text:
Melting and boiling points of transition elements
The melting and boiling points of transition elements increase from scandium ($1530~\mathrm{^\circ C}$) to vanadium ($1917~\mathrm{^\circ C}$). They increase because as we go across the group, we have more unpaired (free) electrons.
But at chromium ($1890~\mathrm{^\circ C}$) however, the melting point decreases even though it has more unpaired electrons than the previous atoms. Why does this happen?

Could it be related to the half-filled d-orbital of Chromium, which is special because it 'borrows' from the s-orbital? (see http://en.wikipedia.org/w/index.php?title=Electron_configuration#Other_exceptions_to_Madelung.27s_rule).

the melting point of chromium is $1907^oC$ not $1890^oC$ - see http://en.wikipedia.org/wiki/Chromium (Check the general properties section on the right side for melting point). This makes it almost the same as Vanadium (which has melting point $1910^oC$).

What I find more peculiar is the massive drop in melting point from Cr to Mn. Mn seems to be the most anomalous, though Cr and Tc also are outliers. Here are graphs for the melting point and boiling point. Can anyone explain the dips satisfactorily?

To answer the question one has to put crystal structure and band structure on same picture.

That would be a part of the explanation. If you can elaborate, please do! Though it is not enough on its own, since some boiling points are also anomalous, and they cannot be due to structural characteristics in the solid.

I don't really like parts of this explanation (was hoping to find a better one here actually), but it's the best I know. I'll build on suggestions by @michielim and @vrtcl1dvoshun.
(Note: I think most of this argument can be transposed to the orbital overlap picture of bonding, but it might be slightly trickier to describe.)

For a backdrop to this answer, I have plotted the melting points and boiling points of the transition metals, so make sure to stare at these for a while before reading on.
From the physicists' "electron sea" point of view of metal bonding, the higher the ionic charge the metal atom can support, the higher the element's melting and boiling points. This explains why group 1 metals such as sodium have quite low melting/boiling points since the metal would be composed of electrons delocalized in a $\ce{M}^+$ lattice. Going towards group 2 and group 3 elements, one can expect to find a $\ce{M}^{2+}$ and $\ce{M}^{3+}$ lattice, and so on. However, this does not mean that, for example, metallic osmium is expected to be made out of $\ce{Os}^{8+}$ cations surrounded by a dense electron sea; as each successive "ionization" of the metal atom is performed, it becomes more unlikely that the next electron can become delocalized since it has to fight against a higher effective nuclear charge relative to the electron before it. Therefore, we can expect some "optimal average charge" for the metal ions in the pure metal. Since melting and boiling points tend to increase towards groups 7 and 8, we can safely assume that this optimal average charge is at least greater than $+3$, and it is likely higher for the heavier elements since they have significantly higher melting/boiling points.
If we can somehow figure out how easily electrons can be removed from their parent atoms to create the electron sea, we may have an argument to explain the trends and apparent discontinuities. First, notice that the $4^{\ce{th}}$ period metals are in general a little bit underwhelming compared to their $5^{\ce{th}}$ and 6$^{\ce{th}}$ period counterparts when looking at the boiling points. This can be attributed to the fact that the $3d$ orbitals are anomalously compact relative to higher $nd$ orbitals. This happens because all first occurrences of a given sublevel (i.e. $1s$, $2p$, $3d$, $4f$, $5g$, etc.) are composed of orbitals whose wavefunctions have no radial node. Since the wavefunction's value is monotonically increasing up to its maximum value, it tapers off quickly. Thus, when a 4$^{\ce{th}}$ period metal atom tries to delocalize its $3d$ electrons, it meets a slight increase in resistance due to its stronger interaction with the nucleus, meaning that the optimal average charge of the ions is slightly lower and leading to lesser interatomic interactions holding the metal together.
Now, for the sudden dips, most visible for $\ce{Cr,\, Mn}$ and $\ce{Tc}.$ As has been suggested, one is tempted to take into consideration the fact that they all have $d^5$ configurations, which represent a half-filled $d$ subshell. It can be argued that they are particularly stable due to exchange energy, though Cann suggests it is not the most adequate explanation and instead forwards the "parallel spin avoidance factor".
In half-filled subshell configurations, there is a maximum in the effective nuclear charge felt by the electrons (compared to the previous elements with no doubly-occupied orbitals) combined with relatively low interorbital repulsions due to the Pauli exclusion principle. The element that follows a half-filled subshell must now put an electron in a previously occupied orbital, creating stronger, intraorbital electron repulsion. These effects combine to suggest that removing an electron from or forcing an electron into half-filled configurations is particularly difficult (notice that $\ce{Cr}$ doesn't follow $\ce{Mn}$ and $\ce{Tc},$ more on that soon). If this is the case, then the delocalization of the third electron (which would create $\ce{Mn}^{3+}$ and $\ce{Tc}^{3+}$ ions) and all those beyond it would be somewhat suppressed. This would effectively reduce the optimal average charge of the ions in the metal, and decrease the strength of interactions.
Finally, $\ce{Cr}$ doesn't show the same electronic behaviour as $\ce{Mn}$ and $\ce{Tc},$ even though it has a $d^5$ configuration, and yet it is still anomalous. What gives? Well, according to Greenwood and Earnshaw (Chemistry of the Elements, 1997), the low boiling point of $\ce{Cr}$ is due to a combination of increased effective nuclear charge and the anomalously small size of $3d$ orbitals. When going from $\ce{V}$ to $\ce{Cr},$ the increase in nuclear charge is enough to pull the already small $3d$ orbitals quite close to the nucleus, disfavouring electron delocalization to the extent that a sudden drop appears. Mo has the same electron configuration as $\ce{Cr},$ yet it does not show nearly as large a dip because its $4d$ orbitals are already significantly more diffuse, and aren't so easily suppressed from delocalizing.

Since writing this post, I have learned that, according to the book Solid State Physics (Ashcroft, Mermin), the Drude model as originally proposed gives no theoretical basis for finding how many electrons per atom in a metal are delocalized, so it has no way to suggest how positively charged the metallic ions are in the lattice.

To make an argument based on electron configuration, one could see that the orbital occupation does affect the melting point:
$$
\begin{array}{lccc}
\text{Element} &   \text{Z}&   \text{Electron Config.}&    \text{Melting Point / K}  \\ \hline
\text{Vanadium}   &   23 & \ce{[Ar] 4s^2 3d^3}  &      2183 \\
\text{Chromium}   &   24 & \ce{[Ar] 4s^1 3d^5}  &      2180 \\
\text{Manganese}   &   25 & \ce{[Ar] 4s^2 3d^5}  &      1519\\
\text{Iron}   &   26 & \ce{[Ar] 4s^2 3d^6}  &      1811 \\
...  &   ... & ...  &      ... \\
\text{Copper}   &   29 & \ce{[Ar] 4s^1 3d^10}  &      1358 \\
\text{Zinc}   &   30 & \ce{[Ar] 4s^2 3d^10}  &      693 \\
\end{array}
$$
This is apparent in both Mn and Zn: the filled s-orbitals and (half-)filled d-orbitals impact the bonding character of the elements. This is also apparent in the 2nd and 3rd-row transition metals, though to a lesser extent due to more diffuse orbitals, greater shielding of the valence electrons, and relativistic effects from heavier nuclei. If you look at @NicolauSakerNeto's graph of the melting points, notice that there is still a dip in the graph when the d-orbitals are (half-)filled (i.e., for elements Tc, Cd, Os, and Hg).
Essentially, melting point boils (haha!) down to bonding:  more unpaired electrons lead to the creation of stronger bonds and thus higher melting points.

But there are 5 unpaired electrons in the d shell of Magnese so, why does it have low melting point?

The half filled d-orbital allows for extra stability compared to the other partially filled states. Read about Hund's Rule and the Aufbau Principal.

That applies to Chromium as well?


text:
How many electrons are transferred in this redox reaction?
I have the following reaction:
$$\ce{CH_4 (g) + 2O_2 (g) \rightarrow CO_2 (g) + 2H_2 O(l)}$$
How do I determine how many electrons are transferred in this reaction? More specifically, I need to know the value of $v_e$ which I am supposed to plug into the Nernst equation, which my textbook explains as the coefficient of the number of electrons transferred in each half-reaction. How does one go about splitting this into two half-reactions?

You can post your "I figured it out" part as an answer to this question, you know :)

Agreed. You should definitely post your answer.

The carbon atom must have an oxidation state (O.S.) of -4 on the left hand side since each hydrogen bonded to it must have an O.S. of +1. On the right hand side, the carbon atom has an O.S. of +4 since each oxygen bonded to it has an O.S. of -2. The carbon's oxidation state increased by 8. This is a hint that 8 electrons were transferred. To verify, examine the oxygen atoms. On the left, we only have free oxygen, therefore the O.S. is 0. On the right, we have 4 total oxygen atoms, each with an O.S. of -2. Thus our net change in oxidation state is 0 - 8 = -8. We can almost conclude that all these electrons were transferred from carbon, but we must check the oxidation states of the hydrogen atoms. On both sides, each H atom has an O.S. of +1, so there is no net O.S. change. Thus we can confirm that 8 electrons were indeed transferred in this reaction.

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What does the signals in a XPS spectrum mean?
I was trying to figure out the different peaks in a XPS spectrum for Molybdenum, but I could not understand them. About XPS, I know that this technique is based on the binding energy of the electrons of different elements

you are quite right, XPS does measure the 'binding energy' of electrons. More specifically, XPS bombards a surface with X-rays and when those X-rays have sufficient energy they are able to sufficiently excite electrons to be emitted from the sample. Each peak in an XPS spectrum corresponds to the energy necessary to emit a given electron.
So what would you expect to see for Mo?

Outer lying electrons (for Mo, we are talking about electrons in the
4p and 4s orbitals) will emit at a lower energy. You should observe
low-energy peaks that correspond to these emissions (approx. 30 - 60
eV).
you should see a cluster of peaks at approx. 230 eV that correspond to the d-orbital electrons.
Finally, you might see discrete peaks at very high energy, approx. 2600 - 2800 eV that    correspond to emission from the 2s and 2p orbitals

In general, outer orbitals emit at lower energy, and inner orbitals emit at higher energy.
Hope this helps!

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text:
If electrolysis splits water, why does only either hydrogen or oxygen (but not both) bubble up on one pole?
If electrolysis splits water, then that means that $\ce{H2O}$ is split into $\ce{H}$ and $\ce{OH}$ or $\ce{O}$.
How come that if a water molecule is split at e.g. the negative pole (anode), only the $\ce{O}$ bubbles up. What happens to the $\ce{H}$ and/or $\ce{OH}$? Do they react with the pole and stay there?

Water, as you may know, has a dualist nature between covalent and ionic bonding; oxygen is the second most electronegative element in the periodic table, while hydrogen's simplistic construction makes it very zen about how it forms bonds (it defines the center point of most electronegativity scales). While the bond between the hydrogens and oxygen of a water molecule are termed covalent, the electrophilic nature of the oxygen makes the bonds border on the ionic, and so water is just as easily thought of as $\ce{2H^+ + O^2-}$, or in more stable terms as $\ce{H^+ + OH^-}$.
In electrolysis, this ionic character of water is exploited in a series of "half-reactions"; at each node, a gas is formed by adding or removing electrons from water to liberate one of its two components, and when that gas precipitates out, what's left remains dissolved in the water as charged ions, which are repelled by the node that created it and attracted to the other node.
When electrons are added to water at the cathode, they are attracted to the partial positive charge of the "hydrogen side" of water, and can then form an orbit around a hydrogen. That liberates the hydrogen; it no longer needs the electron it's sharing with the oxygen to balance its charge, but it still needs a second electron to stabilize its 1s shell, and the best bet is to pair with another hydrogen to form hydrogen gas, which precipitates out. That leaves soluble hydroxide ions in the water, which are repelled by the cathode (and each other, so these ions wouldn't form any compounds) and move toward the anode. It's highly unlikely that both hydrogen atoms would be liberated from a single oxygen, and even if it did, the oxygen, with extra electrons, couldn't bond with another negative-charge oxygen to form gas; it would be much more likely to deprotonate another water molecule forming two hydroxides.
$$\ce{4H2O + 4e^- -> 2H2 (g) (^) + 4OH^- (aq)}$$
At the anode, the positive potential causes water to give up electrons, liberating the hydrogens again, this time as naked protons, which are attracted to water's partial negative pole to form hydronium ($\ce{H_3O^+}$ or simply $\ce{H^+ (aq)}$). The now chargeless but bond-seeking oxygen is left to pair up as oxygen gas, which precipitates out. Hydrogen peroxide is a possible intermediate, but very short-lived, as this peroxide will either find another peroxide and pair up the extra oxygens, or it will give up its hydrogens as protons under the positive voltage potential the same way water would, and become oxygen gas.
$$\ce{2H_2O - 4e^- \to O_2(g)\uparrow + 4H^+ (aq)}$$
... Then, somewhere in the middle, the hydronium and hydroxide meet, and as these proton donors and proton acceptors will do, they reform water:
$$\ce{4H^+(aq) + 4OH^- (aq) \to 4H_2O }$$
You'll notice that only one gas precipitates out at each node; the other atoms are trapped as soluble ions, either protons/hydronium or hydroxide, which cannot form other compounds because their like charges repel them from each other. Instead, they remain dissolved as local acidic and basic solutions. If the potential charge in the nodes is high enough, dipping a litmus strip at each end of the water bath (or dropping in a few mL of pH indicator dye) will yield an acidic reading at the anode and a basic reading at the cathode.


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Source of the no. of isomers in alkanes, and other organic molecules
I would like to ask if there is a source where I can find all the isomers for the alkanes, or other organic molecules. With other words, a list where each molecule of that group is stated in ascending order, and beside there is a value for the no. of isomers it can form.

Search around a bit $\ldots$ There are similar questions about this now. All of them say "it ain't that easy".

Not counting stereoisomers and enantiomers, the number of isomers of the first ten alkanes is as follows:

Methane: 1
Ethane: 1
Propane: 1
Butane: 2
Pentane: 3
Hexane: 5
Heptane: 9
Octane: 18
Nonane: 35
Decane: 75

Source: http://elib.mi.sanu.ac.rs/files/journals/tm/21/tm1121.pdf and Wikipedia articles for these alkanes. The PDF has the number of isomers for the first 30 n-alkanes.
The number of isomers possible with an n-carbon alkane can be calculated by a recursive combinatorial function; however, as the linked paper states, this math isn't for the weak. The number of structural isomers of a 30-carbon alkane is over 4 billion, and a 50-carbon alkane has over 1031 possible isomers. Even organic chemists and petrochemists usually don't bother to identify any one isomer beyond about the pentane mark. Some of the more symmetrical isomers are interesting on their own, since a few of the more compact ones (like the n-methyl types) have markedly different behaviors such as boiling point from longer isomers (for example, tetramethylbutane, an isomer of octane notable for its relatively narrow liquid window). The 2,2-dimethylpropane isomer of pentane is visually interesting because of its isohedral geometry, and likewise the tetraethylmethane/diethylpentane isomer might find interest among a few supremacist groups but is otherwise just another 9-carbon alkane.

There are a lot of "impossible" alkanes included in that numbers. Impossible due to too much crowding in some places. But to calculate this number I do not have the meanest idea. Finding conformers and calculating their energy is a time consuming business even on the fastest computers.

So, there are now ways of calculating the no. of isomers without a computer?

Per the linked article, the original computations were done by hand; however there was an error made in those calculations which was not found and fixed until the calculations were re-run with a supercomputer in the 1980s. Theoretically, any math that a computer can do, a human can do slower. Realistically, when you're not only computing ideal combinations, but also computing the chemical shapes and thus the distortion of bond angles and lengths from the ideal caused by "overcrowding" to determine the feasibility of that isomer, it becomes totally infeasible to do it all on paper.

KeithS's answer is good for the saturated alkanes (generic formula $\ce{C_{n}H_{2n +2}}$). For other organic molecules...
Introducing unsaturation or heteroatoms makes it even more complicated. There are two "possible" structures for $\ce{C3H6}$ - propene and cyclopropane and three for $\ce{C3H4}$ - propyne, allene, and cyclopropene.
While $\ce{C4H10}$ has two isomers, butane and isobutane, $\ce{C4H10O}$ has seven: 1-butanol, 2-butanol, isobutanol, tert-butanol, diethyl ether, methyl propyl ether, and methyl isopropyl ether. For $\ce{C4H8O}$, there are at least 23 valid structures, not including stereoisomers, although not all of them are stable molecules (enols, I am looking at you!).
Beyond maybe 5 or so heavy atoms, the whole effort becomes futile as the number of isomers increases exponentially. Instead, try Googling the formula you care about. ChemSpider will list all known/reported compounds with a given formula. For $\ce{C4H8O}$, there are 42 entries including stereoisomers.


text:
How is this an autocatalytic reaction?
I'm probably missing a simple component of this reaction, but for some reason I not seeing it.
These are two steps that compose a radical process of the Briggs-Rauscher reaction (that are autocatalytic):
$IO_3^- + HIO_2+H^+ \rightarrow 2IO_2 \cdot +H_2O$
$IO_2 \cdot + Mn^{2+} + H_2O \rightarrow HIO_2 + Mn(OH)^{2+}$
This process is autocatalytic in the sense that 2 $HIO_2$ is produced for each one consumed.  However, I'm just not seeing that for some reason.  Like I said, it's probably a simple thing I overlooked. So, I'm looking for an explanation for what's going on here.

In your first reaction, $\ce{HIO2}$ and $\ce{HIO3}$ react to form $\ce{2IO2.}$ and $\ce{H2O}$.
In your second reaction, $\ce{IO2\cdot}$ is being reduced back to $\ce{HIO2}$ by $\ce{Mn^{2+}}$. Since the first reaction produced $\ce{2IO2.}$, this second reaction needs to happen twice for each time the first reaction occurs. The pair of reactions is really:
$$\ce{ IO3- + HIO2 + H+ -> 2IO2. +H2O }$$
$$2\left(\ce{IO2. +Mn^{2+} +H2O-> HIO2 + Mn(OH)^{2+}} \right)$$
When you combine these reactions into an overall reaction, the instances of $\ce{IO2.}$ cancel and you are left with a net gain of $\ce{HIO2}$.
$$\ce{IO3- + H+ +2Mn^{2+} + H2O-> HIO2 + 2Mn(OH)^{2+}}$$
Of course, the net reaction does not hint at the autocatalytic behavior. To emphasize that the reaction is autocatalytic, you could write the overall reaction in a not quite simplest form:
$$\ce{IO3- + HIO2 +  H+ +2Mn^{2+} + H2O-> 2HIO2 + 2Mn(OH)^{2+}}$$


text:
What is the standard heat of formation of the nitrate ion, based on the standard enthalpy of formation of silver nitrate?

Given the standard enthalpy of formation of $\ce{AgNO3(aq)} = \pu{-99 kJ/mol}$ and the standard enthalpy of formation of $\ce{Ag+(aq)} = \pu{+105 kJ/mol}$, what's the standard heat of formation of $\ce{NO3-(aq)}$?

How can I know this if the standard enthalpy of reaction is not known? This is self test 10.1 in Atkins, PC.

Silver nitrate, $\ce{AgNO3}$ is a strong electrolyte. Based on the information given, I expect you are meant to assume that it dissociates completely when dissolved in water:
$$\ce{AgNO3(aq) -> Ag+(aq) + NO3- (aq)}$$
Thus, we can make the assumption that $$\ce{AgNO3(aq)} \equiv \ce{Ag+(aq) + NO3- (aq)}$$
Then, we can state:
$$\Delta_\mathrm{f} H (\ce{AgNO3(aq)}) = \Delta_\mathrm{f} H (\ce{Ag+(aq)})+\Delta_\mathrm{f} H (\ce{NO3- (aq)})$$
$$\implies \pu{-99kJ/mol} = \pu{+105kJ/mol} +\Delta_\mathrm{f} H (\ce{NO3- (aq)})$$
Thus $$\Delta_\mathrm{f} H (\ce{NO3- (aq)})=\pu{-204kJ/mol}$$

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text:
Approximation of electron density of a small molecule
Starting point for Quantum Theory of Atoms in Molecules (AIM) is the electron density $\rho(r)$. It could be determined experimentally (X-ray diffractions). I want to know how can we estimate them ab initio. In AIM
it is often emphasized that the focus is on density rather than a wave function for the later is just a mathematical function while the former is experimentally observable. Are there ways to estimate electron density without first calculating wave function (which will obviously lead to density function).
My interest is to use AIM in quantitative structure-activity relationship. I will typically deal with molecules of around 50 atoms. Usually I will already have the graph (connectivity) of the molecule (eg .SDF file, smile notation etc), how can we incorporate this to speed up electron density calculation?
It would be helpful if brief comment on the computational cost is also included in the answer.

What are you trying to get out of AIM to to use in QSAR? (I ask because the level of accuracy/realism you require will affect what methods are usable.)

I should be able to discriminate molecules with low MIC (minimum inhibitory concentration) from that of molecules having high MIC values.

That's not a thing you can get out of AIM. (As far as I know.)

You are looking for density functional theory. The theory is somewhat involved and you may be interested in A Chemist's Guide to Density Functional Theory or similar text book. A full description of the variety of ways to compute the (ground state) electron density is beyond the scope of an answer and better suited to a textbook.
The general gist is to optimize the energy with respect to the electron density numerically.

can you please give a brief outline (step wise) how to calculate the density. I referred to the book but I still find that they assume density is already given and do further analysis on that.

@DurgaDatta: While DFT does produce a density, it relies on something very much like a wavefunction calculation, and thus is probably not what you are looking for.

@DurgaDatta Additionally, the step by step procedure is going to depend on the software package you choose to use. Ultimately, they will look like "include yadda in the instruction file on line 2 if you want to incorporate diffuse wavefunctions," or "click on Setup Calculation in the Calculation menu".

@Aesin This is not necessarily true. The most common implementations of DFT that are generally used by chemists are indeed based on orbital descriptions, SCFs etc, but this is to reuse existing codebase and easier interpretations for chemists. Using e.g plane waves, pseudopotentials or first order approximations for the KS equation etc one can create different implementations with less chemistry taste.

@Greg: In most DFT techniques I've seen, even when you're using plane waves and pseudopotentials, you're still creating an SCF. It is my understanding that while Hohenberg-Kohn theory doesn't require it, almost all practically used implementations use Kohn-Sham techniques, which do. Your basis set, whether hydrogenic orbital-like, plane-wave, or wavelet-based, does not affect the meaning of the thing you calculate (though will obv. affect the accuracy), though you may have to do more or fewer transformations to produce something you recognise as a chemistry MO.

@Aesin: I agree with what you say. I only pointed out that the likeness is more of a convenience than theoretical necessity. We use orbital or plane wave based wavefunctions to produce "realistic" densities, and easily calculate properties, easy to incorporate the exclusion principle, but one could construct e.g. a numerical density function, and built DFT on it.

Look at SIESTA or DMOL for examples of numerical basis sets. Independently of basis sets, molecular orbitals are always single-electron solutions.

As far as obtaining the density without first computing the wave function--there's almost no way to do it with any accuracy for anything "realistic."  The easiest way to obtain the electron density is to perform an ab initio electronic structure theory calculation.
For example, you can use the GAMESS quantum chemistry program to perform a Hartree-Fock or DFT calculation, and then have it write the resulting wave function to a AIMPAC input deck (WFN file).
The AIMPAC program is the original version of QTAIM, written by the late Richard Bader and his research group. There are more modern versions, such as Todd Keith's AIMALL which is certainly more powerful and user-friendly than the original AIMPAC.
You can probably perform some calculation on fifty or so atoms, assuming a typical organic molecule. It all depends on how fast your computer(s) is and how many basis functions you use.  Constructing molecular graphs with AIM is quite fast.  Where things get expensive is when you start calculating properties of atomic basins, as these are high-accuracy integrations over "irregular" (but well-defined) regions.
Sorry to ruin your day, but the connectivities in your SDF file are probably of little value.  There are no such things as "single"/"double"/"triple" bonds, at least in the context of QTAIM.  Either they are bonded or they are not. Bonding appears when the interaction between two atoms lowers the total energy of the system. QTAIM will give you that information unambiguously.
Sometimes the answers are surprising, but it is guaranteed that they are more "correct" than what we unfortunately call "chemical intuition" and/or simple proximity of two atoms.

Your claims about bonds not being useful is not quite correct. Valence-bond methods can use that information and you can also put that information into your initial guess as in Hueckel approaches.  You could always use NBO to enhance QTAIM.

@Deathbreath 1) No modern electronic structure theory codes will need that bond information to do its job. 2) NBO on top of QTAIM is like painting a mustache on the Mona Lisa. QTAIM doesn't need "enhancement" by arbitrary unitary transformations of wave functions (which themselves have arbitrarily been constrained to be real rather than complex). Also, I'm not sure if NBO even maintains constant density.


text:
I have to write a Python program relating with molecular weight. Any insight?

Write a program that determines the molecular weight of a chemical compound based on the number of hydrogen, carbon, and oxygen atoms in the compound. You should use the following atomic weights:
  \begin{array}{|c|c|c|c|}
\hline
\text{Atom} & \text{Weight}~\pu{gm/mole} \\\hline
\ce{H} & 1.0079 \\\hline
\ce{C} & 12.011\\\hline
\ce{O} & 15.9994\\\hline
\end{array}

I don’t understand the relation between molecular weight and the number of $\ce{H}$, $\ce{C}$, and $\ce{O}$.
The professor has said "there is no need to be able to handle chemical compound names like $\ce{H2O}$ or $\ce{C2H5OH}$. (I plan to discuss how to handle these later in class) Instead, the user should just be asked to specify the number of hydrogen, carbon, and oxygen atoms in the compound.”
I need help to proceed.

It's pretty simple actually. The molecular weight of a compound is the sum of the atomic weights of all its atoms. The same numbers apply to a single atom as for a mole of that atom, but at the single-molecule level the unit of measure is the atomic mass unit or amu (aka the dalton or Da). The number of grams per mole of an element and the number of daltons that a single atom of the element weighs are intrinsically linked due to the definition of what a mole of something is (the amount of a pure chemical substance that weighs the same amount in grams as the mass of one atom or molecule of the substance in daltons).
So, methane (CH4), for instance, has a molecular weight of 1.0079 x 4 + 12.011 = 16.0426 Da. A mole of this gas would weigh 16.0426 grams (which is - interesting science fact - lighter by almost half than that of N2 or O2 gases making up most of the atmosphere, so a balloon of it will float). So, your Python script would simply prompt for the number of hydrogens (accepting integer values only), carbons (ditto) and oxygens (ditto), and plug them into the expression:
total_mass = num_hyd * 1.0079 + num_carb * 12.011 + num_oxy * 15.9994

... before returning total_mass as the answer.
Theoretically, the energy contained as enthalpy in the bonds between atoms contributes to a molecule's mass by general relativity and the first law of thermodynamics. In practice, the amount of energy inherent in even the least stable bonds (which produce the most net energy when they decompose to form more stable ones) is a rounding error when you run it through E=mc2, and so chemists don't bother. Nuclear physicists, who actually change the atoms themselves, do care, because the equation works in reverse for them; they get lots of energy from very small changes in the atomic masses of the reactants vs the products.


text:
How do you determine which entity gains or loses an H+ in an equation?
$$\ce{HCO3- (aq) + H2O(l) <=> H2CO3(aq) +OH- (aq)}$$
I do not understand why this makes an Arrhenius base and why it doesn't form:
$$\ce{HCO3- (aq) + H2O(l) <=> CO3^2- (aq) +H3O+ (aq)}$$
Could someone explain this to me?

Both of these equations are valid, as $\ce{HCO3-}$ is amphoteric (i.e. both an acid and a base).
To answer your question, one has to consider that the $\mathrm{p}K_\text{a}$ value for $\ce{H2CO3}$ is 6.37 and that for $\ce{HCO3-}$ is 10.32. So if we started from "pure" $\ce{HCO3-}$ (let's assume sodium bicarbonate) and add it to water, what would happen?
The $\mathrm{p}K_\text{b}$ for $\ce{HCO3-}$ is of $14 - 6.37 = 7.63$, it is larger than the $\mathrm{p}K_\text{a}$ for the acidic function (10.32). Therefore although both a weak base and a weak acid, it is a stronger base than acid, increasing the pH when added in water.

To sum it up, we know that $\ce{H2CO3}$ is an acid, so for $\ce{HCO3-}$, just look at the $\ce{pK_{a}}$ and the $\ce{pK_{b}}$ values. Pick the greatest one and do it from there.
$\ce{pK_{a}}$ = ${7.6}$
$\ce{pK_{b}}$ = ${6.4}$
So now we know that the entity gains an $\ce{H+}$.
Thanks


text:
synthesizable functional groups
I was told in this question: https://chemistry.stackexchange.com/questions/4789/synthesizability-and-energy-minimization-of-a-structure/4791#4791 to
[Count the] "number of accessible functional groups (alkenes, amines, alcohols, esters, etc) which make it easier to prepare, 5) number of inaccessible functional groups (aromatic rings, polycyclic systems, quaternary carbons, complex or long alkyl groups),"
But when I googled accessible functional groups and inaccessible functional groups no pages had a longer list. I found a SMARTS file which lists some common functional groups. Could someone help me determine which are readily synthesizable?
Thank you

What you are asking amounts to the collected knowledge, wisdom, and intuition of the organic chemistry community. Your best bet will be to find an organic chemist, preferably one with a graduate degree, to work with you on this project. Said organic chemist can apply his/her knowledge/wisdom/intuition to each functional group in less time than it would take you to look up the answer. Doing so frees you to work on the algorithm you describe in your other question.

The beginning of the answer to this question takes me two semesters to get through with my introductory organic chemistry students, and that is just for the common functional groups. After covering the basics of organic structure and reactivity, most texts have 15-20 chapters organized by functional group containing the reactions of and syntheses of said functional groups.
The common functional groups covered in most introductory sequences:

alkanes
cycloalkanes
alkyl halides
alkenes
conjugated dienes and other unsaturated systems
alkynes
aromatic ring systems
alcohols
glycols and other diols
phenols
ethers
epoxides
aldehydes
ketones
carboxylic acids
esters
anhydrides
acyl halides
amides
amines
nitriles
thiols
sulfides
acetals
imines
enamines
heterocycles
nitro compounds
diazo compounds
carbohydrates
amino acids
nucleic acids

If you want to know about every functional group, you will need an advanced, probably graduate level, synthesis course. However, even I have not heard of some of the designations on your list, like semithiocarbazide.

Some good ideas on where to start, if you choose not to consult with an organic chemist.

Check an undergraduate level organic chemistry book out of your local library and get some sense of the subject. In your case, it does not look like you need to know how the reactions work, just that they exist, so you can use the text as a reference source.
Sign up for a free MOOC in organic chemistry. Again, you just want to know that the reactions exist, so you can pick and choose the content from the course.
Acquire a copy of one of the seminal books in organic chemistry, March's Advanced Organic Chemistry. This book is not for the fainthearted but it gives good treatment to the functional groups, including the less common ones.
Find (do not buy - these books are not worth the price tag) a copy of either Comprehensive Organic Transformations: A Guide to Functional Group Preparation by Richard Larock (which is now 14 years old) or the set of Comprehensive Organic Functional Group Transformations, which appears to be many volumes (now 9-10 years old).


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Are there any free electrons in metal vapor?
Metals contain free moving electrons, in both the solid and liquid state. Does this happen when it is in vapor form?

Interesting question. There are free electrons and ions in a metal's gas phase, but even at rather high temperatures, they are relatively rare. The formation of free electrons is given by the reaction:
$$\ce {M_{(g)} → M^+_{(g)} + e^{-}_{(g)}}$$
This is an endothermic reaction for all atoms and molecules, and the energy involved receives a special name: the (first) ionization energy. Comparing a chart of ionization energies for the elements, one finds that almost all elements require over $ 500\ kJ/mol$ (or about $5\ eV/atom$) to ionize. Even at $5750\ K$, the temperature of the surface of the Sun, the average thermal energy of a particle is about only about $0.5\ eV$.
Also, a large part of the electrons that did manage to free themselves from their parent neutral atoms would actually be consumed by other neutral atoms, in a different reaction:
$$\ce {M_{(g)}  + e^{-}_{(g)} → M^{-}_{(g)}}$$
This reaction is slightly exothermic for most elements and molecules, and the energy associated with it also receives a special name: the (first) electron affinity.
Calculating the equilibrium concentrations of each species at a given temperature is possible but is a bit of work. However, for most metals, you can expect to find significantly less than one free electron per ten thousand neutral atoms at $5750\ K$, and the relative amount of free electrons decreases exponentially with temperature.

You are confusing de-localisation with ionisation.               Its significant- its present in traditional fluorescent tubes where mercury is vaporised and shortcircuits  to give arcs which produce UV- in turn producing fluorescence.

@user2617804 I'm considering a thermal plasma in equilibrium, with no applied voltage. Also I'm not sure what you mean about delocalisation.


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Electrolysis: Water to Hydrogen gas?
I recently conducted an experiment where I electrolyzed a solution of sodium chloride. I was using a copper coin as the anode, and I observed a pale blue precipitate in the cup when I was done with the experiment. I then deduced the following:
The chloride ions get oxidized at the anode:
$$\ce{2Cl- -> Cl2 + 2e-}$$
The chlorine gas then reacts with the copper anode to form copper(II) chloride which goes into solution as it is completely soluble.
At the cathode, the hydrogen in the water gets reduced to form hydrogen gas:
$$\ce{2H2O + 2e- -> 2H2 + 2OH-}$$
I know the above half reactions exist from the list of standard reduction potentials.
Next, the copper(II) ions react with the hydroxide ions being formed to produce copper(II) hydroxide, which is both insoluble and pale blue.
However, this source mentions that the hydronium ions in the water get reduced at the cathode:
$$\ce{2H+ + 2e- -> H2}$$
Is this correct? In my understanding, the concentration of hydronium ions ($\pu{1e-7 M}$) would be much too low, and where would the hydroxide ions come from to form copper(II) hydroxide?

I agree that the precipitate is copper(II) hydroxide, but I don't think the oxidation goes through chlorine. I think the copper gets oxidized directly.

In answer to your last statement about the formation of Hydrogen. The actual amount of hydronium ions doesn't matter because it is at equilibrium ($\ce{H2O <=> OH- + H+}$) and so once you use an $\ce{H+}$ you'll simply have more $\ce{H2O}$ dissociate into $\ce{OH-}$ and $\ce{H+}$ (Le Chatlier states) so if you have battery current running through the system you should experience a healthy, visible stream of hydrogen bubbles. I built a similar thing myself back in high school.
The hydroxide ions will come from the water also. As mentioned above, the $\ce{H2O}$ will dissociate into $\ce{H+}$ and $\ce{OH-}$ and you are using both. The $\ce{H2O}$ will keep dissociating and try to keep the $\ce{H+}$ and $\ce{OH-}$ concentrations at $10^{-7} \, \small\text{M}$  as you keep using them.
Feel free to question my knowledge.

Great, thank you, that makes sense. Seeing as the equation for 2H20 + 2e- -> H2 + OH- is on the standard reduction potentials sheet, does this half reaction ever happen in electrolysis?

Yes, absolutely. If it is occurring you should see a bunch of gas bubbles coming off at positive terminal

On a side note you could potentially determine which of the two reactions are occuring at the cathode using the nernst equation along with standard reduction potentials... You'd need to factor in the overpotential but this depend on your setup. Still, might be a good investigation.


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Why do aqueous solutions always have to be electrically neutral?
I was reviewing some analytical chemistry and stumbled upon a section that explained the imperfection of using a salt bridge.
It said that the using dissimilar ions is a problem because in, for example, the case of KCl the K+ and Cl- have different mobility and so you get regions that are rich in K+ and others rich in Cl-. This means that some specific parts of solution (even though it's small) to be not electrically neutral i.e. electrically charged.
So I imagine in theory at least it must be possible to somehow extract these charge-rich areas and put them into a beaker. So what is all this I've been learning about solutions having to be electrically neutral?

Because the electrostatic force is much stronger than you can imagine. Yes, you can get charge-rich areas in a cell. However, the amount of charge held here will be negligible. A rather small amount of charge can overcome the field near the plates and prevent more charge from approaching. The amount is quite small; it's not something that you would notice (and it won't give you a static shock1
1. Everyday static shocks are stronger by orders of magnitude because they deal with a lot of charge accumulated on the surface of an insulator, while here it is a relatively small amount of charge dispersed through the volume of a conductor


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Bonding and Antibonding
So I understand, electrons exist in orbitals, mainly s, p, d, f and that when they bond with one another it will form sigma and pi bonds and that whether it bonds in a certain orientation a molecular bonding orbital or an anti bonding orbital will form.
What I DON'T understand is that when an element bonds it's energy levels will ALWAYS form a series of energy levels that contain bonding and anti bonding orbitals so when they come together, the electrons are always orientated such that these orbitals will form? I also don't really understand what happens when these two atoms meet (the elements that are bonding) how are the orbitals interacting with one another? And what are the orbitals involved?
For those who would like to answer the question feel free to loosely address these questions and refer to fundamental concepts or explain bonding through orbitals briefly.

Orbitals are defined by a set of wavefunctions. Hence, when two orbitals overlap they can either interfere constructively (bonding) or destructively(anti bonding).
Lets take two pz orbitals orbitals overlapping with each other. You must understand that the p orbitals look like a dumbbell. One side of the dumbbell is + and the other side is -.
If - meets - we have constructive overlap. If - meets + we have destructive overlap.
The former gives a bonding orbital ($\sigma$) and the latter an antibonding orbital ($\sigma*$).
Depending on the molecule it is often energetically favourable to be in a bonding state hence bonding orbitals are lower in energy to their anti-bonding counterparts.

A proper answer to your question would require at least a semester of Physical Chemistry. I would like to touch on a couple of issues (very loosely speaking):

We know from quantum mechanics that electrons have certain properties, which are couched in something called a wave function.  Mathematically, wave functions may be complex (i.e. have real and imaginary components)  It is impossible to "see" complex quantities. So there must be something else.
In fact, the only thing that we can "see" is the electron density, and the electron density of an atom is spherical.  The electron density is the square (actually complex modulus) of the wave function, and so a possible set of (perhaps imaginary) wave functions are the "Spherical Harmonics" (s, p, d, ... fuctions)

The only system that has an analytic solution is the Schroedinger equation for a hydrogen-like atom.
So what happens if we "cram" two or more atoms together to make a molecule?  Is there an analytic solution for molecules that gives sigma, pi, etc.?  Nope.  But we can try to make an approximation, called the linear combination of Atomic Orbitals (s,p,d...) to give Molecular Orbitals (LCAO-MO)
The sigma and pi bonds (and anti bonds) are a "basis" or a representation of molecules that we construct from things we know everything about, the atomic orbitals of a hydrogen-like atom.
As a coarse analogy, when we watch TV, the pixels on the screen are a "basis set" which, individually, can be turned red green blue.  Their combination gives rise to the image of whatever you think you see on the screen (State of the Union address, football game, etc.)
Simple fact is: two atoms in separation look like spheres.  If they are pushed together, then their electron density forms a dumbbell shape.  You can not look at the electron density and pick out sigma and pi features.  There is no such thing as "hybridization" to form sp, sp2, sp3...   These are convenient "lies" (linear combinations of primitive atomic orbitals) which sometimes get in the way of something so very simple:   Electrons of Atom A are attracted to Protons in Electron B and vice versa. (This is a slight bend of the truth, because electrons are shared between both.  But it is possible to identify atomic basins, please see Quantum Theory of Atoms in Molecules)
But to follow the conventional explanation: electrons "go" into bonding orbitals (e.g. sigma bonds) first because sigma bonds usually look more like s atomic orbitals.  s atomic orbitals are primarily centered upon the nucleus, and since electrons and protons have an attractive interaction, that is the lowest energy configuration.


text:
Time to effervescent an effervescent tablet
Say I have an effervescent tablet and I put in water.  I thought that if I break the tablet in two halves it would dissolve quicker than with the whole table (without breaking), but it actually takes the same time.
What would be the scientific reason of this?

Are you sure that the time is exactly the same?! Normally the dissolution rate should be a function of the external area of the tablet (i.e. powder with lots of external area dissolves faster than a tablet). If you break the tablet you create some extra area so it should dissolve faster, but maybe the change in area is not that big so you don't notice it. Could you give use dimensions of the tablet?!

michielm thanks for the comment. Why if I break the tablet I create some extra area?

Imagine a whole red apple. Looking at it, all you can see is the red of its skin. Now, cut it in half. Suddenly, you can see not only red, but the white of the insides. The volume of the apple is the same whether it is cut or not, but the surface area is not; there is just as much red skin before or after the cut, but the white insides are entirely new and only appear after the cut. Thus the exposed surface area has increased. How does this affect your experiment? If the surface area of the tablet increases, water can attack it from more places at once, so the tablet is eaten up faster.

Dissolution in general
Dissolution is normally a function of several aspects of the solid and liquid and one of them is the interfacial area between the solid and the liquid. A common relation for the dissolution rate which is not always exact, but shows all trends/dependencies very well is the following:
$$\frac{\mathrm dm}{\mathrm dt}=A \frac{D}{d}\left(C_\mathrm s-C_\mathrm b\right) $$
where $A$ is the interfacial area, $D$ is the diffusion coefficient of your solid molecules in the liquid, $d$ is the thickness of the solute boundary layer and $C_\mathrm s$ and $C_\mathrm b$ are the concentrations of the solid molecules at surface and in the bulk (this term is the driving force).
Dissolution of a whole or broken tablet
To make a simple calculation for the whole vs. the broken tablet let us do the math.  Assuming that we have a disc-like tablet the surface area is 2 times the frontal area $+$ the area of the sides.
$$A=2 A_\mathrm f+A_\mathrm s $$
For the whole tablet
$A_\mathrm f=\frac{\pi}{4}D_\mathrm t^2$ and $A_\mathrm s=\pi h D_\mathrm t$
For the broken tablet
$A_\mathrm f=\frac{\pi}{4}D_\mathrm t^2$ and $A_\mathrm s=\pi h D_\mathrm t  + 2 h D_\mathrm t$
Here I assume that the tablet is cleanly broken in the exact middle of the disc. So you can already see that the surface area is bigger for the broken tablet, but how much will depend on the specifics of the tablet. Rewriting the area we can see how the different aspects of the broken tablet will influence the area:
$$A=\pi D_\mathrm t \left[\frac{D_\mathrm t}{2}+h\left(1+\frac{2}{\pi}\right)\right] $$
From this equation it is clear that if $D_\mathrm t \gg h$ the height term doesn't do much and therefore it will not matter much that the tablet is broken.
For a final check: let's assume some typical numbers now. A typical effervescent tablet is about $D_\mathrm t=2\ \mathrm{cm}$ in diameter and $h=3\ \mathrm{mm}$ thick. This will result in the following surface areas for the broken and the whole tablet: $A_\text{whole}=8.2\ \mathrm{cm^2}$ and $A_\text{broken}=9.4\ \mathrm{cm^2}$.
This means that the difference in dissolution rate will only be about $20\ \%$ which is probably hard to notice due to other effects that disturb the comparison. One such effect is that $d$ depends on the amount of stirring which will be different already when you drop the tablets in the glass in a slightly different way (which is unavoidable, because you have 1 broken and 1 whole tablet).

Technically, it should dissolve faster, just like you thought! It will dissolve faster because there is a greater surface area : solution ratio (i.e. the broken edges, which used to be 'in the middle' of the tablet, are now exposed to the water). In general, the more surface area exposed, the faster it will dissolve.
So why aren't you observing a noticeable increase in speed? How big is your tablet? When you break it in half, are the 'broken edges' small (in surface area) compared to the rest of the tablet? Just for fun, try cutting the tablet horizontally through the middle (like you would cut a bread roll). This should significantly increase the surface area and the speed to dissolve.


text:
How to transform a tertiary alcohol to a carboxyl?
I am trying to transform a lipid with a tertiary alcohol into a carboxylic acid.
More specifically, I want to convert to diplopterol (see below) into something homologous to cholesteryl hemisuccinate (see below).
Any suggestions?
Diplopterol:

Cholesteryl hemisuccinate:

I think you need to specify a bit more information on the specific tertiary alcohol and carboxylic acid that you want because a general tertiary alcohol, $\ce{R1R2R3COH}$, can retain only 1 of the 3 sidegroups and I guess it matters which one you want to keep. Just as a startup: direct oxidation is not an option, because there is no hydrogen available on the central $\ce{C}$

The compound I'm trying to convert is diplopterol: http://www.chemicalbook.com/CAS%5CGIF%5C1721-59-1.gif . I want to end up with something homologous to cholesteryl hemmisuccinate: http://www.goldbandchem.com/1510-21-0.htm

Ok. I assume you want to keep the big molecule and get rid of the methyl groups then

yeah, more or less, see edits to my previous comment...

I am not sure that what you claim to want to do is what you actually want to do.
Cholesterol hemisuccinate is prepared from cholesterol, but not by converting the alcohol functional group into a carboxyl group. The alcohol oxygen becomes one of the oxygen atoms in the ester, but not the carboxyl oxygen. Cholesterol hemisuccinate can be prepared in the lab probably from cholesterol and succinic anhydride in the presence of a non-nucleophilic base (like pyridine or triethyl amine), which can act as a proton shuttle. An acidic workup will be needed to remove the base and protonate the carboxylic acid. The scheme below shows this reaction (sans acidic workup). The alcohol oxygen is highlighted in red.

An identical reaction could be done with your molecule, diplopterol. Dissolve both (1 eq) diplopterol and succinic anhydride (1.1 eq.) in dichloromethane or diethyl ether (or any low polarity aprotric solvent you like). Cool on an ice bath. With stirring, add pyridine (1.1 eq.) dropwise. Let the reaction warm to room temperature, probably about 2 hours. The product will likely precipitate as the pyridinium salt. Check the progress by TLC or some other means. The product will probably be more polar than the reactant. Work up by adding dilute HCl (probably 0.1 M). Extract the aqueous layer with your organic solvent of choice. Combine the organic layers. Wash with brine. Dry over magnesium or sodium sulfate. Remove the solvent by evaporation (rotary being preferred), and purify by chromatography if necessary.
Alternatively, if the product does precipitate at the pyridium salt, collect the salt by filtration, wash the precipitate with ether and then with cold ethanol (will wash away the reactants). Then, add the precipitate to the 0.1 M HCl and extract with ether if necessary. Further purification is probably not needed.

You can try reaction with sodium acetate, anhydride acetic at 120°C for 6h for acetilation product. Because of tertiary alcohol, more hard condicions it´s needed.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Is the activation barrier of a reaction related to the free energy or heat of formation of its reactants?
The energy barrier for activation appears in the derivation of a bimolecular reaction through collision or transition state theory, but it seems that from looking at diagram such as this one  it seems like the barrier to activation would be related to the energy required to break bonds?
Thanks in advance.

They are certainly related, but not necessarily the same. If all bond-breaking and forming would be sequential than the activation energy would equal the bond-breaking energy.
However, it is possible in some reaction mechanisms that the forming bonds 'donate' energy to the break-up and therefore the overall activation energy will be lower. An example is the reaction of hydrogeniodide to hydrogen and iodine show in the figure below. Instead of a full dissocation, two molecules of $\ce{HI}$ line up and the breakup occurs simultaneously with the formation of the new bonds thus allowing a lower energy level.

The only thing you can safely say is that the sum of breaking energies is an upper bound for the activation energy, because single atom level is 'as bad as it gets' (excluding nuclear reactions).


text:
Why does the extent of racemisation in an SN1 reaction depend on the solvent used?
Why is it that the extent of racemisation is greater in an $S_N 1$ reaction caried out using a polar protic solvent but the racemisation is not as extensive if an aprotic solvent is used although the carbocation intermediate is common in both?
eg:- 2-Bromobutane is solvolysed with $\ce{H2O}$ and $\ce{CH3OH}$ but the racemisation is 98% in water and 73% in methanol

What makes you think that 2-Bromobutane will only react in a $S_N1$ mechanism?

Secondary haloalkanes can undergo substitution by both the $S_N 1$ and $S_N 2$ routes. In general these two mechanisms compete with each other. The following diagram is a More-O'Ferrall-Jencks Plot showing the possible mechanism space for a nucleophilic substitution reaction. The reactants are in the upper left. The products are in the lower right. The horizontal axis represents the extent to which the C-LG bond has broken. The vertical axis represents the degree to which the C-Nu bond has formed. The $S_N 1$ pathway is represented by traveling along the perimeter of the plot first to the right and then down. The $S_N 2$ pathway is actually the red diagonal line represented both processes happening simultaneously. There is a third pathway (down and then right), representing nucleophilic attack and then loss of leaving groups. $S_N$ reactions at $sp^2$ centers often follow this route.

Very few reactions stick completely to one of these three extreme paths. For $sp^3$ centers, the mechanism sticks to the upper right of the diagonal depending on a variety of factors like the polarity of the solvent, the nucleophilicity of the nucleophile, the strength of the C-LG bond, steric hindrance at the subsitution center, and how stable the carbocation would be if it formed. The arrows on the plot above suggest the directions that the transition state would move if the nucleophilic attack step became more favored (stronger nucleophile, less polar solvent, less steric hindrance). The end position is not possible for this particular reaction, but such is the result of reusing and image from somewhere else.
In your case, you can tell how close the actual reaction pathway sticks to one of the idealized pathways. Or, you can think of these numbers as the relative likelihood that a substrate will undergo solvolysis by $S_N 1$ or $S_N 2$ mechanisms in that solvent.
In your case, methanol $\ce{CH3OH}$ is less polar than water $\ce{H2O}$, so the $S_N 1$ pathway is less favorable in methanol than in water. However, the $S_N 1$ pathway in methanol is still way more favorable than the $S_N 2$ pathway in methanol (by 73:27).


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Why does molten plastic (PLA) stick to metals (AL)?
I know this is broad so I will narrow:
Why does molten PLA stick to Aluminum?
I ask because I'm looking for a way to prevent/reduce the friction of the plastic inside the aluminum barrel of a 3D Printer. I believe that I can treat the surface of the aluminum with a High Temp enamel however I need to be sure of the physics of why the plastic sticks in the first place. Is it the surface finish, or is the molecular attractiveness? If it is the finish, then I can make it smooth. If it is static attraction then I can charge the barrel etc.

Aluminum reacts with oxygen in the air to produce aluminum oxide, which forms a fine coating on any exposed aluminum surface protecting against further oxidation.
$$\ce{ 4Al + 3O2 -> 2Al2O3}$$
Aluminum oxide is polar, amphoteric, and hygroscopic: it absorbs moisture from the atmosphere. The surface of aluminum oxide is likely coated with $\ce{Al-OH}$ groups. The polar aluminum oxide and aluminum hydroxide surface is attracted to the polar ester functional groups in PLA.
Aluminum surfaces can be rendered nonpolar by silanization. This article may be behind a paywall for you. I apologize. I could not find a good open access reference. The Wikipedia article on silanization is pretty terrible. The surface is rendered nonpolar, which may decrease the interaction with PLA.

So its the oxide layer that is attracting the plastic. Therefore if the oxide layer is removed with vinegar and a polyurethane layer is added immediately afterward, the plastic will not stick as well as it did before?

I don't know if polyurethane sticks to bare aluminum. It might stick to aluminum oxide better. Most polyurethanes are also polar, which means PLA might stick to polyurethane as well.

I did a material test, so far the PLA didn't stick to the polyurethane coated aluminum.

Another common cause why your plastic gets stuck in your hot-end is simple thermal expansion. Try to up the cooling on your hot end to reduce the length of the temp gradient in your metal-barrel hot-end.


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Equilibrium constant
I know that equilibrium constant depends on temperature and pressure. But according to Le Chatelier's principle when we change pressure of reagents the equilibrium is the same. Why?

Good question here - the equilibrium constant is all a little bit abstract, and therefore tough for some of us chemists, at least initially :) I have always thought about the concept of 'equilibrium' in terms of a balloon.
Imagine a balloon, and you've blown the balloon up until it is as full as can be. At this stage the balloon is in one of two 'equilibrium' states:

balloonDeflated $\rightleftharpoons$  balloonInflated

Pressure:
Initially, while blowing up the balloon, you and the balloon are involved in a delicate dance. You want the balloon to be the biggest balloon in the world! But, the balloon doesn't want to be too fat. You and the balloon dance a delicate equilibrium.
So, while you try to blow the balloon up as big as possible, the balloon says: 'No!', until it finally explodes in your face :( Too much pressure from one side. Blown out the equilibrium, and lost the battle.
Temperature:
You've huffed and puffed just as much as you can - (or as much as you can be bothered) - and the balloon is just the right size, determined by the effort you can be bothered putting into it. The balloon size is now in equilibrium with your effort. This is the same with temperature, you can push a reaction faster and further with higher temperatures (more puff).
Equilibrium State
Of course, once you and the balloon have agreed on the optimal size, and you have tied a knot in the balloon, you have reached a happy equilibrium. Yay! Now the fun can begin.
What will happen if you squeeze on one side of the balloon?

roundShape $\rightarrow$ distortedShape

In this analogy, you have added more reagent to the left hand side of the equilibrium. But, by definition, this is no longer an equilibrium, there is more on the left than the right. So, the balloon bulges out to the right, to even up the pressure (i.e. you get more right hand reactants produced to counter the influx of left hand reagent).
I think, the concept of the 'pressure' of reagents is a reference to the concentration of reagents on either side of the equilibrium. A greater concentration on the left, causes a flow to the right and vice versa. Note that this is distinct from the concept of pressure in the normal terms (pounds per square or force).
And this whole process is fluid, a change to one side, affects the other side. Temperature / environmental pressure will affect the initial equilibrium state (your effort / lung power and the potential size of the balloon will affect the initial equilibrium state) and after this, the pressures imposed on either sides of the equilibrium will be countered by a shift to the low-pressure side.

The position of equilibrium depends on temperature and pressure, as Nick Burns describes in his answer. The equilibrium constant depends on temperature only. The equilibrium constant is not a measure of the position of equilibrium, but rather which positions (representing different compositions) along the the reaction coordinate surface are equilibrium positions. Changing the temperature changes the shape of the reaction coordinate surface. Changing the pressure just moves you to a different point on the surface.
Example: consider a hypothetical reaction in which two molecules of gas react to form a third.
$$\ce{A(g) + B(g)<=>C(g)}$$
The reaction quotient $Q$ is simply the ratio of the product of the partial pressure(s) of the product(s) to the product of the partial pressure(s) of the reactant(s). $Q$ can have any value representing different points on the reaction coordinate surface.
$$Q=\frac{P_C}{P_A P_B}$$
The equilibrium constant represents special values of Q in which the reaction coordinate surface has a free energy minimum. For the sake of our example, let's make the equilibrium constant an easy value to work with, say $K=2$.
$$K=Q_{eq}=\frac{P_{C,eq}}{P_{A,eq} P_{B,eq}}=2$$
Since we are talking about gases, we also have the constraint the sum of the partial pressures of the reagents must equal the total pressure of the system. Again, for ease, let us set this value to an easy value to work with, say $P_T = 7 \text{ kPa}$.
$$\frac{P_{C}}{P_{A} P_{B}}=2$$
$$P_A + P_B + P_C = 7$$
Now we have two equations of three variables, which means there are an infinite number of possible values of the partial pressures of the reactions components that fit these two conditions. One is $P_A = 1$, $P_B =2$, and $P_C =4$. Another equally valid combo is $P_A = 2$, $P_B=1$, and $P_C=4$. The position of the equilibrium in this case is better given by a ratio of the sum of the partial pressure(s) of the product(s) to the sum of the partial pressure(s) of the reactant(s):
$$\frac{P_C}{P_A + P_B}=\frac{4}{3}$$
If we change the total pressure to, say $P_T = 4$, we change the allowable combinations of $P_A$, $P_B$, and $P_C$ that can get us to the equilibrium condition of $Q=2$. One allowable combination (of an infinite number) is $P_A=1$, $P_B=1$, and $P_C=2$. The position of the equilibrium is now:
$$\frac{P_C}{P_A + P_B}=\frac{2}{2}=1$$
Le Chatelier's principle suggests that When pressure decreases, the equilibrium shifts toward the side of the reaction with more equivalents of gas.
If we consider changing the situation from $P_A = 1$, $P_B =2$, $P_C =4$, and $P_T=7$ to a situation with $P_T=2$, we can do some math to figure out new values of $P_A$, $P_B$, and $P_C$ that fit this new situation. First, we set the partial pressures of the compoents to $\frac{2}{7}$ of their original value, and calculate Q.
$$P_A =1\times \frac{2}{7}=\frac{2}{7}$$
$$P_B =2\times \frac{2}{7}=\frac{4}{7}$$
$$P_A =4\times \frac{2}{7}=\frac{8}{7}$$
$$Q=\frac{P_C}{P_A P_B}=\frac{\frac{8}{7}}{\frac{2}{7} \times \frac{4}{7}}=\frac{\frac{8}{7}}{\frac{8}{49}}=49$$
Since $C>K$ the equilibrium will shift to decrease $Q$. This can happen by decreasing $P_C$ and increasing $P_A$ and $P_B$. Since there is an reaction equation relating these three components in a 1:1:1 stoichiometric ratio, the change must have the same value vut in different directions:
$$K=2= \frac{\frac{8}{7}-x}{\left(\frac{2}{7}+x \right)\left( \frac{4}{7}+x\right)}$$
Rather than goof up the algebra, I let WolframAlpha do it. The solution for $x$ is:
$$x=\frac{1}{128}\left(-19 \pm \sqrt{681} \right)\approx \{+0.2534..., -1.6106...\}$$
The first answer for $x$ is nonsense, since it would put one partial pressure less than $0$. If we change the partial pressures appropriately, we get new values of
$$P_A = \frac{2}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.5391...$$
$$P_B = \frac{4}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8248...$$
$$P_C = \frac{8}{7}-\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8894...$$
The position of equilibrium is
$$\frac{0.8894...}{0.5391...+0.8248...}\approx 0.65$$
which is less than the value of $1.33$ from when $P_T=7$ and different from the value of $1$, when the partial pressures were different (but still met the criteria than $P_T=2$ and $Q=2$).
Changing the temperature changes the values of $Q$ which define the equilibrium state:
$$K=e^{-\Delta G^\circ/RT}$$


text:
Figuring out the volume of air trapped inside a bottle of water
I am doing a chemistry lab report on figuring out the volume of air trapped inside a bottle.  Here is the procedure:

Fill a larger container or tub about half full with tap water
Fill a 2 L plastic soft drink bottle with tap water.  Put the cap on loosely
Carefully turn the bottle upside down without spilling any water
Put the bottle into the large container of water so that the mouth of the bottle is underwater.  Remove the cap underwater
Feed the flexible tubing under the water so that one end goes inside the bottle
Put your straw into the other end of the tubing.  Do not share straws
When it is your turn, exhale into the straw to collect the air of one normal breath
With a marking pen, mark the volume of air on the soft drink bottle
Figure out the volume of the air trapped inside the bottle.  Record this volume

How might I do number 9? My proposed procedure is:

With a marking pen, mark the new volume of air on the soft drink
bottle
Record this volume, which will be smaller than the volume found in
step 8 as some of the water will have been displaced by the breath
Subtract the new volume of air from the old volume of air to find
the volume of the air trapped inside the bottle
Record this volume

Thanks!

Displacement! Let the container that you have bottle in be filled to the brim inside a bigger container. Then when you breathe into the bottle, the amount of water that is displaced will be the same as the volume of the air.

So could I mark the volume of air with the breath and compare that to the volume of air I marked in step 8?

Yes, you can do that.

So I could simply subtract the smaller value from the larger one?

Yea pretty much. Science works.


text:
How to calculate the formal charge of sulfate ion?
I found a site with this formula if you will,

Formal Charge = [Number of valence electrons on atom] – [non-bonded electrons + number of bonds]

I don't think I am using it correctly in finding the formal charges of each atom in $\ce{SO_4^{2-}}$. $\ce{S}=2$, $\ce{O}=-1$ and the other $\ce{O}=0$ and if you add them together I don't get the overall charge of -2.
 How do you work out formal charges?

Your formula is correct. If $V$ is the number of valence electrons originally assigned to an atom, $N$ is the number of nonbonding electrons, and $B$ is the number of bonds ($\frac{1}{2}$ of the number of bonding electrons to be more precise), then the formal charge $FC$ is:
$$FC = V - (B+N)$$
Sulfate ion has two valid structures that you could draw, one with sulfur having a formal charge of zero and one with sulfur having a formal charge of +2. The structures below (from the Wikipedia page, released into the public domain) follow the annoying but allowable convention of replacing lone pairs with bars.

If the structure on the left (#1) is correct, then the sulfur atom ($V=6$) has six bonds ($B=6$) and no lone pairs ($N=0$). Two oxygen atoms (V=6) have two bonds ($B=2$) and two lone pairs ($N=4$), while the other two oxygen atoms have one bond ($B=1$) and three lone pairs ($N=6$). The sulfur atom and two oxygen atoms have $FC=0$ and the remaining two oxygen atoms have $FC=-1$ for an overall charge on the ion of $-2$.
$$FC_{\ce{S}}=6-(6+0)=0$$
$$FC_{\ce{O_{1,2}}}=6-(2+4)=0$$
$$FC_{\ce{O_{3,4}}}=6-(1+6)=-1$$
If the structure on the right (#2) is correct, then the sulfur atom ($V=6$) has four bonds ($B=4$) and no lone pairs ($N=0$). All four oxygen atoms (V=6) have one bond ($B=1$) and three lone pairs ($N=6$). The sulfur atom has $FC=+2$ and the oxygen atoms have $FC=-1$ for an overall charge on the ion of $-2$.
$$FC_{\ce{S}}=6-(4+0)=+2$$
$$FC_{\ce{O}}=6-(1+6)=-1$$
But, wait! A more fundamental question might be "How can the sulfate ion have two very different structures that puts different formal charges on the sulfur and oxygen atoms?"
Formal charge is a nice bookkeeping mechanism, but has no experimentally valid relationship to the actual charge on any atom in most molecules or ions. Formal charge only equals actual charge on monoatomic species. Formal charge falls into the category of models that we use in chemistry that are 1) helpful, 2) produce the correct answer when used correctly, and 3) completely bogus.  Other models in this category include oxidation number, VSEPR, resonance, and electronegativity. The experimental evidence suggests the real structure of sulfate combines features of both structure #1 and structure #2, but would be challenging to draw using the formalisms we have adopted:

All four $\ce{S-O}$ bonds are equal in length (#2).
The length of the $\ce{S-O}$ bonds is shorter than a normal $\ce{S-O}$ single bond and longer than a normal $\ce{S-O}$ double (#1).
The sulfur atom has a partial positive charge (#2) (note that partial charges, unlike formal charges, have some experimental basis).
The four oxygen atoms have equivalent partial negative charges (#2).
The partial negative charges on the oxygen atoms add to more than $-2$ but not anywhere near $-4$ (#1).

does formal charge of $x$ on an atom really indicates that the charge on it is $x$ in the molecule

Ben Norris' answer is excellent. I'll collaborate with a more visual procedure, so it may provide a different point of view.
The way I've learned it, oxidation numbers can be determined by figuring out where the bonding electrons would be assuming bonds are 100% ionic, while formal charges can be determined by figuring out where the electrons would be if the bond were 100% covalent.
Instead of applying an equation directly, draw the Lewis structure of the compound. Now cut all covalent bonds homolytically, i.e. evenly distributing the two electrons between the bonding atoms. Now count the number of electrons attached to each atom, and subtract the number of valence electrons the free atom has. The number you obtain will be an "electron excess", which is the opposite of the formal charge, so just multiply by -1. Here's your example worked out.


text:
Reduction by thermionic electrons
Certain reduction reactions require electrons as reducing agents. Is the overlapping of orbitals is necessary for electron transfer? I mean if a element $\ce{X}$ is heated and it gives out thermionic electrons then can an oxidizing agent $\ce{Y}$ acquire it and get reduced? If so then will it acquire fast moving electrons or slow moving electrons? What will happen if the inert gas is replaced by $\ce{CO2}$ in an cathode ray tube?


text:
How to find partial pressures of gases only given their percentages?
How do I find partial pressures if I am only given their percentages? Where do I begin?
Example: Assume $\ce{O}=21\%$ , $\ce{N}=78\%$ , $\ce{Ar}=1\%$ by volume. How would I calculate the partial pressures if total pressure is $\pu{1 atm}$.

Pressures of gasses are additive. For example, 1 atm X plus 1 atm Y makes 2 atm total pressure. Because all the components of a mixed gas are at the same temperature and pressure, the fraction of each component by volume is also the same fraction of the total pressure.

Is this generally true or only for ideal gasses?!

Yes, but it works quite well in any "normal" range of densities.


text:
Empirical formula when given percentages?
A binary compound of lead and oxygen contains $90.66 \% \ce{Pb}$. What is the empirical formula for the compound? How do I use the percentage to get to the normal formula, then how do I change the normal formula to empirical?

In these kinds of questions (homework), we like to see what you have tried on this problem so that we can provide a more focused answer. Without knowing where you are stuck, you are likely to get an answer very similar to the example problem in your textbook or your instructor's advice (which obviously weren't helpful or you wouldn't have posted here).

Ah sorry actually I had no idea of how to even begin the question!

You will get the empirical formula straight from this data. In order to convert to a molecular formula, you need more information.
By way of another example:

A binary compound of tin and chlorine contains 62.89% tin by mass. Determine the empirical formula.

Our first goal is to figure out the ratio of moles of tin to chlorine in this compound. Once we know the ratio of moles, we can determine the formula.
What do we know? We know what percentage of tin is in this compound by mass. What is the percentage of chlorine?
In order to get to moles, we need at least a mass of the sample. We were not given that. Since percent by mass is a ratio of masses, it turns out we can choose any mass of sample we want and get the same formula (Try it yourself!).
If we assume that we started with 100.0 g of this stuff, then how many grams of tin would you have? How many grams of chlorine would you have? 100.0 g is a nice starting mass since you are working with percents.
Now, as Ari's answer says, you need to convert mass of tin and mass of chlorine into moles of tin and moles of chlorine by dividing each mass by the atomic molar mass of each element (look it up on your periodic table).
Now that you are in moles, determine the lowest whole number ratio of the two elements by dividing each number of moles by the smaller number of moles. If your numbers are close to whole numbers, you can round. For example, round 1.97 to 2. If your numbers are not close to whole numbers, you may need to do a little more arithmetic:

If you get a number like 1.48, this is close to 1.5 which is 3/2. Multiply both numbers by 3.
If you get numbers like 1.34 (close to 4/3) or 1.67 (close to 5/3) multiply by 3.

Now you have the empirical formula. As Ari's answer says, you need a molar mass of the species to get a molecular formula.
The answer for my example is $\ce{SnCl2}$. Work through the steps and see if you can get to this answer. Then repeat for your lead and oxygen compound.

Can I ask one more thing, I tried it I have the answer for your example and my example but for my question it says the answer is Pb3O4 where as I worked it out to be one : one moles in ratio, so PbO, the answers state that they are three : four atom ratio could you explain that by anychance?

When I went through the maths, I got $\ce{Pb3O4}$. Moles of lead $=90.66/207.2=0.4375$. Moles of oxygen $=9.34/16.00=0.5838$. Divide by the smaller number $0.4375$, and you get $\ce{Pb1O}{1.33}$. 1.33 is too far from one to round (but it is $\frac{4}{3}$ to three sig figs). Multiply by 3: $\ce{Pb}{1\times3} \ce{O}_{1.33\times3}=\ce{Pb3O4}$.

I see ^^ thankyou soo much! your explanation is so clear cut, thankyou for helping me to understand these questions! keep up the great work :)

I'm not going to this question for you. The point of this forum is to educate and not to do your homework.
Imagine you 100 g of compound XY where X and Y are elements.
You know the percentage of X is 70 and of Y is 30.
This means there must be 70 g and 30 g of X and Y respectively.
To convert this to moles you need to do :
$\frac{70}{molar\;mass\;of\;X}$ and similarly for Y.
Then make use of the definition of the empirical formula.
NOTE : To obtain the molecular formula you need the molecular mass.

Just letting you know I did attempt this question before asking and Im not using this site as a homework help, Im genuinely trying to find out how to do empirical formula, and this actually isnt homework its extra questions I am doing doing myself to prepare and brush up on my chem just clearing that up.


text:
What is the titre of a sodium hydroxide-sulfuric acid titration experiment?
So I did for my AS Level exam today a titration experiment, seeing how much sulfuric acid of (I think) about $0.100 - 0.200 \text{ mol dm}^{-3}$, which went in the burette, is needed to react with 0.1M sodium hydroxide, in the conical flask.
We lose a mark for every $0.10\text{ cm}^3$ we are off of the actual titre value that should be obtained by a professional, i.e. our chemistry teacher.
I obtained an average titre of $24.28\text{ cm}^3$
Does anyone know what the titre should be from experience or memory in the lab of this experiment?
I know this isn't a constructive question, so feel free to downvote - I just really want to know so that I'm taken off of the edge I'm sitting on!

There isn't really an answer to your question. The problem is, there is many possible ways in which that lab might have been set up and there's no telling which one you had to sit, not to say that the concentration of sulfuric acid given to you might have been any concentration between 0.100 and 0.200 M.
Also, if you do AS Chemistry, you should also be able to calculate the volume range you should be able to get. With 0.1 M sulfuric acid and 0.1 sodium hydroxide, you expect to use a titre of 12.5 mL of acid against 25 mL of base, to 6.25 mL of 0.2 M acid against 25 mL of 0.1 M base.
Since your titre value seems so off by calculation, I guess that the concentration of the acid could have been about 0.5 M instead, but really, there is no telling.
See a sample marking scheme (I took one from June 2010 and the syllabus doesn't seem to have changed for this May/June sitting):

(a) Accuracy [8]
For the two best titres give:

4 marks for value within 0.2 cm^3 of supervisor
2 marks for value within 0.3 cm^3 of supervisor
1 marks for value within 0.4 cm^3 of supervisor

Concordance [3]
Give:

3 marks if all the ticked values are within 0.2 cm^3
2 marks if all the ticked values are within 0.3 cm^3
1 marks if all the ticked values are within 0.4 cm^3

Average [1]
Give 1 mark if the candidate calculates a correct average (error not greater than 0.05) of all his ticked values.

The second and third mark allocations are really easy to get. As long as you get your readings close to each other and calculate the average titre correctly, you will get 4 marks. I'm not sure where you got this "you lose 1 mark for every 0.1 mL off the supervisor's titre value". As you can see, if you get 2 best titre values withinn 0.2 mL of the supervisors, you get full marks.


text:
Correct definition of terminal alkynes?
My textbook, Organic Chemistry (6th edition) by Brown, Foote, Iverson, and Anslyn, states:

Alkynes in which the triple bond is between carbons 1 and 2 are commonly referred to as terminal alkynes.

Is not this definition incomplete? I can imagine scenarios where we have a terminal triple bond that does not reside between carbons 1 and 2 according to IUPAC rules. E.g., $\ce{CH2=CH-C#CH}$. Here, the terminal triple bond is between carbons 3 and 4 (according to IUPAC rules, doubles bonds take priority when numbering).

You are correct about the absolute numbering of $\ce{CH2=CH-C#CH}$. By the IUPAC priority rules, the alkene has the lower absolute numbering.
However, numbering can also be relative. In your case, the relative numbering 1,2 refer to either of the following:

relative position from the end of the chain, where the chain end is #1
relative position from the functional group you care about (where that group is #1)
suggesting that a reaction occurs on neighboring positions in a molecule.

That being said, the definition given by Brown et al. is imprecise. A better definition is one of the following:

A terminal alkyne is an alkyne in which one (or both) of the two alkyne carbons is at the end of a chain.
A terminal alkyne is an alkyne $\ce{RC#CR}$ in which one or both $\ce{R}=\ce{H}$

I like the second one better, because the first one implies there is only one alkyl functional group in the molecule; there are many polyynes, the simplest being diacetylene with a similar structure but one fewer hydrogen than vinylacetylene.

You're right; the definition in your book is simplistic. A terminal alkyne would be better defined as "an alkyne with the formula $\ce{RC_2H}$", differentiating from an internal alkyne, which would have the formula $\ce{RC_2R'}$ and the added stipulation that neither R nor R' can be a hydrogen. Equivalently stated as in Wikipedia, an internal alkyne has carbon substitutents on all acetylenic carbons, and a terminal alkyne is therefore not an internal alkyne because at least one acetylenic carbon has a hydrogen bonded to it.
The example formula you cite is vinylacetylene, systematic name butenyne or 1-buten-3-yne (here's your IUPAC nomenclature rules favoring double bonds at work; otherwise it'd be ethynylbutene or similar). It is indeed an example of a terminal alkyne, having the exploded formula (C2H3)C2H.

and an internal alkyne is in the center and an external alkyne is between the end and the center. Like for example pent-2-yne or 2-pentyne is an external alkyne but pent-3-yne is an internal alkyne


text:
Pronunciation of "alkene" vs. "alkyne"
Maybe this is better in English.SE, but I think I'm more likely to get better responses here in the "industry" site. This post got me thinking; how do you pronounce "alkyne" (a hydrocarbon with a carbon-carbon triple bond) differently enough from "alkene" (a hydrocarbon with a carbon-carbon double bond) to differentiate the two in spoken conversation? They're obviously spelled differently and so can be separated in the literature, but I can't find a satisfactory pronunciation for either that isn't "al-keen".

English isn't my first language, and in truth I'm always doubtful about the pronunciation of words in chemistry. Sometimes it's possible to find videos where the words are mentioned, for example here or here or here or here (several examples of alkenes and alkynes being pronounced in this last one). That's what I use, al-keen and al-kyne (ky as in kite or sky).

I've always heard it pronounced with a long I sound: "al-kine."

Unfortunately due to the font, your uppercase $I$ in "long I sound" looks confusingly like a lowercase $l$! Just making it clear for anyone else. Maybe I'm the only one who was a bit lost for a couple of minutes.

Perhaps a different way of thinking about the molecule: Very rarely have I heard people using these nouns. Certainly in an organic chemistry world, it is much more likely to talk about the functional group, so you would actually talk about the c-c double bond, or the c-c triple bond, or the carboxyl group... You get the picture.
I know this doesn't really address the pronounciation issue, but think of it this way:
An alkene IS an alkene at the point where there is a c=c bond, the rest of the molecule is kinda unimportant, and it is really this functional group that makes it chemically interesting.
So talk explicitly about the bond of interest, not the whole molecule (at least when you are talking about certain properties you want to exploit that relate directly to this functional group).


text:
What is the maximum pressure that JB Weld can withstand?
JB Weld, a brand name of steel-reinforced epoxy, makes claims of being the "world's strongest bond".
What is the maximum pressure it can withstand, and what is it about the interaction between the steel and the epoxy that makes it so strong?

The MSDS might help people to start thinking about it?! http://www.jbweld.com/wp-content/uploads/2012/08/J-B-Weld-MSDS.pdf

Perhaps this thesis is useful: https://smartech.gatech.edu/handle/1853/19722

This is only a partial answer, addressing the maximum pressure a JB weld can withstand. I am still thinking about the reason for the strength so I will leave it to the 'real' chemists among you for now.
According to Repair Products the JB weld in fully hardened state has the following properties (in psi):
Tensile Strength:   3960
Adhesion:           1800
Flex Strength:      7320
Tensile Lap Shear:  1040

Tensile strength is simply the pressure needed to break the material by pulling it apart. Flex strength is the pressure needed to bend the material which, in case of a perfectly homogeneous sample would equal the tensile strength.
The adhesion is the pressure needed to pull the bond from the surface when pulling perpendicular to the surface (parallel to the bond). The tensile lap shear is also an 'adhesion', but measured when pulling parallel to the surface (perpendicular to the bond).
Since the latter is the lowest it means that this would be the maximum pressure the JB weld could withstand, but only if it is applied exactly perpendicular to the bond. If the direction is in a different direction than you will get some mixture between the adhesion and the tensile lap shear pressures.
What is clear from these numbers is that the bond itself will not break but rather it will detach from the materials it is keeping together.

Project Farm on Youtube has a decent stress test of three different JB Weld variants and some other epoxies here: https://www.youtube.com/watch?v=XObmZIbHOzY .


text:
How hot can Thermal Grease get?
Thermal Grease is used to remove the air gaps between CPU and heatsink. However, computing aside, what would determine the max temp that the grease can handle before it burns or breaks down?

There are different kinds of thermal grease, and the temparature they can handle will vary.

Your best bet is to contact the manufacturer or consult the labeling information for your brand.

Since most thermal greases are ceramic and/or metal suspended in a silicone or organic oil, the temperature the grease remains stable at will likely primarily be determined by the properties of that oil -- either its ease of evaporation, or decomposition temperature (or, I suppose, its flashpoint, but that seems unlikely).
As Ben Norris says in the comment above, your best bet is to contact the manufacturer or check their data sheets: for common CPU uses, most manufacturers will give you stable and extreme temperature ranges which are likely to be fairly conservative (for example, here), and they may be able to help with more esoteric data if you're using it for other purposes.

I tried burning thermal grease; it catches fire pretty easily.

@kurtnelle - That's to be expected; both silicone and other organic oils are organic compounds; get them hot enough, like exposing them to a naked flame, and they'll combust just like any hydrocarbon. Most fuel gas flames burn at between 1000-1500 degrees, with some of the crazier ones like acetylene going up to 2500. That's well above the flash point of just about any organic substance, as well as a few metals.

However, without true combustion as an ignition source, very few things on the surface of the Earth can sustain those kinds of temperatures through other means. You could likely put a dollop of silicone thermal grease on a cookie sheet and stick it in the oven at 450*F for an hour and it wouldn't burn (it would likely melt). The circuitry that grease protects by conducting heat away from it would melt down at a third that temperature.


text:
Better understanding the pyrolysis reaction
I'm no chemist, but much of my work revolves around biochar.  I've always understood it with the following rough approximation:
$$
\ce{CH2O (dry plant matter) + heat -> black carbon + H2 + CO + CH4 + hydrocarbons}
$$
I've got two questions:

Is the above a decent approximation?  In what ways is it misleading (as all approximations always are)?
A colleague recently told me that methane is only formed in this reaction when there is water vapor present -- he said that it mixes with CO at high temperature to form the methane.  Is this true?  Will one not get any methane from oven-dried biomass that is exposed to heat in the absence of oxygen?

The presumption here is that the biomass is purely sugars: glucose- and xylose-based starches and cellulose.  The essential components of lignin, which make up about 20% of the biomass, is phenols and propane. You may need to rebalance based on those additional components.

Thanks for the answer!  But could you take it down a notch?  What about that equation is misleading in terms of resultant products?  Under what conditions will methane form?

The amount of methane you get will depend on moisture, temperature, the original biomass, and other contaminants.  There are also some other complications: the stuff left in the bottom of your fireplace consists largely of potassium hydroxide (KOH). 'HCs' might include the phenols and propane, but might also simply be bits of partially combusted product or reactions with nitrogen to create NOx.  If carbon starts out bound to 2H and 1O, the extra hydrogen has to come from somewhere.  Water would work, but it might also come from H2 or ammonia.

The equation above is a decent approximation of one group of reactions. Another one given here:
$$
\ce{C_{c}H_{h}O_{o} -> z CO + x H2 + w H2O + v C_{n}H_{m}O_{p} + u C_{(s)}}
$$
This is only the pyrolysis part of the process. There's a host of reduction reactions, I only give the ones that produces methane:
$$
\ce{CO + 3 H2 <-> CH4 + H2O} \\
\ce{C_{(s)} + 2 H2 <-> CH4}
$$
As you can see, methane is formed when there's hydrogen in the gas phase. Besides the pyrolysis there's two reactions that generate free hydrogen:
$$
\ce{CO + H2O -> CO2 + H2} \\
\ce{C_{(s)} + H2O <-> CH4}
$$
So indeed, you need water vapor to generate the hydrogen for your methane. If I recall correctly, methane is only a minor component in pyrolysis gas most of the time. By caloric value, the biggest component is either carbon monoxide or tar (long, complex hydrocarbons).
Last but not least you have a host of oxisation reactions (both in the gas phase and on the solid stuff) that power the pyrolysis - I won't go into them here.

my Latexs rusty! how do I get proper equation arrows?

See here for information on how to use mhchem to typeset chemical formulas and the like.


text:
Why does nucleophilic substitution of a haloalkane with ammonia produce ammonium, not a hydrogen halide?
For example, $\ce{CH3CHBrCH3}$ reacts with $\ce{:NH_3}$ to make $\ce{CH3CH(NH3^+)CH3 +:Br-}$,$$\ce{CH3CHBrCH3 \rightarrow CH3CH(NH3^+)CH3 +:Br-}$$then reacting again with $\ce{:NH_3}$  to make $\ce{CH3CH(NH2)CH3 +:Br^- +NH4^+}$:
$$\ce{NH_3 + CH3CH(NH3^+)CH3 +:Br- \rightarrow CH3CH(NH2)CH3 +:Br^- +NH4^+} $$
Question: why does the last step not produce $\ce{CH3CH(NH2)CH3 +HBr}$? That is, why is Ammonia a better nucleophile than the halide ion, despite having a lower partial negative charge (the halide has a $-1$ charge)?
Edit: a suggestion made by a teacher is that my proposed step does in fact happen, it's just that it occurs as an intermediary to the last step.  Apparently, it's very favourable to have a Hydrogen Halide and Ammonia acid-base reaction. Is he correct, and if so, why is this?

The answer to this is quite similar to one I gave recently for another set of reactions with hydrogen halides as possible products.
Your proposed last step is:
$$\ce{CH3CH(NH3^+)CH3 +Br- \rightarrow CH3CH(NH2)CH3 +HBr}$$
Here, you have a weak acid $\ce{(CH3CH(NH3^+)CH3)}$ and a very weak base $\ce{(Br^{-})}$ reacting with each other to turn into a weak base $\ce{(CH3CH(NH2)CH3)}$ and a very strong acid $\ce{(HBr^)}$. This is highly unfavourable from a thermodynamic sense, because you're trying to inject a lot of chemical potential into the molecules at no cost. Though your teacher suggests it might happen, I wouldn't be very comfortable saying that, if only for the fact that there's another, easy option.
It is possible that the previous argument is slightly circular, so it's best to explain why $\ce{HBr}$ is such a strong acid / $\ce{Br^-}$ is such a weak base. As you mention, $\ce{Br^-}$ has a negative charge, and one might expect that to mean it is a better base. However, that affirmation is only useful as a rule of thumb when comparing substances of different structure, being strictly valid solely when looking at a single sequences of related structures (such as $\ce{PO_4^{3-}}$/$\ce{HPO_4^{2-}}$/$\ce{H_2PO_4^{-}}$). The strength of a base is determined by several factors, such as charge localization, bond strength between base and a proton, steric hindrance, and strength of interaction of the protonated/deprotonated species with the solvent (if it exists). There are very strong neutral bases and very weak negatively charged bases ($\ce{SbF_6^-}$) known.
$\ce{Br^-}$ is a poor base because, while it has a net negative charge, that extra electron is bound to a somewhat electronegative atom, thereby stabilizing the charge. Not only that, but the bromide anion is quite large, so the charge is "spread out" over a significant volume and would attract a proton less, in some sense. Meanwhile, though $\ce{NH_3}$ has no extra charge, and even though the central nitrogen atom is more electronegative than bromine, the non-bonding pair of valence electrons in $\ce{NH_3}$ is highly localized, especially due to the fact that second period p-block elements are anomalously small. This makes $\ce{NH_3}$ an attractive target for a proton.


text:
Ratio of concentrations
I'm studying for my chemistry final and having a hard time with this question on one of the practice exams. Here's the question:
At $20$ degreees Celsius, Henry's Law constant for Ar is half of that for Kr. If a gaseous Ar/Kr mixture containing 20% molar fraction of Kr is placed over water at $20$ degrees Celcius, what is the ratio of concentrations of Ar to Kr in water?

Since there is a 20% molar fraction of Kr, that means the mixture contains 4 parts Ar to 1 part Kr.
4 parts * 1/2 gas constant = 2 times more Ar than Kr.

this would make a good answer if you expand on it.

Henry's Law is mathematically expressed by $$p=k_Hc$$ where $p$ is the partial pressure above the liquid, $k_H$ is the Henry constant for the specific gas at the given temperature, and $c$ is the concentration of gas in liquid. If you set up one equation for each gas and divide the two, you get $$\frac{p_{Ar}}{p_{Kr}}=\frac{k_{H, Ar}}{k_{H, Kr}}\frac{c_{Ar}}{c_{Kr}}.$$ You are given both the ratio of partial pressures and the ratio of the Henry's Law constants; the ration of concentrations should be easily calculated from that point.


text:
Strongest chloride complex as predicted by HSAB theory
Why won't $\ce{Ni^2+}$ form a complex with $\ce{Cl-}$, while the ions $\ce{Cu^2+}$ and   $\ce{Co^2+}$ form the complexes $\ce{[CuCl4]^2-}$ and $\ce{[CoCl4]^2-}$?
According to the HSAB theory, $\ce{Cl-}$ is considered a hard/borderline base. So I figured the strongest complex would be the one in which the center ion is the hardest acid. The $\ce{Co^2+}$ ion is the largest and the least electronegative. So it should be the softest acid (although they are all considered borderline). It forms a complex with $\ce{Cl-}$. $\ce{Ni^2+}$ should be the hardest acid, and yet it does not form a complex with $\ce{Cl-}$ at all. Why is that?
References:

http://www2.chemistry.msu.edu/courses/CEM812/HSAB_Theory.pdf
R. Pearson, Chemical hardness and density functional theory, J. Chem. Sci., Vol. 117, No. 5, September 2005, pp. 369–377

Are all of the metals in your complexes supposed to be in the $+2$? oxidation state? If so, the charge on your complexes should be $2-$. If not, then all the metals need to be $+3$, which is very unlikely for copper.

The link to your handout explains why $\ce{Fe^{3+}}$ (and it is $\ce{Fe^{3+}}$ in $\ce{FeCl4^{-}}$) is a harder acid than $\ce{Co^{3+}}$ or $\ce{Cu^{2+}}$ or $\ce{Ni^{2+}}$.

Hard acids (in context, HA) are characterized by (s,f blocks, left side of d block in higher OS's)

Iron is the furthest to the left of the four metals you mention. Thus, it has fewer electrons in its $d$ shell to balance the positive charge:

$\ce{Fe^{3+}}$ is $\ce{[Ar]}3d^5$
$\ce{Co^{3+}}$ is $\ce{[Ar]}3d^6$
$\ce{Ni^{2+}}$ is $\ce{[Ar]}3d^8$
$\ce{Cu^{2+}}$ is $\ce{[Ar]}3d^9$

and

Low electronegativity (χ) of the acidic atom. A value in the range 0.7-1.6 is typical of hard acids;

Iron has the lowest electronegativity of the four metals:

$\ce{Fe}$ 1.83
$\ce{Co}$ 1.88
$\ce{Ni}$ 1.91
$\ce{Cu}$ 1.90

Iron is the hardest Lewis acid of the metal cations you list.
As for $\ce{Ni^{2+}}$, it does form a complex $\ce{NiCl4^{2-}}$ with $\ce{Cl-}$. Here is a Google search for it.