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Hexagonal unit cells
I know 3 types of unit cells, the simple cubic, the body centered and the face centered.
But what is a hexagonal unit cell, and why does it have a hexagon base while others are all cut into a cubic? And how can I calculate the number of atoms in every primitive hexagonal unit cells?

Welcome to chemistry.SE! I've taken the liberty to edit your question some, while retaining the original meaning.

The cubic primitive, cubic body-centered and the cubic face-centered are just the three possible Bravais lattices of the cubic crystal system.
In total, we know seven crystal systems, namely (in order of increasing symmetry) triclinic, monoclinic, orthorhombic, tetragonal, trigonal, hexagonal and cubic.
Among these, the triclinic crystal system has the lowest symmetry ($a \ne b \ne c;\ \alpha, \beta, \gamma \ne 90°$), the cubic crystal system has the highest symmetry ($a = b = c;\ \alpha = \beta = \gamma = 90°$).
A careful look at illustrations of the hexagonal closest packing ($7+3+7$ atoms in $ABA$ layers) in a hexagonal prism and a bit of thinking gives the number of atoms:

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

The total count thus is $\mathbf{6}$.
These diagrams of the hexagonal closest packing, do not represent the the unit cell as the most simple unit from which the crystal is build by symmetry operations.
Cutting the hexagonal prism (along the axis through the atoms centering the top and bottom faces) into three pieces gives identical rhombohedra. These can be transferred into each other by rotations along the axis described above.

So how many molecules are in each hexagonal unit cells? And why I saw many people cut hexagonal unit cells into three structures that look like body centered unit cells. Check this out http://www.youtube.com/watch?v=vXtO8afu8wg

@user40003 Did that answer your questions?

I havent't got time to read it yet. I will do it probably tmr or tonight!Thanks for your answer though

So when we usually say hexagonal unit cells, we refer to the hexagonal structure one or the one has been cut into rhombohedra

@user40003 The unit cell of the hexagonal crystal cystem is a rhombohedron.


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What is the cost of 1.00 g of calcium ions as provided by this brand of dry milk?

A student determined the $\ce{Ca^{2+}}$ ion content in a sample of nonfat dry milk. For complete titration of the $\ce{Ca^{2+}}$ ion present, a sample weighting $1.483 ~\mathrm{g}$ required $41.33 ~\mathrm{mL}$ of $1.183 \times 10^{-2} ~\mathrm{M}$ EDTA solution. Calculate the mass of the $\ce{Ca^{2+}}$ ions in the titrated sample.

So I did:
$41.33 ~\mathrm{mL} = 0.04133 \mathrm{L}$
$(1.183 \times 10^{-2}~\mathrm{mol\, L^{-1}})\times (0.04133 \mathrm{L}) = 4.89 \times 10^{-4} ~\mathrm{mol}\, \ce{Ca^{2+}}$
$(4.89 \times 10^{-4} ~\mathrm{mol}\, \ce{Ca^{2+}}) \times 40.078 ~\mathrm{g\, (mol\, \ce{Ca^{2+}})^{-1}} = 0.0196 ~\mathrm{g}\, \ce{Ca^{2+}}$
My question is:
A $25.06$ oz box of the powdered milk analyzed above costs $5.46$.

Calculate the cost of $\pu{1.00 g}$ of $\ce{Ca^{2+}}$ ion as provided by this brand of dry milk.

How do I do that?

Carrying the units through the calculation is a good decision! Nice example for a showing progress in homework question.

First you calculate the value $v$ of the powder per gram:
$$ 25.06~\mathrm{oz} = 710.44~\mathrm{g} \; \hat{=} \; 5.46~\$ \Rightarrow v=0.007685~\mathrm{\$\, (g\, Powder)^{-1}}$$
Then you calculate the mass of powder ($m$) that contains one gram of calcium ions:
$$ m=\frac{1.483~\mathrm{g\, Powder}}{0.0196~\mathrm{g\, \ce{Ca^{2+}}}} = 75.66~\mathrm{\frac{g\, Powder}{g\, \ce{Ca^{2+}}}} $$
Finally, you multiply the two results for the price $p$:
$$ p = v\times m = 0.58~\mathrm{\$\, (g\, \ce{Ca^{2+}})^{-1}}$$

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Does supersaturated caffeine polymerize?
I tried preparing a supersaturated solution of caffeine by mixing anhydrous caffeine powder with water in a thin Turkish tea glass and heating it up over a candle (desperate times...).
This worked as expected and a lot more caffeine dissolved into the clear solution after I slowly swung it back and forth over the flame. The water didn't boil, no visible vapors formed on the surface, and I was holding the glass with my fingers, so I would say that the temperature was well below 100°C.
After the solution cooled down, instead of crystals precipitating, I found the glass filled with a fibrous cotton-like white mass. I pulled it out (it mostly stuck together) and dried it up on sheets of paper. Now I have white flakes which are flexible instead of brittle (still feel very much like cotton).
This would seem an awful lot like some polymer forming - except the substance tastes a lot like crystalline caffeine, and melts and evaporates much like caffeine - with the vapors having the same distinctive acrid (but not burnt) smell.
I'm pretty sure the answer to the polymer question is no, but in case I'm right I'd be very curious as to what's actually happening.

Google Images "caffeine crystals"  They are beautiful fibrous masses.  The solvent can influence crystal shape by making various faces grow faster or slower.  Benzil slowly crystallized from hot optically resolved pinene is slender needles.  Crystallized from room temp supersaturated mixed xylenes with a needle seed crystal, it is single crystal blocks.

so why would the crystals be flexible? also, why would they not adhere at all to the glass, and not precipitate, but rather form as clouds in the solution?

The crystals are flexible given aspect ratio. 1 cm of thin spaghetti is rigid. 30 cm easily flexes.  Dust in the water affords multiple nucleation centers. Bottled water is free of particles. Wrap the hot liquid in a rag for much slower cooling.  Try recrystallizing caffeine from vodka, http://lxsrv7.oru.edu/~alang/onsc/solubility/allsolvents.php?solute=caffeine (alcohol vapors are flammable).

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How do I convert benzoic acid to m-nitrobenzyl alcohol using 2 steps?
The question asks to carry out the following conversion in not more than 2 steps:
Benzoic acid to m-nitrobenzyl alcohol
I first used nitration (conc. $\ce{HNO3}$ /conc.$\ce{H2SO4}$) and then I'm got stuck, as reducing the $\ce{-COOH}$ group would also reduce $\ce{-NO2}$ group (using $\ce{LiAlH4}$).
I tried to reduce $\ce{-COOH}$ group using $\ce{LiAlH4}$ to $\ce{-CH2OH}$ and then proceed to nitration but got confused whether $\ce{-CH2OH}$ is o/p or m-directing. I think it should be o/p directing.

Then $\ce{-COOH}$ is meta directing. Does that solve your question? (think about a reducing agent with selective affinities)

@SatwikPasani I can't think of such a reagent which can reduce acid but not nitro...

Use the meta-directing effect of $\ce{-COOH}$: nitrate benzoic acid first.
Reduce m-nitrobenzoic acid with $\ce{LiAlH4}$. The nitro group is not attacked, but when you calculate the equivalents of  $\ce{LiAlH4}$ needed, don't forget that you are reacting an acid with a hydride.

Edit 1:
Nitroalkanes or nitroalkyl-substituted arenes can be converted to the corresponding amines by $\ce{LiAlH4}$. This is usually not the case for nitroarenes. Here, catalytic hydrogenation, e.g. over $\ce{Pd/C}$ is the method of choice.
Edit 2:
Let's play, change the order of events and first reduce benzoic acid to benzyl alcohol. The hydroxymethyl group isn't m-directing, but what will happen to it under nitration conditions ($\ce{H2SO4}$ and $\ce{HNO3}$)?
Under these pretty acidic conditions, elimination of water from the starting material furnishes benzyl cations, which might react with benzyl alcohol in different ways, such as (multiple) Friedel-Crafts alkylations or etherification. I'd dare to say that the result is a mess.

Wikipedia says that it will reduce nitro group. See reduction to amines. http://en.wikipedia.org/wiki/Reduction_of_nitro_compounds

@Awesome Irrelevant in your case. Check out the link given in that article.

Oh right...Thanks. But please do tell whether $\ce{-CH2OH}$ is o/p directing or m-directing.

Isn't the hydroxymethyl substituent like a methyl group with a hydroxy group dangling from its far end? In electrophilic aromatic substitution, it's not m-directing. But would it survive treatment with sulfuric acid/nitric acid?

According to wikipedia $\ce{LiAlH4}$ and similar metal hydrides will reduce nitroarenes to azo compounds. http://en.wikipedia.org/wiki/Reduction_of_nitro_compounds#Reduction_to_amines

http://pubs.acs.org/doi/abs/10.1021/jo00956a011
Required is a reduction that acts upon the carboxyl as such.  Consider borane-THF, or its equivalent sodium borohydride and BF3·Et2O.  Next best is borane-dimethyl sulfide, given the odor.  Aliphatic nitro groups are reduced.

Use nitration by mixed acid and then $\ce{SOCl2}$ to replace the $\ce{OH}$ of carboxylic group. Then use $\ce{NaBH4}$ to reduce acid chloride to alcohol.

this is a three stage process not a two stage

bon is right. 2 steps, not three. Please study the question well before answering. That way the quality of the Q and A will be maintained.

You were initially on the right track:
After nitration with conc. HNO3 and conc. H2SO4, perform reduction with Diborane (B2H6), under mild conditions as it does not easily reduce functional groups like nitro, halo, ester, etc.
Voila, you have m-Nitrobenzyl alcohol.

If a reducing agent does not attack an ester, I’m having a hard time imagining it attacking an (even more stabilised) acid.


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Are the electrons also distributed for the alpha decay?
Consider this reaction of alpha decay:
$$ \ce{^{222}_{88}Ra -> ^{218}_{86}Rn + ^4_2 He} $$
My question is, whether or not the electrons also get distributed in the reaction like the nucleons?

The equation for the $\alpha$-decay omits the charges.
Since the $\alpha$-particle actually is $\ce{^4_2He^{2+}}$, the resulting nucleus is initially left with a surplus of two electrons. These are emitted too, probably on the same timescale.
So, the answer is yes: the $\alpha$-decay of a neutral nucleus yields another neutral nucleus, a double-positively charged $\alpha$-particle and two electrons.


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Why does the cathode ray tube only start glowing at low pressures?
I am considering the cathode ray tube of the set up, when the cathode rays were discovered:

Voltage = very high
Gas = any
Atmospheric Pressure = changing

At different pressures, there are different observations:

1 atm: the tube does not glow.
0.1 atm: the whole tube lightens up
0.01 atm: the tube glows at the anode end (Glowing is shown as there is a detector coating of ZnS.)

Note: I don't have this set-up, this is just taken by me from theory.
Why does, when I am just reducing the pressure from 1 atm to 0.01 atm, the tube at first not glow, then it glows completely, and only at low pressure it glows just at one end?

At 1 atm. it is difficult create stable plasma, because breakdown voltage is very high.
Your electrodes are located at least 10-15 cm I think, and it is simply you do not have enough electrical potential between them to create initial plasma channel.
Then pressure goes down, voltage is fixed. When you reach breakdown voltage you get your glow (the Townsend discharge happens just as glow discharge becomes visible)-normal glow discharge. What happens? You ionized your gas, you have free electrons and ions, which are floating in different directions: ions are floating to negatively charged cathode and electrons flow to positively charged anode.
Because charge carriers are moving in space. They are rubbing against each other, part of the energy is dissipated and goes into heat and other is emitted and you see your glow. Ions are moving relatively slow, electrons, on contrary,  move very fast comparing to ions, because they are light weighted. So this fast free electrons, accelerated by a sufficiently strong electric field, give rise to electrical conduction through a gas by avalanche multiplication caused by the ionization of molecules by ion impact.
Then you continue to reduce your pressure more and more, less and less gas molecules can be ionized by free electron flow, less and les collisions are happening (less energy is dissipated), so the intensity of glow is going down, and finally it stops, because the breakdown voltage increased and overcome your applied voltage. (see Paschen Curve, lower is a pressure higher is the breakdown voltage.)
This is generally how it works.

Cathode rays travel in a straight line.
This was confirmed with a Maltese cross covered with phosphor that illuminates with radiation. As voltage is applied between the cathode and the anode, the cathode rays emitted struck the metal Maltese cross used, creating a very sharp image of the same cross at the end of the tube.
Crooke's tubes need to be evacuated, as gas particles within would scatter the electrons.

Image 1 shows the Crookes tube apparatus. The gas pressure inside is reduced and eventually disappears at very low pressures.
In image 2, even though evacuated, the rays that come from the cathode still persist, proving that the rays are emitted from the cathode and different from light as they travel in straight lines.

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How do I derive the Nernst equation in the case in which R is absent?
Does anyone know what the intermediate calculations are to obtain equation 1.4.14? Taken from Fundamentals and Applications (Bard) page 31
To be more precise, how to obtain (1.4.14) from $\frac{nFAm_O}{nFAm_R}=\frac{i_l-i}{i}$

You have to use equation 1.4.12, i.e. $E = E^{0^{\prime}} + \frac{RT}{nF} \ln \frac{C_{\mathrm{O}} (x=0)}{C_{\mathrm{R}} (x=0)}$, as basis and substitute equations 1.4.11 and 1.4.13 into it, then you arrive at 1.4.14 by using the basic property of logarithms that $\ln (ab) = \ln (a) + \ln(b)$.

Thank you for the comment Philipp. With this clarification, the derivation is simple:
$E=E_0+\frac{RT}{nF}ln\frac{C_O(x=0)}{C_R(x=0)}$ (1.4.12) where
$C_O(x=0)=\frac{i_l-i}{nFAm_O}$ (1.4.11) and $C_R(x=0)=\frac{i}{nFAm_R}$ (1.4.13).
Than, combining equations (1.4.11) and (1.4.13) to (1.4.12), we obtain:
$E=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{nFAm_O}}{\frac{i}{nFAm_R}}=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{i}}{\frac{nFAm_O}{nFAm_R}}=E_0+\frac{RT}{nF}(ln\frac{i_l-i}{i}- ln\frac{nFAm_O}{nFAm_R})$=
$E_0-\frac{RT}{nF}ln\frac{m_O}{m_R}+\frac{RT}{nF}ln\frac{i_l-i}{i}$


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What's the mechanism of the reaction of (E)-methyl 3-(furan-2-yl)acrylate in acidic methanol?
In the organic chemistry textbook by Clayden et al. they mention that the reaction of (E)-methyl 3-(furan-2-yl)acrylate in acidic methanol yields dimethyl 4-oxoheptanedioate without providing the reaction mechanism and leaving it to the reader to figure it out:

I've tried for over an hour now but didn't get to this product. I assumed that since the reactant already has the right number of carbon atoms I won't have some kind of dimerization to consider, only reactions involving one molecule of (E)-methyl 3-(furan-2-yl)acrylate. I think, the first step should be a protonation at the furan 3- or 5-position with a subsequent attack of methanol on the carbonyl group that was formed followed by an acetal cleavage. The problem is that I don't know how to get the second ester group and how to get rid of all 3 $\ce{C=C}$ double bonds? Unfortunately, I don't have access to ChemDraw on this computer so I can't provide some example mechanisms I came up with.

As Uncle Al already pointed out, the mechanism starts with protonation of the ester carbonyl oxygen. This yields a resonance-stabilized cation whose charge can be delocalized into the furan ring, where it is subsequently captured by nucleophilic attack of methanol in $\alpha$ position to the ring oxygen.

The cyclic acetal thus formed undergoes [1,5]-H shift which is driven by rearomatization of the furan ring. The reactive 2-methoxyfuran is protonated at the 3 position, which yields a resonance-stabilized acetal cation. During aqueous workup, this cation is captured by water, which results in ring opening. The protonated ester carbonyl group is deprotonated by the ketone enolate, and after subsequent keto-enol tautomerization, the final product is eventually obtained.

In the top part I would start by pushing electrons from the furan oxygen. That gives me the same intermediate. Then I don't really like that 1,3-H shift. You can drive forward by rearomatisation and get rid of the exocyclic double bond too, so 1,5-H shift, to give a 2-methoxyfuran which is even more reactive and will open wide by the standard mechanism, starting by pushing electrons from the furan or the methoxy O and protonate the 3-position

@K_P I was also sceptical about that 1,3-H shift, because of an apparent lack of "driving force" for this reaction step. Your suggested 1,5-H shift, which is driven by rearomatization and also leads to the acetal cation, makes much more sense. I have edited the mechanism accordingly.

Unfortunately I cannot agree with the currently accepted answer. This 1,5-hydride shift that was invoked seems to have a very strained transition state (the cyclic, planar system means that the "1" and "5" atoms are quite far apart) and with all due respect, I am quite skeptical as to whether it would really work. Furthermore furan does not enjoy a particularly large aromatic stabilisation and "rearomatisation" as a driving force is not too convincing. Furan, in all its reactions, has a very pronounced tendency to lose its aromaticity. For example even nitration with the mild reagent acetyl nitrate ($\ce{AcONO2}$) leads to a 2,5-addition product and you actually need to add a second reagent (pyridine) to eliminate $\ce{AcOH}$ and rearomatise. For all intents and purposes furan reactivity is much closer to that of an enol rather than that of benzene.
Instead of a 1,5-hydride shift, proton transfers should suffice, especially since there are carbonyl groups all over the molecule (i.e. the proton transfer is simply a tautomerism and doesn't go through any particularly unstable intermediates).
One way of thinking about this question would be to try and do it in reverse. 1,4-Dicarbonyl compounds are known to cyclise to form furans in acidic solution, so this should simply be the microscopic reverse of that reaction. The key cyclisation step involves the attack of an enol oxygen onto a protonated carbonyl group. So, in this reaction, the fragmentation step should be the reverse: the ejection of an enol oxygen from a (hemi)acetal.
I would propose this mechanism instead.

So far, this is the same as in the currently accepted answer. However the usage of the furan lone pair to push electrons saves time in drawing all the resonance forms (this is what was meant by K_P's comment also).
Protonation of the furan oxygen (not drawn) brings us to the start of the next step:

As described previously this fragmentation is the reverse of the cyclisation reaction used to generate furans. Following tautomerisation of two of the enols, most of the molecule is already in place and the remaining question is how to generate the second ester functionality. The issue is that the carbon that should become the ester has an extra hydrogen on it. Instead of having a hydride shift it makes much more sense to move it around by tautomerisation!

The ketene acetal can be protonated and attacked by methanol at the carbonyl carbon, but that would be unproductive as there is no way of getting rid of one of the methyl groups on oxygen (unless we invoke a $\mathrm{S_N2}$ reaction on the methyl group with methanol acting as the nucleophile). So, it probably needs to wait until water is added during the reaction workup, at which point it is hydrolysed to give the desired product.

This commentary may be a bit late in the game but some historical perspective may be appreciated. This mechanism question was around when I was a graduate student in the 1960's. The reaction was reported in 1877 by Marckwald.1 He conducted the reaction in 95% ethanol (water present) in the presence of HCl. The diacid of ketodiester 7 was formed by saponification and its composition was confirmed by combustion analysis. The mechanism is no different with methanol and unless methanol is scrupulously dried, the keteneacetal mechanism will not prevail nor will a [1,5]-hydride shift survive. Another point about the mechanisms provided thus far is that enols of ketones and esters, if so regenerated, rapidly tautomerize.

The reaction is catalytic in HCl. Esterification can be effected at any stage of the process, but here, it has been accomplished for convenience in the first step (1 ---> 2). The sequence 2 ---> 4 is a redox reaction in that the double bond is reduced to an alkane while C5 of the furan is oxidized, all through the aegis of a proton. Furans bearing α-alkoxy groups are highly susceptible to acid hydrolysis. Protonation at the β-position leads to facile introduction of water (4 ---> 6). The enol ether 6 is readily protonated causing collapse to Marckwald's keto diester 7.
To distinguish between the mechanism presented here and the keteneacetal mechanism presented above, workup in the presence of H2O18  while the reaction mixture is still acidic, will incorporate O18 into the ester carbonyl by rapid deuterolysis of the keteneacetal but not into the esters by the mechanism presented here. Incorporation of O18 into an ester is expected to be much slower than into a keteneacetal.

Marckwald, W. Chemische Berichte, 1877, 20, 2811.  DOI: 10.1002/cber.188702002140

Starting furan is also a bis-enol ether hiding a terminal aldehyde and an internal ketone.  Seven skeletal carbon atoms including that conjugated olefin.  Product has seven skeletal carbon atoms and three carbonyls in the right places.  The double bond is reduced and what would be the terminal aldehyde is oxidized to the acid, appearing as the methyl ester.
Protonate the ester carbonyl.  1,3 shift of the double bond affords a carbenium ion next to the furan ring and =C(OH)(OR), the enol of the ester.  That is downhill in energy for the alpha-cation conjugating to the electron-rich aromatic ring.  1,3 shift of the ring double bond leaves the ring oxygen an enol ether (exocyclic double bond)  and moves the charge into the ring.  1,3 shift moves the charge alpha to the ring oxygen (again downhill in energy - resonance contributors), where it is captured by methanol to make an acetal.
Move the endocyclic unsaturation into the acetal. I see some ugly here, so I hope somebody does better.  Work it up in aqueous acid.  Protonate the endocyclic double bond to put the charge alpha to the captured MeO.  1,2-hydrogen shift to put the charge where it is needed, captured by water.  Now hydrolyze everything to the parent carbonyls.

Thanks for your answer. I'm not sure that I understood every part of your mechanism - especially the sentence "Move the endocyclic unsaturation into the acetal" is a bit unclear to me - but it did set me on the right track and provided me with some ideas I hadn't considered, e.g. 1,2-hydrogen shift. I have figured out a mechanism which looks (still a little odd but) mostly good to me and I will post it when I get access to ChemDraw again. Your answer was very helpful (+1 for that) but if you want me to mark it as the accepted answer you'll have to put in the chemical formulas.

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text:
How does high-frequency electrolysis of water work?
I have read that combining the DC current with a high-frequency AC current, the electrolysis of water speeds up. Is this true? In that case, how is less energy wasted as heat? Or does it simply catalyze the process?

Do you have a source for that? Also, what do you mean by combining DC and AC? I assume you mean a rippled DC current, that is one that is changing its voltage by way of an overlaid AC without changing polarity?

First of all, have a look at the wikipedia page on electrolysis of water.

I also like this review: Zoulias et al.: A Review on Water Electrolysis, TCJST, 4 (2) (2004) 41-71
Specifically they list a number of actually existing installations (context: renewable energy) and their actually achieved efficiency.

Speed up does not necessarily have anything to do with higher efficiency. In electrolysis it is often the other way round: if you want to squeeze out the maximum free energy, you need to do the reaction infinitely slowly (despite thermodynamics having a dynamic name, it looks at infinitively slow processes).

Thus, speeding up usually means that you find a way to put more power through your system. The big issue is to find a way of doing this without loosing (too much) efficiency.

Pulsed/modulated DC: looking through a few papers I liked this one: Shimizu et al.: A novel method of hydrogen generation by water electrolysis using
an ultra-short-pulse power supply, Journal of Applied Electrochemistry (2006) 36:419–423, DOI 10.1007/s10800-005-9090-
They aim at avoiding the diffusion controlled situation by having the pulses short enough so that no depletion zone occurs.
Look at these diagrams:

So they report one setting where the pulsed electrolysis is actually more efficient than DC in their cell.
There's more research going on on this, however the papers I found reported increased efficiency compared to pure DC electrolysis, but the absolute efficiencies are around 10%.   However, compare their numbers with the 80% efficiency cited by the review for an industrial alkaline electrolysis. Note that one big difference is the voltage that is applied: for the DC it is around 1.85 - 2.05 V, so much less overvoltage. Note also that when they say that higher voltage speeds up ion transport, then this overvoltage is converted to heat (ions face friction in the medium when they travel) and thus basically lost.

Another line that looks real is that if you go to higher temperatures, a (small) part of the energy can be supplied by heat. As heat is cheap, this may help. One point one has to be aware though, that the efficiency calculations may be done with respect to the electric energy only (neglecting the heat input) and thus look artificially nice (like efficiencies of condensing boilers calculated against the lower heating value).

I found a bunch of nonsense claims in the internet, about the resonance frequency of water helping to split bonds.

The first thing to realize here is that there is no one resonance frequency of water. With suitable energy, you can excite rotational, vibrational and electronic states (I left out translation - there transition energies minute). At room temperature you can say as a rule of thumb that most molecules will be in some excited rotational state, but in the vibrational and electronic ground states. Excitation energies for rotation are in the far infrared or microwave energy/frequency region. Widely used e.g. in the microwave oven at 2.45 GHz ($\approx$ 12 cm). Actually, the whole region is full of bands where water absorbs. Note that microwave heating of water does not cause electrolysis. Vibrational transitions are around 2.9 μm = 105 THz = 3500 cm⁻¹ and 6μm = 50 THz = 1635 cm⁻¹ with lots of combinations and overtones throughout the near infrared region. Quite exceptionally, the visible region is basically free of water absorption. Electronic transitions (breaking of bonds) need energies in the UV, and here we meet bands that lead to photodissociation, e.g. at 166nm (taken from Wikipedia). That corresponds to 1.8 PHz = $1.8 \cdot 10^{15}$ Hz. Compare this to the kHz and MHz where your link claims dissociation.

This doesn't mean that the pulsed DC cannot help, nor that impedance spectoscopy won't give important information. But resonance frequencies in the kHz range are electrical LC-circuit resonances depending on cell and electrode geometries and electrical double layers etc. But neiter on vibrations nor breaking of the bonds of the water molecule.

To give the "method" you ask about some real world numbers,

at the very end of the Wiki page the energy efficiency for industrial water electrolysis is cited as usually between 50 and 80 %.
The paper then proposes to burn the gas in an internal combustion machine. As such a stationary process could be adjusted so that the engine is at its maximum efficiency, we may assume 1/3 or 35% efficiency here.
we then need a generator to convert the mechanical energy into electric energy. Fortunately, that step is rather efficient. Say, 95 %.
A fuel cell would be more efficient than the combustion - generator combination: ca. 40 - 60 % according to Wikipedia.
Unfortunately, also battery charging is not 100% efficient. Let's assume 80–90% (taken from Wikipedia on Li-ion batteries) For batteries that are charged with higher current (or current density) efficiency is less. Example would be lead-acid batteries as used in cars. Wiki quotes efficiencies between 50 and 80 %.

Taking these numbers together, I conclude that after going once through the cycle of the proposed "perpetuum mobile", 8 - 24 % of the energy are retained in a "useful state" while 76 - 92 % became heat. With fuel cell, we may be able to "boost" the energy efficiency to 43%.

Useful general knowledge (in addition to the law of energy conservation)

The US patent system is different from e.g. the German patent system in that here in Germany the patent application has to have commercial/industrial usability. This includes a technical argument why it works (according to the physical laws). A perpetuum mobile would be rejected on these grounds (of course the inventor could prove his case with a prototype). US patents do not have this technical check.

generators (mechanic -> electric conversion) are sources of current, while batteries (galvanic cells) are sources of voltage.

Without being a chemist, I thought that the efficiency of electrolysis is measured with respect to some ideal amount of energy for a perfectly efficient operation, that is, without taking at all into account the energy potential of the generated gasses. (Which I would presume to be huge compared to the energy needed to produce them via electrolysis.) Consequently, it appears to me illogical to compare the efficiency of the electrolysis process against the efficiency of burning the gases in an internal combustion machine in order to see if there is a net gain, no?

@MikeNakis: That ideal amount of energy typically  accounts only for the energy content of the produced gas (chemical energy/enthalpy;  pressure and/or temperature may or may not be included but for hydrogen or hydrogen + oxygen they are very small compared to the chemical energy = your potential). So 45 % efficiency in diagram (b) above means that a bit less than half of the electric energy they put into the cell ends up as chemical energy of the producs; the losses (heat) are in fact not negligible but a bit higher than the energy stored chemically in the produced gas.

"illogical to compare the efficiency of the electrolysis process against the efficiency of burning the gases in an internal combustion machine in order to see if there is a net gain" That would indeed be illogical - if only because under the law of energy conservation, there cannot be a net gain. But the OP linked to a cycle of hydrolysis -> (gas storage) -> combustion ->  electrical generator -> (battery storage) -> hydrolysis again. And for that you can calculate how much energy is left after one round. (Or, how much energy you have to "pay" for that storage option).

hmmm, I can't make sense, but that's almost certainly because of my incomplete understanding of the processes involved and even the terminology being used.  But thank you for trying.  I think it will be highly inefficient to try and gain an understanding of these matters by exchanging comments.  I will try to find someone to explain it to me in a conversation.  Once again, thanks.  Cheers!

I agree that, if there were any improvements made by using pulses or AC in hydrolysis, it couldn't have anything to do with resonant frequency. Logically, since DC had no frequency, and it does break the bonds, it does so in a non-resonant way, therefore, it would appear pointless to attempt any kind of frequency modification. That said, perhaps the pulsing could improve the results, but mechanically (acoustically?), as in creating more surface area for hydrolysis to occur (simply making DC electrolysis more effective).  Don't know.

With regards to surface area (of the electrodes), increasing it helps, but increasing it by several or hundreds of orders of magnitude, say, by the addition of a catalyst (the catalyst acting as trillions of nanoscopic electrodes) should vastly  improve the reaction. Funny, there's a study showing just that, here:
http://www.labmanager.com/news/2016/12/new-approach-to-water-splitting-could-improve-hydrogen-production#.WFmrT33fRLF

From the site that you linked:

A battery would be used as a source of electrical energy which would separate the hydrogen/oxygen into gases. The gasses would then fuel an internal combustion engine, which would power a generator to continously recharge the battery as well as deliver useable mechanical energy. If this sort of motor can be made to work, the energy crisis on this planet will be over forever.

Sounds like somebody proposes a perpetual motion engine of the first kind. I'm
very much impressed! The rest on that page is techno-babble of the same quality.
Edit This doesn't mean that there's might not be serious research on electrolysis using something else than constant DC, e.g. PWM (pulse width modulation).
A cursory search (there might be other sources) gave:
K. Mazloomi, N. Sulaiman, S. A. Ahmad, N. A. Yunus,
Analysis of the Frequency Response of a Water Electrolysis cell,
Int. J. Electrochem. Sci., 2013, 8, 3731-3739, PDF

Yes, what you cited there actually says its a perpetual motion engine, which if of course impossible. I'm just curious if it is true that you need less energy to break the bonds if your're using a rippled DC current. This is not the only source I've found. I'm not educated on any of these fields, so it's difficult to know what everyone is doing. Some people claim to use a high frequency current. I believe this might be just a homemade transformer with a transistor on to keep the flow running. They claim that this requires less energy per liter of $\ce{H2}$ than by using a continous DC flow.

It seemed to me that this source is written by someone without any knowledge of physics, and is merely a summary of something else that person has read, which is why I'm curious if it's true or not. In big scale physics, like bridges, the resonating frequency can make a huge bridges break. Does this apply to molecule bondings?

@FriendofKim: bonds breaking with wave of suitable energy: well in a way photochemistry does just that. Have a look at http://en.wikipedia.org/wiki/Visual_cycle Light of suitable frequency is absorbed, light that is (too) far from the absorption maximum isn't. The absorbed energy leads to isomerisation. You can say that a π -> π* corresponds to breaking the double bond (not the single bond, though).

Aha, so these using a rippled DC source wouldn't be able to create a resonance like the wind does with a bridge. It would be logical, though, that there is a way to do electrolysis in a more efficient way. A lot of the energy is going directly into heat. We "just" need to figure out how to use more of that energy to separate the water molecules.

@FriendofKim: you get the electric LC resonance due to the capacity of the double layer. But thats many (9) orders of magnitude away from dissociation energy. The pulsed DC has some points, but not breaking or weakening bonds.

Among the "scientific" mumbo jumbo I have read here, most commenters have missed a few key facts.

Most gains in electrolysis are made at extremely high voltage and extremely low currents. This sets up a charged capacitance state where the hydrogen and oxygen are pulled to the near breaking point to their respective charges.

Concerning frequency, it's not about energy absorption. It's about creating a base frequency within one or more atoms within the molecule, then hitting it with a brief secondary frequency pulse to cause a shattering effect much like when an opera singer shatters a crystal glass.

The idea here would be to add an insulator to the positive and negatively charged plate such as sealing them in a thin layer of plastic or other such noncondictive material. Then setting up a base pulsed DC current, preferably one that is a resonant of the natural base frequency of that atom. Then intermittently utilizing a secondary higher or lower frequency to initiate a shattering effect on the molecule.


text:
Is there a name for the pair (c, R), where c is the stoichiometric coefficient of a reactant R in some chemical reaction?
I am programming a simulation, and I need a good name for a data structure that stores pairs $(c, R)$, where $c$ is the stoichiometric coefficient of a reactant $R$ in some chemical reaction. Is there a name for these pairs $(c, R)$ in chemistry?

In chemistry, often quantities get denoted by an index $i$, which then stands for some compound ($i=\text{A, B, C, ...}$).
Examples:

The stoichiometric coefficient $\nu_i$
The concentration $c_i$
The amount of substance $n_i$

The notion of grouping together the stoichiometric coefficient with the substance it uses is a novel one for me, and will most likely not have a name.
I suggest you use a name such as "coefpair" or the like and document your code accordingly.


text:
How do ligands and coordination complexes work in this case with copper (II) chloride?
What happens when Ammonia  is added to an (aq) solution of copper II chloride?
$\ce{CuCl2 + NH3 ->\; ?}$
I know that ligands bond with metals to form ion but I'm not sure about how they work in a reaction.

When solid $\ce{CuCl2}$ is dissolved in water, aquo complexes are formed.
$\ce{CuCl2_{(s)} + H2O -> [Cu(H2O)6]^2+ + 2 Cl-}$
Addition of ammonia results in a ligand exchange reaction to give tetrammine complexes.
$\ce{[Cu(H2O)6]^2+ + 4 NH3 -> [Cu(NH3)4(H2O)2]^2+ + 2 H2O}$

I get how NH3 as a ligand replaces (H2O)6 but how did you come uo with the products? [Cu(NH3)4(H2O)2]^2+ + 2H2O Thanks.


text:
What is the structure of a lipid and why is this example non-polar?
Lipids usually consist of 3 fatty acids and 1 glycerol. I looked up what they look like
So how is it not a polar molecule?  The $\ce{-OH}$ side has negative charge, and the hydrocarbon side chain has a positive charge.

How do you get that the side chain has a positive charge?

See Klaus's answer: Lipids are a class of biomolecule that are usually nonpolar or ambiphilic - they have a polar region (usually small, like the OH on cholesterol) and a nonpolar region (usually large, like the hydrocarbon part of cholesterol).

Why was his question downvoted?

Lipids usually consist of 3 fatty acids and 1 glycerol.

That's wrong. You're giving the definition of fats, which are just a subclass of lipids.
The structure in your question is the skeleton of cholesterol - without the stereochemistry.


text:
Energy required to break the "delocalised electron cloud" of benzene
I learned that when benzene undergoes hydrogenation, $\pu{208KJ}$ of energy is given off and thus $\pu{152KJ}$ less energy than what would be given off if Kekule's structure was correct. My confusion is that my book says that this $\pu{152KJ}$ of energy is required to break the delocalized electron cloud - this does not make sense to me.
How is the energy discrepancy of value the energy required to break the delocalized electron cloud in the benzene?

Let's sum it up :)

When cyclohexene is hydrogenated, $\mathrm{-120\, kJ\cdot mol^{-1}}$ is released.
When 1,3-cyclohexadiene is hydrogenated, $\mathrm{-232\, kJ\cdot mol^{-1}}$ is released. That's pretty close to 240, so it looks that the energies for the hydrogenation of the double bonds sum up. So far, we're cool.

Based on Kekule's model of alternating single and double bonds in benzene, we would now assume that the energy for hydrogenation is $\mathrm{ 3 \times -120 = -360 \, kJ\cdot mol^{-1}}$.
But as your textbook says it's only $\mathrm{-208\, kJ\cdot mol^{-1}}$.
We must conclude that Kekule was almost right, but the bonding situation in bezene is special. The energy of the molecule is much lower than that of a hypothetical cyclohexatriene. And when it's lower in energy, less is released upon hydrogenation. The difference of $\mathrm{152\, kJ\cdot mol^{-1}}$ is attributed to resonance stabilization.
Edit 1 If that's still a bit unclear with the signs of the energies, we can do some reverse experiments in mind.
Let's take cyclohexane and rip adjacent hydrogen atoms off to generate unsaturated molecules.

We need $\mathrm{120\, kJ\cdot mol^{-1}}$ to make cyclohexane.
We need $\mathrm{232\, kJ\cdot mol^{-1}}$ to make 1,3-cyclohexadiene.
If benzene would be cyclohexatriene, we would need $\mathrm{360 \, kJ\cdot mol^{-1}}$.

But we need less energy to make benzene, because it is more stable than a hypothetical cyclohexatriene. Actually, it is even more stable than 1,3-cyclohexadiene!
Edit 2

Benzene's actual structure is stable by 152 KJ w.r.t. Kekule's structure. That 152KJ is used for breaking down the delocalised electron cloud of benzene.

Guess what? The teacher is just repeating with other words what we have already worked on our own.
Imagine that you want to convert benzene to the hypothetical Kekule structure.
It's a process that would "consume" energy, $\mathrm{152\, kJ\cdot mol^{-1}}$ to be precise.
Why is that? It's because benzene has the delocalized $\pi$-system that aromatic molecules use to have. If you want to overcome that and push the stable molecule uphill and turn in into the less stable (= not having the delocalisation) cyclohexatriene, you have to do some work.

but why is $152KJ$ = the amount of energy required to break the delocalised electron cloud of benzene?

@Eliza I'm not sure that I understand, but did you consider the +/- signs for the energies?

@Eliza Did the penny drop now? ;)

Hmm... now I completely understand how the actual benzene structure is more stable than kekule's predicted structure :) However, I still have this confusion, I'll just quote my teacher "Benzene's actual structure is stable by 152 KJ  w.r.t. Kekule's structure. That 152KJ is used for breaking down the delocalised electron cloud of benzene"... The second sentence is what I have problems with :/

What is resonance stabilization... Many of my chemistry books refer to this but do not actually say what it is...

@Eliza Now you really have to read a bit on your own online about aromaticity :) We can't do that here all morning ;)

@Eliza - 152 kJ/mol is the amount of energy required to break the delocalised electron cloud of benzene, because we predict that 1,3,5-cyclohexatriene, which assumes NO delocalisation has 152 kJ/mol higher energy than benzene. So hypothetically to go from benzene to 1,3,5-cyclohexatriene you need 152 kJ/mol

@Brian: So, the energy of 152KJ hypothetically is the energy required to go from a delocalised pi-system as in the benzene to form 3 double bonds as in 1,3,5-cyclohexatriene (no delocalization) ?

@Klaus Warzecha : I guess aromaticity is a very huge chapter in chemistry :P I just started learning about the basics of benzene.. a long way to go :)

@Eliza A journey of a thousand miles begins with a single step :)


text:
The significance of the dissociation constant of an indicator
In the chapter acid-base, I learned about indicators and how they are actually weak acids.
This is the equation given in our book:
$\ce{HIn + H_2O <=> H_3O^+ + In^-}$
$K_{in}= \left(\frac{[\ce{H_3O^+} ][\ce{In^-}]}{[\ce{HIn}]}\right)$
When $\ce{HIn}$ and $\ce{In^-}$ are of different colors
My question:

Further in my book, the concentration of $\ce{H_3O^+}$ at which $\ce{[HIn]=[In^{-}]}$ is calculated. Apparently, this is the concentration of $\ce{H_3O^+}$ at which the indicator changes color and thus this can be used to identify end point of a titration. However, how is this the point where the indicator "changes" color, I mean isn't the indicator changing color throughout the titration as the concentration of $\ce{H_3O^+}$ is changing continuously as the alkali is added to the acid for example
Why do different indicators have different $pK_{in}$ i.e. "change color" at different concentration of $\ce{H_3O^+}$

Well, yes and no. You want $\ce{[HIn] = [In^-]}$ for a two colored indicator like methyl orange which is red in acidic form and yellow in the ionized form. Thus the solution would most orange when $\ce{[HIn] = [In^-]}$. However for phenolphthalein which is colorless in the $\ce{HIn}$ form and pink in the $\ce{In^-}$ form having $\ce{[HIn] = [In^-]}$ would overshoot the endpoint.

1. It all depends on the $\text{p}K_{\text{in}}$ of the indicator.
If we look at the Henderson-Hasselbalch equation:
$$\text{pH} = \text{p}K_{\text{a}} + \log \left( \frac{[\ce{A-}]}{[\ce{HA}]} \right) $$
You can see that when $\text{pH}=\text{p}K_{\text{a}}$, then logarithm part must equal 0. This occurs when we have equal parts of the protonated and unprotonated state, because when $[\ce{A-}]=[\ce{HA}]$, the ratio between them will be 1, and $\log 1=0$.
Every time you increase $\text{pH}$ by 1, to balance the equation, the logarithm must also be 1 higher, and this happens when the thing you take the logarithm of becomes 10 times larger. This means that for a small change in pH, we have a large change in the ratio between protonated and deprotonated forms.
Typically, with indicators, you say that the colour change happens when $\text{pH} = \text{p}K_{\text{a}} \pm 1$, because outside of that interval, you will quickly have almost purely $\ce{A-}$ or $\ce{HA}$.

If you follow this far, let's try an example with an indicator with a high $\text{p}K_{\text{in}}$

You put an indicator with $\text{p}K_{\text{in}}=10$ into pure water, which has a $\text{p}K_{\text{a}}=7$. From these values, we can see that water is a stronger acid than the indicator. This means that $[\ce{H3O+}]$ just in pure water will be high enough, that the indicator is largely protonated:
$$\text{pH} = \text{p}K_{\text{in}} + \log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right) $$
$$7=10+\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)$$
$$\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)=-3$$
Which means $[\ce{HIn}] = 1000*[\ce{In-}]$ - try plugging in 1/1000, 1/100, 1/1, 100/1 etc. into the logarithm on your calculator to see why this is so!
Anyway, we see that when the indicator has a $\text{p}K_{\text{in}}$ significantly higher than the $\text{p}K_{\text{a}}$ of the solution, it will mostly be protonated. Why? We will see this in part 2 of the answer.

Another example. Let's see what happens if we have a low $\text{p}K_{\text{in}}$ indicator

You put an indicator with $\text{p}K_{\text{in}}=3$ into pure water ( $\text{p}K_{\text{a}}=7$). This time the indicator is a stronger acid. This means that $[\ce{H3O+}]$ is not high enough to keep the indicator largely protonated:
$$\text{pH} = \text{p}K_{\text{in}} + \log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right) $$
$$7=3+\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)$$
$$\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)=4$$
Which means $10000*[\ce{HIn}] = [\ce{In-}]$ - Statistically, every time you find a protonated indicator molecule, you will have found 10000 deprotonated indicator molecules!

Titration scenario: acetic acid titrated with NaOH, using an indicator with $\text{p}K_{\text{a}}=10$

Acetic acid has a $\text{p}K_{\text{a}}=4.7$. The acetic acid is a stronger acid than water, so it will have increased the concentration of $[\ce{H3O+}]$. To raise the $\text{pH}$, we need to lower this concentration. As we add $\ce{OH-}$ to the solution, the following reaction occurs:
$$\ce{H3O+ + OH- <=>> 2H2O}$$
We are removing $\ce{H3O+}$ from the solution. But the acetic acid was only partially deprotonated, and what happens if the remove $\ce{H3O+}$ from the following reaction?
$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
Well, by the principle of Le Chatelier, acetic acid and water reacts further, to replenish the $\ce{H3O+}$. This goes on until the acetic acid is deprotonated to a large degree, and then the $\ce{OH-}$ very quickly raises the $\text{pH}$ of the water, since there is no acid to react with anymore.
When the $\text{pH}$ gets into the area of the $\text{p}K_{\text{in}}$ of the indicator, the colour change starts to happen. It happens swiftly, because we are not using very much of the indicator. This means that it doesn't take a lot of $\ce{OH-}$ to do to the indicator what we did to the acetic acid.
2. Indicators have different equilibrium constants because they have different acid/base energetics
Indicators have different values for $pK_{in}$ for exactly the same reasons as acids have different $\text{p}K_\text{a}$. It depends on the stability of the bond to the hydrogen that can be lost as a proton. If the bond is very unstable (as in HCl), the proton will lose its share of the electrons in the bond, and become free very easily, and the compound will have a low $\text{p}K$.
For a compound such as acetic acid, protons will become free, but protons from $\ce{H3O+}$ can also transfer back to the acetic acid. The more $\ce{H3O+}$ that exists in the environment, the larger the chance, that a deprotonated acetic acid molecule will collide with it and receive the proton.
For the indicator with a high $\text{p}K_{\text{in}}$, if there is a lot of $\ce{H3O+}$, then the indicator molecules will be re-protonated as fast as they can lose their protons. Thus, they overall stay protonated.
The equilibrium constants $K_{\text{in}}$, $K_\text{a}$, etc. basically tell you how likely the molecule is to shed its proton.


text:
What are angular and radial nodes?
Nodes are the points in space around a nucleus where the probability of finding an electron is zero. However, I heard that there are two kinds of nodes, radial nodes and angular nodes. What are they and what information do they provide of an atom?

Note that nodes are properties of orbitals and wavefunctions (and thus of electrons) and not exactly of atoms.

1. How to get the number and type of nodes for an orbital
As you said, nodes are points of zero electron density. From the principal quantum number $n$ and the azimuthal quantum number $\ell$, you can derive the number of nodes, and how many of them are radial and angular.
$$\text{number of nodes}=n-1$$
$$\text{angular nodes}=\ell$$
$$\text{radial nodes}=(\text{number of nodes})-(\text{angular nodes})$$
So each type of orbital ($s, p, d$ etc) has its own unique, fixed number of angular nodes, and then as $n$ increases, you add radial nodes.

Examples: First shell

For the first shell, $n=1$, which means the number of nodes will be 0.

Examples: Second shell

For the second shell, $n=2$, which yields 1 node.

For the $2s$ orbital, $\ell = 0$, which means the node will be radial
For the $2p$ orbital, $\ell = 1$, which means the node will be angular

Examples: Third shell

The third shell, $n=3$, yielding $3-1=2$ nodes.

The $3s$ orbital still has $\ell = 0$ meaning no angular nodes, and thus the two nodes must be radial
The $3p$ orbital still has one angular node, meaning there will be one radial node as well
The $3d$ orbital has two angular nodes, and therefore no radial nodes!

2. The difference between radial and angular nodes
Radial nodes are nodes inside the orbital lobes as far as I can understand. Its easiest to understand by looking at the $s$-orbitals, which can only have radial nodes.

To see what an angular node is, then, let's examine the $2p$-orbital - an orbital that has one node, and that node is angular.

We see that angular nodes are not internal countours of 0 electron probability, but rather is a plane that goes through the orbital. For the $2p_\text{z}$-orbital, the angular node is the plane spanned by the x- and y-axis. For the $2p_\text{y}$-orbital, the angular node is the plane spanned by the z- and x-axis.

Commenting on the last paragraph. Angular nodes 1) do have zero electron probability density and 2) are not necessarily planes.

The accepted answer has nice pictures, but has a couple of small factual inaccuracies in the last paragraph. In particular, it is not necessary for an angular node to be a plane (even though it frequently is). Here's a bit more maths to illustrate what's going on.
Atomic orbitals, which are one-electron wavefunctions, are split into two components: the radial and angular wavefunctions
$$\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_{lm}(\theta,\phi)$$
so-called because they only have radial ($r$) and angular ($\theta$,$\phi$) dependencies respectively. If either of these two components are zero, the total wavefunction is zero and the probability density there (given by $\psi^*\psi$) is also zero.
A radial node occurs when the radial wavefunction is equal to zero. Since the radial wavefunction only depends on $r$, this means that each radial node corresponds to one particular value of $r$. (The radial wavefunction may be zero when $r = 0$ or $r \to \infty$, but these are not counted as radial nodes.)
Analogously, an angular node is simply a region where the angular wavefunction is zero.* In the case of the p-orbitals, this is a plane. However, angular nodes do not necessarily have to be planes, as we'll see later.
The number of radial and angular nodes is dictated by the forms of the wavefunctions, which are derived by solving the Schrodinger equation. For a given orbital with quantum numbers $(n,l)$, there are $n-l-1$ radial nodes and $l$ angular nodes, as previously described.
An example
Let's take a look at one of the hydrogen 3d wavefunctions with $(n,l) = (3,2)$. $a_0$ is the Bohr radius of the hydrogen atom. We would expect $3-2-1 = 0$ radial nodes and $2$ angular nodes.
$$\mathrm{3d}_{z^2} = \underbrace{\left[\frac{4}{81\sqrt{30}}a_0^{-3/2}\left(\frac{r}{a_0}\right)^2 \exp{\left(-\frac{r}{3a_0}\right)}\right]}_{\color{blue}{\large \text{radial: }R_{32}}} \underbrace{\left[\sqrt{\frac{5}{16\pi}}(3\cos^2\theta - 1)\right]}_{\color{red}{\large \text{angular: }Y_{20}}}$$
All the individual terms in the radial wavefunction can never be zero (excluding the cases of $r = 0$ or $r \to \infty$ as I described earlier). Therefore, this orbital has no radial nodes. Surprise, surprise.
The angular nodes are more interesting. The angular wavefunction vanishes when $3 \cos^2\theta - 1 = 0$. Since $\theta$ takes values between $0^\circ$ and $180^\circ$, this corresponds to the two solutions $\theta = 54.7^\circ, 125.3^\circ$. Both of these solutions are angular nodes. This is how they look like:

The dotted lines are the angular nodes. They are not planes, but rather cones. They correspond to one particular value of $\theta$, which in spherical coordinates is the angle made with the positive $z$-axis; I have marked these angles on the diagram.
If you can obtain the forms of the wavefunctions, then it is easy to find the radial nodes. Mark Winter at Sheffield has a great website for this; just click on the orbital you want on the left, then "Equations" near the top-right corner.

Footnote
* If we stick to complex atomic orbitals, which are simultaneous eigenfunctions of $H$, $L^2$ and $L_z$, then the dependency on $\phi$ is always of the form $e^{\pm im\phi}$, which can never be zero, so angular nodes never arise due to $\phi$-dependency. However, radial and angular nodes are most commonly discussed in the context of real atomic orbitals, obtained by linear combination of the spherical harmonics. These do have angular nodes that depend on both $\theta$ and $\phi$, but they are often much simpler to express in terms of Cartesian coordinates $(x,y,z)$ (obvious examples being the $\mathrm{2p}_x$ and $\mathrm{2p}_y$ orbitals). The $\mathrm{3d}_{z^2}$ orbital is an exception that was deliberately chosen as an example because its angular nodes are not planes.

For the angular nodes would it be appropriate to say that theta and phi can form a line in 3D space? This line would also pass through the origin. Hence, all points along this line (towards the selected side of the origin, changing sides would change theta and phi) would have an angular wave function equal to zero. So this line forms a node.  I don't see how a plane is formed. Could you please illustrate how a plane could be formed as a node?

The set of all points with $\theta = \pi/2$ forms the xy-plane.

First of all, thats a wonderfully succinct explanation. Just one doubt: why is r=0 not counted as a radial node?

@satan29 sorry that this reply is so late. I am not sure myself and can only speculate. First of all, the solution $r = 0$ is a trivial one, so isn't very interesting to begin with. However, there is another issue that I can think of: $r = 0$ doesn't correspond to a "surface" or a "region" of space, it refers to a single point. If you assume that a node means some region of space, then $r = 0$ won't meet that definition.

@orthocresol, If $Z_{eff}=1$, then $Z_{eff}^{\frac32}=1$. But you have written in the radial wave function $a_0^{-\frac32}$ How is that? Would you explain?

@DhamnekarWinod First of all, your question is not clear. What is the significance of this quantity $Z_\text{eff}^{3/2}$, and why should I care whether it’s 1 or not? Are you reading a different form of the radial wavefunction, which doesn’t contain any $a_0$’s, which suggests that it’s expressed in atomic units? In that case, $a_0$ is set to 1 and there would not be any discrepancy. The form of the wavefunction I wrote doesn’t use atomic units.

@orthocresol, Haven't you  used the formula and other relevant information for 3d orbital given here $\rightarrow$ https://winter.group.shef.ac.uk/orbitron/atomic_orbitals/3d/3d_equations.html If not, which formula you have used? Let me know

@DhamnekarWinod Yes, the formulas on that website use atomic units. It is exactly the same as in my answer, except that $a_0$ is taken to be unity, so on that website you won't see any terms of $a_0$. The specific term $a_0^{-3/2}$ that I have here isn't related to the $Z$ term; it's merely a normalisation factor to make sure that the entire wavefunction integrates to $1$. You can find the forms I used in e.g. Griffiths, Introduction to Quantum Mechanics 2ed. Sorry, but I'm not keen on explaining this further in the comments. If you have any more questions, please ask a new question.

If you vibrate a piece of rope then it is quite easy to show that nodes will appear, and when standing waves are produced the nodes are fixed in space and time. Between the nodes the rope oscillates up and down. The n$^{th}$ harmonic has $n-1$ nodes. The more nodes there are, the greater is the energy is required to produce them. In a vibrating rectangular membrane nodal lines are produced and in a vibrating disc nodal rings. The nodal lines, where the amplitude is zero separate normal vibrational modes.
Similarly with atoms and molecules. Nodes are points where the wavefunction crosses zero, and its amplitude is zero. Where the wavefunction gradually approaches zero (at the origin or infinity) are not considered nodes. Nodes naturally appear in all solutions of the Schrodinger equation, even for simple systems such as a particle in a box. Not all wavefunctions have nodes, the lowest energy one does not, (e.g. the S orbital in atoms, zero point vibration and zero rotation in molecules, lowest MO in a molecule). The larger the number of nodes a wavefunction has the greater the energy eigenvalue. Some pictures of wavefunctions and nodes has been given in the answer by @Brian.
There appears to be no special significance to a node, it arises purely out of the solution of the Schrodinger equation and is a consequence of the solutions (hence boundary conditions) we choose to obtain quantisation. Without nodes it is difficult to see how wavefunctions describing different energy eigenvalues could be constructed.
Briefly extending the question to include bond formation. In forming bonds from atomic orbitals the parity of the wavefunction is important. Upon interchanging coordinates, i.e. inversion operator, odd parity (ungerade, u) produces -1 itself and even parity (gerade, g)leaves the orbital indistinguishable. The node in a p orbital causes it to have odd parity, as s orbital even. The symmetry (parity) determines whether two orbitals can combine to form a bonding orbital or anti-bonding one, e.g. $\sigma$ or $\sigma ^*$ or $\pi$ or $\pi ^*$. Anti-bonding orbitals always have nodes. Nodal planes are also important in Diels Alder addition and similar reactions as here symmetry is important. In spectroscopic transitions in molecules, the nodal structure of the molecular orbitals  determine, via symmetry, which transitions are allowed or forbidden.


text:
How do I work out the amount of H+ ions in a sodium hydroxide and water solution?
I know the $pH$, pressure, temperature and concentration of the solution. Is there a method I can use to work out the amount of $\ce{H+}$ ions?

i.e the dissociation of the solution

The pH can be directly related to the concentration of $\ce{H3O+}$:
$$\text{pH}=-\log{[\ce{H3O+}]}$$
$$\Downarrow$$
$$[\ce{H3O+}]=10^{-\text{pH}}$$


text:
How is an aqueous Ag+ the same as [Ag(H2O)2]+?
$\ce{Ag^{+}_{(aq)}}$ can be written as $\ce{[Ag(H2O)2]+}$ but how would you know that it bonds with 2 water molecules?
I don't get when ligands randomly form 6, 4, 2 bonds.

There is no way to absolutely predict the co-ordination number in any complex entity, without knowing say for example, magnetic or geometric properties of the complex. All we can do is give a probable configuration based on intelligent guesses. The co-ordination number depends on the characetristics of the ligands and the central ion.
Silver ion, $\ce{Ag+}$ usually forms complexes with Co-ordination Number two. This is partly because the silver ion is small sized, making it difficult to accommodate higher co-ordination numbers around itself owing to increased ligand-ligand repulsion, and partly due to the low charge on silver ion which makes it difficult to hold more than two ligands together. Therefore, in most of its complexes, silver ion co-ordinates with two ligands to form a linear entity. But to confirm this, you need to study solvated silver ions spectroscopically, magnetically or geometrically (diffraction), without which the above arguments can only give you, at best, a good guess.
Other examples include $\ce{[Ag(NH3)_2]+}$, and the polymeric $\ce{[Ag(en)]ClO4}$, both of which have co-ordination number of the silver ion 2.

Cu is directly above Ag in the periodic table, so surely Cu2+ is smaller in size than Ag+, both from the charge and electron shell perspectives. If true, it'd be hard to argue Cu2+ can have six ligands in a complex while a larger Ag+ can have only two due to steric hindrance


text:
Why does HCl ionize in water?
Why does the Hydrogen proton "prefer" to join the $\ce{H2O}$ molecule instead of remaining bonded to the chlorine atom?

Chlorine is a very large atom compared to hydrogen, which means the overlap between the hydrogen $1s$-orbital and the chlorine $sp^3$ orbital is poor, which translates into a low bond energy.
Additionally, chlorine is very electronegative, which helps pull the H-Cl bonding pair onto the chlorine atom. The chloride anion solvation in water also plays a big part.
It's simply because it's energetically favourable. Maybe someone knows in greater detail than I do.

Your answer is already more descriptive than I expected. I was wondering if there was a better answer than simply "being energetically favourable". Saying that water is a stronger base than the chloride ion is just a way to mask that. So is mentioning HOMO/LUMO orbital energies. Of course, all answers boil down to potential energy minimization (which itself boils down to entropy maximization), but the difference is how much feel you can get for why it happens, and how much you can extrapolate to other systems from it. Perhaps someone can still find something to add.


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Insoluble Group I/ammonium salts
Most solubility references (e.g. this one) say that there are "few" insoluble salts of alkali metals.
Are there any exceptions?

Though I don't know the details, some types of uranium ions can precipitate sodium and ammonium cations when added to water as their respective diuranates. If I recall correctly, a qualitative method to identify sodium in aqueous solution is to add potassium uranyl sulphate in the absence of any other non-group 1 cations. I'm curious what other insoluble salts exist, especially for the heavier alkalis.

https://chemistry.stackexchange.com/questions/82328/are-there-any-sodium-salts-that-are-insoluble-in-water

The link given is dead. Please update.

Lithium carbonate has poor water solubility.  Cesium triphenylcyanoborate is insoluble in water.

They would advertise $\ce{NaBPh4}$ as a reagent to determine $\ce{K+}$ gravimetrically.  Old-timers may remember the feared Kalignost.  $\ce{XBPh4-}$, for X=$\ce{Rb+, Cs+, NH4+,Tl+}$ is similarly insoluble.
Thanksfully, now there is methods like AAS, and we plague the kids with that in Instrumental Analysis/Analytical Chemistry classes instead.

Perhaps the simplest case involving limited (<0.1 molar) solubility is lithium fluoride, whose solubility is only about one part in 750 by mass at 25°C (about 0.05 molar).

There are some cations whose Group I/ammonium salts are not soluble in water. For example, (thanks Nicolau) the diuranate of ammonium is insoluble in water.
Additionally, (as I just found) the hydrides (H-) of alkali metals are not soluble, it seems, in any room-temperature liquid solvent. They hydrolyze violently (http://goo.gl/Dl8Bd4), and the hydride ion has never been observed in solution.

I'm not sure whether it's right to say that substances that undergo solvolysis aren't soluble. Technically, they do dissolve, they just react immediately afterwards.


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Determining the lowest rank non-vanishing multipole moment
I have been told that the multipole moments of rank 0,1,2,3,...n have respectively 1,3,5,7,...2n+1 components which transform like the spherical harmonics of rank 0,1,2,3,...n. How do I use the character tables to determine the lowest rank non-vanishing multipole moment for the following molecules: $\ce{H2O, BF3, CF4, CO2, SF6}$?


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Why does glass shatter when heat is applied to it?
When I poured hot tea into a cup, the cup shattered after about three seconds. The cup was made of glass. It was pretty thin. How did the glass and the tea react in terms of chemicals. This is a bit odd because wasn't heat applied to sand in order to make glass, shouldn't the glass break up into the reactants that made it?

This is not a chemical process.
Objects typically expand when heated. Think in railway profiles in the summer, think in loosening a stuck nut by heating it with a blowtorch.
However, when the material is under stress and a bad heat conductor, unevenly heating will result in cracking.
While it is a pity for the cup, the breaking of glassware to thousands of rounded (!) pieces upon uneven or punctual (mechanical) impact may be exactly what you want under different circumstances. Think in tempered (toughed) glass used in windows of railroad passenger cars.
On a side note, there's the Bologna Bottle. In the making, these bottles are slowly cooled on the outside - but very rapidly on the inside, which maximizes the stress.
As a consequence, you might use the bottle as a hammer to drive a nail into something, but beware if you drop something in the bottle. Here's a video to demonstrate the effect. (Be warned that there's a lot of screaming kids in it.)

Similar to the bologna bottle, see also https://en.wikipedia.org/wiki/Prince_Rupert's_Drop


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What aspects of a molecule's chemical makeup determine which portions of it will react?
Is there a specific way to know how two or more arbitrary molecules will react?
An example that prompted this question: $\ce{(HOC6H4)2CMe2 + 2 NaOH → (NaOC6H4)2CMe2 + 2 H2O}$ (from here). How can you know how those molecules will react? Is it just a bunch of substitutions, with certain atoms in the reacting molecules reacting better than others? What factors would determine which atom in the molecule would react with the other molecule?

A full answer to this question is like the table of contents for an entire course in synthetic chemistry.

@matt_black True :) That's why I didn't go into great detail ;)

Let me answer with a question (or two):
How do you identify in an electric circuit which parts are just the wiring and which are functional parts?
How's that for a mechanical engine? Can you tell cogwheels and cranks from "just the basic structure"?
It's all about pattern recognition in the first to identify the functional groups.
Next step is to know how they react under certain conditions. As a rule of thumb, a group with a high electron density will probably react with something more positively charged (think plus and minus attract each other).
But it takes quite a while and learning to predict exactly what will go on exactly in the reaction. But that not different from any other scientific/technical discipline.
I tried to keep this as general and simple as possible ;)
Edit
Let me work you comment in:

Generally molecules don't contain ions though

Says who? :D Molecules may contain ionic units, think in zwitterions of amino acids, think in all the mesoionic sydnones, münchnones and whatnot. Well, no. Forget about the special cases ;) But if it has more than one atom and is electrically neutral in total it is a molecule. At least in the less rigid sense of the definition.
But yes, the electronegativity is a good point.
Forgive me that I'm not going into detail why the example with fluorine and sodium is a not the best one, but in principle, you're right: Such a setup would allow to form a new $\ce{C-C}$ bond between a carbon that is negatively polarized (the one bearing the sodium) and another one that is positively polarized (the one bearing the halogene).

Generally molecules don't contain ions though, so you mean something like electronegativity? So if a molecule were to contain fluorine, and a second one sodium, would it (probably) react with each other, instead of with any other elements in the molecule, because of the difference in electronegativity, and therefore the increased electron density (the fluorine pulling electrons towards itself)? (Just trying to explain it to myself with the limited stuff I've learned in class.)

That's surprisingly simple then (it's still complicated, but it uses the same mechanisms involved in more "basic" chemistry)  Thanks a lot!

@Goocholplex I tried to keep it as simple as possible without being wrong ;) But if you found it helpful and inspiring please consider to "accept" the answer. I'd appreciate that :)


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Photoelectric effect and electron loss
I've read that when a certain amount of UV light is shone on a metal surface, electrons are ejected (the photoelectric effect).
Are these electrons from the metal atoms themselves? and in case they are - what happens to the metal atoms seen as they have lost an electron? (Are they chemically different?)

You may find http://en.wikipedia.org/wiki/Photoelectric_effect and especially http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html illuminating.

It is NOT the amount of light (number of photons), it is the energy of each photon (its frequency).  When the energy of a photon exceeds the vacuum work function of the surface, an electron is unbound.  There obtains a charged surface and a free electron.  Unless the surface is part of a circuit (e.g., grounded or connected to the cathode), the next electron out must overcome the vacuum work function plus the electrical potential.
The electron is emitted from the conduction band of the bulk metal surface.  The positive charge is delocalized across the surface if it is of constant radius of curvature.   Surface electric field divergence is inversely proportional to radius of curvature.  Unless the surface is very smooth you can get cold cathode emission or electrical breakdown of the encapsulating medium.


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Is pitchblende the same as Uraninite?
Different websites say different things on the matter.
 For example:
 This Website   Holds that pitchblende is also known by the name Uraninite.
This Website   Holds that pitchblende is a type of Uraninite.
My question is, which website is correct? I have always known the first site to be a reputable source, so I assume that this is a contentious topic.

If you search online for Uraninite then you find various websites saying that uraninite is called pitchblende.
On wikipedia you find :

Uraninite is a radioactive, uranium-rich mineral and ore with a
  chemical composition that is largely UO2, but also contains UO3 and
  oxides of lead, thorium, and rare earth elements. It is most commonly
  known as pitchblende

even the word pitchblende will automatically redirect you to the uraninite wikipedia page .
Another trustworthy website is mindat.org : uraninite information can be found on this link:http://www.mindat.org/min-4102.html (It seems that there are various names associated to the mineral):

Re-named in 1845 by Wilhelm Karl von Haidinger for its composition
  containing uranium. Originally recognized before the element uranium
  was known and called "schwarz beck-erz" by Franz Ernst Brückmann in
  1727. Subsequently described with several names including "pseudogalena" and "pitch-like zinc-blende" by Johan Gottschalk
  Wallerius (Vallerius) in 1747. Named pechblende by Axel Cronstedt in
  1758. Named "uranerz" by Martin Klaproth after he discovered the element uranium in this mineral. Many spelling variations of the above
  names known.


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what are reduction oxidation reactions?

in the picture, the product is a neutral material, but in the illustration of the molecular structure, the compound is shown in its ions, why we can't show it in its ions even in the equation or if we can't, why in this molecular structure it is shown as ionic although it is neutral?

Your questions involve a lot of chemical concepts.
An oxydo-reduction is reaction where electrons are exchanged between species including or not molecular rearrangement.
You totally can write down any salt in the following way, emphasizing ion charges:
$\mathrm{NaCl}_{(s)} \equiv \mathrm{Na^+Cl^-}_{(s)}$
As long as you specify the phase, because when writing down an ionic form the chemist might assume that it is solvated into water. To underline it you may write the following:
$\mathrm{Na^+Cl^-}_{(aq)}$
Further more solid generally have numerous possible distinct phases, therefore you may need to specify which crystalline structure the salt have.
The following equation is explicit:
$2\mathrm{Na}^{(0)}_{(s)} + \mathrm{Cl_{2(g)}} = 2\mathrm{Na^+Cl^-}_{(s)}$
And underlines that an electron is withdrawn from metallic sodium in order to produce chlorine. Anyway, this equation is global balance this it hides many details of the complete and complex chemical process, especially the formation of the crystalline structure. And it does not provide any information on conditions of reaction.

Simply, Oxidation and reduction is the reaction in which one of the species donates eletron and another species accepts the electron .
As for the reaction involved in formation of Na Cl(s) , the reaction mechanism is as
Na will release an electron from its outermost orbit to attain octet state forming Na+ cation.Likewise, the Cl atom will accept the electron forming Cl- anion to attain octet state. As a result, a ionic bond will be formed between them which is indeed a very strong chemical bond and finally after multiple bonding between them a crystalline structure will be formed.
Similarly even it is neutral it is presented in ionic form in its molecular structure mainly to specify that there is ionic linkage between these two atoms


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Is pyrite (FeS₂) an ionic or a covalent compound?
I have searched all over the web and found a lot of diverse explanations, but none of them are concluding exactly whether $\ce{FeS2}$ (solid - pyrite) is a covalent or an ionic compound.
From electronegativity, it should be covalent as the $\Delta\chi=0.7$ which is less than $1.5$ and thus said to make covalent bonds and therefore be a covalent compound.
From the definition of ionic bonds, which are bonds between a metal and a non-metal element (whereas covalent bonds are bonds between non-metal elements), it should be an ionic compound.
Does someone know which of those is 'true', or better, if there is another, more detailed explanation?

http://books.google.com/books?id=L-1K9HmKmUUC&pg=PA246&lpg=PA246&dq=FeS2+pyrite+ionic+covalent&source=bl&ots=zVuxXXU_k-&sig=BAfe1lWsyxO4YjjU7tLDIJxYVFk&hl=en&sa=X&ei=XXMKU5jfHYm5iAeN7YGYCQ&redir_esc=y#v=onepage&q=FeS2%20pyrite%20ionic%20covalent&f=false

May be of interest (unfortunately behind a paywall) http://pubs.rsc.org/en/content/articlelanding/2014/sc/c3sc52977k#!divAbstract

You seem to have fallen into the trap of thinking that ionic and covalent bonds are fundamentally different. They are not - they are just two ends of a spectrum, which has an arbitrary division somewhere in the middle into an ionic and covalent regime. This is explained in the answers to this question.
In the case of pyrite we have a relatively hard cation, with a small ionic radius and charge of +2, and a rather large anion, with a charge of -2. Therefore there will be a significant degree of covalent character in the Fe-S bonds due to the polarising effect of the cation on the anion. This is confirmed by experimental results and theoretical calculations which suggest that the charge on Fe is about +2/3 and the charge on S is about -1/3. This is significantly less than the expected charges of +2 and -1 from a purely ionic model and so indicates that there is significant sharing of electrons. This is supported by measurements and calculations of the electron density, which show significant electron density between the atoms.
Reference: http://pubs.rsc.org/en/content/articlepdf/2014/sc/c3sc52977k

Your definition of ionic bonding is incorrect as non-metallic polyatomic cations (e.g ammonium) form ionic compounds. Ionic bonding is the electrostatic attraction of cations and anions forming a regular giant lattice. Transition metal compounds do not have simple close packing ionic lattices as they prefer coordination compounds due to the energies of d-orbitals.
Here is the pyrite structure:
Note that the iron has octahedral geometry and sulfur has tetrahedral. These are the most common geometries in coordination chemistry.

but couldn't you simplify Iron(II) Sulfide(or pyrite) to Fe^2+ ions and S^2- ions ionically bonded like how you can simplify Iron(III) Oxide into Fe^3+ ions and O^2- ions in ionic bonds?


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How to calculate the start concentration for a first order reaction?
The half-life of the first order reaction, A → products, is 33.8 s. What was the original concentration, [A]O in mol/L, if after 2.5 minutes, the concentration of A is 0.0796 mol/L? Round your answer to three significant figures.
I think the formular is 1/(A)t = kt + 1/(A)0
So I tried 1/33.8s = (0.0796)
so it = 1/0.0796.
I know this is not correct, but I think I'm using the right formula.  I looked back at my notes and nothing is really making any sense, but this formula.

For a first order reaction $\ce{A ->[k]P}$, the rate constant is given by
$$k[A] = - \frac{d[A]}{dt} $$
So, at any time $t$, the concentation $[A]_t$ is
$$[A]_t = [A]_0 \cdot e^{-kt}$$
At the half-life time $t_{1/2}$, $[A]_t = \frac{1}{2}[A]_0$, isn't it?
Would you agree that we can calculate $k$ using $k = \frac{ln 2}{t_{1/2}}\ $?
Knowing $k$ and $[A]_t$, calculation of $[A]_0$ doesn't seem impossible now ;).


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How could I create a machine that as rapidly as possible creates greenhouse gases from air?
For a very large-scale project, I am trying to create a greenhouse gas demonstration device, which rapidly produces greenhouse gases from atmospheric air using only electricity, or any chemical reactions or chains of chemical reactions that can attain their input energy from electricity. At the intake would be a very powerful fan which rapidly forces air through the system, to speed up the process. How would it be done?
UPDATE:
I was thinking nitrogen oxides would be best, considering the atmosphere is approximately 78% nitrogen and 21% oxygen. But what would be the fastest way to produce them, assuming price is not really an issue?

Could you specify which greenhouse gases you are interested in? As you can see by the answer already given, there is quite a range of gases that are (or can be) considered greenhouse gases and I'm guessing you have some particular ones in mind?!

Start a fire. That will convert the oxygen in air into CO2.

I was looking at oxides of nitrogen. They are rather effective greenhouse gases, and the air is 78% nitrogen, 21% oxygen already, making both elements abundantly available for reactions. I just need to find a reaction pathway that will very quickly convert them into nitrogen oxides.

Start with the Haber process. Convert the ammonia from the Haber process to nitric oxide by oxidation with Pt. Use nitric acid to produce ammonium nitrate. Heat the ammonium nitrate. You will need a lot of money though. A lot.

Keep on saving the world, but stay away from chemistry.

Current concentrations of $\ce{CO2}$ and $\ce{CH4}$ in the atmosphere are around 390 ppmv and 1.75 ppmv, resp. These concentrations and the wish to obtain large quantities of both gases rapidly seem mutually exclusive, unless you use (at least one) additional carbon source.
Taking the options for interconversion into account, $\ce{CH4}$ in pressurized bottles would be an obvious choice ;).

I was thinking more along the lines of nitrogen oxides, considering the atmosphere is 78% nitrogen and 21% oxygen anyway

The most powerful Greenhouse Effect gas of all is water.  Run a vaporizer to 100% humidity

http://disc.sci.gsfc.nasa.gov/featured-items/images/fig2_air_infrared_spectrum
to 50 microns
http://earthobservatory.nasa.gov/Features/Tyndall/Images/absorption.gif
to 15 microns, near IR
The Greenhouse effect incremental champ is a tank of $\ce{SF6}$ (insulation and arc quench for high voltage electrical switches and transformers).  Note that industrial grade is often contaminated with $\ce{SF4}$ that is extremely toxic.  However, your job is already being done for you,
http://www.afeas.org/greenhouse_gases.html
http://en.wikipedia.org/wiki/Greenhouse_gas
http://www.eia-international.org/explosion-of-super-greenhouse-gases-expected-over-next-decade
HFC-23, by-product of HCFC-22 manufacture. is 14,800 times $\ce{CO2}$.

I would think that carbon dioxide and methane would be your options due to their abundance. However, they are very difficult to harvest because of their non-reactivity. There are research groups working with metallo-organic frames/cages that capture carbon dioxide. However, making these takes time and significant funds. In addition, they are not that great when capturing carbon dioxide from atmospheric air. You have to take into account that despite of their abundance, the amount of $\ce{CO2}$ and $\ce{CH4}$ in the atmosphere is still small. I think the frame works will first be installed on factories funnels which emit large quantities of carbon dioxide.

You could, however, produce large quantities of sulfur hexafluoride. May be easier.

Nevermind the comment, I just remembered that it can asphixiate people. It's non-toxic though.

Not sure what you are specifically looking for, but I would look into the Solvay Process (At least part of it). An old well known way to produce CO2 from an inorganic source (Which you have tagged). Its fairly simple, the Solvay process is a production process for creating Sodium Bicarbonate, One of the steps is to create CO2 from limestone. This is done from heating up limestone to about 1000 degrees C, which when I did the Solvay process in chem lab we used a bunsen burner but could be done with a good electric burner/furnace.
If your measuring CO2 in PPM in the experiment then this would work.

The issue is this process does not use just air

Try an electric air compressor with extremely high performance in the PSI column. liquefy the air and separate the gases by boiling points. That is how it is done now.


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Van der Waals nonspecific interaction definition
Every definition I see of Van der Waals interactions claim that they are short-range, nonspecific interactions between two chemical species. What does "nonspecific" mean in this context?

I would assume that it means that all elements and molecules have them. I may be wrong though

The definition of the van der Waals force that I like (and apparently so too do the folks who have contributed to the Wikipedia article on the topic) is much more *specific about what is and is not a van der Waals force. This definition originates in the International Union of Pure & Applied Chemistry Compendium of Chemical Terminology
The van der Waals force is the sum of the attractive or repulsive forces between molecules (or between parts of the same molecule) other than those due to covalent bonds, the hydrogen bonds, or the electrostatic interaction of ions with one another or with neutral molecules or charged molecules.
Thus, a van der Waals force is a catch-all term to include all other very weak interactions between atoms and molecules:

Dipole-dipole interactions between permanent dipoles
Dipole-induced dipole interactions
Induced dipole - induced dipole interactions (London dispersion)
$\pi-\pi$ stacking
Halogen bonding
Repulsive forces derived from atoms trying to be inside another atom's van der Waals radius (Steric strain and torsional strain)

Your answer is not full, it does not answer why  van der Waals force is "nonspecific" forsce.

It's nonspecific in that it is any force that is not one of those specific three forces.

Don't forget chalcogen bonding ($\sigma$-hole bonding), $\pi$-hole bonding (snicker), pnictogen bonding, $n\rightarrow\pi$* interactions, hydrogen bonding... the list goes on and on.

Yes, the IUPAC definition is incomplete since newer interactions are emerging.

There are 3 forces that could be named as van der Waals' force (see wikipedia article)
What is common between them? The common between them is that these interactions are electrostatic interactions between dipoles of different nature (permanent dipoles, induced dipoles). And electrostatic interaction is not specific: there is no space specificity, or chemical specificity.
An electrostatic interaction between dipoles according to Coulomb law decays as $$\sim 1/r^6$$, that is why van der Waals' forces (Keesom force,Debye force,London dispersion) force are short-range.

I vaguely remember London and Debye forces proportional to $\frac{1}{r^6}$.

See here for an excellent description of the distance dependence of van der Waals interactions.

My mistake, I've corrected.


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What chemicals cause the smell of a septic tank?
I am wondering what chemicals exactly contribute to the smell of a septic tank system. I read about $\ce{H2SO4}$, $\ce{NH3}$, but I am not sure. Is methane also a factor or is that odorless?

According to the Wikipedia article Septic tanks, as septic tanks are generally anaerobic environments, pungent smells from the formation of hydrogen sulfide $\ce{H2S}$, which is sometimes referred to as 'rotten egg gas'. As you stated, ammonia $\ce{NH3}$ can be produced - also with a distinct smell.
Also, according to the Wikipedia article, fermentation could occur, producing the odourless methane and carbon dioxide as well as phosphates.
Incidentally, $\ce{H2SO4}$is the chemical formula for sulfuric acid.

While it is true that $\ce{NH3}$ and $\ce{H2S}$ can be produced (larger amounts collect usually only in abnormal conditions), my nose suggests that they are not the main cause of the typical smell (with the caveat that gas mixtures as well as different concentrations can have totally different smell from the compounds). My guess would be leaning towards small amounts of thiols and volatile amines along the lines @KlausWarzecha suggests. Also do not forget butyric acid.

Some of the "fragrances" are formed in the degradation of different amino acids.
Amaterasu has already described that hydrogen sulfide is an important component in the septic tank perfume. It is formed in the degradation of the sulfur-containing amino acids methionine (1) and cysteine (2).

In the anaerobic degradation of methionine, the initially formed methylmercaptan ($\ce{CH3SH}$, 3) is methylated to dimethylsulfide ($\ce{CH3-S-CH3}$, 4). Further microbial transformations of 3 and 4 eventually yield $\ce{H2S}$ .

The degradation of the amino acid tryptophan (5) furnishes 3-methylindole (skatole, 6), the aroma of feces.
Just for the completion of the sensoric enjoyment: Other amino acids, such as ornithine (7) and lysine (8) form 1,4-diaminobutane (9) and 1,5-diaminobutane (10), respectively upon decarboxylation (= loss of $\ce{CO2}$).

The trivial names of these diamines are putrescine and cadaverine, which suggests another odor than that of roses and lemons ;)
I'm however not sure if the last two compounds are actually formed in septic tanks.

The smell coming out of a septic tank depends on (1) what goes into the septic tank, (2) how long the material has resided in the tank, (3) where the septic tank is located, (4) the current weather/climactic conditions, and (5) how well the tank is maintained.  The point being, at any given time or location the smell of a septic tank can change due to the emission of a different combination of chemical compounds.
Interestingly, if hydrogen sulfide is overwhelmingly the dominant compound, that likely means the septic tank is being fed sulfur water. This compound is has a very distinct smell of rotten eggs.
Relative to other malodorous compounds, hydrogen sulfide and ammonia have a relatively high concentration.  The high concentration is certainly an issue from the air pollution standpoint,  however, these compounds have a relatively high odor detection threshold compared to some of the other malodorous compounds.
Fresh waste may have odors dominated by indole and skatole which, while not very high in relative concentration, have lower odor detection thresholds.  Because the odor detection threshold is low, a very small quantity of these compounds gives fresh waste its distinctive odor.  Other compounds include the methly sulfides, butyric acid, valeric acid, and phenol.  (It is worth noting indole is a highly unusual compound in that at high concentrations it smells like fecal matter, but at low concentrations it smells like flowers.)
Most people do not notice the odor of a septic system until something goes wrong.  If something is wrong then an unpredictable release of many oddball malodorous organics may be caused by the incomplete digestion of the fresh waste by microorganisms living in the septic system.  If the septic system is operating correctly, then the contents has time to be thoroughly digested by the microorganisms living in the system.  The system should be releasing compounds in the most broken down forms, which would be water, carbon dioxide, methane, hydrogen sulfide, and ammonia.  If the septic system is functioning properly, the odor from the system should be minimal.
-Also, as a side note some septic systems prevent the compounds emanating from fresh waste from coming into a house through the use of traps.  These traps may be in floor drains.  Make sure all traps draining into a septic system are periodically flushed with water.
Methane has no odor.  This may be confusing because malodorous compounds, like dimethylsulfide, are added to natural gas, which is mostly methane, to give its unique smell.  This smell has the purpose of leak detection.
Sulfuric acid and ammonia are used for clog removal and cleaning.  If someone who cannot tell a clog from a backed up septic system tries to throw these chemicals down the drain as opposed to paying someone to pump the system, then all these chemicals may do is kill the micro-organisms.  If the microorganisms die then the system may emit the odor of the chemicals as well as undigested fresh waste.


text:
Temperature in the Maxwell-Boltzmann Distribution
I am attempting to generate the probability for the specific speed of a molecule using the Maxwell-Boltzmann distribution, but I cannot decide on which temperature I should use in the equation.
For example, I have a very, very cold liquid (let's say 70 K) at room temperature (we'll call that 296.15 K), but would I use the temperature of the liquid in the equation, or should I use the temperature of the air around the liquid?

from wiki (MB distribution) "Particle in this context refers to either gaseous atoms or molecules, and the system of particles is assumed to have reached thermodynamic equilibrium." Therefore, liquid with different temperature than environment is not very suitable for Maxwell-Boltzmann distribution, though it might be possible for a very crude estimates (temperature gradually varing from 70K to 296.15K)

I thought that Maxwell Boltzmann distributions were applicable in systems in equilibrium. In this case, you certainly are not in equilibrium. The calculation would bequite complex. I dont think you can calculate that with just temperature. I second what user26143 said. However, please let me know if you find something that counters this notion. I am interested in it.

I was indeed working on a crude estimation, and the liquid is certainly not going to make it all the way up to room temperature under the circumstances that I'm in

The problem is what is your system? If you are considering the liquid itself, then you direct apply 70K to the equation. But if you include the room, then you have an inequilibrium system and you cannot do it.

If you are assuming a pseudo-steady state condition (where the bulk of the liquid is in thermal equilibrium with itself for the timescale of interest), and you want to study the bulk molecules of the liquid, then you want to use the temperature of the liquid.
Now, if you are going to be studying the surface of the liquid, using a Maxwell-Boltzmann distribution is much more questionable - it definitely isn't strictly correct, at least on time scales of any substantial length, but you might be able to get away with it for rough approximations.
If you gave some more details as to the specific application, I could probably give you a better answer.


text:
Reaction of ethene with ammonia and sodium
Is the following sequence of reactions correct?
$$\ce{CH2CH2 ->[][\ce{NH3},\,\ce{Na}] NaHC=CHNa}$$
$$\ce{NaCH=CHNa + 2CH3CH2Br-> CH3CH2CH=CHCH2CH3}$$

Have a look at the Birch reduction.

@tschoppi If it wasn't ethene but methylpropene and a branched hydrocarbon, I'd suggest sulfuric acid or hydrofluoric acid to achieve the alkylation. But the double bond would be gone then.

@KlausWarzecha Well, after seeing ($\ce{NH3}$, $\ce{Na}$) I was thinking Birch and nothing happening afterwards... I didn't realize that the goal of this might be "alkylation of ethene" ;)

@tschoppi No problem!

No, because for ethene, $\mathrm{p}K_\text{a} = 44$. You won't be able to abstract even one proton.
Maybe, you were thinking in ethyne ($\ce{HC#CH})$? In that case, you are right! The bis-alkylation would work with sodium amide as a base!
But is the simultaneous abstraction of both protons the most plausible reaction then?


text:
how to add a methyl group to a straight carbon chain
I mean like how can I convert ethane into propane, ethene into propene and like this further on

Direct ethene to propene (ETP) conversion seems possible at 400 °C on mesoporous silica (M41) loaded with nickel ions.

Note that I didn't know that either until now and just found it in a cursory search!

Why would you want an expensive process to convert cheap molecules into other cheap molecules?  Brominate ethane, make the organo-lithium, displace methyl iodide. Epoxidize ethylene, react with methyl lithium, eliminate to the olefin. Silly.
Doing something cheap, easy, and fast on a huge scale with unwanted methane from oil wells to form a high energy density liquid for pipeline conveyance will make you rich. Methanol is marginal - and still can't be done.  Perhaps methane plus air into titanium-silicalite (plus hydrogen peroxide, or cobalt- or vanadium-doped plus air) then that mixed product through ZSM-5 into aromatic gasoline.  That could be exciting in various ways.
http://dx.doi.org/10.1016/j.apcata.2008.04.020
http://chemelab.ucsd.edu/methanol/memos/final.html

Because I am in high school and I have to it on paper not in reality, this is on of the higher order thinking skills question that my school have to me.

A. M. Schwartz, M. Petitjean, "[6.6]Chiralane: a remarkably symmetric chiral molecule," Symmetry: Culture and Science 19(4) 307-316 (2008). It is a molecule whose sense of chirality cannot be named, even in principle, http://www.mazepath.com/uncleal/schwart3.png  When CAS and Agnew Chemie cannot name a bunch of chiral centers, I say "publish it."  They said "no."  It does not impact my income.

Folks, I know you were likely just joking with each other, but someone reading your comments later might misconstrue them.

Who reads organic chemistry? "8^>) Employment is EPA-expelled offshore. September 1969, Michigan State, 1200+ Majors Organic enrollment. June 1973, 15 BS/Chem graduated.  Wanting it more than anything else is a good start.

Firstly, you cant go ethane to propene, it should be propane.
Chemically, you can do it by following steps:

halogenate with $\ce{Cl}$ in UV.
then use $\ce{KCN}$ to make nitrile - it will replace $\ce{Cl}$.
hydrolyse it using $\ce{KOH}$.
then it will be a carboxylic acid.
reduce it using $\ce{LiAlH4}$ and then, you have got propane.

You can go for ethene the same way, but there will be complications.

You haven't really done that, have you?  Photochlorination of ethane will yield ethylchloride (btw suspect of being cancerogenic) plus butane (from the termination step of the radical reaction) plus chlorinated butane(s).  $S_N1$ reactions of alkyl chlorides with sodium or potassium cyanide in DMSO do work in principle. Here, the boiling point of ethylchloride is 12 °C, the melting point of DMSO is 19 °C.   Propionitrile to propionic acid looks ok, but what about the reduction?  The reaction of propionic acid with $\ce{LiAlH4}$ will yield 1-propanol.