molecular dynamics or monte carlo techniques. In our group we use 1000 to 5000 explicit water molecule but we can't do it for large reaction.
Another method is implicit solvation model: Here electrostatic potential is applied to account the solvent effect but this procedure is so expensive that now that potential is replaced by the solvent induced reaction field energy.
In our group we are using a hybrid approach that involves communication of several software at different level of theory. In my quantum chemistry class i used spartan and qchem and they have some implicit solvation but i didn't use GAMESS.
From GAMESS manual i found that they have solvation module. They are using QM/MM approach. That's actually a very good approach. Please see page 9 of this file:  http://www.sdsc.edu/~kimb/gamesstut.pdf
In this following file at page 3 & 4 you will also find the list of keywords that are needed to start a solvation model run
http://www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf


No, substitution by an alkyl group produces a small downfield shift in both proton and carbon nmr. Therefore, in a typical hydrocarbon a secondary carbon will be downfield of a primary carbon.  The assignments in your figure are correct.

A possible explanation for the downfield shift of a carbon atom when a proton is replaced with another carbon atom is as follows.  Hydrogen is more electropositive than carbon, therefore, in a C-H bond a significant amount of the electron density in the bond resides around the carbon.  When we replace the electropositive hydrogen with a relatively more electronegative carbon, then there is less electron density around the central carbon atom.  This reduction in electron density deshields the carbon nucleus and shifts it downfield.  Every time another hydrogen is replaced by carbon a further downfield shift occurs.


pKa and pH behaviour of a drug lignocain

If I understand your question correctly, this should answer it:

An "uncharged" base with a high pKa is protonated to form an ionic species:
$$\ce{B + HX <=> HB^+ + X^-}$$
$$Ex:\ \ce{NH3 + HCl <=>> NH4^+ + Cl^-}$$

A negatively-charged base with a high pKa is protonated to form an uncharged species:
$$\ce{B^- + HX <=> HB + X^-}$$
$$Ex:\ \ce{OH^- + HCl <=>> H2O + Cl^-}$$


Sodium permanganate from manganese dioxide and sodium hypochlorite

I performed this reaction, and the answer is yes, concentrated sodium carbonate can be used in place of sodium hydroxide, though the latter works better. The reaction goes under normal pressure, but it requires heating. The rate of the reaction also depends on the quality of manganese dioxide - freshly prepared works best. You will notice the formation of permanganate as soon as you boil the reaction mixture, but complete conversion of dioxide may take several minutes to hours. If you want a complete conversion, better use the sodium hydroxide.


Is the fact that table salt is composed of Na & Cl a chemical or physical property?

The problem rises when we are mistaken about what a property is.

  The different types of matter can be distinguished through two components: composition and properties. UCDavis

Do a search about chemical property, and you get:

  a property or characteristic of a substance that is observed during a reaction in which the chemical composition or identity of the substance is changed. Dictionary.com

  A chemical property is any of a materials properties that becomes evident during a chemical reaction; that is, any quality that can be established only by changing a substance's chemical identity. Simply speaking, chemical properties cannot be determined just by viewing or touching the substance; the substance's internal structure must be affected for its chemical properties to be investigated. However a catalytic property would also be a chemical property. Wikipedia

  Any characteristic that can be determined only by changing a substance's molecular structure.
   Boundless

What's a physical property?

  A physical property is any property that is measurable whose value describes a state of a physical system. The changes in the physical properties of a system can be used to describe its transformations or evolutions between its momentary states. Physical properties are often referred to as observables. They are not modal properties. - Physical property in wikipedia.

Basically, a physical property is what that doesn't deal with the composition of matter; but other factors, such as it's physical state. "Physical" isn't at its literal meaning.

Table salt is never pure $\ce{NaCl}$; because simply there's nothing pure in the nature. It's mainly composed of Sodium Chloride crystals. But if, and only if, we consider it to be only made from $\ce{NaCl}$, we are talking about its chemical composition and therefore, it must be a chemical property.

I understand that sometimes there are wrong questions for us to answer, so if composition is to be a property, it must be a chemical one.


Can chloride attack the carbonyl bond in formaldehyde in the reaction with hydrochloric acid?

We can both immediately agree on the first step of the reaction between formaldehyde and hydrochloric acid which is the protonation of the carbonyl oxygen. The mechanism of formaldehyde oligo- or polymerisation then suggests the attack of the oxygen atom of another formaldehyde moiety to form a hemiacetal. This would be a nucleophilic addition onto a π system.

One possible course this species can now take is the back reaction to reform formaldehyde. Another one is a subsequent attack of the newly-formed hydroxy moiety of the hemiacetal function onto another carbonyl group to form a full acetal. The latter reaction creates a product which is inert enough to not undergo back reactions.

You are indeed correct that chloride, which is also a nucleophile, can equally attack the π $\ce{C=O}$ bond of formaldehyde. And this attack would indeed result in chloromethanol. However, this reaction is also reversible as is the attack of oxygen. Furthermore, this molecule has no real way of stabilisation: chloride will always be a good leaving group while a formal alcoholate is not. Thus, while the attack of chloride may occur it is not productive and it is rapidly reversed.

Therefore, we do not see any chlorinated product — which would decompose on isolation at the latest anyway.


Why does caffeine powder cause apple cider (or any carbonated drink?) to foam?

I don't need to know why you're mixing uppers (caffeine) with downers (alcohol) but if the powdered caffeine is removing the fizz from your cider then you might try dissolving it in hot water first and then adding the hot caffeine solution to your cider.  In dissolved form, the caffeine cannot provide the nucleation sites for the formation of CO2 bubbles that MaxW and vapid pointed out.

The solubility of caffeine in water is 200mg/ml at 80 oC so you should be able to dissolve an adequate dose of caffeine in a small enough quantity of hot water to avoid both excessive warming of your cider and precipitation of the caffeine upon cooling and dilution.

It behooves me to mention that recommended caffeine administration should not exceed 200mg every 3-4 hours in adults, but since you're adding it to alcoholic beverages, I doubt you give a monkey's.  Here's the drugs.com link anyway:

https://www.drugs.com/monograph/caffeine-caffeine-and-sodium-benzoate-injection-caffeine-citrate.html

And solubility data here:

http://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/c0750pis.pdf


How to calculate the average size of cubic crystals of KCl powder?

For a cube, the surface area is given by the formula $6a^2$ where $a$ is the length of one side. The volume is given by $a^3$. Surface area per volume is
$$\frac{6a^2}{a^3}= \frac{6}{a}.$$

We know the total surface area to be $\pu{1620 cm^2}$. We can calculate the total volume from the mass of the $\ce{KCl}$ and its density.

\begin{align}
  V &= \frac{m}{\rho}\\
  \rho(\ce{KCl}) &= \pu{1.98 g//cm^3}\\
  V &= \frac{\pu{0.5 g}}{\pu{1.95 g//cm^3}}\\
    &= \pu{0.3 cm^3}
\end{align}

An average crystal should have the same ratio as the total. So we set the two equal and solve for the length.

\begin{align}
  \frac{6}{a} &= \frac{\pu{1650 cm^2}}{\pu{0.3 cm^3}}\\
  a &= \frac{6 \times \pu{0.3 cm^3}}{\pu{1650 cm^2}}\\
  a &= \pu{9E-4 cm}
\end{align}


Note that $C_\mathrm{2v}$ is a subgroup of $D_\mathrm{2h}$. This means that you're just overlooking some of the symmetry elements. In addition to the $C_2$ axes penetrating the lobes "end to end", you also have a $C_2$ axis perpendicular to the orbital plane through its center.

Your $C_\mathrm{2v}$ guess is also inconsistent because that point group only has one $C_2$ axis, but you said yourself that you already found two.


Molecular mass has NO pertinence to BP, water and ammonia.  Look up the BP of perfluorocyclohexane, hexafluorobenzene, and the hydrocarbons.  If I want to volatize iron, ferrocene.  If I want to volatize iron more, $\ce{Fe(hfac)_3}$ where the ligand is the bidentate enolate anion of 1,1,1,5,5,5-hexafluoro-2,4-pentanedione.  Big MW.  What are the MW and BP of $\ce{UF_6}$?  $\ce{-Si(CH3)_3}$ also confers volatility.

Boiling points are more about polarizability, intermolecular van der Waals and such. Bigger molecules are overall electronically floppy.  Benzene, toluene, xylenes – nice BP steps, but not because of MW. Look up $\ce{SF_6}$ bp.  $\ce{I(CF3)_7}$ boils near 0 °C.  Calculate that MW, and its vapor’s mass/liter at STP compared to dry air at 29 g/liter.


Which is the major product of bromination of 3-(dimethylamino)benzamide, ortho or para?

Here's the 3D structure of the substrate:

As you can see, the $\ce{-N(Me)2}$ group is clearly larger than the $\ce{-CONH2}$ group. Also notice that the $\ce{-N(Me)2}$ group is planar with the benzene ring. This is because this structure is the most favourable for $\mathrm{p-\pi}$ conjugation 1.
Your guess that A would not form is correct. The large size of the $\ce{-N(Me)2}$ group would also point to the fact that C is the major product. For a better explanation of the major product, we can take a look at the 3D structures of B and C:

You can clearly see that in B, the bromine atom pushes the dimethylamine group out of it's stable configuration, where it is coplanar with the benzene ring. This requires more energy and also produces an unstable product. Therefore, C is the major product in this reaction.
References:

Novikov, V. P., et al. “Molecular Structure and Conformation of N,N-Dimethylaniline by Gas-Phase Electron Diffraction and Quantum-Chemical Calculations.” Russian Journal of General Chemistry, vol. 74, no. 8,2004, pp. 1247–53. doi:10.1007/s11176-005-0146-9.