The average disintegration rate is given by $\bar n=\sum_{n=0}^{N_0} nP(n)$ where $P$ is the binomial distribution. If $M$ is the average number of events in time $t$ then $\displaystyle M=N_0(1-e^{-\lambda t})$. For times smaller than the lifetime $1/\lambda$  the $\displaystyle e^{-\lambda t} \to 1-\lambda t$ then $M=N_0\lambda t$ and as the disintegration rate is $M/t$ this leads to $\displaystyle \frac{dN}{dt}=-\lambda N$. Thus provided the measuring time-bins are small compared to the lifetime this rate law should hold.

As the decay proceeds the number of counts becomes small, perhaps $10^6$ at short times to just 1 or 2 at long times then of course the signal to noise becomes huge, zero counts in some time bins for example and 1 or 2 counts in others. (This becomes an issue in analysing data, see Anal. Chem., 2003, 75 (16), pp 4182–4187).

The expected variance is, as you quote, $V=\sigma^2=N_0e^{-\lambda t}(1-e^{-\lambda t})$ and can be rewritten as $V=Me^{-\lambda t}$ where $M$ is the mean number of events recorded in time $t$. As the time bins are normally short so that $\lambda t <<1$ then $\sigma=\sqrt{M}$ so that $\displaystyle \frac{\sigma}{M}=\frac{1}{\sqrt{M}}$


Existence in an oxidative atmosphere means many commercial products may be, and often are, exposed to oxygen.  Many things, and almost all drugs, react with oxygen.  The products of these reactions (oxidations, normally) are different molecules that may be less effective than the original drug.  As such, the efficacy of a drug that is beyond its expiration date is presumably lower, meaning dosing instructions become invalid.  Similarly, water content in the atmosphere may alter reactivity of drugs.

In general, studies to determine expiration dates take large amounts of time, and are hence expensive.  The standard for such an expiration would normally be a remaining purity level of the drug in question.  For example a company might study their drug for a year and note it is still 99% unchanged.  This means it meets a pretty good standard after one year.  From a marketing perspective they just label the drug as expiring after one year.  This allows them to be out of the legal liability because they are disclosing actual data indicating the efficacy of the drug over a period of time, and if a consumer takes the drug outside the expiration date, they are using the drug counter to instructions.  It also tends to generate extra sales for people who keep old drug stocks around, and must buy new ones to guarantee efficacy.  The actual drug may be still therapeutically useful even if "expired".

Specifically related to zinc oxide, I see no pathway for further degradation.  Wikipedia reports Anusol actually contains zinc sulphate (monohydrate), which I expect to be similarly robust.  However, the other but the qualities of the product ( presumably an ointment or lotion ) may have degraded such that its application may be less effective.  Additionally the Anusol plus contains a drug called Pramocaine Hydrochloride.  This compound is likely to oxidize overtime.


Why do distilled alcoholic beverages taste different from each other?

Two different aspects come into play here. For once, most of the whiskey aroma stems from aging the fresh destillate in oak barrels previously used to store wines, e.g. sherry. Colour an aroma thus, at least in part, stem from a  slow extraction process using the fresh destillate as a solvent.

But thinking in other beverages, particular the clear fruit spirits, there's more in it. The key to a beverage that doesn't smell and taste just like ethanol but has a cherry aroma is the sloppyness of the destillation :) In the lab, a destillation or rectification is performed to isolate a particular compound from a mixture at a sharp boiling point.

In the production of alcoholic beverages, the destillation typically aims to obtain three fractions over different boiling point ranges.

At lower boiling points, methanol, acetone and ethylacetate are removed.
The upper range mostly furnishes long chain alcohols.

These two fractions aren't meant to be digested ;)

The central fraction, however, will contains all the fragrances, ethanol and water.


Molecular formula for 4-methyl-1-pentanol?

Actually, the formula you arrived at is $\ce{CH3C5H12OH}$ = $\ce{C6H16O}$.  In any case, in your "2n+2" step, you need to reduce that number of hydrogens by 2.  One is eliminated where the methyl group is attached, and a second is eliminated where the $\ce{-OH}$ group is attached.  So removing 2 hydrogens from $\ce{C6H16O}$, gives you the desired $\ce{C6H14O}$.


What undergoes reduction in a zinc-copper voltaic cell?

Copper cannot undergo reduction because, in the list of components of the cell you listed, it is only present in its most reduced state, Cu(0). Thus, it cannot be copper that is reduced. See here for additional hints about your system.

You may have been confused by the fact that, in the historical (and widely taught) Daniell cell, copper is indeed reduced: it is present as Cu(+II) in a solution of copper sulfate.


When is Serpeck's process preferred over Bayer's process?

From this source

Baeyer's process: This method is mainly applied when ferric oxide is present as chief impurity.
Serpeck's process: This process is used when silica is present in considerable amount of bauxide ore.

I couldn't find any information on the efficiency , but I suspect both are efficient in their own way. We cannot compare these two processes as they are used for two different impurities.


What is the pKa of the hydronium, or oxonium, ion (H3O+)?

The controversy surrounding the $\mathrm{p}K_\mathrm{a}$ of hydronium mostly arises from the definition of $K_\mathrm{a}$ or lack thereof.

There is no IUPAC definition of $\mathrm{p}K_\mathrm{a}$ or $K_\mathrm{a}$.

The closest IUPAC defined term is the standard equilibrium constant, which can be denoted as $K^\circ$ or just $K$.

Physical chemistry texts such as Levine and respected works such Bates's "Determination of pH — Theory and Practice" define $K^\circ_\mathrm{a}$ of an acid in water as:

$$\frac{a(\ce{A-})a(\ce{H3O+})}{a(\ce{HA})a(\ce{H2O})} \tag{1}$$

Where $a$ is activity.

Substituting that the acid is $\ce{H3O+}$:

$$\frac{a(\ce{H2O})a(\ce{H3O+})}{a(\ce{H3O+})a(\ce{H2O})} = 1 \tag{2}$$

and of course $-\log(1) = 0$.

The number $-1.74$ that some quote for the hydronium $\mathrm{p}K_\mathrm{a}$ comes from:

omitting the activity of water from the denominator of the $K^\circ_\mathrm{a}$ definition (equation $(1)$); and
taking the concentration of water (about $55.5~\mathrm{M}$) as the $K_\mathrm{a}$ of $\ce{H3O+}$.

With that, one obtains the $\mathrm{p}K_\mathrm{a}$ as $-\log (55.5) = -1.74$.  For example, Levine even has the $-1.74$ value in a figure comparing the $\mathrm{p}K_\mathrm{a}$'s of various acids, but has a footnote explaining that the value for $\ce{H3O+}$ is based upon the alternative $\mathrm{p}K_\mathrm{a}$ definition.

However, revisiting the analysis that $K^\circ_\mathrm{a} =1$ is probably the most authoritative paper on this issue:

New point of view on the meaning and on the values of Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water, which insists that

$$\ce{H2O + H3O+ <=> H3O+ + H2O}$$

"does not correspond to an actual chemical process" and therefore "it is not legitimate" to extend the concept of $K^\circ_\mathrm{a}$ to $\ce{H3O+}$ in water.

The article goes on to say that only by studying the $K_\mathrm{a}$ of $\ce{H3O+}$ in another solvent such as ethanol can $\ce{H3O+}$ be compared to other acids.

The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in ethanol is $0.3$ and $\mathrm{p}K_\mathrm{a}$ values are $1.0 \pm 0.3$ units lower in water than in ethanol, so the article suggests a $\mathrm{p}K_\mathrm{a}$ of $-0.7$ for $\ce{H3O+}$ in water, for the purpose of comparison to other acids.


Does the salt in the ocean act as a binding agent on sand?

Good observations!

Seawater contains salts in small amounts such as such as $\ce{LiCl}$ and $\ce{MgCl2}$ that are quite hygroscopic, which could keep the sand wet longer.

There is also dissolved organic matter (http://www.eoearth.org/view/article/154471/) in seawater that might act as a hydrating agent or as a wetting agent, helping to hold onto the water and to bond it to the sand.

You might try an experiment comparing the appearance of (washed) sand after being wet with plain water, with $\ce{NaCl}$ salt water of the same concentration as seawater, and with seawater itself to see if there is a difference. Though this seems like a casual test, this could be of significance in cases where it's necessary to drive on a sandy shore.


Term symbol for atomic nitrogen: why can't S be -1/2?

Why not go one step further and make all the electrons have spin down? Then you should get $S = -3/2$, no?

In fact, what's happening is: you are confusing the projection of the spin quantum number, $M_S$, with its magnitude, represented by the spin quantum number $S$.

The $\pm1/2$ that you are summing are the individual projections of the electron spins, $m_s$. When you add up projections, you get a projection $M_S$. This projection can indeed be $-1/2$, and it can indeed be $-3/2$ as well, but the term symbol does not depend on $M_S$; it depends on $S$.

To obtain the total spin quantum number $S$ you have to use the Clebsch-Gordan series to couple different sources of angular momenta together. In this case we would first couple the spins of two electrons

$$S_{12} = s_1 + s_2, s_1 + s_2 - 1, \cdots, |s_1 - s_2|$$

Note that for an electron, it has spin $s = 1/2$. Again, the projection of the spin can be $+1/2$ or $-1/2$, but that is not what we are interested in. We are interested in the magnitude of its spin, which is only $1/2$, not $-1/2$.1 If we plug $s_1 = s_2 = 1/2$ into that equation above we get

$$S_{12} = 1, 0$$

and now if we couple this to the spin of the third electron we get

$$S = S_{12} + s_3, S_{12} + s_3 - 1, \cdots |S_{12} - s_3|$$

Again, $s_3 = 1/2$ only. But $S_{12}$ can take on two values so we will have to evaluate this sum two times, one time for each possibility:

$$\begin{align}
S_{12} = 1 &\Longrightarrow S = \frac 32, \frac 12 \\
S_{12} = 0 &\Longrightarrow S = \frac 12
\end{align}$$

Note that the absolute value in the Clebsch-Gordan series forbids the possibility that $S < 0$. This makes sense since a magnitude of a spin cannot be negative.

In the end your two possibilities are $S = 3/2$ or $S = 1/2$. Due to Hund's first rule the $S = 3/2$ term is the ground state and the $S = 1/2$ terms are excited states.

Next, the $L$ in the term symbol is not equal to 1 just because the p orbitals have $l = 1$. Note that one is a small $l$ and the other is a big $L$. The big $L$ in the term symbol is obtained again by coupling the individual values of $l$ for each orbital. In this case all three p orbitals have $l =1$ and you would have to work through the Clebsch-Gordan series again.

$$L_{12} = l_1 + l_2 , l_1 + l_2 - 1, \cdots , |l_1 - l_2|$$

$$L = L_{12} + l_3, L_{12} + l_3 - 1, \cdots, |L_{12} - l_3|$$

and if you set $l_1 = l_2 = l_3$, should reach the conclusion that $L$ can take on the values of $3,2,1,0$. However you have to be careful about this, because the $L = 3$ value is forbidden by the Pauli principle, which is not taken into account by the Clebsch-Gordan series. A term with $L = 3$ would necessitate the existence of a microstate with $m_L = +3$, which would imply that all three electrons are in the p orbital with $m_l = +1$, which is not possible.

1 Technically, the spin is $\displaystyle \sqrt{\frac{1}{2}\left(\frac{1}{2} + 1\right)}\hbar$, but whatever. The correct phrasing is that the quantum number representing the magnitude of the spin can only be $1/2$.


How to draw resonance structures for CH3COCN

Here are the significant resonance structures for pyruvonitrile.


Iterative method for calculating pH of a weak acid / base

This method of repeating iterations is interesting, though I would suggest something perhaps a little more straightforward (and rigorous) than doing something until you are "acceptably close."

This method I learned is called a RICE diagram.

Reaction

$$\ce{PO4_{(aq)}^{3-} +H2O_{(l)}<=>HPO4_{(aq)}^{2-} +OH^{-}_{(aq)}}$$

Initial (concentrations)

$$[\ce{PO4^{3-}}]=0.20~~~~~[\ce{HPO4^{2-}}]=0~~~~~[\ce{OH-}]=0$$

Change (in concentrations)

$$[\ce{PO4^{3-}}]=-x~~~~~[\ce{HPO4^{2-}}]=+x~~~~~[\ce{OH-}]=+x$$

Equilibrium (concentrations)

$$[\ce{PO4^{3-}}]=0.20-x~~~~~[\ce{HPO4^{2-}}]=x~~~~~[\ce{OH-}]=x$$

We can write the reaction expression as:

$$K_\mathrm{b}=\frac{[\ce{HPO4^{2-}}]\cdot[\ce{OH-}]}{[\ce{PO4^{3-}}]}$$

We substitute in the equilibrium concentrations and the $K_\mathrm{b}$:

$$2.22\times10^{-2}=\frac{[x]\cdot[x]}{[0.20-x]}$$

And solve for $x$:

$$2.22\times10^{-2}=\frac{x^2}{0.20-x}$$

$$4.44\times10^{-3}-2.22\times10^{-2}\cdot x=x^2$$

$$0=x^2+2.22\times10^{-2}\cdot x-4.44\times10^{-3}$$

Use the quadratic formula:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

$$x=\frac{-(2.22\times10^{-2})\pm\sqrt{(2.22\times10^{-2})^2-4(1)(-4.44\times10^{-3})}}{2(1)}$$

$$x\approx-7.87\times10^{-2},~5.65\times10^{-2}$$

Obviously the negative solution is not applicable to our problem, so $x=5.65\times10^{-2}$.

Since $x=[\ce{OH-}]$, $[\ce{OH-}]=5.65\times10^{-2}$.

We know that:

$$[\ce{H+}]\cdot [\ce{OH-}]=1\times10^{-14}$$

so:

$$[\ce{H+}] \cdot [5.65\times10^{-2}]=1\times10^{-14}$$

$$[\ce{H+}] =\frac{1\times10^{-14}}{5.65\times10^{-2}}\approx1.77\times10^{-13}$$

If you have access to a calculator with a numerical solver function, I would recommend that instead of using the quadratic formula, but if not, this is how it is done.


How does lead toxicity from jewelry happen vs. occupational exposure?

Metallic lead is comparatively inert, but after exposure to air it quickly develops a film of lead compounds, including litharge, $\ce{PbO}$, basic lead sulfate, $\ce{Pb2SO4(OH)2}$, etc. Though these are not very soluble, sweat and bacterial action can convert them to much more soluble compounds.
Lead is also very soft, so touching an article made of lead leaves a residue on the fingers, similar to the way using an aluminum-handled tool leaves the hands covered in gray aluminum and its compounds.
The danger is usually not immediate, but rather the accumulation of toxic lead in the body. However, lead worn next to the skin is steadily abraded. Also, children (and pets) put things in their mouths or even swallow them. Stomach acids do dissolve lead, and there is high mortality in waterfowl from swallowing lead pellets.
This is why lead has no place in play things (or perhaps in most products, since it will almost invariably leach out to poison the environment).
That said, this is an abstract comment, and not to be construed as a way to "safely" use lead in toys.


Why is the ionization energy for Hydrogen non-zero?

The ionization energy (IE) of an atom or molecule describes the minimum amount of energy required to remove an electron (to infinity) from the atom or molecule in the gaseous state.

With ionization energy, an electron is not "kicked out" by other electrons, but rather it is "the energy required for the electron to 'climb out' and leave the atom."

Since the electron "is drawn inwards by positive electrostatic potential," it would make sense to infer that the more "tightly drawn" the electron is to the nucleus, the more energy it would require to "climb out."

More of this can be found on the wikipedia page.


The idea of Markivnikov or anti-Markovnikov addition fundamentally isn't related to the mechanism by which a reaction took place, rather, the term is a way of describing the observed products obtained from certain kinds of addition.

The original statement of  Markovnikov addition is as follows, and applies specifically to the addition of hydrogen halides ($\ce{HX}$) across an alkene:

  In the addition of hydrogen halides to unsymmetrically constituted [unsaturated] hydrocarbons, the halogen atom becomes attached to the carbon bearing the lesser number of hydrogen atoms.

The statement was a purely empirical one (i.e. it was based on the results of experiment, rather than any consideration of the mechanism/pathway that the reaction had taken). It can also be expanded to consider any addition of a polar reagent ($\ce{RX}$) across alkenes and alkenes. The following definition is taken from the IUPAC Gold Book (a standard reference for definition of chemical terminology):

  In the heterolytic addition of a polar molecule to an alkene or alkyne, the more electronegative (nucleophilic) atom (or part) of the polar molecule becomes attached to the carbon atom bearing the smaller number of hydrogen atoms.

Thinking about the situation mechanistically, the Markovnikov rule basically just says that the less electronegative atom in the polar reagent $\ce{RX}$ ends up bound to the end of the alkene (or alkyne) most able to stabilise a developing positive charge (i.e. the most hindered end of the alkene/alkyne).

In terms of labelling the products of reactions, as discussed in your question, its perfectly valid to describe the products as those of Markovnikov addition (as stated previously, the term itself is saying nothing directly about mechanism, even if implied).


Sometimes authors use buzz words in their paper in order to make their work novel or fancier than it should be. However, such newly coined terms never get popular and vanish with the author into oblivion. No, cis and trans can never mean cathode or anode. It refers to the chambers in which the electrode is place.
I looked up the Oxford's Latin Dictionary:
cis has three meanings

(in time) before, within
(in space) on the near side of, on this side of
(expressing motion) to the near side of

trans

to the other side of

Now look at the cell configuration in your link. The very first lines of the paper you linked says

DNA translocation through protein or synthetic solid state nanopores
is a two stage process that involves capture of a DNA coil by the pore
including its delivery to the pore and insertion of one of its ends
into it, followed by its motion through the nanochannel. The process
may be driven by voltage $\Delta$V typically a few hundred millivolts
applied between two macroscopic electrodes positioned on both sides of
the partition separating the cis where DNA is initially placed and the
trans to where it translocates through the pore chambers, as shown
schematically in Fig. 1


Can a s orbital overlap with any p orbital to form a sigma bond?

$\ce{Px}$, $\ce{Py}$, and $\ce{Pz}$ orbitals can all overlap with an $\ce{S}$ orbital to form a bond.  It simply depends upon along which axis (X, Y or Z) the bond will be directed. These concepts of overlap do not change between Molecular Orbital theory and Valence Bond theory.


What type of polyester is most commonly used for clothing?

Based on several sources (shown below), it appears that polyethylene terephthalate (PET) is in fact used in the production of fabrics, including textiles. PET is marketed under names such as Dacron, Terelene, or Lavsan by various companies depending on your location. As to if it is the only textile polyester, I was unable to be completely certain. However, it appears to be a major textile polyester at the very least, as 39 million megagrams were produced in 2008 for fabrics alone.

Though it was not part of your question, I thought that I should include the following, as I found it interesting:

As it turns out, not only is PET used as a fabric, but it also makes a very nice fabric that is highly resistant to wrinkling. This is due to the presence of an aromatic ring within the monomers, which makes it quite strong/stiff; especially in this fibre form. These wrinkle-resistant fabrics are appropriately known as "permanent press fabrics."

To find out more information, I have included some sources below:

https://en.wikipedia.org/wiki/Polyester
https://en.wikipedia.org/wiki/Polyethylene_terephthalate
https://www.britannica.com/science/polyethylene-terephthalate


Why is gold unreactive when only one electron is in the outer shell?

First off, gold does react. You can form stable gold alloys and gold compounds. It's just hard, mostly for reasons explained by the other answer

The reason bulk gold solid is largely unreactive is because the electrons in gold fall at energies which few molecules or chemicals match (i.e., due to relativistic effects).

A nice summary of some work by Jens K. Norskov can be found here:
http://www.thefreelibrary.com/What+makes+gold+such+a+noble+metal%3F-a017352490

  In their experiments, they distinguished between gold atoms' ability to break and form bonds and the ease with which they form new compounds, such as gold oxides. The two qualities are related: To make a compound, gold atoms must bond with other atoms, yet they cannot do so until they have sundered their bonds with neighboring gold atoms.

I think this is a nice succinct explanation. You always have this trade-off in reactions, but in gold, you don't get much energy in the new compound formation, and you're losing the gold-gold interactions.

You can, of course, react gold with aggressive reagents like aqua regia, a 3:1 mix of $\ce{HCl}$ and $\ce{HNO3}$.

If properly done, the product is $\ce{HAuCl4}$ or chloroauric acid.


How to connect structures in Avogadro at certain angles?

Did you try a (Universal Force-Field) geometry optimization? I would:

Select all atoms
Deselect (hold Ctrl + click) the 4 (6? 8?) carbons of interest
Click Extensions -> Molecular Mechanics -> Ignore Selection
Click Extensions -> Optimize Geometry

Alternatively, you could optimize a few more atoms around/in the bridge while the others are constrained. The G.O. algorithm is pretty robust and handles many systems quite well. A simple example of hand-drawn cyclohexane:

I got a result like this by making 2 nanotubes with chiral index (4,4), drawing 6 bonds and optimizing the bridge:

If this is ineffective, I would resort to:

Change the 4 carbons to different elements e.g.

$\ce{O        H}$

$\ce{N        S}$

(like the arrangement you have in the picture, so you can identify them later)

Click File -> Save As... and save the file as an .xyz
Calculate the O-N and H-S distance by $r_{ij} = \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$ from the coordinates in the .xyz file.
Ideally, Avogadro saved the .xyz coordinates aligned in z (which is default when you draw molecules without rotating the initial view).
(Even if not,) re-write the coordinates of the atoms aligned in x and z, e.g. $\ce{O}$ at $(x, y, z)$; $\ce{N}$ at $(x, y-r, z)$.
Or (yikes) write a quick program to generate possible systems with the fixed $r$ you calculated, name them system-000.xyz, system-001.xyz, system-002.xyz, etc., and view the "trajectory" frames until you find one pleasing. (Viewing can be made much easier with VMD, e.g. vmd system-*.xyz).

Avoiding Avogadro: If you use Molden and load your system (as an .xyz, .car, .pdb etc.), you can use the ZMAT-editor to manually change bond angles and dihedrals:

Finally, if all else fails, it may be time to consider a full quantum-mechanical treatment of the system (if you're familiar with High Performance Computing) with a free (for academia!) program like Orca (where one can constrain certain atoms during the optimization). An example input would be:

! hf 6-31g* opt  # or copt, for cartesian optimization

%pal
nprocs 128   # num. processors
end

%scf
maxiter 2000 # max self-consistent field iterations
end

%geom
Constraints # atoms to keep fixed starting at index 0
{C 61 C}
{C 62 C}
{C 290 C}
{C 291 C}
{C 292 C}
end
end

*xyz 0 1  # charge=0; mult=2S+1 = 1 (closed shell)
  C      -1.306743      4.170005      1.012072
  C      -0.042414      3.718873      0.686759
  C       0.210647      2.367809      0.785537
  C      -2.267754      3.271952      1.435683
  C      -2.920909      0.889649      1.948837
  C      -4.153277      1.187877      2.498723
  C      -4.989183      0.155126      2.896328
  C      -4.571713     -1.152339      2.735382
  C      -3.332221     -1.380404      2.172864
....... etc.
*


Why is allyl chloride more reactive towards substitution than alkyl chloride?

But how allyl halide is more reactive than alkyl halide?

Reactivity towards $\ce{S_N1}$
The key to reactivity towards $\ce{S_N1}$ is the stability of the formed carbocation. Allyl system stabilises the carbocation through overlap with the vacant p orbital (@gsurfer999 has shown the resonance structures in his answer below). However, note that any allyl halide wouldn't be better at $\ce{S_N1}$ than any alkyl chloride.
In fact, the tertiary alkyl chloride's rate of $\ce{S_N1}$ is faster than that of allylic chloride which is secondary at one end. (due to $\ce{+I}$ effect of 3 methyl groups as well as hyperconjugation)
Visual aid for stabilisation through hyperconugation (from Clayden):

For such comparison, here's a useful table again from Clayden:
Reactivity towards $\ce{S_N2}$:
In $\ce{S_N2}$ the most important factor is the substrate. The reaction proceeds through a single transition state with trigonal bipyramidal geometry and $\ce{\sim sp^2}$ hybridisation of the electrophilic carbon. The p-orbital makes two partial bonds, one with the nucleophile and another with the leaving group. Thus, it's electron deficient.  In such a case, the allyl system provides the additional electron density through conjugation.
However, again note that primary alkyl chloride is better at $\ce{S_N2}$ than allylic system due to least steric hindrance!


Why is the iron(II) on the heme molecule depicted as having 6 electrons in its 5 d non- and anti-bonding orbitals?

I've seen that there are consistently 6 electrons in these orbitals — which would be consistent with iron(0) but not iron(II) as I understand it.

Iron is in the 8th group (or group VIIIb in older terminology). Thus, in its ground state it has eight electrons; two of which are in the 4s orbital. However, practically always once you oxidise a transition metal and enter coordination chemistry, the orbital energies shift relatively to each other and it becomes more stable to have all the energies in the d-orbitals. Thus, the typical configuration for iron(II) is not $[\ce{Ar}]\,\mathrm{3d^4\,4s^2}$ but $[\ce{Ar}]\,\mathrm{3d^6\,4s^0}$.

  I am under the impression that there must be 18 electrons, as suggested by the 18 electron rule (which I am aware is a loose guideline).

I think, you’re not aware of the looseness of the 18-electron-rule guideline. Most of the time it does nothing except for stating a maximum that can occur. However, there are still many complexes that violate it; e.g. anything octahedral with zinc(II) (22 electrons), anything with copper(II) (21), octahedral nickel(II) (20) etc. It is much more valuable when going to neutral carbonyl complexes; e.g. iron typically forms pentacarbonyliron $\ce{[Fe(CO)5]}$ due to the 18 electron rule. But in the context of standard cationic complexes. you can safely ignore it.


Diffusion Phenomena : Why does diffusion occur?

The short answer
They don't.
The long answer
Molecules are very fast. For example, a single molecule in the air around you can be as fast as 500 km/h. The reason that you do not feel them is because they are too small.
Also, you need to understand that there are very, very, very many molecules in a "small" volume.
So, until this point, you have many molecules which travel very fast.
Suppose a gas is divided to high and low concentration areas. The area with high concentration would have more molecules, and let's say 50% of them is travelling towards the area of low concentration (they all move randomly, and in a straight line unless collision happens).
Let's say, 50% of the gas molecules in the area of low concentration is also travelling to the area of high concentration.
Then, you would have a net movement of molecules from high to low concentration.
The keyword here is net. Some molecules do go from low to high concentration. However, they are outnumbered.
After they reach equilibrium, there is still gas molecules travelling between the two regions, just that they are equal in number.
Extra
Of course, because my explanation being not very formal and strict, there would be some errors in my explanation, but I hope that you understand it.


Note the image of acetic acid (top) and the image of the carboxylic structural isomer methyl formate (middle).  As you can see the bonding location of the proton and the methyl group switch between the structural isomers.  Note the image of formic acid (bottom).  If you switch a hydrogen and a hydrogen you still get formic acid.  Yes, structural isomers of carboxylic acids make esters, but formic acid does not have a structural isomer, so the rule does not apply.

 Acetic Acid
 Methyl Formate
 Formic Acid


Why does sweetening milk need less sugar as compared to tea?

It's all in your taste buds. Lactose in milk is a form of sugar. It is naturally a little sweet. Tea is slightly bitter so it might seem you need to add more sugar to make it sweet.


Reason for the stronger acidic property of phenol than alcohol

In fact I find a more simple reasoning with resonance structure. When phenol loses the $\ce{H+}$ the phenolate ion is stabilized due to the resonance effect, as shown below:

The energy of the dissociated form is lower and so phenol has more chance to be in the solution dissociated with the phenolate ion. Aliphatic alcohols are not stabilized by resonance so they are not very prone to dissociation.


Are there any free phosphorus-31 NMR spectral libraries?

There are quite a few it seems, a couple of examples are below:

University of Wisconsin Phosphorus NMR data, starts with a shifts overview.

Stanford's Magnetic Resonance Laboratory site P NMR at the SMRL, provides comparisons and a protocol for testing.

A general NMR site is NMR Wiki - has many links to many sources.


Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?

There is a reason for everything.

  Does Bent's rule have any utility?

YES! It wouldn't be there if there was not. But I will get back to this at the end of this post.

Let me go through the points raised by your teacher first:

  Coulombic considerations can be used to rationalize bond angles, strengths, and lengths without the use of Bent's rule.

This might be true to some extent, but that only means that coulombic considerations give the same result as a complete orbital analysis. Boltzmann, Planck, Einstein, Schrödinger, Pauling, etc. worked the field for a reason - classical mechanics cannot be applied to molecular systems. Any theory that simplifies it beyond quantum mechanics, is a model to understand or comprehend - not a model to predict.

Orbitals are not real so why invoke orbitals when simpler Coulombic considerations are available? (Occam's razor)

Occam's razor would suggest picking molecular orbital or valence bond theory, as it states that from competing hypotheses the one with the fewest assumptions should be selected. Coulombic consideration is purely based on assumptions, it is a rather crude approximation to the correct picture.
The statement, that orbitals are not real is clearly wrong. There are many books that explain how one can comprehend this. Teaching that orbitals are not real is, with all due respect, just plain wrong.

The Coulombic effect is also real as opposed to orbitals.

The coulombic effect is not more real than orbitals. Orbitals should be used to explain the coulombic effect.

But let's talk about Bent's rule. The IUPAC goldbook states:

  In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.

It is important to know, that Bent's rule is, like Lewis rule of two (or eight, known as octet rule), a rule (not a law) based on observation. Hence its character is more or less one of a guideline. The key paper was published in Henry A. Bent, Chem. Rev., 1961, 61 (3), 275–311.

Bent's rule originates from valence bond theory (VB). Early forms of VB theory proposed that a molecule can entirely be described by a set (linear combination) of resonance Lewis structures, i.e. every bonding orbital is a directed localised orbital, which will be occupied by two electrons and the same applies to lone pairs. The advantage of this approach is that it is easily comprehensible. The disadvantage is, that it is quite inflexible. The reason for this was the use of either pure $\ce{s}$ and $\ce{p}$ functions or fixed hybrid orbitals $\ce{sp^3,~sp^2,~sp}$, which only allowed a subset of angles. A direct quote from Bent's original paper states this as follows:

  For atoms that satisfy the octet rule, and it is with such atoms (e.g., carbon, nitrogen, oxygen, fluorine) that this review is primarily concerned, the four valence-shell orbitals of an atom may be thought of as compounded from one spherically symmetrical $\ce{2s}$ atomic orbital and three mutually perpendicular dumbbell-shaped $\ce{2p}$ orbitals, designated $\ce{2p_$x$, 2p_$y$, 2p_$z$}$, ($x, y, z$ indicate mutually orthogonal axes). The $\ce{2s}$ orbital is lower in energy than the $\ce{2p}$ orbital; hence the ground state of a carbon atom, for example, is written $\ce{ls^2~2s^2~2p^2}$ (not $\ce{ls^2, 2s~2p^3}$. However, in a molecule such as methane, which is known to have the shape of a regular tetrahedron, the $\ce{2s}$ orbital of carbon may be considered to mix (or “hybridize”) with the three $\ce{2p}$ orbitals to form four equivalent orbitals that point to the corners of a regular tetrahedron and make angles with each other of $109^\circ28’$. These are called hybrid orbitals. Each contains one-quarter $\ce{s}$ character and three-quarters $\ce{p}$ character, abbreviated $\ce{s^{$1/4$}p^{$3/4$}}$ or $\ce{sp^3}$, or $\ce{te}$ (for tetrahedral) (223). In ethylene, on the other hand, the $\ce{2s}$ orbital of carbon may be considered to mix with but two of the three $\ce{2p}$ orbitals to form three equivalent hybrid orbitals that lie in a plane and make angles with each other of $120^\circ$; these three hybrids, which contain 33 per cent $\ce{s}$ character, are abbreviated $\ce{s^{$l/2$}p^{$2/3$}}$ or $\ce{sp^2}$, or $\ce{tr}$ (for trigonal); perpendicular to the plane of these three orbitals is the axis of cylindrical symmetry of the remaining pure $\ce{2p}$ orbital. The geometry of acetylene is often explained by assuming that the carbon $\ce{2s}$ orbital mixes with but one of the three $\ce{2p}$ orbitals, forming two $\ce{di}$ (for digonal) hybrids that point in diametrically opposed directions; these are abbreviated $\ce{s^{$1/2$}p^{$1/2$}}$ or $\ce{sp}$.

The original approach allowed already good results. However, Bent already draw the conclusion, that hybridisation is connected to molecular geometry:

  Molecular geometry provides then a direct clue to interorbital angles and
  (table 1) the distribution of $\ce{s}$ character. For example, from the bond angles in ammonia ($106^\circ46'$) (284) and water ($104^\circ27'$) (134, 284), it is inferred that the nitrogen atom of $\ce{NH}$, devotes slightly more $\ce{s}$ character to its bonding orbitals than does the oxygen atom of $\ce{H2O}$.

In conclusion to this (unfortenuately I ran out of time, so this may be extended soon) he suggested to use fractional linear combinations of hybrid orbitals to add more flexibility to the VB approach. Therefore Bent's rule is an extension to VB theory, that refers to the linear combination of atomic orbitals, that eventually form hybrid orbitals.

The formulated rule was later merely a result of the approach to flexibilize this theory. It is therefore a very important part of theoretical chemistry.

Of course there are now many more modifications to VB theory, and it is nowadays considered complementary to molecular orbital theory, but this would certainly exceed the scope of this question.


How can one distinguish an IR spectroscopy alcohol band on an organic molecule versus the alcohol bands from water in the sample?

The main working rule of a good FTIR spectrum is that everything must be bone dry! The absorbance in IR is additive just like Beer's law, i.e. your spectra is the sum of water traces in your sample as well as your suspected alcohol. There is no simple way to deconvolute it. Dry your product thoroughly in a vacuum dessicator for at least a day and re-run the IR. If your big -OH band vanishes, it means your product was not dry. If it persists, your product likely has an hydroxide group.

KBr must be oven dried well before use as well.


Your textbook is correct. Glass surface may appear smooth, but in reality it is really rough and irregular if viewed under a microscope. Here you will find some pictures. Have you ever seen an insect walking on a window? It is possible because of those irregularities on the glass surface. Insect legs have tiny hooks which graple to all these lumps, cracks and holes on the glass.


Why do noble metals like silver not form natural cubic crystals?

Native silver crystals can be found in cubic, octahedral, or dodecahedral forms. As you mention, silver often is found with gold, copper and other metals, and with gold, tends to hexagonal stacking, rather than cubic crystals.

Silver may also form pseudomorphs, in which the crystal shape is from its progenitor or its surroundings. For example, dyscrasite, $\ce{Ag3Sb}$, or acanthite, $\ce{Ag2S}$ may be replaced by silver in the form of the antimony or sulfur minerals.


Which compound or compounds are responsible for color change in blue butterfly pea flower tea?

OK, that was simple. This

https://www.google.de/search?q=Clitoria+Ternatea+blue+color

gave this

https://www.ncbi.nlm.nih.gov/pubmed/14568080

which lead me to this

https://en.wikipedia.org/wiki/Anthocyanidin

and I guess that's the culprit. I judge from that abstract linked above that the actual plant can contain a wide mixture of those anthocyanidines.


Who needs high pressure?  Ammonium can be amalgamated into mercury like an alkali metal.  As with metals generally, there are not really any ammonium molecules in this phase; rather, the ammonium cations are intermingled with mercury cations and delocalized valence electrons.

Nor is all that high a pressure needed for unamalgamated ammonium, relative to metallizing other materials.  Bernal and Massey (1954) reported that a pressure of 250 kbar would be sufficient to metallize an ammonia-hydrogen mixture, which may be compared with the rocky material deep in Earth's mantle remaining nonmetallic even at megabar pressures.

Ammonium is easy to metallize because it has, indeed, an alkali-metal electronic structure.  The tetrahedra structure common to ammonium and methane, $\ce{CH4}$, is most stable with eight valence electrons -- there are four bonding orbitals, corresponding to the four bonds in methane, and four antibonding orbitals.  In methane the neutral molecule matches this optimal electron count, but in ammonium the neutral molecule would have to include a ninth valence electron in its antibonding orbitals, like an alkali metal atom having one to many electrons for a stable noble gas configuration. Either this extra electron breaks down the proposed ammonium molecule (which is what we ordinarily see) or, under moderately high pressure or in a metallically bonded matrix like the amalgam, it goes into the metallic bonding.


What's up with all that magic? (A chapter formerly known as Introduction)
The hunt for for the holy grail of density functional theory (DFT) has come a long way.[1] Becke states in the introduction of the cited paper:

Density-functional theory (DFT) is a subtle, seductive, provocative business. Its basic premise, that all the intricate motions and pair correlations in a many-electron system are somehow contained in the total electron density alone, is so compelling it can drive one mad.

I really like this description, it points out why we use and need DFT, and as it also points out the flaws, that every computational chemist has to deal with: How can something with such a simple approach be correct?
Something that is often forgotten about DFT is, that in principle it is correct. It's the implementations and approximations, that make it incorrect, but usable. Becke states this in the following quote:

Let us introduce the acronym DFA at this point for “density-functional approximation.” If you attend DFT meetings, you will know that Mel Levy often needs to remind us that DFT is exact. The failures we report at meetings and in papers are not failures of DFT, but failures of DFAs.

I sometimes here that the abbreviation DFT is often used in the wrong context, since we are not talking about the theory itself any more, but about the implementations and approximations of it. One suggestion I heard was, that it should rather be used as density functional technique.
With that in mind I would like to state, that I absolutely agree with the previous answer by user1420303 and it's subsequent comment by Geoff Hutchison. Since you asked for a somewhat more practical approach, I like to offer the advice I usually give to students new in the field.
Old is bad, isn't it?
Some of the functionals are now around for about thirty years. That does not make them bad, maybe even the opposite. It shows, that they are still applicable today, giving reasonable results. One of my personal favourites is the conjunction of Becke 1988 and Perdew 1986, often abbreviated as BP86.[2] It's a pure functional which is available most modern quantum chemical packages.[3] It performs usually well enough for geometries and reasonable well for energies for simple systems, i.e. small organic molecules and reactions.
The magical functional B3LYP was one of the first hybrid functionals, and it was introduced by Gaussian's very own developers.[4] A lot of people were surprised how well it worked and it quickly became one of the most popular functionals of all time. It combines Becke's three parameter functional B3[5] with Lee, Yang and Parr's correlation functional.[6] But why are we surprised it works? The answer is quite simple, it was not fitted to anything. Frish et. al. just reworked the B3PW91 functional to use LYP instead of PW91. As a result, it heavily suffers or benefits from error compensation. Some even go as far as to say: “It is right for the wrong reasons.”[7-9] Is it a bad choice? No. It might not be the best choice, but as long as you know what you are doing and you know it is not failing for your system, it is a reasonable choice.
One functional is enough, is it?
Now that we established, that old functionals are not out of fashion, we should establish something very, very important: One is never enough.
There are a few things, where it is appropriate to do most of the work with one functional, but in these cases the observations have to be validated with other methods. Often it is best to work your way up Jacob's ladder.[10]
How do I start?
It really depends on your system and what you are looking for. You are trying to elucidate a reaction mechanism? Start with something very simple, to gain structures, many structures. Reaction mechanisms are often about the quantity of the different conformers and later about suitable initial structures for transition states. As this can get complex very fast, it's best to keep it simple. Semi-empirical methods and force fields can often shorten a long voyage. Then use something more robust for a first approach to energy barriers. I rely on BP86 for most of the heavy computing. As a modern alternative, another pure density functional, M06-L is quite a good choice, too.[11] Some of the popular quantum chemistry suites let you use density fitting procedures, which allow you to get even more out of the computer. Just to name a few, without any particular order: Gaussian, MolPro, Turbomole.
After you have developed a decent understanding of the various structures you obtained, you would probably want to take it up a notch. Now it really depends on what equipment you have at hand. How much can you afford? Ideally, more is better. At least you should check your results with a pure, a hybrid, and a meta-hybrid functional. But even that can sometimes be a stretch.[12]
If you are doing bonding analysis, elucidation of the electronic structure, conformation analysis, or you want to know more about the spectrum, you should try to use at least five different functionals, which you later also check versus ab initio approaches. Most of the times you do not have the hassle to deal with hundreds of structures, so you should focus of getting the most accurate result. As a starting point I would still use a pure functional, the worst thing that could happen is probably, that is reduces the times of subsequent optimisations. Work your way up Jacob's ladder, do what you can, take it to the max.[13]
But of course, keep in mind, that some functionals were designed for a specific purpose. You can see that in the Minnesota family of functionals. The basic one is M06-L, as previously stated a pure functional, with the sole purpose of giving fast results. M06 is probably the most robust functional in this family. It was designed for a wide range of applications and is best chosen when dealing with transition metals. M06-2X is designed for main group chemistry. It comes with somewhat built in non-covalent interactions and other features. This functional (like most other though) will fail horribly, if you have multi-reference character in your system. The M06-HF functional incorporates 100% Hartree-Fock exchange and was designed to accurately calculate time dependent DFT properties and spectra. It should be a good choice for charge transfer systems. See the original publication for a more detailed description.[14]
Then we have another popular functional: PBE.[15a] In this initial publication an exchange as well as a correlation functional was proposed, both pure density functionals, often used in conjunction.[15b] I don't know much about it's usefulness, since I prefer another quite robust variation of it: PBE0, which is a hybrid functional.[15c,d] Because of its adiabatic connection formula, it is described by the authors as a non-empirical hybrid functional.[15d]
Over the years there have been various developments, some of the are called improvement, but it often boils down to personal taste and applicability. For example, Handy and Cohen reintroduced the concept of left-right correlation into their OPTX functional and subsequently used it in combination with LYP, P86 and P91. Aparently, they work well and are now often used also as a reference for other density functionals. They went on and developed a functional analogous to B3LYP but outperforming it.[16]
But these were obviously not the only attempts. Xu and Goddard III extended the B3LYP scheme to include long range effects. They claim a good description of dipole moments, polarizabilities and accurate excitation energies.[17]
And with the last part in mind, it is also necessary to address long range corrections. Sometimes a system cannot be described accurately without them, sometimes they make the description worse. To name only one, CAM-B3LYP, which uses the coulomb attenuating method.[18] And there are a couple of more, and a couple of more to come, head on over to a similar question: What do short-range and long-range corrections mean in DFT methods?
As you can see, there is no universal choice, it depends on your budget and on the properties you are interested in. There are a couple of theoretical/ computational chemists on this platform. I like BP86 as a quick shot and answer questions relating to MO theory with it, shameless self-promotion: Rationalizing the Planarity of Formamide or Rationalising the order of reactivity of carbonyl compounds towards nucleophiles. And sometimes we have overachievers like LordStryker, that use a whole bunch of methods to make a point: Dipole moment of cis-2-butene.
So I picked a functional, what else?
You still have to pick a basis set. And even here you have to pick one that fits what you need. Since this answer is already way longer than I intended in the first place (Procrastination, yay!), I will keep it short(er).
There are a couple of universally applicable basis sets. The most famous is probably 6-31G*. This is a nice ancient basis set that is often used for its elegance and simplicity. Explaining how it was built is easier, than for other basis sets. I personally prefer the Ahlrichs basis set def2-SVP, as it comes with a pre-defined auxiliary basis set suitable for density fitting (even in Gaussian).[19]
Worth mentioning is the Dunning basis set family cc-pVDZ, cc-pVTZ, ... . They were specifically designed to be used in correlated molecular calculations. They have been reworked and improved after its initial publication, to fit them to modern computational standards.[20]
The range of suitable basis sets is large, most of them are available through the basis set exchange portal for a variety of QC programs.
Sometimes an effective core potential can be used to reduce computational cost and is worth considering.
*Sigh* What else?
When you are done with that, consider dispersion corrections. The easiest way is to pick a functional that has already implemented this, but this is quite dependent on the program of you choice (although the main ones should have this by now, it's not something brand new). However, the standalone DFT-D3 program by Stefan Grimme's group can be obtained from his website.[21]
Still reading? Read more! (A chapter formerly known as Notes and References)

Axel D. Becke, J. Chem. Phys., 2014, 140, 18A301.
(a) A. D. Becke, Phys. Rev. A, 1988, 38, 3098-3100. (b) John P. Perdew, Phys. Rev. B, 1986, 33, 8822-8824.
Unfortunately this functional is not always implemented in the same way, although the differences are pretty small. It basically boils down as to which VWN variation is used in the local spin density approximation term. Also see S. H. Vosko, L. Wilk, and M. Nusair, Can. J. Phys., 1980, 58 (8), 1200-1211.
P. J. Stephens, F. J. Devlin, C. F. Chabalowski, and M. J. Frisch, J. Phys. Chem., 1994, 98 (45), 11623–11627.
Axel D. Becke, J. Chem. Phys., 1993, 93, 5648.
C. Lee, W. Yang, and R. G. Parr, Phys. Rev. B, 1988, 37, 785–789
Unfortunately the B3LYP functional suffers from the same problems that are mentioned in [3].
The failures of B3LYP are known and often well documented. Here are a few recent papers, but there are many, many more. (a) Holger Kruse, Lars Goerigk, and Stefan Grimme, J. Org. Chem., 2012, 77 (23), 10824–10834. (b) Joachim Paier, Martijn Marsman and Georg Kresse, J. Chem. Phys., 2007, 127, 024103. (c) Igor Ying Zhang, Jianming Wu and Xin Xu, Chem. Commun., 2010, 46, 3057-3070. (pdf via researchgate.net)
Just my two cents, that I am hiding in the footnotes: “Pretty please do not make this your first choice.”
John P. Perdew and Karla Schmidt, AIP Conf. Proc., 2001, 577, 1. (pdf via molphys.org)
Yan Zhao and Donald G. Truhlar, J. Chem. Phys., 2006, 125, 194101.
Note, that not always full recomputations of geometries are necessary for all different functionals you apply. Often single point energies can tell you quite much how good your original model performs. Keep the computations to what you can afford.
Don't use an overkill of methods, if you already have five functionals agreeing with each other and possibly with an MP2 calculation, you are pretty much done. What can the use of another five functionals tell you more?
Y. Zhao, N.E. Schultz, and D.G. Truhlar, Theor. Chem. Account, 2008, 120, 215–241.
(a) John P. Perdew, Kieron Burke, and Matthias Ernzerhof, Phys. Rev. Lett., 1996, 77, 3865. (b) The exchange functional was revised in Matthias Ernzerhof and John P. Perdew, J. Chem. Phys., 1998, 109, 3313. (c) Carlo Adamo and Vincenzo Barone, J. Chem. Phys., 1999, 110, 6158. (d) Kieron Burke, Matthias Ernzerhof, and John P. Perdew, Chem. Phys. Lett., 1997, 265, 115-120.
(a) N. C. Handy and A. J. Cohen, Mol. Phys., 2001, 99, 403-12. (b) A. J. Cohen and N. C. Handy, Mol. Phys., 2001, 99 607-15.
X. Xu and W. A. Goddard III, Proc. Natl. Acad. Sci. USA, 2004, 101, 2673-77.
T. Yanai, D. P. Tew, and N. C. Handy, Chem. Phys. Lett., 2004, 393, 51-57.
(a) Florian Weigend and Reinhart Ahlrichs, Phys. Chem. Chem. Phys., 2005, 7, 3297-3305. (b) Florian Weigend, Phys. Chem. Chem. Phys., 2006, 8, 1057-1065.
The point where the use of more basis functions does not effect the calculation. For the correlation consistent basis sets, see a comment by Ernest R. Davidson, Chem. Phys. Rev., 1996, 260, 514-518 and references therein. Also see Thom H. Dunning Jr, J. Chem. Phys., 1989, 90, 1007 as the original source.
DFT-D3 Website; Stefan Grimme, Jens Antony, Stephan Ehrlich, and Helge Krieg,
J. Chem. Phys., 2010, 132, 154104.
Have fun and good luck!


Equilibrium constant when adding more of a reactant

You're correct.  The equilibrium constant will not change.  The equilibrium constant is only defined at equilibrium.  Your analysis of the situation was flawless.


Interesting question. I actually had to look up the definition of petrichor.  After getting a little nostalgic about rainfall, I decided to go ahead and forage around for an answer for you.  I found a nice summary article here for those that are interested, and then a really nice bit of history here with a link to a great video about how rainfall helps to produce aerosols, which can contain the petrichor compounds.

Your question regards the reactions in the soil that give the petrichor smell, so let me get to that:

The main constituents of petrichor are volatile plant oils and geosmin. Oxidation of the fatty acids produced by plants can lead to some of the volatile compounds that constitute petrichor. Geosmin is a metabolite of actinobacteria and is produced through a biological pathway.

https://en.wikipedia.org/wiki/Geosmin - a brief intro to Geosmin
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3013058/ - A free paper highlighting the biosynthesis of geosmin


Here is some additional info from web

Wood's metal, also known as Lipowitz's alloy or by the commercial names Cerrobend, Bendalloy, Pewtalloy and MCP 158, is a eutectic, fusible alloy with a melting point of approximately 70 °C (158 °F). It is a eutectic alloy of 50% bismuth, 26.7% lead, 13.3% tin, and 10% cadmium by weight.

I have experience using the material for fixturing of difficult to secure /  shaped parts for machining. It can also help in fixturing/holding - then later melted away.

The eutectic composition will privide lowest Tm and get the info from a phase diagram.

Do not add mercury for safety reasons.


The difference in behavior between the two methods is likely how much "wiggle" the molecule is allowed to have, or how much additional kinetic energy you allow the molecule to have when searching for the minimum energy conformation. The equatorial conformation is the global minimum conformation, but the axial conformation is a local minimum. Additionally, between the two conformations, there is another minimum called the twist-boat. You can read about them on this Wikipedia article or you could check out an introductory organic chemistry book from a library. Molecular dynamics tend to stop when a minimum is found. In other words, if a small alteration to the geometry increases the potential energy, the calculation stops. Monte Carlo calculations on molecules are designed to examine the evolution of conformation with time, and so geometry moves out of energy minima are allowed.

Unsubstituted cyclohexane has four different conformations: Chair (relative energy 0 kJ/mol), half chair (+45 kJ/mol), twist boat (+23 kJ/mol), and boat (+30 kJ/mol). A molecular dynamics calculation may not allow large enough steps to get over the activation barrier to get out of a chair conformation. The following energy diagram shows the conformational space of cyclohexane.

The energy difference between equatorial and axial methylcyclohexane is 7.6 kJ/mol (1.6 kcal/mol on this table), which is small compared to the difference between the chair and the half-chair conformation. The energy diagram below approximates the energy steps between equatorial and axial methylcyclohexane.

Any computational method that allows potential energy steps large enough to get to the half-chair will certainly isomerize the axial conformer to the equatorial conformer. Presence or absence of solvent molecules in the simulation may be irrelevant.


Method 1: Assume the compound is ionic

For binary compounds consisting of a metal (but not a transition metal) and a nonmetal, the best method of determining oxidation numbers is to assume that the compound is ionic and determine the charge on the corresponding ions. The oxidation number of a monatomic ion is equal to its charge. This a good assumption because most compounds of this type are ionic, and it will work on many compounds that might be covalent. To make this work, you need to memorize some typical charges for ions:

Alkali metals - Li, Na, K, Rb, Cs - are typically 1+
Halides - F, Cl, Br, I - are typically 1-
Alkaline earth metals - Be, Mg, Ca, Sr, Ba - are typically 2+
Chalcogenides - O, S, Se, Te - are typically 2-

These are typically sufficient to deduce charges for most binary ionic compounds.

Let's do some examples.

  Definitely ionic: $\ce{Li3P}$

First, we assume that this compound is comprised of lithium ions and phosphide ions. The lithium ion is $\ce{Li^+}$, and there are three of them for a total of $3+$. Thus the charge on phosphorous must be $3-$. The oxidation number of lithium is +1, and the oxidation number of phosphorous is -3.

  Definitely covalent: $\ce{BeH2}$

Beryllium hydride is covalent, but this approach still works. If we assume this compound is a beryllium cation, $\ce{Be^2+}$, and two hydride anions, we can work it out. The hydride anions must be $\ce{H^-}$ to make the compound neutral. Thus, the oxidation number of beryllium is +2, and the oxidation number of hydrogen is -1.

Method 2: Based on structure

If we know the structure, we can use an alternative method. This method works best for covalent compounds, and does not work well for ionic lattices or network compounds. For each atom, we can assign the oxidation number as follows:

Each bond to a less electronegative atom contributes -1 to the oxidation number
Each bond to a more electronegative atom contributes +1 to the oxidation number
Each bond between two atoms of the same electron contributes +0 to the oxidation number

Let's look at some examples:

  $\ce{H2O2}$

The structure of $\ce{H2O2}$ is $\ce{H-O-O-H}$. Each oxygen atom has one bond to hydrogen (-1) and one bond to oxygen (+0) for an oxidation number of -1. Each hydrogen has one bond to oxygen (+1) for an oxidation number of +1.

  $\ce{HCN}$

The structure of $\ce{HCN}$ is $\ce{H-C#N}$. The hydrogen atom has one bond to carbon (+1), for an oxidation number of +1. The carbon atom has one bond to hydrogen (-1) and three bonds to nitrogen (+3) for an oxidation number of +2. The nitrogen atom has three bonds to carbon (-3), for an oxidation number of -3.

  $\ce{NaC2H}$

This compound is an ionic compound with one of the ions having covalent bonds. The structure is $\ce{Na^+ ^-C#C-H}$. The charge on sodium is 1+, so its oxidation state is +1. The first carbon atom has a formal charge of 1- and three bonds to another carbon atom (+0). The formal charge is added to the contributions from the bonds to get an oxidation state of -1. The other carbon atom has an oxidation state of -1 because of one bond to hydrogen (-1) and three bonds to the first carbon (+0). The oxidation state of the hydrogen atom is +1.

Your compound $\ce{Mg3C2}$

Either approach will work for $\ce{Mg3C2}$. If we treat it as ionic (a valid assumption based on results reported in this paper by Hongzhe Pan, et al.), there are three $\ce{Mg^2+}$ cations in the formula unit, and carbon needs to have a corresponding charge to make the compound neutral.

In the same paper by Pan, et al., the structure of a $\ce{Mg3C2}$ monolayer is described as a honeycomb kagome lattice. Below is a representation of part of this lattice:

.

In this lattice, each magnesium is bonded to two carbon atoms, and carbon is more electronegative than magnesium. Each carbon atom is bonded to three magnesium atoms.

The similar compound $\ce{Mg2C3}$ is a little trickier. If we use the ionic method, we can get an average oxidation state for each carbon atom of $\dfrac{+4}{3} \approx +1.333$. An atom cannot have a fractional oxidation state, so the total +4 must be distributed unequally over the three carbon atoms. If we look at the structure, we can see that one carbon atom (the one in the middle) is different than the other two.

$$\ce{Mg^2+ ^2-C=C=C^2- Mg^2+}$$

Using the formal charges on the carbon atoms, you can determine the oxidation states.


Is there a way to turn alcohol into a solid gel?

If your alcohol is ethanol, calcium acetate works very well.  I use calcium acetate to make gel in petri dishes, and sprinkle metal salts on it for colored flame demonstrations.  Calcium acetate is also used in some canned alcohols used for warming/cooking foods.

3 grams of calcium acetate dissolved in 10 ml of water will make a saturated solution (some may not dissolve, which is OK).  This is mixed with 75 ml of 95% ethanol, which will then form a colloidal gel identical to that used in canned alcohol warmers.  I do not know if this gel is suitable for your purposes, but it works well for a heat source.  If used as a heat source, a pinch of salt should be added to the aqueous solution before mixing with alcohol to make the flame more visible.


Can the titrant and the unknown material be flipped?

The titration reaction does not know or care whether one reagent is in the buret and one is in the conical flask (Erlenmeyer flask, it is not a beaker in your figure). And certainly you can standardize/titrate potassium permanganate with ferrous ammonium sulfate solution to determine the concentration of permanganate ion. Ferrous ammonium sulfate hexahydrate is a primary standard in analytical chemistry.

If you are extraordinary careful, you can fill the iron (II) solution in the buret and have permanganate in the conical flask. The end-point will be slightly vague (deep purple to colorless). You really have to be extra-cautious. The question is why bother with this painful eye-stressing procedure? Always take the permanganate in the buret.

A dark solution can be used in the conical flask. For example, in iodometric titrations, we start with a deep brown solution, which fades, to nearly a colorless solution upon titration with thiosulfate. In order to confirm the end-point, you would put a drop of starch before your expected equivalence point. Again, the color change is from deep blue to colorless.


Do hydroxymethyl-substituted cyclopropylmethyl cations undergo ring expansion?

The following takes place-

And then H3O+ can be assumed to give H+ and H2O.
Try it out and you'll get to the product.

If the positive charge is shifted to carbon 1, it would form a more stable alkene since it is more substituted. But after formation of C1-C4 double bond, no further reaction will occur since after expansion + charge will be on a double bonded carbon which would be highly unstable. Since Benzene is more stable than the product which will be formed after shifting of + charge to Carbon 1, and since heat is also applied, there will be enough energy to form more stable benzene and thus the reaction takes place in such a manner.


Urine typically is yellow, but in combination with blue dyes, a green colour can be obtained.

Noteworthy compounds are methylene blue, thymol, Cimetidine (antacid), or Mitoxanthrone (used in cancer therapy).

I've mentioned these just for the sake of completeness - I think it is irresponsible and stupid to spike food with anything that doesn't belong there.

Hiding a webcam in a shelf and running zoneminder probably isn't legal either and will get you in trouble with the privacy officer, but at least that doesn't mess around with the health of anybody else.


Reaction of copper (II) ion reacting with sodium hydroxide

Yes, the reaction is correct however it may be preferable to either include all or no spectator ions.
Use either a net ionic equations (omit the Na+), molecular equation (include the copper compound full) or complete ionic equation.

For example, if the compound is copper sulphate the equations would be (without states)

Net ionic
$$\ce{Cu^2+}+\ce{2OH^-->Cu(OH)_2}$$
Molecular equation
$$\ce{CuSO_4}+\ce{2NaOH->Cu(OH)_2}+\ce{Na_2SO_4}$$
Complete Ionic Equation
$$\ce{Cu^2+ +SO_4^{2-}}+\ce{2Na^+ +2OH^-->Cu(OH)_2}+\ce{2Na^+ +SO_4^2-}$$

Edit

In the case where you don't know what copper compound it is, use a net ionic equation. As a side note, copper hydroxide is not white. You might want to check your calculations.

If you're wondering what it is, check what other compounds will emit green.


which are the possible ways to detect impurity in drugs?

There are a lot of types of impurity. Basically, much as any plant that's growing where you don't want it is a weed, any material in your sample that isn't what you want to have in it is an impurity.
Since you asked a question in simple terms I will attempt to answer it in equally simple terms:

Byproduct contamination - The theoretical reaction is usually not the only possible one, especially in organic chemistry such as pharmaceutical work. Areas of the reaction which didn't have just the right mixture of reactants, or wasn't quite at the right temperature, could have produced different products. Additionally, devations from the ideal when adding reactants can result in unreacted quantities of those reactants.
Various forms of chromatography, such as HPLC, CC, GC, MS, etc can detect and even separate these various byproducts from the ideal product. How effective any one of these techniques is depends on the type and combination of the products and byproducts. One of the most basic aspects of any drug research chemist's job is to isolate their product from all of the byproducts of their reaction.

Structural isomer contamination - Two compounds are structural isomers if they have the same molecular formula (number and type of atoms), but have different molecular structures. Tetramethylbutane, for instance, is based on a 4-carbon alkane, to which two additional methyl groups each are added to the carbons in the "2" and "3" positions of the alkane. The resulting molecule has all the same carbons and hydrogens as the totally linear octane molecule, but its structure causes very different behaviors (for instance, it has a much narrower range of temperatures at which it is a liquid).
Your average petrochemist doesn't care too much about the difference between any of the structural isomers of these hydrocarbon alkanes; their primary reaction of interest, combustion, only cares about the required stoichiometric ratio of fuel to air, and the resulting rate and heat of combustion. However, such isomers pose problems in pharmacology for the above reasons; a reaction that was theorized to produce a molecule with one arrangement may in fact produce an entirely different isomer, with similar physical properties as the desired molecule, but much different reaction chemistry including biochemical effects. Chromatography methods based on molecular weight or physical properties like boiling point or solubility may or may not be able to tell you whether you have the right isomer, but other methods might show a difference in the physical properties of the chemical allowing separation.

Chirality and Racemization - Even if two isomers have all the same atoms, and each atom is bonded to the "correct" adjoining atoms, the molecules can have different shapes because the atoms are bonded to different relative positions of their adjoining atoms. These "spatial isomers" or "stereoisomers" have many forms, from structures that exhibit the same "mirror-image" symmetry that your left and right hand do (enantiomers), or that have certain functional groups transposed by rotation around a "chiral center" (diasteromers, including cis-trans and most detro-levo isomers). Again, like structural isomers, these stereoisomers can cause problems for pharmacologists because the different isomers can have different reactive properties in the complex biochemical soup that is the human body.
Sometimes these chiral isomers can be separated from each other; enantiomers and other spatial isomers are often "optically active", meaning they rotate polarized light in opposing directions depending on their chiral arrangement. Certain "chiral resolving agents" can also separate these isomers based on a differing affinity for one isomer over another, and these agents are often used as "stationary phases" of column chromatography including HPLC that can separate and thus "purify" the desired isomer from its chiral mirror. However, a few troublesome spatial isomers will "racemize"; that is, they will readily rearrange between the two stereoisomers when given the right environment, and so any attempt to produce a pure sample of one enantiomer is ultimately pointless.


Can a network covalent solid be amorphous?

The most obvious example is amorphous carbon. As mentioned in a comment, another good example is glass (although, whether it's truly "solid" past the glass transition temperature).

Really, a "network" simply means "solid with covalent bonding" and of course it's easy to imagine many possible amorphous network solids.

Vulcanized rubber might qualify too, but it would depend on whether there were many non-interlinked macromolecules. Usually macromolecules would be considered molecular solids due to the high number of non-bonded packing interactions. If the amount of cross-linking is high, classifying as a network solid makes sense.


When can a molecule be considered freely rotating at room temperature?

I'm not a transition-state physical chemist, but I think a good approach to this problem is transition-state theory, specifically the Eyring equation:

$$k = \kappa \frac{k_b T}{h} e^{\frac{-\Delta G^{\ddagger}}{RT}}$$

This equation tries to predict the rate constant $k$ from an assumed pseudo-equilibrium between the transition-state and the starting material, which are assumed to differ in energy by $\Delta G^{\ddagger}$.  $k_b$ is Boltzmann's constant, $h$ is Planck's constant, $T$ is temperature, and $\kappa$ is the "fudge factor" or transmission coefficient, the fraction of transition states that react productively instead of converting back to starting material.

In the conformational reactions you are talking about, I think a good starting place would be to assume the eclipsed conformations are the transition state, although I don't know if this has been shown to be 100% true experimentally or computationally.   As a first approximation, let's assume there isn't a huge entropy term in $\Delta G^{\ddagger}$, so that it is well-approximated by the energy level differences you found.  If we also assume that $\kappa$ is 1.0, then we can just multiply the $e^{\frac{-\Delta E}{RT}}$ you've already calculated by $\frac{k_b T}{h} = 6.2\times10^{12}\mathrm{~s^{-1}}$ at 298 K to get (very) approximate rate constants.

Ethane: $6.55 \times 10^{-3} \rightarrow 4.1 \times 10^{10}\mathrm{~s^{-1}}$
Butane: $4.79 \times 10^{-4} \rightarrow 3.1 \times 10^{9}\mathrm{~s^{-1}}$
$\ce{H2O2}$: $4.29 \times 10^{-5} \rightarrow 2.0 \times 10^{8}\mathrm{~s^{-1}}$
Ethene: $2.09 \times 10^{-44} \rightarrow 1.3 \times 10^{-31}\mathrm{~s^{-1}}$

That shows that rotational equilibrium of everything except ethene should be very rapid, but that the half-life for ethene equilibration much greater than the age of the universe ($4 \times 10^{17} \mathrm{~s}$).

Suppose we want to find an energy difference that should lead to a rate constant that is neither very fast nor very slow, say we'd like $k\approx\mathrm{1~hr^{-1}}$.  Inverting our very very crude approximation gives:

$$\Delta E = - RT \ln{\left(\frac{k h}{\kappa k_b T}\right)}$$

Plugging in $k=\mathrm{1~hr^{-1}=2.7\times10^{-4} s^{-1}}$ gives (if I did my math right) 93 kJ per mol.  Some well-known atropoisomers do have estimated energy differences in that range.


Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?

TL;DR Fluorine is electronegative and can support the extra negative charge that is dispersed on the six X atoms in $\ce{SX6}$, whereas hydrogen cannot.

First, let's debunk a commonly taught myth, which is that the bonding in $\ce{SF6}$ involves promotion of electrons to the 3d orbitals with a resulting $\mathrm{sp^3d^2}$ hybridisation. This is not true. Here's a recent and arguably more understandable reference: J. Chem. Educ. 2020, 97 (10), 3638–3646 which explains this. Quoting:

The natural ionicity, $i_\ce{SF}$, of each $\ce{S-F}$ bond [in $\ce{SF6}$] is 0.86, indicating a rather ionic σ bond. Each fluorine has an average charge of $−0.45$, resulting in a sulfur center of charge $+2.69$. [...] In summary, the electronic structure of this system is best described as a sulfur center with a charge somewhere between $2+$ and $3+$; the corresponding negative charge is distributed among the equivalent fluorine atoms. Shown in Figure 12 is the orbital occupation of the sulfur center, $\ce{3s^1 3p^{2.1} 3d^{0.19} 5p^{0.03} 4f^{0.01}}$. The minimal occupation of d-type orbitals eliminates the possibility of $\mathrm{sp^3d^2}$ hybridization.

If not via d-orbital bonding, how does one then describe the structure of $\ce{SF6}$? I'll present an LCAO-MO answer. Here's a "simple" MO diagram (I won't go through the details of how to construct it). It's actually fairly similar to that of an octahedral transition metal complex, except that here the 3s and 3p orbitals on sulfur are below the 3d orbitals.

Just for the sake of counting electrons, I treated the compound as being "fully ionic", i.e. $\ce{S^6+} + 6\ce{F-}$. So sulfur started off with 0 valence electrons, and each fluorine started off with 2 electrons in its σ orbitals. I've also neglected the π contribution to bonding, so the fluorine lone pairs don't appear in the diagram.
You'll see that, for a total of six $\ce{S-F}$ bonds, we only have four pairs of electrons in bonding MOs. The other two pairs of electrons reside in the $\mathrm{e_g}$ MOs, which are nonbonding and localised on fluorine. If we want to assign a formal charge to sulfur based on this diagram, it would be +2, because there are only actually four bonds. We could perhaps use Lewis diagrams to represent it this way:

The "hypervalent" resonance form contributes rather little and does not rely on invoking d-orbital participation; see Martin's comment on my answer below for greater detail about the resonance contributions. I am guessing that its existence can be mostly attributed to negative hyperconjugation, although I'm not 100% sure on this. The trans and cis resonance forms are not equal, so their contribution is not the same, but the contribution from each individual trans resonance form has to be the same by symmetry. Overall, the six fluorines in $\ce{SF6}$ have to be equivalent by the octahedral symmetry of the molecule. You could run a $\ce{^19F}$ NMR of the compound and it should only give you one peak.
(An alternative way of looking at it is that two of the $\ce{S-F}$ bonds are "true" 2c2e bonds, and that the other four $\ce{S-F}$ "bonds" are in fact just a couple of 3c4e bonds, but I won't go into that. For more information on multi-centre bonds, this article is a nice introduction: J. Chem. Educ. 1998, 75, 910; see also refs. 12 and 13 in that article.)
Right from the outset, we can see why $\ce{SH6}$ is not favoured as much. If we use the same framework to describe the bonding in $\ce{SH6}$, then those "correct" resonance forms that we drew would involve $\ce{H-}$. I'll leave it to the reader to figure out whether $\ce{F-}$ or $\ce{H-}$ is more stable.
Alternatively, if you want to stick to the MO description, the idea is that in $\ce{SH6}$, the relatively high energy of H1s compared to F2p will lead to the nonbonding $\mathrm{e_g}$ orbitals being relatively higher in energy. All things being equal, it's less favourable for a higher-energy orbital to be occupied, and $\ce{SH6}$ would therefore be very prone to losing these electrons, i.e. being oxidised.
In fact, if we do remove those four electrons from the $\mathrm{e_g}$ orbitals, then it's possible that these six-coordinate hydrides could form. But obviously we might not want to have a $\ce{SH6^4+}$ molecule on the loose. It'll probably lose all of its protons in a hurry to get back to being $\ce{H2S + 4H+}$. Is there anything better? Well, there's the species $\ce{CH6^2+}$, which is methane protonated twice. It's valence isoelectronic with $\ce{SH6^4+}$, and if you want to read about it, here's an article: J. Am. Chem. Soc. 1983, 105, 5258. While it's hardly the most stable molecule on the planet, it's certainly more plausible than $\ce{SH6}$.
Now, just to come back to where we started from: d-orbital participation. Yes, there is an $\mathrm{e_g}$ set of d orbitals that can overlap with the apparently "nonbonding" $\mathrm{e_g}$ linear combination of F2p orbitals, thereby stabilising it. It is true that some degree of this does happen. The issue is how much. Considering the fairly large energy gap between the $\mathrm{e_g}$ orbitals, this interaction is bound to be fairly small, and is nowhere near enough to justify a $\mathrm{sp^3d^2}$ description of it; Martin's comments contain more details.


Confused by notation of atomic number Z and mass number A on periodic table of elements

Periodic tables of elements (PTEs) are often abused by designers. Books are more trustworthy as long as they are written by scientists. Long story short, the second notation $(\ce{^{12}_{6}C})$ is the correct one.
There is an easy to remember AZE notation: $^A_Z\ce{E}$.

I suspect the PTE you were looking at lists standard (averaged) atomic weights of the elements $A_\mathrm{r}$ rounded to the nearest whole number so it may appear as if those were the mass numbers $A$, probably something like this:

  * The atomic weights listed on this Table of Elements have been rounded to the nearest whole number. As a result, this chart actually displays the mass number of a speciﬁc isotope for each element. An element's complete, unrounded atomic weight can be found on the It's Elemental website: http://education.jlaborgﬁtselemental/

Note that a good periodic table usually includes a legend which deciphers and justifies designer's choice.
Even better example is the Periodic Table of the Elements by NIST.
Despite atomic number $Z$ also located in the upper left corner of the cell, its location is typographically literate as both $Z$ and $A_\mathrm{r}$ have distinct place, different typeface and the legend unambiguously denotes which is which:


I'm going to go against the grain here:  the Maxwell distribution does describe the distribution of molecule speeds in any (3-D) matter, regardless of phase.

Suppose we have a system of $N$ molecules with masses $m_i$, positions $\vec{r}_i$, and velocities $\vec{v}_i$ (with $i = 1, ..., N$).  Assume that the total energy of this system is of the form
$$
E(\vec{r}_1, \dots, \vec{r}_N; \vec{p}_1, \dots, \vec{p}_N) = U(\vec{r}_1, \dots, \vec{r}_N) + \sum_{i = 1}^N \frac{\vec{p}_i^2}{2 m_i },
$$
i.e., a potential energy depending on the molecules' positions and a kinetic energy.  According to Boltzmann statistics, the probability of finding this system within a small volume of phase space $d^{3N} \vec{r} \, d^{3N} \vec{p}$ is
$$
\mathcal{P}(\vec{r}_1, \dots, \vec{r}_N; \vec{p}_1, \dots, \vec{p}_N) d^{3N} \vec{r} \, d^{3N} \vec{p} \propto e^{-E/kT} d^{3N} \vec{r} \, d^{3N} \vec{p} \\= e^{-U(\vec{r}_1, \dots, \vec{r}_N)/kT} \left[ \prod_{i=1}^N e^{-p_i^2/2m_i k T}\right]d^{3N} \vec{r} \, d^{3N} \vec{p}
$$
To find the probability distribution of finding molecule #1 with a particular momentum $\vec{p}_1$, we integrate over all other configuration variables (i.e., $\vec{r}_1$ through $\vec{r}_N$ and $\vec{p}_2$ through $\vec{p}_N$).  Because the $E$ is the sum of a contribution from the positions and a contribution from the momenta, the Boltzmann factors can be split between these integrals, with the result that
$$
\mathcal{P}(\vec{p}_1) \, d^3\vec{p}_1 \propto e^{-p_1^2/2 m_1 kT} d^3\vec{p}_1 \left[ \int e^{-U(\vec{r}_1, \dots, \vec{r}_N)/kT} d^{3N}\vec{r}\right] \left\{ \prod_{i = 2}^N \int e^{-p_i^2/2 m_i kT} d^3 \vec{p}_i \right\}
$$
These integrals are nasty (especially the one over all $N$ position vectors), but they're just constants with respect to $\vec{p}_1$, which means that the can be folded into the proportionality constant:
$$
\mathcal{P}(\vec{p}_1) \, d^3\vec{p}_1 \propto e^{-p_1^2/2 m_1 kT} d^3\vec{p}_1.
$$
A similar logic applies to every other molecule in my system.  In other words, the probability of finding a molecule with a momentum $\vec{p}$ doesn't depend at all on how they interact with each other, assuming that their interaction energy is dependent only on their collective positions.  Since these molecules obey the same momentum distribution, they must also have the same velocity distribution, and in particular they obey the Maxwell speed distribution.

So the answer to the question asked in your title, "Does the Maxwell-Boltzmann distribution apply to gases only?" is "no";  it applies to all phases of matter, in the sense that it describes the distribution of particle speeds and energies.  However, the question you ask at the bottom of your post, "Can I use a Maxwell-Boltzmann distribution to interpret reaction rates at the molecular level in liquids and/or solids?" may also be "no";  the connection between reaction rate and activation energy is not straightforward if the medium is dense, as was pointed out by @porphyrin in their answer.


When heat is increased, does water move upward in a straw due to increased capillary action or increased pressure?

It almost certainly was due to thermal expansion of water and the air above it. Consider that the thermal expansion coefficient of water near room temperature is ~$0.000214$ K$^{-1}$. [Though it is not linear over a broad range, assume for the small temperature rise in your car that it is.]

If the water was cold (e.g. ~$277$ K, maximum density) and it warmed to ~$300$ K ($80$o F), a temperature change of ~$23$ K, it would have expanded by a factor $(300 \ce{K} - 277 \ce{K}) \cdot 0.000214 \ce{K^{-1}} = 1.005$.

Assume you have a one-liter bottle filled completely with water when you start; by the time you return it's $1.005$ L. In the whole bottle, the change would hardly be noticeable, but the narrow straw greatly magnifies the rise of liquid. If the straw is $6$ mm in diameter, the volume of liquid is given as: $$V=\pi\cdot r^2\cdot h$$

$$5 \mathrm{mL} = 5 \ce{cm}^3$$

$$r = 0.3 \ce{cm}$$

$$h = \frac{5 \ce{cm}^3}{\pi \cdot (0.3 \ce{cm})^2} = 18 \ce{cm}$$

So the water would have risen about $18$ cm in the straw!

This exercise only took into account the water's expansion... for any air above the water in the bottle, and for other (approximately) ideal gases, $PV=nRT$ and thermal expansion would be far greater.

To some extent, the bottle itself expands, decreasing the rise of the liquid, so these calculations are not very accurate. Most likely, air expansion contributed more to the "mass ejection" than did that of water... it depends on whether you view your container as half-full or half-empty.


Is this calculation of the amount of required hydrogen gas correct?

Kedar's calculation may be right. But it must be realized that the reaction will be rather slow. Even with a $1 M$ solution, the reaction is slow. The reaction of $1.5$ $\pu{g}$ metallic zinc with such a diluted $\ce{HCl}$ solution ($0.1 M$) will last more than one hour. Anyway, in order to inflate such a balloon a higher pressure than the atmospheric pressure must be applied. If not, the balloon will keep its original dimension. How to increase the pressure of $\ce{H2}$ in such a balloon for inflating it, if the hydrogen is produced by the reaction of zinc on $\ce{HCl}$ ?


Can an organic molecule have a triplet ground state?

As Martin has mentioned, carbenes are a good starting point if you are looking for organic compounds with a triplet ground state.

In these carbenes, the HOMO is not twofold degenerate, as your question suggests. Instead, there are two singly occupied orbitals with different energy. The energy gap, though, is typically small, and since there is an energy cost associated with pairing electrons in the same orbital, the electrons prefer to stay apart.

$\ce{CH2}$ itself has a triplet ground state, but is very difficult to observe; most methods of generating carbenes actually produce singlet carbenes, and $\ce{CH2}$ is so reactive that singlet $\ce{CH2}$ reacts with other stuff before it decays. Not all that interesting, eh.

There are some examples of persistent triplet carbenes, though.1 This one, reported by Hirai and Tomioka in 1999,2 has a half-life of $\pu{940 s}$ in degassed benzene at $\pu{20 ^\circ C}$:

In general, simple alkyl and aryl carbenes adopt a triplet ground state. These particular aryl groups on the carbene provide a large degree of kinetic stabilisation, which prevents simple dimerisation of the carbene (to form an alkene); the para alkyl groups also hinder reaction at that position. A further computational study of this molecule has been carried out by the Schaefer group, which supports the analysis above.3

References

Hirai, K.; Itoh, T.; Tomioka, H. Persistent Triplet Carbenes. Chem. Rev. 2009, 109 (8), 3275–3332. DOI: 10.1021/cr800518t.
Hirai, K.; Tomioka, H. A Triplet Carbene That Can Almost Be Bottled. J. Am. Chem. Soc. 1999, 121 (43), 10213–10214. DOI: 10.1021/ja991387c.
Woodcock, H. L.; Moran, D.; Schleyer, P. v. R.; Schaefer, H. F., III. The Almost Bottleable Triplet Carbene:  2,6-dibromo-4-tert-butyl-2′,6′-bis(trifluoromethyl)-4′-isopropyldiphenylcarbene. J. Am. Chem. Soc. 2001, 123 (18), 4331–4335. DOI: 10.1021/ja003552q.


Preparing a 0.030 M solution of potassium ferrocyanide trihydrate?

It is quite trivial. Just calculate how many moles of potassium ferrocyanide trihydrate you need to measure out (n = C x V), then multiply with molar mass to see how many grams you need to measure out. After measuring the sample on analytical balance, quantitavely transfer it to the volumetric flask. Add some water and dissolve the sample. After homogenizing the solution, add water carefully to the full mark. You can't actually calculate the exact molarity of the prepared solution: you need to determine it experimentally, by performing a standardization with silver-nitrate standard using chromate salt as an indicator.


How to know which reduction and oxidation reactions occurred?

Here, Fe(III) will be reduced to Fe(II), and Copper(0) will be oxidized to Cu(II).
The standard potential for Cu2+/Cu is +0.34 V, and for Fe3+/Fe2+ it is +0.77 V.
Nernst's law gives the potentiels for a 0.5 M solution. It is +0.34 - 0.009 V = 0.33 V pour the Copper electrode, and 0.77 - 0.018 = 0.75 V for the Fe3+/Fe2+ electrode.
With this choice, you get nearly your results.
It you had chosen another system of half-reactions, you will get results which are far away from your measurements.


Water is not, in fact, colourless, but is slightly blue. As you can see below, in the visible range, there is slightly more absorption toward the long wavelengths, so more blue light makes it through than redder light. However, the absorption is very low in the visible range so the blue colour is faint and not apparent unless looking through a significant thickness of water. The reason things like $\ce{Cu^2+}$ solutions appear so vividly blue is because they also absorb red light, but very strongly, so a good deal more blue light than red passes through the solution. (see here)

I'm not sure exactly what you mean by transparent here, but if you mean why isn't water opaque, because water doesn't absorb much visible light (as described above), and doesn't contain any non-uniformities (like particles or the fat globules in milk) that can cause scattering, photons can travel through water in a straight line and most make it through without being absorbed, making water colourless (when looking through a small quantity) and transparent, much as air is.

Source


Common etchant for Stainless Steel 304L and Titanium Grade 2

Common etchant for stainless steel is $\ce{FeCl3}$ with $\ce{HCl}$ (up to 3%) heated at 45°C.

The etching solution for $\ce{Ti}$ looks fine.


What is the relationship between ortho, meta and para-dimethyl benzene?

These compounds do not have a stereochemical relationship. Stereoisomers have the same connectivity of atoms (constitution) with different arrangement in space. The (1,2), (1,3), and (1,4) in the names of these compounds indicate that this is not the case. These are not stereoisomers of any kind.

Not that the qualifying text about having to break bonds to interconvert is unhelpful and irrelevant. All isomers require bond-breaking to interconvert.

So, what then are the relationships between these three compounds? Let's look at a decision flowchart for relationship between structures:

  Do they have the same formula?

Yes.

  Do they have the same connectivity?

No. Thus, they are constitutional isomers. Constitutional isomerism is not a type of stereoisomerism. The two main types of stereoisomerism are enatiomerism and diastereomerism.

  It would be worth contacting your instructor and asking for clarification, noting that these are not examples of any kind of stereoisomerism. It could be an honest mistake in the wording of the question. As someone in the teaching profession, my experience is that most instructors are happy when a student understands the content at a level high enough to catch mistakes.


If you have two components in a mixture, say water and ethanol, and you heat the liquid to the boiling point of that mixture (at the given pressure), you may find that the composition of the resulting vapor and the remaining liquid are different. In that case, you have a non-azeotropic mixture and the composition of vapor and liquid will change as you continue boiling.

For the case of water/ethanol, at some point, the composition of vapor and liquid will match (and no longer change, no matter how long you boil, until all liquid has evaporated). That composition/mixture is called an azeotrope.

The liquid mixture of oxygen/nitrogen is an example for a system that does not show an azeotrope.


Decomposition of potassium permanganate under prolonged heating

Potassium manganate is produced by reacting manganese dioxide with potassium hydroxide and oxygen gas :

$\ce{2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O}$

So I would suspect that the reverse process will occur when further heating. (Of course when there is no water available you might get other products, maybe potassium oxide instead of potassium hydroxide)


Why polyethylene (PE), wool (polyamide) and cotton (cellulose) burnt with different smells, while they just product CO2 (and N2)?

Things are more complex than you suppose

There are two key reasons why these compounds burn with different smells.

The first is that your assumption about what the substances contain is wrong. Biological fibres contain plenty other than carbon and hydrogen. Often they will have lots of nitrogen and sulphur and various other elements and these often produce a variety of pungent chemicals that provide strong odours even in low concentrations. Plus artificial polymers will contain small amounts of other ingredients to modify their properties and these will often contain elements that give more complex combustion products some of which may be smelly.

The second is that burning these compounds in normal conditions is far from a perfect reaction yielding just the obvious oxidation products like carbon dioxide (which would be odourless). Even pure PE (and most of the PE you get will not be pure as it will contain various fillers and additives with other elements in it) will not burn cleanly to produce water and carbon dioxide). In normal conditions the burning will produce a variety of hydrocarbons and particles some of which will have odours. This is fairly obvious from the soot produced (which you would not get at all if the burning was completely efficient in producing carbon dioxide). The variety of those "imperfect" combustion products will be very different for different starting materials. So even compounds with roughly the same chemical composition (but different structure) may burn to give different combustion side products and, hence, will smell different.


Professor Jonathan M Goodman gave an explanation on the InChi mailing list:
https://sourceforge.net/p/inchi/mailman/inchi-discuss/thread/B7D99356-8B55-43BE-8C2A-B8CC3EB37799%40cam.ac.uk/#msg37289766
In summary, the given class of molecules represents a special case where the current InChi algorithm (1.06) produces less than useful results in order to being able to handle all other cases satifyingly. Prof Goodman hints at work being done in organometallics which may affect this behaviour. So, it's not a bug in the implementation of the algorithm, it is the algorithm itself that needs improvement.
The linked message also contains a smaller minimal case, confirming that it is the hetero-ion together with the keto-enol configuration causing the phenomenon.


Soap is a fatty acid salt of an alkali metal, typically sodium or potassium. The fatty acid itself is always insoluble in water, and therefore acidification of the soap solution produces two layers, a lower aqueous layer with the alkali salt of the acid and an upper layer of fatty acid laced with aqueous material. This can be treated with an organic solvent to dissolve the fatty acid and expel the aqueous material, giving two immiscible layers.
I suspect that your soap is a combination of fatty acid soap plus SLES.
Sodium laureth sulfate is not a soap, it is a surfactant, and it is so constructed so as to be more soluble in water than sodium laurate (which is a soap) or sodium lauryl sulfonate (which is a surfactant) by addition of typically 3 -OCH$_2$CH$_2$O- (ethylene glycol) groups. This complicates the precipitation and extraction; another variable is the solvent: while petroleum ether is organic, it has poor solvent power, and long chain acids are not well dissolved. That's why you have to do the extraction many times.
I visualize your experiment to go this way:

When you add the concentrated HCl to the sodium laureth sulfate, you insolubilize the HLES (the acid form), but it is not easily separable. A variable here is whether there is excess acid and how much. Another variable is that the precipitation is incomplete.

You add petroleum ether to extract the HLES and reject water and excess acid, leaving, hopefully, mostly NaCl in the aqueous phase. You do this again, extracting more HLES.

However, HLES is somewhat soluble in water (because it contains 3 -OCH$_2$CH$_2$O- groups), and is quite a good emulsifier for organics, and pet ether has a lot of low mw molecules, so before the last of the HLES is extracted, you find that the HLES has attracted a lot of water and a lot of pet ether. But this combination cannot contain all the water, so it is immiscible with the lower aqueous layer; it contains too much water to be miscible with the pet ether, so it is a third phase. Excess acid in step 1 complicates things because acids increase the solubility of ethoxylated surfactants.

You get something like this:

So instead of dealing with a soluble/insoluble situation, you have a significant but incomplete precipitation of HLES, and what doesn't precipitate out manages to emulsify with some of the pet ether and create the third phase.
A possible way to improve your separation is to add NaCl to the mixture after adding the HCl to maximize the salting-out effect. Another adjustment is to minimize excess acid (add just enough acid to precipitate the HLES - measure pH with paper) or add NaOH to adjust pH to near neutral (pH 6-8), when you get the third phase. This minimizes acid solubilization of the HLES, but keep the aqueous layer saturated with NaCl by adding solid NaCl if necessary.
You could also substitute toluene for the pet ether, but there is somewhat more hazard. Diethyl ether would be the extractant of choice many years ago, but alas, we must emphasize safety.