The glassware shown is a drying tube. The bulb holds cotton or glass/quartz wool while the tube part holds a drying media such as anhydrous calcium sulfate to keep moisture out of the vessel during moisture sensitve reactions.


A typical example for the application of models and experimental methods of chromatography for geologic materials (e.g. soil, sediment) is the $K_\mathrm d$ model, which can be used for understanding the long-term behaviour of contaminants in the subsurface environment (e.g. the rooting zone of crops).

The parameter $K_\mathrm d$ is the partition coefficient, i.e. the ratio of the concentration of a contaminant sorbed on a specified solid phase (e.g. soil) to the concentration in a specified liquid phase (e.g. groundwater):

$$K_\mathrm d=\frac{\text{concentration in solid phase (per kg)}}{\text{concentration in liquid phase (per l)}}$$

The partition coefficient $K_\mathrm d$ can be used to calculate the retention factor $k$:

$$k=\frac\rho\varepsilon K_\mathrm d$$

where
$\rho$ is the dry bulk density of the solid phase, and
$\varepsilon$ is the total porosity of the solid phase.
Typical values for soil are $\rho=1.4\ \mathrm{kg\ l^{-1}}$ and $\varepsilon=0.3$.

This parameter is related to the retardation factor $R_\mathrm F$:

$$R_\mathrm F=\frac1{k+1}$$

The retardation factor $R_\mathrm F$ corresponds to the ratio of the distance travelled by the contaminant to the distance simultaneously travelled by the mobile liquid phase:

$$R_\mathrm F=\frac{v_\text{contaminant}}{v_\text{mobile phase}}$$

For example, if the annual precipitation exceeds the evaporation by $300\ \mathrm{mm}$ and the porosity of the soil is about $\varepsilon=0.3$, the drainage water travels with an average velocity of $v_\text{mobile phase}=100\ \mathrm{cm\ a^{-1}}$ downward through the soil.

Assuming a mean value for the partition coefficient of plutonium in all soils of

$$K_\mathrm d=7.4\times10^2\ \mathrm{l\ kg^{-1}}$$

the corresponding retention factor is

$$\begin{align}
k&=\frac\rho\varepsilon K_\mathrm d\\[6pt]
&=\frac{1.4\ \mathrm{kg\ l^{-1}}}{0.3}7.4\times10^2\ \mathrm{l\ kg^{-1}}\\[6pt]
&=3.5\times10^3
\end{align}$$

and the retardation factor is

$$\begin{align}
R_\mathrm F&=\frac1{k+1}\\[6pt]
&=\frac1{\frac\rho\varepsilon K_\mathrm d+1}\\[6pt]
&=\frac1{\frac{1.4\ \mathrm{kg\ l^{-1}}}{0.3}7.4\times10^2\ \mathrm{l\ kg^{-1}}+1}\\[6pt]
&=2.9\times10^{-4}
\end{align}$$

Therefore, the migration rate is

$$\begin{align}
v_\text{contaminant}&=R_\mathrm F\cdot v_\text{mobile phase}\\[6pt]
&=2.9\times10^{-4}\times100\ \mathrm{cm\ a^{-1}}\\[6pt]
&=0.029\ \mathrm{cm\ a^{-1}}
\end{align}$$

i.e. a downward migration of $1\ \mathrm{cm}$ takes about 35 years.


Looking for a radio-transparent epoxy/resin

You might not be able to find a home-usable thermosetting resins such as epoxy or RTV silicone rubber that meets your needs: low dielectric loss (dissipation factor), moderate hardness and stability at microwave cooking temperatures.

Bakelite (urea-formaldehyde thermosetting resin) might be usable, as it's hard and temperature resistant, though its dissipation factor is ~0.005, a tiny bit higher than ideal. However, one-off molding of Bakelite is non trivial, requiring pressurized and temperature controlled molds.

A table of electrical properties of some plastics shows polystyrene, polypropylene and some fluorinated plastics such as polytetrafluoroethylene with dissipation factors <0.001, but these are less hard than Bakelite.

Glass-filled plastics such as glass-epoxy laminate (printed circuit board, PCB, material) and RO4000® hydrocarbon ceramic laminates might be better choices. However, instead of casting a part, you'd need to machine the part from one or more PCB's laminated together. This might actually be easier, for a one-off part, than casting.


Why does the colour of the same compound (e.g. magnesium nitride) differ?

Minor differences in color are likely either due to impurities or due to intrinsic crystal defects such as f-centers. For example, $\ce{ZnO}$ is often a shade of yellow, rather than pure white, due to those defects. In alkali halides, h-centers are similar defects giving rise to colors. You can easily create color centers in $\ce{NaCl}$ with a spark coil in just a few minutes.

Another possibility is that there are mixtures of different allotropes, or polymorphic forms. An extreme example is transparent diamond vs. black graphite. BTW, diamonds having varied colors due to impurities (e.g. nitrogen) and to intrinsic defects are useful in studying quantum phenomena, as well as gems.


Reversible and Irreversible adiabatic expansion

The change in internal energy $U$ is
$$\Delta U=Q+W$$
where $Q$ is amount of heat transferred to the system and $W$ is work done on the system.
Since the process is adiabatic, no heat is transferred into or out of the system, i.e. $Q=0$ and thus
$$\Delta U=W$$

The reversible expansion is performed continuously at equilibrium by means of infinitesimal changes in pressure $p$.
The corresponding infinitesimal pressure-volume work $\delta W$ is
$$\delta W=-p\,\mathrm dV$$

For an ideal gas, the internal energy $U$ only depends on amount of substance $n$ and temperature $T$. Since $n$ is constant in a closed system, the change in internal energy $U$ is
$$\mathrm dU=C_V\,\mathrm dT$$
where $C_V$ is heat capacity at constant volume.

Equating $\mathrm dU$ and $\delta W$ yields
$$C_V\,\mathrm dT=-p\,\mathrm dV$$
and since $pV=nRT$ for an ideal gas
$$\begin{align}
C_V\,\mathrm dT&=-\frac{nRT}V\,\mathrm dV\\
\frac{C_V}T\,\mathrm dT&=-\frac{nR}V\,\mathrm dV
\end{align}$$

Integration from the initial state ($T_1,V_1$) to the final state ($T_2,V_2$) yields
$$\begin{align}
\int\limits_{T_1}^{T_2}\frac{C_V}T\,\mathrm dT&=\int\limits_{V_1}^{V_2}-\frac{nR}V\,\mathrm dV\\
C_V\int\limits_{T_1}^{T_2}\frac1T\,\mathrm dT&=-nR\int\limits_{V_1}^{V_2}\frac1V\,\mathrm dV\\
C_V\ln\left(\frac{T_2}{T_1}\right)&=-nR\ln\left(\frac{V_2}{V_1}\right)\\
\ln\left(\frac{T_2}{T_1}\right)&=-\frac{nR}{C_V}\ln\left(\frac{V_2}{V_1}\right)\\
\frac{T_2}{T_1}&=\left(\frac{V_2}{V_1}\right)^{-nR/C_V}\\
&=\left(\frac{V_1}{V_2}\right)^{R/C_{\mathrm m,V}}
\end{align}$$
where $C_{\mathrm m,V}$ is molar heat capacity at constant volume.

Since $C_{\mathrm m,p}-C_{\mathrm m,V}=R$, the exponent can be rewritten
$$\begin{align}
\frac{T_2}{T_1}&=\left(\frac{V_1}{V_2}\right)^{R/C_{\mathrm m,V}}\\
&=\left(\frac{V_1}{V_2}\right)^{C_{\mathrm m,p}/C_{\mathrm m,V}-1}\\
&=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\\
\end{align}$$
where $\gamma$ is the ratio of the heat capacities $\gamma=C_p/C_V=C_{\mathrm m,p}/C_{\mathrm m,V}=c_p/c_V$. For a monoatomic ideal gas, $C_{\mathrm m,p}=\tfrac52R$ and $C_{\mathrm m,V}=\tfrac32R$, and thus $\gamma=\tfrac53$.

Note that, in the given question, the volumes $V_1$ and $V_2$ are unknown. However, the values for the pressures $p_1$ and $p_2$ are given. Since $pV=nRT$ for an ideal gas,
$$\begin{align}
\frac{T_2}{T_1}&=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\\
T_2\cdot V_2^{\gamma-1}&=T_1\cdot V_1^{\gamma-1}\\
\frac{p_2V_2}{nR}\cdot V_2^{\gamma-1}&=\frac{p_1V_1}{nR}\cdot V_1^{\gamma-1}\\
p_2\cdot V_2\cdot V_2^{\gamma-1}&=p_1\cdot V_1\cdot V_1^{\gamma-1}\\
p_2\cdot V_2^\gamma&=p_1\cdot V_1^\gamma\\
\frac{p_2}{p_1}&=\left(\frac{V_1}{V_2}\right)^\gamma
\end{align}$$

In contrast to the reversible expansion, an irreversible expansion is not performed continuously at equilibrium by means of infinitesimal changes in pressure. In the limiting case, the value of the pressure $p$ abruptly changes from $p_1$ and $p_2$. The following expansion proceeds in one step against a constant pressure $p_2$. Hence,
$$\delta W=-p_2\,\mathrm dV$$
and thus
$$\begin{align}
C_V\,\mathrm dT&=-p_2\,\mathrm dV\\
\int\limits_{T_1}^{T_2}C_V\,\mathrm dT&=\int\limits_{V_1}^{V_2}-p_2\,\mathrm dV\\
C_V\left(T_2-T_1\right)&=-p_2\left(V_2-V_1\right)
\end{align}$$
Since $pV=nRT$ for an ideal gas,
$$\begin{align}
C_V\left(T_2-T_1\right)&=-p_2\left(V_2-V_1\right)\\
&=-p_2\left(\frac{nRT_2}{p_2}-\frac{nRT_1}{p_1}\right)\\
&=-nRT_2+nRT_1\frac{p_2}{p_1}\\
C_VT_2+nRT_2&=C_VT_1+nRT_1\frac{p_2}{p_1}\\
T_2\left(C_V+nR\right)&=T_1\left(C_V+nR\frac{p_2}{p_1}\right)\\
T_2\left(C_{\mathrm m,V}+R\right)&=T_1\left(C_{\mathrm m,V}+R\frac{p_2}{p_1}\right)\\
\end{align}$$


What does it mean if a functional group has "higher priority" in terms of IUPAC naming of organic compounds?

Whether trying to figure out if a double bond is "E" or "Z", or if a chiral center is "R" or "S", the Cahn–Ingold–Prelog priority rules (see the section "Assignment of priorities") are used to prioritize the groups.  In the case of a double bond there are 2 substituents (a lone pair of electrons can be a substituent) at each end of the double bond.  Priorities (1 and 2) are assigned to the substituents at each end using the Cahn–Ingold–Prelog rules provided in the link above.  The first rule used to set priorities is based on the atomic number of the atoms attached to the double bond.  If there is a tie (the 2 attached atoms at one end of the double bond have the same atomic number), then we move out to the next atoms and compare these.  The process is continued until the tie is broken (eventually the tie must be broken if the molecule being analyzed is really capable of being described as "E" or "Z").  After the priorities have been assigned, if the two priority-1 substituents at opposite ends of the double bond are on the same side of the double bond, then the molecule is the "Z" (zusammen, together) isomer, otherwise it is the "E" (entgegen, opposite) isomer.

Example:

image source


Is there a consensus on whether amber can seal in air bubbles for millions of years?

There is at least a fair degree of consensus (there will always be naysayers) that the results that were performed back in the 80s and 90s by Berner and Landis were false positives that could not be confirmed by further investigations. A good historical account of this can be found for instance in Nick Lane's book "Oxygen" (first published in 2002). Quoting from this book:

  The overall consensus, which has now persisted for more than a decade,
  was that the air trapped in amber could not have been ancient [...]

That "could not have been" has an air of finality about it, and you can probably take this book's word as accurate. However, all of the amber in the world has not been examined, so I would say chances may be slim but not none that some will be found containing traces of an ancient atmosphere. Probably part on the answer depends on how old the amber is, and how it was preserved. A very different question is whether gases in the amber can be regarded as representative of the atmosphere of origin, or useful. Perhaps reactions with the amber (oxidation) during burial have altered it beyond recognition.


Does hybridization take place in other atoms apart from central atom in a molecule?

Yes, all atoms with p or d orbitals can hybridize in a molecule, not just the central atom.  Take acetone for example, the carbonyl carbon is $\ce{sp^2}$ hybridized and the carbonyl oxygen is also $\ce{sp^2}$ hybridized.  Another example is methyl chloride, the methyl carbon is $\ce{sp^3}$ hybridized and the chlorine is also $\ce{sp^3}$ hybridized.  As a final example consider dimethyl ether, each of the carbon atoms is $\ce{sp^3}$ hybridized and the central oxygen is also $\ce{sp^3}$ hybridized.


Let's start with the specific examples. For the conversion of ethene to ethane, you need a reaction partner to provide the extra hydrogen, e.g. dihydrogen:
$$\ce{H2C=CH2 + H2 -> H3C-CH3}$$
In the reaction, a single bond and the pi-bond of the double bond are broken, and two single bonds are formed. If you look up the four bond dissociation energies (two broken and two formed), you get a fairly good estimate of the reaction enthalpy.
$$\ce{ADP + phosphate -> ATP + H2O}$$
I'm leaving out the charges and the protonation state of phosphate to make things simpler. If we look a bit closer, we can write:
$$\ce{RO3P-OH + H-OPO3R' -> RO3P-OPO3R' + H-OH}$$
So when ATP is formed, we are not only making the phospho-anhydride bond (and a bond within water), but also breaking two bonds.

Lastly, when teaching cellular respiration, we speak of energy being stored in the phosphate bonds when ADP is converted to ATP and in photosynthesis, of the energy from sunlight being stored in the bonds of glucose.

Is this line of thinking correct, if not why?

You could say the energy is stored in the reaction, and when the reaction runs in the other direction (hydrolysis), it gets released again. Similarly, glucose is a source of free energy only if oxygen is available as reaction partner. Breaking glucose into atoms is not a source of free energy.

Let's look at the first statement It takes energy to break bonds.

Is this true of spontaneous processes?

Here is the reaction this refers to:
$$\ce{molecule -> free atoms}$$
The entropy of reaction is expected to be positive (more particles on product side). If the reaction were also exothermic, these molecules would not be stable under any condition (Gibbs energy of reaction would be negative at any temperature). So bonds of molecules that actually exist are expected to have a positive bond dissociation energy.
The other questions
The answers to the rest of the questions follow from the information above.


Of course! Rb is more electropositive than hydrogen, so hydrogen is released.

As you mention, $\ce{RbOH}$ is a bit more expensive than $\ce{NaOH}$. At present, $\ce{RbOH}$ is ~US\$500/kg and $\ce{NaOH}$ is ~US\$500/tonne, three orders of magnitude cheaper.

You might use $\ce{CsOH}$ or $\ce{FrOH}$, too. The Fr will even illuminate the reaction with Cherenkov radiation.


Why is phosphorus (v) oxides molecular formula P4O10 and not P2O5

There is no clear rule about how to name a compound which is actually existing as a dimer, a trimer... or an octamer, as long as it does not make a difference in a reaction equation, which is usually the case.

It does not matter in an equation if you count sulfur as a single atom entity (S) or as its true form (S8). You will find the same quantity in both cases.

The same stands for P2O5/P4O10: the true form is usually P4O10 (depending on the solvent), but the 2 forms give the same result at the end.


Adulteration of turmeric with Metanil yellow

I hope to keep this short.

Metanil yellow is an azo dye synthesized from the coupling of metanilic acid and diphenylamine, as described here. Here's a sketch:

Now, the thing with most azo dyes like metanil yellow is that when they are taken internally, liver enzymes or intestinal flora will usually reduce the azo dye to the components that were used in the diazo coupling. (See this article for more details.) For metanil yellow, the metanilic acid is poorly absorbed from the intestinal tract (and is thus easily excreted), and your remaining worry is the diphenylamine, which is a suspected mutagen and carcinogen (apart from the usual toxicity associated with arylamines). (In short, the metanil yellow itself is not your worry, but the metabolite diphenylamine sure is.)

By way of contrast, the natural coloring agent in turmeric is the phenolic compound curcumin:

As it turns out, both compounds are in fact acid-base indicators, though they have different transition ranges; for metanil yellow, the pH transition range is 1.2-2.3 (red to yellow), while for curcumin, the transition range is 7.8-9.2 (yellow to red-brown). (Data taken from the Handbook of Acid-Base Indicators.) I'm not quite sure where the purple/violet color would come from, but the key point is that natural turmeric would be reddish in an alkaline environment, while metanil yellow will be reddish in an acidic environment; however, vinegar (pH ~ 2.4) is not sufficiently acidic to display the color change for metanil yellow, and thus a stronger acid like hydrochloric acid is needed to display the color change.


What is the liquid which has the most hydrogen atoms per volume?

Hydrazine has more hydrogen atoms per mL than water
Some simple calculations give the moles of hydrogen per mL in some possible alternative liquids. Taking into account density and molecular mass we get the following results:
Water 0.056 mol/mL 2 Hydrogens -> 0.11 mol H/mL
Hydrazine 0.0313 mol/mL 4 hydrogens -> 0.125 mol H/mL
symmetric dimethylhydrazine 0.0138 mol/mL 8 hydrogens -> 0.111 mol H/mL
methyl hydrazine 0.0191 mol/mL 6 hydrogens -> 0.115 mol H/mL
So, while some other liquid at STP hydrazine derivatives beat water, it looks like hydrazine has notably more hydrogens per unit volume. This is probably the record.


Problems regarding Kc,Kp and rate of reaction

Well, there are two cases:

$a + b = c + d$; and
$a + b \neq c + d$

For case 2, it's fairly simple to tell how the equation was balanced because your $K_C$ will come in a different unit, but for case 1, because $K_C$ is dimensionless we depend on the convention that Chemical Reactions are balanced on the lowest whole number. So in the case of the example given, we would balance it as:

$$\ce{2H2 + O2 <=> H2O}$$

and $K_C$ would be defined on that basis.


Grignard Reagent in THF vs in Diethyl ether

THF has a number of differences with diethyl ether:

Oxygen non-bonding doublets are more available for bonding with magnesium because the 2 ethyl groups are not freely rotating around (you may think of THF as diethyl ether with its arms tied in its back).
THF boils at 66°C (ether 34.6°C). Because the reaction is usually carried out so to maintain a gentle reflux, the higher boiling temperature of THF will speed up the reaction, and solubilize reactants/products better.

As @Alchimista pointed out, THF is just one of the possible solvents you can use if ether fails or gives poor yields. You can also use higher boiling ethers like dimethoxyethane.

Note: this is a collective answer asked by other users. I have been harassed. Call the police :-)


Mathematical properties of free Gibbs energy

n is discontinuous. The least amount you can change it by is ~1 /6.022E23 of 1 mole. You'll have to be more specific about what type of (mathematical) continuity you are referring to, but it doesn't seem useful to ask about its fine grained continuity since it is a macroscopic concept.  Statistical (quantum) Thermodynamics is used both for microscopic systems and as the basis for macroscopic quantities such as Gibbs Free Energy. You need to be careful: GFE is one meaning of Thermodynamic Potential, but not the only one. I don't have any idea what you mean by G's coercivity.


Molecular Orbital Hybridization vs Delocalization for positive charge stabilization

Basicity is the ability to donate a pair of electrons.
In the case of pyridine, the lone pair is in a sp2 orbital. This orbital does not participate in any resonance, and is smaller than the sp3. This means the electrons will be closer to the nucleus due to stronger attraction, making the pyridine a weaker base, since it's harder for it to donate these electrons.

Tertiary carbocations are more stable, that is correct, but you should look closely on what that means. A tertiary carbon is carbon atom that is to 3 other alkyl groups , not just any other atom. Alkyl groups are weak electron-donating groups, therefore stabilizing the carbocation, because they increase the electron density in the electron-poor center.
That is why according to Markovnikov's rule, the electrophile, in this case the proton, will bond to the less substituted carbon - in order to create a more stable secondary carbocation, which the nucelophile (the chloride) will attack.


There are many things to consider and improve your code. Many components
in an ab initio electronic structure calculations need supervision to achieve a
good performance. I am no expert on those calculations, but I will point you to some ideas and references. On the theoretical side, you could calculate E_ab for all shell pairs first. For more clarification on the order of operations and their scaling look at
Figure 3 in Multi-electron integrals. The article is written by a well-recognized researcher in the field that wrote a monumental book where you can find all needed expressions.
On the computational side, I would say that maybe Numba is not able to optimize your code. Writing it with Numba is not a guarantee of speed. I just took a glimpse at the code and, without going deep in the code, I saw many if statements. Only a straightforward observation. As general advice, it is all about how data is moved
and the flow of operations. Avoid allocations, try to rewrite as much as possible.
Profile your code to find the bottlenecks. Try to write it as vector or matrix operations to be able to use the large registers of the processor. There are many tricks to do these operations fast, but sadly they are buried elsewhere.
I would like to end with a reference to the recent literature on the topic. I just came up with an article that I managed to find again now: Fast Evaluation of Two-Center Integrals over Gaussian Charge Distributions and Gaussian Orbitals with General Interaction Kernels.


Radiation in space and its effects on chemical reactions?

It depends....

First let's define the "problem space".

Chemical reactions are typically driven by a few eV (electron volts). "Ionizing radiation" can be keV or MeV which is vastly excessive. In fact that amount of energy is so massive that it would "rip up" the chemical bonding in any material. Only solid metals would be somewhat stable since the 3D lattice locks the atoms into place. $\ce{NaCl}$ for example could decompose to Na metal and $\ce{Cl_2}$ gas.

Any sort of chemical reaction in space has to have a container of some sort which would mitigate the penetration of the ionizing radiation somewhat. But it surely won't stop all of the ionizing radiation. There are nuclear experiments that are conducted a mile underground to try to remove cosmic particles.

The primary source of ionizing radiation in our solar system is the sun. So more distance from the sun is your friend.

"Performing chemical reactions in space" sort of implies that a human is doing the reactions. A biological entity is going to be more effected than any vat of chemicals. So in an environment which humans or plants could safely operate, then performing chemical reactions won't be a problem.


Is the following molecule aromatic or antiaromatic?

It turns out this question has some subtleties.  The analysis presented below, based on Huckel MO theory, indicates that:

1.
The compound as presented by the OP is not expected to be aromatic and would be anti-aromatic by the Huckel model -- meaning in practice it would probably distort out of full conjugation to avoid this calamity.

2.
The dication, however, does appear to be aromatic.  Part of the reason is that while we ordinarily suppose that five-membered rings must be negatively charged in an aromatic system, a phenanthrene-like accumulation of three such rings can become aromatic with a positive charge instead.

Same formula, but ...

The OP suggests a comparison with the aromatic compound corannulene, which like the compound in question has the formula $\ce{C20H10}$ and twenty conjugated pi electrons.

Let's compare the Huckel energy levels for these two compounds, keeping in mind that an aromatic system will have (1) a stabilized highest occupied energy level and (2) a wide energy gap between this level and unoccupied states (figure made by the author):

Occupying the lowest available orbitals in corannulene on the left (including two degenerate pairs between $\beta$ and zero that are close together, thus appear as thick bars), we see the desired effect for aromaticity: stabilized highest occupied orbitals, wide gap to the unoccupied ones).  Not so on the right with the OP's compound: here there are nine bonding orbitals so putting in twenty pi electrons means having a mixture of unoccupied and occupied states at the nonbonding level where there is a degeneracy.  This is a classic antiaromatic situation (similar, actually, to triangulene which us invoked in another answer), and thus the answer to the original question is the compound we are examining will not be aromatic.  It will form a diradical or  find a way to get out of antiaromaticity instead.

Thrown for a loop

What happened?  In multicyclic systems aromatic coupling is not just a matter of having the right number of electrons.  Hidden in the energy calculations is a topological difference.  In corannulene five of the conjugated atoms are in an interior closed loop surrounded by the rest of the molecule, while the OP's compound lacks such an internal loop.  This topological change can alter aromatic coupling, which is sensitive to how loops of conjugated atoms are connected to each other.  For this particular combination of molecular formula and pi-electron count, corannulene has a good topology for forming an aromatic system.  The OP's compound does not.

Making lemonade out of lemons

Many Huckel-antiaromatic systems become aromatic if one places the proper charge on them, adjusting the number of pi electrons and thus occupied pi orbitals.  The classic case is cyclootatetraene which goes from an antiaromatic/distorted nonaromatic neutral molecule to an aromatic dianion when we combine it with potassium.  What can be done in this direction for the OP's compound?

According to the Huckel energy levels we ought to either fill the degenerate orbitals as in cyclooctatetraene, or empty them.  We see that in this case the latter is what works.  The dication, as opposed to the dianion, gas a stabilized highest orbital and a wide gap up to what are now unoccupied degenerate states.  In other words, the hallmarks of aromaticity.

But where can these positive charges reside?  We can make the seven-membered ring aromatic with one positive charge, but shouldn't the benzene rings be neutral and the five-membered rings negative like the cyclopentadienyl anion?

We crack open our topology textbook again and we discover that three separate rings are not connected in the same way as the accumulated five-membered rings in the OP's molecule/dication.  We'd better check the Huckel orbitals inherent to a cumulated five-membered ring system like the one in the OP's molecule, having the formula $\ce{C11H7}$ (this figure also by the author):

The seemingly similar phenanthrene system is a fourteen-pi-electron aromatic system.  Here, however, with fewer input atomic orbitals we see the widest gap, thus likely the greatest stabilization, occurs with only ten pi electrons.  This means the eleven-carbon system will have a positive overall charge.  The chain of three "cyclopentadienyl" rings generates an aromatic cation and thus can take on the second positive charge in the dicationic, predicted aromatic form of the OP's compound.


If brine is either unpalatable or has too high a freezing point for you, other options include a solution of glycerol in water. Glycerol is perfectly safe to ingest in modest quantities -- it's used as a food additive and a medicine -- although it tastes a little weird (a bit like extra-sour sweet and sour sauce).

Although it's mostly been replaced by ethylene glycol now, glycerol used to see a lot of use as an antifreeze, and is still used in domestic applications like ice cream makers. Apparently it's best used in a 60-70% solution, when it can get a freezing point below -40°C.

The paper linked below puts the optimum point at -46.5°C with two-thirds glycerol, one-third water by weight.

"Freezing Points of Glycerol and Its Aqueous Solutions", Leonard B. Lane, Ind. Eng. Chem., 1925, 17 (9), pp 924–924, DOI: 10.1021/ie50189a017


If 10ml of 40% sucrose is 15.7 grams, then the mass of 150 ml of 40% sucrose is

$$ 15 \times 15.7 grams$$

You can not determine the mass of 150 ml of 20% sucrose from the information given.

You could calculate from this sucrose solution density table:
http://homepages.gac.edu/~cellab/chpts/chpt3/table3-2.html

Note that the table shows the premise "I know is the mass of 10 ml of 40% sucrose, which is about 15.7 grams" to be incorrect.


What will the final product of this synthetic sequence be?

Potassium carbonate is not going to give you hydrolysis of a tertiary bromide. It will however close the halohydrin to form an epoxide (1). This intermediate undergoes Lewis acid catalyzed rearrangement to form the spiro compound that is choice (D).

Note that 1 or the carbocation could undergo certain elimination reactions to form (B) and (C). These are probably minor products.


Understand that ammonia typically adopts a trigonal pyramidal structure with a bond angle of $107^\circ$, which assumes $\mathrm{sp^3}$ hybridisation, rather than trigonal planar, which would assume $\mathrm{sp^2}$. Therefore, the lone pair is in a similar orbital as the other three $\ce{N-H}$ bonds, allowing for ammonia to act as a Brønsted base (accept $\ce{H+}$) and as a Lewis base (in coordination complexes).

The two states interconvert rapidly at room temperature.

The nitrogen atom is simply too small to allow for a bond angle of $90^\circ$ (unhybridised). The heavier homologue phosphorus, on the other hand, can allow this angle. Thus, phosphane $\ce{PH3}$ is much less basic than ammonia (even though phosphonium $\ce{PH4+}$ exists — but that is much more acidic than ammonium and can be used as a source for phosphane).


Relative levels of the two most prominant glutamate species at pH 4.7

The two species are related. They are conjugated acid and base to each other. Assuming one species $\ce{A^2-}$ with concentration $x \,\pu{mol/L}$, the second species $\ce{HA-}$ with concentration $y\,\pu{mol/L}$. There is a third species $\ce{H2A}$ with concentration $z \pu{mol/L}$.
$$\frac{10^{-4.7}\times x}{y}=10^{-9.67} \tag{1}$$
$$\frac{10^{-4.7}\times y}{z}=10^{-2.19} \ \ \tag{2}$$
Based on above two equations, the two most prominant species are $\ce{HA-}$ and $\ce{H2A}$ and ratio should be $y/z$. I utilized definition of $K_\mathrm a$ instead of $\mathrm pK_\mathrm a$ to get the above two equations. $10^{-4.7}$ is simply $[\ce{H+}]$ at $\mathrm{pH} = 4.7$
Therefore the final answer is:
$$\frac{y}{z} = 10^{2.51}$$


I can answer the chemistry part of your question... the candle-making part is better suited to a candle making forum.

There is no easy way of converting the other 90% of (older) pine-sol into pine oil and newer pine-sol has 0% pine oil. Best to purchase pure pine oil.

Pine oil is cyclic:

The other pine-sol ingredients are esters and other linear compounds. So using pine-sol to make pine oil would be ridiculous.

I am not certain how much pine oil is needed for a candle to be scented though. It may just be enough. For that you will have to just experiment... or ask on a candle forum.


Explain why these ions are more strongly hydrated in an aqueous solution?

I believe your reasoning is correct (more or less). Solvation is influenced by charge to size ratios. Higher the charge to size ratio, the easier it is for the species to be solvated by water molecules.

Apply this idea to both your cases. (note: in the second case, I don't get your argument about molar mass. The idea is much simpler, magnesium's atomic radius is smaller than sodium's and it has a higher positive charge. What can we conclude from this?

Also, reconsider the first case: a positive ion would not be attracted to the positive end of a dipole.


Gibbs energy: the term −TdS. Where does the reversible heat go?

The way to think about microstates is as alternative configurations of a system which satisfy constraints consistent with the observed macroscopic properties of the system. The collection of such microstates forms an ensemble.

The misconception lies in the idea of "concentrating" kinetic energy, and in thinking that reducing the number of available microstates changes the available energy. It doesn't necessarily do so. It is necessary to be careful, since the idea that "concentrating kinetic energy reduces the entropy" is not generally correct. In general you should consider all available degrees of freedom. An ideal gas only has translational degrees of freedom (kinetic energy) so in this case this restriction to kinetic energy applies.

In addition, during an isothermal process performed on an ideal gas, the internal energy does not change, and neither does the average kinetic energy ($K=\frac32 N \mathrm{k_B}T$), but the number of available microstates may change (for instance with an expansion), as reflected in the dependence of entropy on volume. The change in entropy reflects a change in the probability distribution over the ensemble.

To illustrate what is meant with "dispersion" (and by extension entropy) envision a maze in which the system wanders. As the maze gets larger the system gets more lost in that maze, it has more choices where to go.

If you use a particle in a box model, assuming quantum like behavior, then the gaps between energy levels become smaller as you increase the size of the container, that is, with expansion. This results in spreading probability of occupation (dispersion) over more states. Now, irrespective of how the isothermal expansion is performed - reversibly or irreversibly (assume a free expansion) - the entropy cost for the system is the same. However, in the case of the spontaneous expansion the surroundings remains at constant entropy, whereas its entropy is reduced in the reversible case. In addition (and this is a key point), no work was performed in the free expansion.


Predicting bond-strength of metal carbonyls

As the π-antibonding orbital of CO is filled by electrons from the metal, they weaken the C-O bond compared with free CO.  The metal-carbon bond is strengthened in kind. That's what the books says.


Mustard oil test for distinguishing between primary aliphatic and primary aromatic amines?

No, aromatic primary amines cannot be distinguished from aliphatic amines using mustard oil test as it reacts with carbon disulfide to give phenylisothiocyanate.

Aniline is reacted with carbon disulfide in the presence of mercury chloride to give the isothiocyanate. The reaction is given in this site.

$$\ce{C6H5NH2 + S=C=S ->[HgCl2] C6H5NCS + HgS + 2HCl}$$.

Aliphatic amines also give the same products,  isothiocyanate, mercuric sulfide and hydrochloric acid and as such this test cannot be used to distinguish between aliphatic amines and aromatic amines.

Alternatively, it is also formed in the presence of ammonium hydroxide and lead nitrate.


Why does blood get further oxidized than Iron oxide?

There are two misconception behind your question:

You assume you can directly compare the colors of different compounds and derive oxidation states e.g of a heme (porphyrin) comparing to iron-oxide. No, it is not true.
You assume this whole story has anything to do with copper. No, the blueish-green is nothing to do with copper, it is a (porphyrin) degradation product.

The color of transition metal compounds mostly can be derived from the crystal field (ligand field) around the central metal ion. In inorganic iron-oxo/hydroxo compounds have different structure and hence different field than heme compounds. In heme, the N coordination of porphyrin ring is generating a very strong planar field, and the oxygen coordinates as molecule on the top of it. Charge transfer to the ring can also be an important factor in heme derivatives. As you see it is a whole complicated mass, and not much direct connection with different inorganic salts you see in lab.

The colors you see are coming from oxo-hemoglobin (which you have in your healthy blood), and most probably several steps of enzymatic degradation of heme. The porphyrin ring is oxidized and cut open by an oxygen on the the iron. Several colorful compounds are produced, including the iron complexes themselves and biliverdin (green), bilirubin (yellow). These are the same colors you see under your skin, of you hit yourself hard and have some bleeding in your tissues. First the injured area is deep red/blue, than go to greenish, finally yellowish patch. This also happens continuously in your body, when your blood cells get old, and heme is enigmatically degraded so the whole thing can be recycled:  http://www.elmhurst.edu/~chm/vchembook/634bilirubin.html


Why weren't the electrons attracted to doubly charged Helium particles in the Rutherford experiment?

The gold foil experiment conducted by Rutherford (or more appropriately by Geiger and Marsden under Rutherford's direction) involves a stream of $\alpha$ particles (which are helium nuclei) bombarding a thin gold foil.

Why didn't some electrons go with the alpha particles because of electrostatic attraction? Some probably did. The experiment was equipped with a fluorescent zinc sulfide screen to detect alpha particles ($\ce{He^{2+}}$ ions). This screen might have also detected $\ce{He+}$ ions (but not necessarily in a way that was different than the detection of alpha particles), but probably not free electrons or helium atoms. The experiment was only set up to detect alpha particles, and so that's all that Rutherford, Geiger, and Marsden observed. We'll never know how many (if any) alpha particles picked up electrons in their experiment. We could, however, repeat the experiment, replacing the fluorescent screen with a mass analyze to detect at least the difference between $\ce{He^{2+}}$ and $\ce{He+}$.

More importantly, we known that many alpha particles passed through (and some were deflected) unchanged. Why didn't these particles pick up electrons? The answer to this question is another question: What is holding those electrons in the gold foil?

Gold atoms have 79 protons holding down those electrons. The first ionization energy of gold is 890.1 kJ/mol and the second is 1980 kJ/mol. However, these values are for isolated gold atoms. Gold (like all metals) is held together by metallic bonds with delocalized electrons occupying a valence band. Each electron is not being held by just one gold nucleus. Once the first few electrons are ejected or snatched by the alpha particles, the gold foil will have a net positive charge. It will be much harder to remove negatively charged particles from the positively charged foil.


Why can you not make a compound with a quaternary alpha carbon using malonic ester?

Double alkylations of the $\ce{CH2}$ centre of malonic esters do work!

Have a look at this marvel by R. P. Mariella and R. Raube, published in Org. Synth., 1953, 33, 23

EDIT 1

The use of sodium metal in a properly dried alkanol (in the case of ethanol, refluxing over magnesium ribbon is a good choice) was (and still is) an excellent and cheap method to generate sodium alkoxide in situ:

\[\ce{2Na + 2ROH -> 2NaOR + H2}\]

I don't know if nowadays commerically available sodium ethoxide and/or sodium hydride were that much used back in the days when the procedure was published. Fact is that both these can rot away in humid air when somebody leaves the bottles and cans open for too long, and this happens frequently in university labs.

With sodium metal, you're on the safe side: a freshly cut piece of metal is always good to go.

EDIT 2

For the example given above, no further direct alkylation at that position is possible, and this is exactly, what ron pointed out in his
comment to the question:

  Perhaps the author was trying to say that by using just the malonic ester synthesis you can't prepare a compound with 3 (potentially different) alkyl groups on the alpha carbon.

He also suggested the viable solution, which I had outlined in another question on this site. Upon acidification and heating, $\beta$-ketoesters and 1,3-diesters undergo decarboxylation, which makes them great synthons to introduce $\ce{-CH2CO-CH3}$ and $\ce{-CH2COOH}$, respectively.

Converting a 1,3-diester to the corresponding carboxylic acid this way means that the new structure now again has an $\alpha$-proton that can be abstracted. The resulting enolate can be alkylated.

In the case of a carboxylic acid, this is usually done in a solvent like tetrahydrofurane or 1,4-dioxane, using two bases. For the deprotonation of the acid, $\ce{NaH}$ is sufficient, while the abstraction of the $\alpha$-proton, a strong base like $\ce{LDA}$ is used. (Exactly this is outlined in the textbook for the alkylation of an ester).


How do scientists study liquid tungsten?

One way is to use electristatic or electromagnetic forces to hold the liquud in place.  This abstract refers to a "non-contact method" and an "electrostatic levitate", which us enough to reveal the basic strategy.  Unfortunately the article is behind a paywall, so you have to put up to get details on the good stuff.

Industrially, metals like tungsten are not processed as liquids.  They are ... see here.


Electroplating an alloy from separate metal solutions

In short, yes, it is possible to electroplate alloys, but it isn't as simple as just mixing the ions you want in the ratio you want.
In the simplest approximation, the alloy deposition would behave as two simultaneous, separate depositions at the same voltage and the composition could be controlled by changing the voltage, assuming each component has a different current-voltage response.
In the real world, unfortunately, there are numerous interactions between the ions in solution, the metals on the electrode, etc. and the outcome becomes quite complex to predict.
Here is a decent open-access review of the chemistry and applications of this type of process.


See the Wikipedia article on octane rating. 95% octane is the same as an octane rating of 95, and it means the petrol has the same knocking characteristics as a mixture of 95% iso-octane and 5% heptane.


The method shown in OP's question to find a meso-compound is the way most people used. However, I usually used the Cahn-Ingold-Prelog (CIP) assignments of chiral centers for this purpose. It is a fact that the mirror image of (R)-chiral center is corresponding (S)-chiral center. For example, bromochlorofluoromethane is a chiral compound and the morror image of its (R)-stereoisomer is its (S)-stereoisomer as depicted in the diagram below:

When (S)-isomer is rotated $180^\circ$ vertically, the resultant structure would not be superimposed on (R)-isomer, as shown in the diagram. This is to justified the above mentioned fact.

Now, let's assign the compound in hands with CIP system. I'd call it (2 R, 3 S)-butane-2,3-diol (it could also be (2 S, 3 R)-butane-2,3-diol, based on the numbering). Either way, carbon #2 has three different groups attached to it, which are identical to those on carbon #2. Based on the given assignments on carbon #2  and #3, if you'd put a mirror between these carbons, you must see the mirror image of one part (carbon #2) on other side (carbon #3) as depicted in the diagram. Therefore, the compound has plane of symmetry (the mirror) and hence, it is the meso-isomer.

To conclude this fact, I included two enanthiomers of this compound ((2 R, 3 R)-butane-2,3-diol and (2 S, 3 S)-butane-2,3-diol), none of which gives mirror images when a mirror placed in between two chiral carbons (see the diagram).


IUPAC nomenclature of cycloalkadienes with substituents and alkenes on the same carbons

The question is based on a false premise:

Generally we try to give just enough information to be able to reconstruct the original compound, but not too much.

This isn’t true. The prime aim of IUPAC nomenclature is clarity and the construction of unambiguous names.

Three logicians walk into a bar
The bartender asks: “Do you all want a drink?”
The first logician says: “I don’t know.”
The second logician says: “I don’t know.”
The third logician says: “Yes.”

IUPAC nomenclature doesn’t work like that. Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information or hidden logic involving other structures.
The fact that “1-methylcyclohexa-2,4-diene” would actually be called 5-methylcyclohexa-1,3-diene does not change the preferred name for 1-methylcyclohexa-1,4-diene.
Nevertheless, there are a few exceptional cases where locants are omitted, but only when there is no ambiguity. For preferred IUPAC names, the nomenclature rules explicitly stipulate when locants are omitted. In most cases, if any locants are essential for defining the structure, then all locants must be cited for the structure.
Also note that your third example is clearly wrong even when following your unorthodox logic. Your name “1,4-dimethylcyclohexadiene” could mean 1,4-dimethylcyclohexa-1,4-diene as well as 1,4-dimethylcyclohexa-1,3-diene or even 1,4-dimethylcyclohexa-1,2-diene, which are all correct IUPAC names.


What is the chemistry of this household cleaning recipe?

Baking soda is sodium bicarbonate, and vinegar is dilute acetic acid. Common salt (table salt) is sodium chloride. Sodium bicarbonate will react with acetic acid to form sodium acetate (in solution), carbon dioxide gas, and water. Sodium chloride will not participate in the reaction. The reaction equation would be:

$$
\ce{NaHCO3(s) + HCH3COO(aq) -> NaCH3COO(aq) + CO2(g) + H2O(l)}
$$


How to determine the total pressure of a gaseous mixture of xenon tetrafluoride and hydrogen?

You are correct up to the last part. The total pressure would be the sum of all the partial pressures in the reaction vessel.

We know that $$P=\left(\frac{RT}{V}\right)n$$ and therefore $n$ varies directly with $P$ as $T$, $R$, and $V$ are constant. For the sake of the problem (it won't matter) we can just say that $x=\left(\frac{RT}{V}\right)$. Therefore, $P=xn$ and the starting amounts of substance are equal to $\frac{2}{x}\pu{ mol}$ $\ce{XeF4}$ and $\frac{10}{x}\pu{ mol}$ $\ce{H2}$.

From the balanced equation and the starting materials and stoichiometry, we know that $\frac{2}{x}\pu{ mol}$ $\ce{XeF4}$ would react with $\frac{4}{x}\pu{ mol}$ $\ce{H2}$ to form $\frac{2}{x}\pu{ mol}$ $\ce{Xe}$ and $\frac{8}{x}\pu{ mol}$  $\ce{HF}$ to make $\frac{10}{x}\pu{ mol}$ product.

However, you only reacted $\frac{4}{x}\pu{ mol}$ $\ce{H2}$ of the $\frac{10}{x}\pu{ mol}$ total, so there is $\frac{6}{x}\pu{ mol}$ unreacted $\ce{H2}$ left in the reaction vessel.

$$\begin{multline}
  \frac{10}{x}\text{ amount of substance of product }
+ \frac{6}{x}\text{ amount of substance of unreacted }\\
= \frac{16}{x}\text{ total amount of substance}
\end{multline}$$

From $P=xn$, the fact that this occurred instantly, and $T$, $R$, and $V$ are constant, we then know that the final pressure is $\ce{16 kPa}$.


The top of the p block is unusual in the atoms being so small they can readily form strong multiple bonds, because (from a VB point of view) they can get close enough for good side-by-side p orbital overlap.  Hence they prefer to form multiple bonds if they can, and both N and O can do that to form diatoms.  So extended chains of either N or O atoms are generally less stable than a collection of N2 or O2 molecules.  Or if you have extra capping atoms around, like H, then a collection of N2/O2 plus some H2O/NH3, et cetera.

Carbon is the exception because it has four valence electrons, and a quadruple bond to form C2 is too high energy.  (From a VB point of view, there are no valence atomic orbitals left to overlap to form the 4th bond.)  The next smallest molecule it can form with itself with lots of double bonds is buckminsterfullerene (C60).  So without another element around, a long chain of O atoms can form a more stable collection of O2 molecules, while a long chain of C atoms can't form anything more stable at all.  (When other elements are around, all bets are off: long chains of carbon tipped with hydrogen are notoriously unstable to CO2 and H2O if there's plenty of oxygen around.)


Solvent effects on acidity of ethanoic acid

Ethanoic acid has $\mathrm{pK_a}$ values of 4.76, 9.63 and 12.3 in water, methanol and DMSO respectively (source here and here). Comparing this with the $\mathrm{pK_a}$ values given in the question we can see that ethanoic acid is most acidic in water and least acidic in DMSO. $$\ce{pK_{a}(H2O) < pK_{a}(CH3OH) < pK_{a}(DMSO)}$$

In general the acidity of a species in different solvents is affected by three factors:

The dielectric constant of the solvent. Solvents with larger dielectric constants favour charge separation because the force pulling the charges together as they are separated is lower. This is evidenced by the Born equation which estimates the Gibbs free energy of solvation of gaseous ions: $$\Delta G = -\frac{N_Az^2e^2}{8\pi\epsilon_0r_0}\Bigg(1-\frac{1}{\epsilon_r}\Bigg)$$
The strength of solvation of the conjugate base of the acid.
The strength of solvation of the conjugate acid of the solvent.

Dielectric constant: The dielectric constants of water, methanol and DMSO are 80.1, 32,7 and 46.7 respectively (source). This gives some support to the greater acidity in water but contradicts the observed acidity in methanol and DMSO. This suggests that dielectric constant may not be the major factor that determines the relatives acidities.

Anion solvation: As you suggested in the question, water and methanol are protic solvents but DMSO is aprotic and so the ethanoate ion will be more strongly solvated in water and methanol, shifting the acid-dissociation equilibrium in favour of dissociation. This is considered to be the main factor which explains the greater acidity of ethanoic acid in protic solvents as opposed to aprotic solvents. When comparing water and methanol the strength of anion solvation will be broadly similar because both form $\ce{O-H\bond{~}Anion}$ hydrogen bonds.

Cation solvation: The difference in the stability of the conjugate acid of the solvent may be the factor that causes the difference between water and methanol. The $\mathrm{pK_a}$ of $\ce{CH3OH2+}$ is approximately -2.5 (several sources, none giving exact answers) and the $\mathrm{pK_a}$ of $\ce{H3O+}$ is -1.7. Therefore protonated methanol is less stable than protonated water and so this disfavours dissociation of the acid in methanol. This page claims that this can be attributed to oxygen $p$-orbital donation into $\ce{C-H}~\sigma^*$ orbitals but I don't have enough knowledge of MO theory to verify this (maybe someone else can).

Overall the greater acidity of ethanoic acid in water than methanol seems to be down to the greater dielectric constant of water and the greater stability of the solvent cation formed.

This page says some good things about the subject (some of my ideas were developed from reading this).

Additional reading: This paper looks to have some detail on the anion solvation aspect but I can't access the full text so I'll leave it as a link.


Preparing an inert atmosphere reaction vessel with only an input?

Her setup was wrong. There needs to be an outlet, the simplest solution in this case being a second needle with nothing connected to it, puncturing either septum to act as a vent to the atmosphere. After a while, depending on the volume of the flask and amount of nitrogen flow, most of the atmospheric air is removed.

This procedure is, however, quite inefficient, and likely unsuitable for particularly sensitive reactions. While most oxygen and moisture is quickly removed, it becomes exponentially slower to remove the last remnants. A much quicker and more reliable method is to perform vacuum/nitrogen backfill cycles. Three or four cycles will fully replace the interior atmosphere with nitrogen gas of the same purity as the input. This can be performed using a single needle and a Schlenk line (or any system connected to nitrogen and a vacuum pump with a intermediating stopcock), which essentially switches the same needle between a gas inlet and a gas outlet.


Side products during the electrolysis of Copper Sulfate?

There are a few key factors that I expect are contributing here:

$\ce{CuSO4}$ concentration – If your $\left[\ce{Cu^2+}\!\right]$ in solution is low, then that will disfavor deposition, as permeakra noted in a comment. You want at least $25$ to $100~\mathrm{g\over L}$ of copper in the electrolyte.
Cell voltage – $15~\mathrm V$ is an exorbitantly high voltage for copper electrodeposition. In order to avoid gassing at the cathode, you'll want to operate down in the $1.5$ to $2.0~\mathrm V$ range, tops.
Electrolyte conductivity – you don't mention anywhere that you've added other constituents to try to increase the conductivity.

If this is a home experiment, I would try dissolving the $\ce{CuSO4}$ in table or pickling vinegar instead of plain water. That will give you more charge carriers in solution, and also decrease the $\mathrm{pH}$, both of which should help.
If this is a school or work experiment with access to proper lab equipment and chemicals disposal services, I would use $1\%$ to $10\%~\ce{H2SO4}$ as the base electrolyte. That will really get your conductivity up, and your cell voltage down.

Finally, on the methods side, I would recommend choosing your voltage in order to initially target a current density in the vicinity of $\sim\!10~\mathrm{mA\over cm^2}$, as a value where I would expect to see little to no $\ce{H2}$ evolution. You can always nudge the voltage up from there as needed.

  Is $\ce{H2}$ generation happening as a result of the weak sulfuric acid in solution?

Almost certainly not.

  Can you calculate the products of simultaneous electrochemistry reactions?

Not with much certainty. Most of the time the best one can do is make qualitative arguments about what reactions might be happening.  Detailed modeling of all reactions occurring at an electrode is still prohibitively difficult, as far as I know.


Distinguishing between octahedral and tetrahedral holes

To visualize holes in closest packing of spheres, you need 6 balls (marbles, table tennis balls, etc) of equal size. 3D closest packing can be constructed from 2D layers of closest packed spheres as such:

The light blue layer is on the bottom and the dark blue layer is on top. The holes, tetrahedral or octahedral, are in between layers.

To find a tetrahedral hole, take three balls and arrange them in a triangle to form part of the lower layer. Then place one sphere in the center on top (you have to keep the lower level together somehow or the structure will collapse). Your tetrahedral hole is in the center of the tetrahedron formed by the 4 balls (marked T in the picture).

To find an octahedral hole, you need a helper or glue. Take three balls and arrange them in a triangle. Take another three balls and arrange them in a triangle. Place the two triangles on top of each other, twisted against each other by 60 degrees. This is a octahedron on its side, and in the center is the octahdral hole.

In the picture above, these positions are marked by T and O. You just have to imaging that they are in between the layers, and then you might be able to count nearest spheres. The picture below shows the arrangement of the nearest spheres, only.

If you don't have 6 balls handy, but have two hands handy, you can also do the following. Place thumb, pointer and middle finger together so that the finger tips form a triangle. Point at the center of the triangle with the pointer of the other hand. That's the tetrahedral hole. Now, form a triangle with the finger tips of your other hand. Bring the 6 finger tips together with a little twist so that they "lock in". At the center of the 6 finger tips, you have the octahedral hole.

So, whether you are a visual or a haptic learner (not that that would be a thing), this will do the trick, I hope.

  There are some images of octahedral and tetrahedral holes

Actually, I only found images of the spheres representing the atoms, no images of the shape of the space in between. If you squeeze some blue play dough between two layers of marbles, here is what you get (see also https://leancrew.com/all-this/2019/06/cheesegraters-and-crystals/):

Within a layer, each indentation is surrounded by six others. The arrows point at the six locations with the largest voids (in the center of triangles formed by the sphere and two of its nearest neighbors in that layer). If you look closely, each indentation has three holes, the contacts with the layer below. The voids take on different shapes depending what is underneath. If a sphere is directly underneath, it is a tetrahedral hole. The shape of this is a tetrahedron with "dented in" faces (far right). If a center of a triangle is directly underneath, it is an octahedral hole. The shape of this is a cube with "dented in" faces (near right; cube and octahedron are complementary archimedian solids: the centers of the faces coincide with the other's vertices and vice versa. A tetrahedron is complementary to itself).

If you shrink the spheres around the atoms to points, you get a space-filling arrangement of tetrehedra and octahedra called a tetrahedral-octahedral honey comb.

Teaching this material

Stefanie Lenzer, Bernd Smarsly and Nicole Graulich recently published a paper in J Chem Ed on using clear spheres to explore crystal packing and voids.


Le Chatelier's principle with pure solids and liquids

Your first statement is right.
You just need a certain amount of solid or pure liquid for the reaction to occur, just like how putting in more solid in a saturated solution doesn't increase dissolution, while you need solid in there to call it an equilibrium between aqueous and solid phases.
However, the amount of solid or liquid does change the kinetics of the reaction: the rate at which it proceeds to equilibrium.

But if you put in some solid A, the Q is 0, so some B and C will form. It's the same reason why putting in a sugar cube in water increases aqueous sugar concentration: there is an equilibrium between solid and aqueous components, it's just that the constant means the maximum amount of aqueous concentration.


Can a reaction of the form A + B → AB ever be endothermic?

A bond does not simply form because two atoms are close enough. You also have to bring them into the right electronic state to do so. Which takes various amounts of energy. So yes, an A+B -> AB reaction can be endothermic.

Also beware of mistaking the abscissa in those energy diagrams as a distance: It's usually just a symbolic "reaction progress", or it is a line on a multidimensional hypersurface describing the energy potential with respect to all angles, distances, electronic parameters etc. in the reaction.


What is the nature of the mysterious Kw, the ionic product of water?

The solution product for the reaction $\ce{H2O -> H+ + OH-}$ is a constant, $10^{-14}$, which we call $K_w = [\ce{H+}][\ce{OH-}]$. To make any aqueous solution, we put a solute into water. Because the dissociation constant (which is basically an equilibrium constant) of a substance only changes with temperature, $K_w$ remains the same. Thus, for any aqueous solution at a given temperature, $K_w$ is constant.

You need to treat $K_w$ as an equilibrium constant. For example, if I have a reaction $\ce{A -> 2B + C}$, $K_{eq} = \frac{[\ce{B}]^2[\ce{C}]}{[\ce{A}]}$; let's say I measured at equilibrium $[\ce{B}] = 1$ M, $[\ce{C}] = 2$ M, and $[\ce{A}] = 0.1$ M . This means $K_{eq} = 20$. Now, if I increased the concentration of $[\ce{A}]$ to $5$ M, $K_{eq}$ wouldn't change: the concentrations would have to adapt to my change. This is the same for water: once I determine $K_w$, it is the same for all concentrations of $\ce{H+}$ and $\ce{OH-}$.


The system doesn't "want" anything.  This is a major misconception with Le Chatelier's principle.

Think about the following reversible reaction, everything in the gas phase:

$$\ce{A + B <=> C}$$

We'll simplify the rate law of the forward reaction to

$$\text{rate} = K_\mathrm{fwd}[\ce{A}][\ce{B}]$$

The reverse rate law is

$$\text{rate} = K_\mathrm{rev}[\ce{C}]$$

If the volume is doubled, what happens to the rate of the forward reaction?  It is quartered.  What happens to the forward reaction?  It is halved.  So, the reaction "shifts to reactants".

If the volume is halved, what happens to the forward rate?  It is quadrupled.  The reverse rate is doubled.  The reaction "shifts to products".

What happens when an inert gas is added?  Since the inert gas appears in neither the forward nor the reverse rate law, nothing happens.

I'm afraid you have learned some misconceptions.

Le Chatelier's principle was developed before much of what is studied in a general chem course today was known.  It has been used to manipulate the outcome of reactions.  All of the predictions made by Le Chatelier's principle can be made by understanding how reactions work at the particle level.  Le Chatelier's Principle is from the macroscopic perspective and leads many students to generate misconceptions.


Do crown ethers have to have two carbons between the oxygens? If so, why?

To be a crown ether, the compound in question must be an ether, i.e. must contain oxygen atoms which are bonded to two carbon atoms neither of which is bonded to another oxygen atom. Thus, if you had just one carbon between two subsequent oxygens you would no longer be dealing with a crown ether; if you wanted, you could call it a crown acetal. But it would have vastly different properties; most notably it would break down in aquaeous solution under acid catalysis.

To the best of my knowledge, there is no formal restriction other than that given in the first paragraph for crown ethers. 20-crown-5, which would have three $\ce{CH2}$ groups between each ether oxygen, would be a valid crown ether in my humble opinion. However, most cations that are coplexed by oxygen prefer five-membered rings for various reasons, resulting in many complexing agents — $\ce{en, dmg, m2g, phen}$ to name just a few examples — having exactly two other atoms between the ligating atoms. Thus, practically all crown ethers one hears of have two carbons between the oxygens.


Can Fe and Ca react and become unusable in some foods?

Studies have indicated that calcium inhibits iron absorption in humans. However, numerous dietary factors influence iron absorption, and the effect of calcium may not be as pronounced when calcium is served as part of a whole diet. Calcium does not have any significant long-term effect on iron absorption. In addition, milk and milk products have not been found to affect iron absorption.
References:

Lönnerdal B. Calcium and iron absorption--mechanisms and public health relevance. Int J Vitam Nutr Res. 2010 Oct; 80(4-5):293-9. doi: 10.1024/0300-9831/a000036.
Grinder-Pedersen et.al., Calcium from milk or calcium-fortified foods does not inhibit nonheme-iron absorption from a whole diet consumed over a 4-d period, The American Journal of Clinical Nutrition, Volume 80, Issue 2, August 2004, Pages 404–409, doi: 10.1093/ajcn/80.2.404
https://www.dairynutrition.ca/nutrients-in-milk-products/calcium/calcium-and-iron-absorption-is-there-an-interaction


I would love to find numbers, but alas, I have failed. If anyone can point me towards energy differences between the orbitals I am lacking, please feel free!

There is no a priori physical reason why lithium, sodium, potassium and calcium give flame colours but beryllium and magnesium would not. For all these elements (and hydrogen), the principal mechanism works in the same way: An electron is thermally excited into a higher orbital, and when it relaxes back to its ground state it releases a photon of exactly that wavelength. The flame colour is always due to gaseous atoms (not ions), so the electrons are always excited from an s-orbital.

The strongest lowest-energy excitation and relaxation should always be $n\mathrm{s} \rightarrow n\mathrm{p}$, i.e. from one shell’s s-subshell to that shell’s p-subshell. For sodium, this energy difference corresponds to $589~\mathrm{nm}$ or $2.10~\mathrm{eV}$, for lithium it is $671~\mathrm{nm}$ or $1.85~\mathrm{eV}$, for potassium $767~\mathrm{nm}$ and $1.61~\mathrm{eV}$ and for calcium $657~\mathrm{nm}$ and $1.89~\mathrm{eV}$.[1]

We can see that the energy difference rises from potassium to calcium, even though they are neighbours and the excitation is from the same orbital to the same different one. This is because calcium’s higher nuclear charge contracts the orbitals, stabilises them and thus lowers their energy — compare with the ionisation energy which also typically rises across a period. This effect is, of course, more pronounced when going across the second or third period when compared to the fourth; thus, I would expect an even greater energy difference when going from sodium to magnesium and even greater than that from lithium to beryllium.

The best explanation I have for the colourlessness of magnesium and beryllium is that their excitations are too close to ultraviolet to be properly observed due to strong contractions of the orbitals, but without numbers that is a weak explanation at best.

Notes:

[1]: I would be grateful for confirmation that these values I found are correct.


Enthalpy of Reaction and Efficiency of Electrochemical Energy Devices

The efficiency is  $\eta=\displaystyle \frac{w}{\Delta H}$ where $w$ is the useful work output, and $\Delta H$ the reaction enthalpy. This can be written as $\displaystyle \eta=\frac{(\Delta H-Q_{in})}{\Delta H}$ and $Q_{in}$ is the heat added where $Q_{in}=T\Delta S$ when operated reversibly giving the efficiency equation you quote. In effect the efficiency accounts for heat losses or gains relative to the reaction energy available. Gibbs was first to realise that because heat from the surroundings can be added to or removed in a reaction the work done can be greater (or less than) than $\Delta H$. In your definition the efficiency can therefore be $\gt 100$%.


Which make HCO3- to show two pH values at two scenarios?

The problem is that both titration curves are wrong: at the first equivalence points, the $\mathrm{pH}$ does not equal $\mathrm{p}K_\mathrm{a1}$ and the $\mathrm{pOH}$ does not equal $\mathrm{p}K_\mathrm{b1}$. As well, the two provided dissociation constants for carbonate ion, in the OP's upper figure, are also wrong.
The figure below shows two titration curves: 1) for $\pu{10 mL}$ of $\pu{0.1 M}$ "carbonic acid", with $\pu{0.1 M}$ $\ce{NaOH}$ titrant (black rising curve) and 2) for $\pu{10 mL}$ of $\pu{0.1 M}$ sodium carbonate solution, with $\pu{0.1 M}$ $\ce{HCl}$ titrant (red falling curve):

The $\mathrm{pH}$ of the starting $\pu{0.1 M}$ carbonic acid is 3.67 and the $\mathrm{pH}$ of the starting $\pu{0.1 M}$ sodium carbonate is 11.65. At the first equivalence point, for both titration curves, the $\mathrm{pH}$ is approximately 8.34. The $\mathrm{p}K_\mathrm{a}$ values are as shown in the figure and are from (or derived from) the reference at the bottom.
The next two figures show Excel spreadsheet $\mathrm{pH}$ calculations. The first one shows that $\mathrm{pH} = 8.338$ at the first equivalence point, for the neutralization of sodium carbonate solution with $\ce{HCl}$:

At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate plus $\pu{0.05 M}$ $\ce{NaCl}$. Ignoring ionic strength effects, the $\ce{NaCl}$ does nothing to the $\mathrm{pH}$.
The next one shows that $\mathrm{pH} = 8.339$ at the first equivalence point, for the neutralization of carbonic acid with $\ce{NaOH}$:

At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate. In both cases, the $\mathrm{pH}$ is approximately 8.34 and is consistent with this answer.
The two $\mathrm{p}K_\mathrm{a}$ values for carbonic acid are from D.G. Harris, In Quantitative Chemical Analysis, 7th Edition; Appendix G; W. H. Freeman and Company, ©2007. The two $\mathrm{p}K_\mathrm{b}$ values for carbonate ion are obtained from the two $\mathrm{p}K_\mathrm{a}$ values for carbonic acid via $\left(\mathrm{p}K_\mathrm{b1} = 14.00 - \mathrm{p}K_\mathrm{a2}\right)$ and $\left(\mathrm{p}K_\mathrm{b2} = 14.00 - \mathrm{p}K_\mathrm{a1}\right)$.


How to find out unpaired electron in S2 molecule?

The molecular orbital energy-level diagram of $\ce{S2}$ is given below:-

(source)

We can see  how close the energy the 3s and 3p atomic orbitals are because their energy separation will determine whether the  $\ce{π_3_{p}_{x,y}}$,  or the $\ce{σ_{3}_{p}_{z}}$ molecular orbital is higher in energy. Because the ns–np energy gap increases as the nuclear charge increases , the $\ce{σ_{3}_{p}_{z}}$ molecular orbital will be lower in energy than the $\ce{π_3_{p}_{x,y}}$ pair.

Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons. Ten valence electrons are used to fill the orbitals through $\ce{π_{3}_{p}_{z}}$and $\ce{π_3_{p}_{y}}$, leaving 2 electrons to occupy the degenerate $\ce{π^{⋆}_{3}_{p}_{z}}$ and $\ce{π^{⋆}_{3}_{p}_{y}}$ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of $\ce{S2}$ is $\ce{(σ_{3}_{s})^{2}(σ^{⋆}_{3}_{s})^{2}(σ_{3}_{p}_{z})^{2}(π_{3}_{p}_{z,y})^{4}(π_{3}_{p}^{⋆}_{x,y})^{2}}$ with 2 unpaired electrons at the antibonding $\ce{π^{⋆}}$ orbital and thus is paramagnetic. The bond order is (8 − 4) ÷ 2 = 2, so we predict an $\ce{S=S}$ double bond.


How would one correct for a blank in a potentiometric titration?

Even though the other compounds should theoretically not interfere with your titration, a method/matrix blank like you've proposed is always a good idea.

What I would do is simply subtract the blank potential value from that of the sample for each volume addition, then plot and find the inflection point as you would usually do (i.e. from the derivative plot, etc.).

If you are correct in your assumption that the other compounds won't interfere, then the position of the inflection point should be the same as that calculated without the blank, within experimental error (i.e. within a standard deviation or two or three, etc. of a set of replicate titration values.).  If you are incorrect and there is some interference, then the blank subtraction will correct for it and the blank corrected value is what should be used/reported.


Which silver halide is the most soluble?

The solubilities of silver halides decreases down the periodic table:

$\ce{AgF}\ : K_{sp} = 205$
$\ce{AgCl}: K_{sp} = 1.8\times10^{-10}$
$\ce{AgBr}: K_{sp} =5.2\times10^{-13}$
$\ce{AgI}\ \ \ : K_{sp} =8.3\times10^{-17}$

The rationale for this trend is typically described using the concept of hard/soft acids and bases.  (See J. Chem. Educ., 1968, 45, page 581 and also page 643 for early discussions on HSAB theory.)  Briefly, ions with similar size and high polarizability will lend themselves to more covalent interactions, and therefore will have decreased solubility in water.  HSAB classifies ions as either hard or soft, with soft being large and polarizable and hard being small and highly charged.  H-H and S-S interactions are strong while H-S or S-H interactions are weaker.  Ag is classified as soft and the halides go from hard (F-) to soft (I-).


Epoxide ring opening via decarboxylation

You are correct about protonating the epoxide of the glycidic acid 1 (by some other molecule of glycidic acid). Ring opening gives stable, tertiary cation 3 and NOT cation 6, which is destabilized by the carbonyl group. Collapse of 3 gives CO2 and the enol 4 which tautomerizes to the aldehyde 5.


Does excessive sulfuric acid as catalyst affect the synthesis of aspirin?

Did the addition of excessive sulfuric acid affect this in any way? What could have caused salicylic acid crystals to be produced instead of aspirin?

The intended reaction was the formation of an ester from a a phenol and an anhydride.

However, with an excess of sulfuric acid, acidic hydrolysis of the product according to an $\mathrm{A_{Ac}2}$ mechanism takes place.


What difference between pi bond and bent bond? Is it possible to replace sigma-pi bond model and use only bent bond instead?

In the famous words of George Box, "All models are wrong. Some are useful."
The bent bond and sigma/pi descriptions are two apparently distinct models for representing the underlying structure of electron density of molecules. Both are oversimplifications, but the key question (as with any model) is which is better for accounting for all of the known molecular structures and predicting the results of observations we have not yet made.
There are examples that support both. Bent-bonds have been cited famously as better predicting the conformation of propene, but the sigma/pi gives more intuitive results for reactions like the Diels-Alder. As in the case of the VB vs MO theory debate, both models can be elaborated with greater complexity to the point that they give indistinguishable results, but the sigma/pi approach has largely taken over because in its basic form it is more intuitive and gives accurate predictions in most cases.
The general concept of a bent bond remains useful for highly strained systems such as cyclopropanes, where the electron density of a single bond does not appear to align with the axis connecting the atoms.
To answer your final question,

If it not, then how can we consider which chemical result will
be bent bond, and which will form pi bond?

it is rare that the sigma/pi model will fail you in modeling a molecule with a double bond, and since it is the prevailing model pedagogically, you would do best to use it preferentially and only invoke bent bonds for strained single-bond systems.


How to determine the reaction of a strong electrolyte?

$\ce{HMnO4}$ is s a strong acid (compare oxidation state with perchloric acid).  Its anion is then a weak base.  The salt is completely dissociated in water solution.


How do I calculate the density of propane assuming an ideal gas?

As you have already figured out that we have to use the ideal gas law $PV= nRT$, which is you have modified as $$\frac{n}{V}= \frac{P}{RT}$$

To find the density, we can further modify the equation as,
\begin{align}
\frac{m}{MV} &= \frac{P}{RT}\\
\frac{\rho}{M}&= \frac{P}{RT}\\
\rho &= \frac{PM}{RT}\\
\end{align}

Now we just have to replace the parameters with their respective values.

The value of temperature should be converted to Kelvin as $(0 + 273)~\mathrm{K} = 273~\mathrm{K}$.

Likewise, the value of gas constant $R$ should be chosen as  $0.8206~\mathrm{L~atm~K^{-1}{mol}^{-1}}$.
\begin{align}
\rho &= \frac{1~\mathrm{atm} \cdot 44.09562~\mathrm{g~mol^{-1}}}{{0.8206~\mathrm{L~atm~K^{-1}{mol}^{-1}} \cdot 273~\mathrm{K}}}\\
&=1.9683~\mathrm{g/L}\\
&=0.0019683~\mathrm{g/cm^3}\\
\end{align}


van der Waals coefficients for helium and neon

The question intregues me so I looked up the $a$ and $b$ values in Wikipedia on the web page Van der Waals constants. This doesn't answer the question, but adds more data to focus the question.

$$\begin{array}{cc}
\hline
\text{Element} & a\text{ / }\mathrm{L^2 \ bar\ mol^{-2}} & b\text{ / }\mathrm{L\ mol^{-1}}\\
\hline
\ce{He} & 0.0346 & 0.0238 \\
\ce{Ne} & 0.2135 & 0.01709 \\
\ce{Ar} & 1.355  & 0.03201 \\
\ce{Kr} & 2.349  & 0.03978 \\
\ce{Xe} & 4.250  & 0.05105\\
\ce{Rn} & 6.601  & 0.06239 \\
\hline
\end{array}$$

A plot of the Van der Waals $a$ and $b$ values.

So is the b value for He high, or is the b value for Ne low ?!?

I also looked up for each element what was listed for the covalent radius and the Van der Waals radius on each elements individual Wikipedia webpage.

$$\begin{array}{cc}
\hline
\text{Element} & Covalent\text{ (pm) } & Van~der~Waals \text{ (pm)}\\
\hline
\ce{He} & 28 & 140 \\
\ce{Ne} & 58 & 154 \\
\ce{Ar} & 106±10  & 188 \\
\ce{Kr} & 116±4  & 202 \\
\ce{Xe} & 140±9  & 216 \\
\ce{Rn} & 150  & 220 \\
\hline
\end{array}$$


Why do Hydrogen and Helium have molar volumes higher than an ideal gas?

As hydrogen and helium's molecular weight are too less, the intermolecular attractions are also too less. So the a/V^2 is negligible.  So the v.d.o equation becomes P (V - nb) =RT . By which, it can be said as the pressure of these elements are too less and from Boyle's law , the volume is higher.


Silicon is very oxophilic and $\ce{SiCl4}$ irreversibly hydrolyzes in water as follows:
$$\ce{SiCl4_{(l)} + 4 H2O_{(l)} -> 4 HCl_{(aq)} + Si(OH)4_{(s)}} $$

The release of $\ce{HCl}$ creates hydrochloric acid and lowers the pH.


You might consider chlorhexidine gluconate. It is a cationic surfactant used as a skin cleanser and disinfectant prior to surgery, and is an ingredient in certain mouthwash products, so it must have low toxicity. Drug stores sell scented and unscented solutions for hand washing -- choose the unscented.


What is the name for a reaction with a solid product rising from its container?

In pyrotechnic chemistry, they might call it a pharoh's snake (or serpent) reaction. To a degree, (or rather 330 °C) pyrolysis of the sugar is occurring (starting at 170 °C).

In cooking, they would call it caramelization, although the term is also applicable in general chemistry. Less specifically, you can call it a "browning reaction"... note- caramelization (non-enzymatic) should not be confused with a Maillard reaction (enzymatic). More specifically, you could call it dehydration or decomposition of sugar with sulfuric acid. Somewhere in-between you might find the term carbonization.

You may also find the terms carbon soufflé and carbon foam interesting.


How to define custom force in quantum chemistry packages?

The short answer is not. You cannot define a constant force for an atom or set of atoms in a system.

If you still want to modify force constants, this is the way to do it:

Gaussian, you can modify through the keyword GEOM(IHarmonic=n), where n is part of: $\frac{n}{10^6}$ which indicates the respective force applied to the system. Its units are $\frac{Hartree}{Bohr^2}$. To more info, please refer to this link.
If you want to introduce a specific force to an atom or set of atoms in MM, the best way to do it is by parameterizing your own system into your desire environment. The most important part is that you have to have in mind that the right part of the principal ec. of any MM describes the parameters of the specific force field of your desire system or set of systems under specific conditions (i.e. aminoacids in water like OPLS)

$$\vec{F}=\frac{\partial V}{\partial x}$$

Moreover, when you talk about forces in QM you are talking about the forces acting on electrons. Not on atoms.
But, if you still want to know a way to know how the forces of electrons (From DFT) and, at the same time, nuclei are, I recommend you Car-Parrinello. Please, refer to this link
If you have a particularly interest on simulating systems in QM with specific forces acting on electrons, the best way to do it (As far as I know) is to develop your own density functional.

Update

I Gaussian input file was requested to figure out how the keyword and input is used: Geom(iharmonic=n). I did not find the respective file, instead, I give a try twice by crating two input files with different values of n. The used Gaussian input files were:

chk=Test.chk
#p opt b3lyp/6-31g geom(iharmonic=n)

Test

0 1

H
H 1 0.9
O 2 1.2 1 120.

and the used values of n were 2 (On the left) and 20 (On the right). The respective Energy and RMS profile of both simulations are:

Also, in Gamess, you can use these keywords:  Statpt, Force and its related keywords like Zmt. More info, Please see this link at page 2-121 and 2-131.


Why did the alpha-particles in Rutherford's experiment not collide with the electrons?

Protons and neutrons have a mass approximately 1836 times greater than an electron, ignoring relativistic effects and nuclear binding energies. There are 4 of these baryons in an $\mathrm{\alpha}$-particle. There is no way a puny electron could do much to deflect an $\mathrm{\alpha}$-particle that is almost 7500 times more massive. This is like if a 150-lb human tried to block a 747 (the 747-100's maximum takeoff weight is only about 6400 times greater than the person's).


How does LeChatelier’s principle (which makes qualitative predictions) relate to the quantitative changes observed when an equilibrium is disturbed?

How does this qualitative principle relate to the known quantitative changes that occur when an equilibrium is disturbed by a change in temperature, concentration or pressure?

Equilibrium constant K and reaction quotient Q
One way to quantitatively describe equilibrium processes is to define an equilibrium constant K and a reaction quotient Q. The equilibrium constant is defined by an equilibrium constant expression which contains all concentrations or partial pressures that are affected by the reaction.
For example, if we look at the following reaction:
$$\ce{A(aq) + B(aq) <=> C(aq) + H2O(l)}$$
The equilibrium constant expression for this reaction is:
$$K = \frac{\ce{[C]_{eq}}}{\ce{[A]_{eq}}\ce{[B]_{eq}}},$$
with $\ce{[X]_{eq}}$ designating the equilibrium activity (or fugacity for gas state) of a species. At a given temperature, the equilibrium constant will have the same value when the reaction is at equilibrium, no matter what the starting conditions.
The reaction quotient is defined in the same way as the equilibrium constant, but using $\ce{[X]_{t}}$ (the current activity) instead of the activity at equilibrium $\ce{[X]_{eq}}$. For the example reaction above, it is:
$$K = \frac{\ce{[C]}_t}{\ce{[A]}_t\ce{[B}_t},$$
Q can have arbitrary values from zero to infinity, depending on what reaction mixtures is found at a given time. Q will change over time if there is a net forward or reverse reaction.
Criterion for equilibrium based on Q vs. K
When Q equals K, the reaction is at equilibrium because the current activities $\ce{[X]_{t}}$ plugged into the equilibrium constant expression match K. Any net reaction in the forward direction would typically increase Q, while any net reaction in the reverse direction would typically decrease Q. This relies on the assumption that the activity of reactants decreases and that of the products (of the species included in the equilibrium constant expression) increases as the reaction goes forward (and vice versa as it goes in reverse).
Change in temperature
When, after the reaction has attained equilibrium, the temperature changes (through an external cause) and consequently the equilibrium constant changes, Q no longer equals K. Concentrations that had been constant over time will change now. In the example, we are changing the temperature at $t = 5$ (arbitrary units) so that the equilibrium constant doubles. The only way to make Q equal K again at this temperature is to have a net forward reaction (this will increase product activities and decrease reactant activities in the reaction quotient expression). An example is shown below:

How temperature affects the equilibrium constant is given by the van't Hoff relation. Stated in qualitative terms, the equilibrium constant of exothermic reactions decreases with an increased temperature, while that of endothermic reaction increases.
Change in activity
If we change the activity of a single species (for example by adding or removing solute) in a reaction that has already attained equilibrium, Q will change. If the temperature is maintained, K will remain constant, and Q will no longer be equal to K. As a result, a net reaction will be observed.
For example, if we double the activity of A in the reaction above, Q will be halved, and the reaction has to move in the net forward direction to attain equilibrium again:

After equilibrium is attained, Q will have the same value as it did before the external manipulation, but the set of activities will be different (necessarily, because there is more matter now).
Limitations and Criticism

Are there any limitations of Le Chatelier's principle as it is stated by modern sources?

The main limitation is that there are multiple non-equivalent ways the principle is stated, so different ways of defining the principle might lead to different counterexamples.
Historically, Le Chatelier wrote (Comptes-rendus de l’Académie des sciences, 99 (1884), p.786-789.) :

Tout système en équilibre chimique stable soumis à l’influence d’une cause extérieure qui tend à faire varier soit sa température, soit sa condensation (pression, concentration, nombre de molécules dans l’unité de volume) dans sa totalité ou seulement dans quelques-unes de ses parties, ne peut éprouver que des modifications intérieures, qui, si elles se produisaient seules, amèneraient un changement de température ou de condensation de signe contraire à celui résultant de la cause extérieure.

This google-translates into:

Any system in stable chemical equilibrium subjected to the influence of an external cause which tends to vary either its temperature or its condensation (pressure, concentration, number of molecules in the unit of volume) in its totality or only in some of its parts, can only experience internal modifications, which, if they occurred alone, would bring about a change in temperature or condensation of a sign contrary to that resulting from the external cause.

Wikipedia states the principle as:

When any system at equilibrium for a long period of time is subjected to change in concentration, temperature, volume, or pressure, (1) the system changes to a new equilibrium and (2) this change partly counteracts the applied change.

Chemistry OpenStax has two versions:

[in text]if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.
[Overview]An equilibrium system subjected to a disturbance will shift in a way that counters the disturbance and re-establishes equilibrium.

ChemPrime by Moore et al states:

if a system is in equilibrium and some factor in the equilibrium conditions is altered, then the system will (if possible) adjust to a new equilibrium state so as to counteract this alteration to some degree.

Limitations of the principle are discussed, for example, in this letter in the Journal of Chemical Education, and in the literature cited within. Another article in the same journal discusses how applying Le Chatelier's principle leads a convenience sample of educators to incorrectly answer questions such as :

The reaction
$$\ce{2NO2(g) <=> N2O4(g)}$$
is at equilibrium in a syringe. If the volume is decreased at constant temperature by moving the plunger, will the concentration of NO2(g) be higher or lower than the original concentration when equilibrium is re-established? Give reasons for your answer.

The misconception is that while Le Chatelier's principle correctly predicts that the reaction will form product after the pressure is increased, this does not completely offset the increase in partial pressure from lowering the volume.
Some authors argue that it makes sense not to teach Le Chatelier's principle to avoid such confusion, and instead treat equilibrium in a quantitative matter, calculating the new equilibrium state rather than predicting it qualitatively.


Use of osmosis to measure molecular weight of macromolecules

It has to do with the fact, that osmotic pressure of any solution is dependent only on the relative number of particles in the solution (in case of dilute solutions), irrespective of their nature. Hence osmotic pressure is called a colligative property.
Therefore, you take a certain known weight of the substance you want to calculate the molecular mass of, prepare a solution and measure the osmotic pressure. Then you compare this to the known osmotic pressure of standard solution which tells you the relative number of particles required to  create such a pressure. From this and the knowledge of the weight you used to prepare the solution, you can get the required molecular weight (weight of Avogadro's number of particles). In practice, certain complications, like dissociation and association might complicate the matter a bit.
This method is widely used to calculate the molecular weight (especially) of polymers. See Berkeley and Hartley method.


Remember that $\ce{HCl}$ (aq) and $\ce{H2O}$(l) are liquids while the oxygen and chlorine are gases . As the number of moles of gases on the left  hand side (one mole of oxygen) is smaller than that of the right hand side (two moles of chlorine), decreasing the pressure makes the equilibrium position move to the right hand side, therefore the concentration of chlorine increases.


Can molten aluminium be used to reduce iron oxide (rust) into iron?

Short answer: Yes.

Take a look at the Ellingham Diagram.

The line for aluminium oxide is lower than the line for the all the iron oxides. This indicates greater stability of aluminium oxide, or in other words indicates aluminium's greater affinity for oxygen. Thus, based on thermodynamics alone one would expect aluminium to be able to reduce iron oxides.

Based on this figure, one would also expect this reaction to be spontaneous at all temperatures.

Important caveat, the ignition temperature remains critical. Just as a piece of paper will not combust unprovoked, even though the reaction is spontaneous, the thermite reaction needs to reach its ignition temperature.

Additionally, of interest to you would be this paper. Given below is a DSC curve from it

The endothermic peak ca. 660 corresponds to the melting of aluminium, while the exothermic peaks correspond to the thermite reaction.

The activation of the process is reported as 145 kJ/mol, and according to he authors in the study the product is controlled by the diffusion of $\ce{Al}$ in $\ce{FeAl2O4}$


Is perchloride the same as perchlorate ion?

Short answer: Yes, sometimes "perchlorate" can be called "perchloride" in older literature, but because of the classical definition of a "perchloride", it's so confusing and technically wrong that the $\ce{ClO_4^-}$ anion should be called either "perchlorate" or "chlorate(VII)".

In older naming systems, pre-IUPAC, a "per-ide" is quite simply "any substance containing an unusually high proportion of the named element". So, a "perchloride" is any substance with more chlorine than is "normal" for that mixture of elements. This name was typically given to metal halide salts where the metal has multiple possible oxidation states; the highest oxidation state of the metal forms the metal "per-X-ide". for instance, titanium perchloride aka titanium tetrachloride, $\ce{TiCl_4}$ (IUPAC name titanium(IV) chloride), as differentiated from the trichloride and dichloride salts of titanium, and iron perchloride aka ferric chloride (that name's rarely used as iron only has two stable chlorides, and so the traditional "ferric" and "ferrous" descriptors work fine).

However, the perchlorate anion isn't actually a "perchloride" by that definition; it's more a "peroxide", as there's only one chlorine (which is relatively "normal") that has been "superoxidized" by four oxygens (not normal).

This use of "perchloride" as a synonym for "perchlorate" is more the result of a bad simplification of the traditional naming system, where salts of "-ic" acids are "-ides". While the rule follows for the hydrohalic acids (HF, HCl, HBr, HI) and most oxoacids (which, when they end in "-ous", form "-ites", as in hypofluorous acid, chlorous acid and sulfurous acid), the highly oxygenated oxoacids like chloric, perchloric, sulfuric, nitric, phosphoric and boric acids form "-ates", not "-ides", to mirror the low-oxygen "-ous" acids and their "-ite" anions/salts.

When not talking about the perchlorate anion, you sometimes hear about organic molecules that have had all their hydrogens replaced by halogens. Perchlorocarbons and perfluorocarbons are the chlorine and fluorine derivatives of hydrocarbons (for instance, the dry-cleaning chemical perchloroethylene, "Perc", and the perfluoroethylene monomer of polytetrafluoroethylene, aka Teflon).

"Peroxide" has found a home in the IUPAC system as the official name of the $\ce{(O_2)^{2-}}$ anion, composed of two single-bonded oxygens (often as the related "hydroperoxide", which is an OOH functional group bound to a carbon or to hydrogen). However, because each chlorine anion in a "perchloride" is typically bound to whatever the chlorine is oxidizing, the multiple chlorides are not a single anion and so aren't named as such in IUPAC.


Sameer. The key to understanding this is that the two asymmetric carbons are indistinguishable because of the symmetry in the molecule. When you do your inversions on them, the molecule looks different, and should be an enantiomer, but after you do the 180 deg rotation, you have essentially switched them over (now the molecule looks the same as the first molecule, but only because you can’t differentiate the two asymmetric carbons).

Imagine that you switched one of the terminal methyl groups with an ethyl, then try the same trick. Because you have broken the symmetry of the molecule, the 180 deg rotation doesn’t get you back to the start. Hence, this meso isomerism relies on the mirror plane of symmetry (or a point of inversion) existing in addition to multiple asymmetric centers.


How well can triglycerides be resolved using TLC?

If you work with normal phase TLC, you may lower the polarity of your mobile phase, and instead of mixing ethyl acetate with petrol ether, you may give diethyl ether / petrol ether (like 5:95, v/v) a try.

Some of the TLC manufactures even provide databases, pointing towards publications were compounds of interest were successfully separated.  Often, these references refer to a separation on TLC-scale.  CAMAG's CCBS Database is one noteworthy example.


Easily removable material that sticks to skin

I have an engineering solution for you that bypasses the need to find a chemistry solution.

  I'm thinking more along the lines of a sticky polymer sheet that readily sticks to the skin

You want an adhesive bandage. Adhesive bandages are thin, stick to skin, and are easy to remove. In my experience, the less expensive generic brands are both thinner and stickier, so go with a generic brand. If you are worried about the thickness, do not use the central part with the embedded gauze pad. Only use the wings. Use a strong glue to fix your motor to the back of the adhesive wing.


Can ICE tables give two solutions? What does this mean physically?

In other words, is there a scenario where I could end up with two possible [solutions]?

No, there is always a single solution. The reaction quotient Q assumes values from zero (no products) to infinity (no reactants). It varies monotonically with the extent of reaction (the x in the ICE table). So there is always a solution (because Q covers the full range of possible equilibrium constant), and it is always unique (because for a given set of starting values, every value of Q is unique - follows from the monotonic variation).

The only exception is a process where none of the concentration vary with x (when all species are pure solids or liquids). In this case, Q takes a single value (except for running out of reactant or product). If that value is equal to K, x can take a range of values. If Q is different from K, there is no solution, and there won't be any equilibrium (i.e. the reaction would go to completion).


Change of potential difference during electrolysis while adding salt

Tungsten lamps change resistance as they warm up. At room temperature, the resistance of a small flashlight lamp (e.g. #222) might be 10 ohms when brightly lit, but only 1.5 ohms at room temperature.

At first, the tungsten filament acts like a short circuit, so most of the voltage of the battery is dropped across the series resistance of the electrolytic cell. Once enough current goes through the lamp, though, its resistance abruptly rises and more of the voltage is dropped across the filament.

If you use an LED instead of a tungsten-filament bulb, the change in voltage would be more linear as you add salt. Note that an LED will be effectively an open circuit until a minimum voltage is reached (~1.6 V for a red LED, ~3 V for blue or white LED), but then current increases linearly until ~20-50 mA, at which point some may burn out.


Conductometric titration of Strong Acid And Strong base

You correctly identify that the number of cations do not change in solution, so from a charge perspective it appears that there should be no significant change in the conductivity.

However, consider two things:

Sodium ions are much heavier and larger than free protons and so should travel through water more slowly. Charge moving more slowly
means lower conductivity.
In addition to their smaller size and lightness, Protons can also "hop" from water molecule to water molecule and so can move through the solution much more quickly and easily. Other water molecules do not act as obstacles

Point 2 is called the Grotthuss Mechanism.


Preference of certain positions during chlorination of an alkene

In your first example, propene plus chlorine, one of the Cl atoms bonded to a "CH" and the other, that you described as bonding to the end, bonded to a "CH2," so it's not just CH groups getting all of the chlorines.  Instead, it's the carbons at the ends of the double bond, however many hydrogens they might be attached to.

I think part of your misunderstanding lies in the way that you're writing/imagining the alkene.  The propene molecule has a shape, and the geometry of the molecule plays a major role in its reactions.  The double bonded section is more or less flat, and the CH3 is spread out in space.  The chlorine atoms attach to the ends of the double bond, wherever that may be.  The mechanism for this attachment is beyond the scope of your question, but if you're curious, you can find more information here.  At the very least, know that the chlorines do not attach to the carbons at the same time; there is an intermediate compound formed.

(If my drawing isn't clear, you can try building a model of the molecule.)

It would require the breaking of a C-H or C-C sigma bond in order for the chlorine to get to the CH3 carbon, and that's unlikely.  As I mentioned above, the C=C section is planar, so these carbons are open on the sides for attack from a halogen.  A bond still has to break for a new C-Cl bond to form, but in this case it is the weaker C-C pi bond that breaks.  With easy physical access and a lower energy barrier, the reaction goes forward.


What is exactly ortho effect in Reimmer-Timmer reaction?

I think that the coordination step is responsible for the ortho selectivity. The phenoxide anion has a large negative charge on the oxygen atom, so the electron-poor carbene will tend to go there and the conjugated attack will result in the ortho product.


Reaction involved in Combustion of Nitrocellulose-Nitroglycerine(NCNG) solid propellant?

Nitrocellulose is complex stuff. It is generally considered to be the fully nitrated version of cellulose, in which every available alcohol functional group is converted to a nitrate:

$$\ce{[C6H10O5]n +3HNO3 -> (C6H7O2(ONO2)3]n +3H2O}$$

In principal it probably contains some small amount of partially and/or non-nitrated units.

If we consider nitrocellulose to be fully nitrated, then the combustion reaction is:

$$\ce{ 2C6H7N3O11 + 9/2O2 -> 12CO2 + 3N2 + 7H2O}$$

For nitroglycerin $(\ce{C3H5N3O9})$:

$$\ce{2C3H5N3O9  -> 6CO2 + 3N2 + 5H2O + 1/2O2}$$

It will depend on the ratio of nitrocellulose to nitroglycerin what the overall $\ce{CO2}:
\ce{N2}:\ce{H2O}$ mole ratio is. I'll leave you to figure out the mass ratios using the molecular masses of the products.