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How does the choice of metal (oxide) catalyst affect the range of unsaturated compounds that can be hydrogenated?
In the hydrogenation of unsaturated compounds with hydrogen gas and a catalyst, the choice of palladium on carbon is able to hydrogenate alkenes and alkynes, but is unable to hydrogenate aromatic compounds. Using rhodium or Raney nickel, however, allows one to hydrogenate aromatics. How does the choice of catalyst affect what kind of unsaturated compounds can be hydrogenated?

@Georg I think he wants a mechanistic explanation.

@Georg: Oh, I assure you I did. Unfortunately I seem to not be reading the right references, since a lot of them just mention "catalyst X works for hydrogenating substrate Y". Now, anything else you can offer, good sir?

The currently accepted hydrogenation mechanism for metal like palladium is the Horiuti-Polanyi mechanism. (image)  This mechanism works with good yield for alkenes and alkynes.  For an aromatic compound, the activation energy is much higher because the aromaticity of the target molecule is being broken by the first hydrogenation.  Indeed, 1,3-cyclohexadiene is very unstable and will revert to benzene and hydrogen instead of continuing to cyclohexane.
Addition of a Lewis Acid makes palladium an acceptable catalyst for the hydrogenation of benzene, however.  An intermediate compound with the benzene and the LA is first formed, and hydrogen is added to this intermediate.  This paper discusses the mechanism of hydrogenation in the presence of a Lewis Acid.
Rhodium catalysts work slightly differently.  They generally have ligands attached that help stabilize the intermediates.  You can see a simplified mechanism here and there are papers discussing specific catalysts here and here.  Basically, in the case of benzene, the relatively unstable 1,2-cyclohexadiene is held in place on the catalyst until it has been completely hydrogenated.

In that Horiuti- Link there is a big error. No D in the upper product, althouh the reaction is +D and -H ! In general, JM asked for heterogenous katalysts, and there is little to no knowlegde besides the phenomenology of results. The problem is the microscopic structure of the Pd or Pt or Ni preparations.

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If nothing sticks to Teflon, how does Teflon stick to frying pans?
The most notable characteristic of polytetrafluoroethylene (PTFE, DuPont's Teflon) is that nothing sticks to it. This complete inertness is attributed to the fluorine atoms completely shielding the carbon backbone of the polymer.
If nothing indeed sticks to Teflon, how might one coat an object (say, a frying pan) with PTFE?

Why do the fluorine atoms have to be on both sides?

""This complete inertness is attributed to the fluorine atoms completely shielding the carbon backbone of the polymer."" This is wrong, any atom will shield the carbons. Because the fluorine diameter is small, this shielding is even lower for PTFE than in other similar polymers. The true reason is the low polarsisability of the fluorine substituents. So, all other molecules try to "stay among themselves" resulting in minimum interface area.

@Soandos, care to elaborate?

It has to be so common a question that the answer is actually given in various places on Dupont's own website (Dupont are the makers of Teflon):

“If nothing sticks to Teflon®, then how does Teflon® stick to a pan?"
  Nonstick coatings are applied in layers, just like paint. The first layer is the primer—and it's the special chemistry in the primer that makes it adhere to the metal surface of a pan.

And from this other webpage of theirs:

The primer (or primers, if you include the “mid coat” in the picture above) adheres to the roughened surface, often obtained by sandblasting, very strongly: it's chemisorption, and the primer chemical nature is chosen as to obtain strong bonding to both the metal surface. Then, the PTFE chain extremities create bonds with the primer. And thus, it stays put.

While this is probably the best answer available publicly, it also doesn't seem very satisfying from a scientific viewpoint.

Before putting Teflon on a pan, the pan is scratched hardly leaving some tiny holes where the hole opening is smaller than the hole inner size (like a bottle). When Teflon is cast into that, it physically can't get out. That's also why when you scratch a Teflon pan, things start sticking to it.

they spray a primer first on the substrate to be coated (the frying pan) , then they spray the fluoropolymer powder on it. the primer is the tricky part. i think the idea is they make another copolymer of ptfe and another monomer that dont have all its C atoms saturated with F. this copolymer sticks to ptfe because it contains ptfe and at the same time can stick to other substances more readily vs pure ptfe polymers due to other c atoms not fully saturated with Florine.this copolymer is mixed with a polyamide or polyimide or other forms of plastic that would stick after melting to the frying pan. so to sum it up . the polyamide plastic sticks to the aluminum of the pan. then the copolymer of ptfe and perfluoro polymer sticks to the polyamide and finally the top coat of ptfe is applied over that copolymer and sticks to it . the idea here is this copolymer that is half ptfe and half perfloropolymer that sticks to ptfe above and other adhesives below. its a gradient from sticky substance up to a completely non stick pure ptfe. at least this is what i understood from reading some of the patent information about the primer used by chemours(dupont). for more detailed information you can refer to these patents.

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During the development of aspartame, why was the methyl ester chosen?
One frequent argument against the safety of the synthetic sweetener aspartame is that it can be hydrolyzed, with one of the hydrolysis products being methanol, which is known to be toxic. Of course, usually not much aspartame is added in food anyway to yield a sufficient amount of methanol or its oxidation product formaldehyde, but this does beg the question: why wasn't, say, the ethyl ester chosen to be developed as an artificial sweetener? Is there a structure-activity relationship between the alcohol used to esterify the dipeptide and the sweetness of the product?

You mention the irrelevance of the toxicity ba methanol yourself. So, the inventor then either did not think about it or the methyl is essential for taste and/or degree of sweetnes. I think it is the latter, because nobody would stay with the methyl ester today, if that was not a necessity. Aspartame use is limited by chemical stability, so eg the presumably much more stable  t-Bu ester might be commercially relevant. That this is not done, speaks for itself.

This question raises more important issues than just the technical "why methyl ester," so I'll address those too.
The easiest explanation for their focus on the methyl ester is that the ethyl ester just isn't nearly as sweet. This report says it is approximately $10\times$ less sweet (see Table VI on page 2689 and the entry '$\ce{Asp-Phe-OEt}$' with "$++$" being defined on pg. 2684). It's an easy economic choice between the two, assuming the only difference is sweetness/g, but you're also worried about the perceived safety concerns of the methyl derivative.
A toxicology class I once took can mostly be summed up by a cliche phrase, "the dose makes the poison." Even too much of a good thing is by definition a bad thing, for example too much oxygen, water, or even the sun can lead to death. We also say that too little of any compound won't cause measurable harm.
It's correct to say that aspartame will be metabolized into amino acids and methanol. Methanol itself is toxic in high doses, in the same manner ethanol is, by acting as a CNS depressant. This acute poisoning can cause respiratory failure or other medical emergencies such as kidney failure. Ethanol poisoning occurs in microorganisms too: it has been proposed that some yeast produce ethanol as a means of eliminating competition. This would have been potentially deadly for a prehistoric mammal too, if it were not for an enzyme, which metabolize ethanol. This enzyme is called alcohol dehydrogenase (ADH), which leads to biochemical pathways that allow the safe removal of ethanol. However, methanol is a different case. ADH instead make formaldehyde, which is then converted to formic acid by ALDH. Formic acid, in large quantities, is used in the animal kingdom as a weapon, most notably in bee venom and from the bites of a fire ant. In low quantities it is fairly benign. Formaldehyde is just a nasty chemical. Both are produced in the body as a natural response to methanol and it's easy to see how too much of either couldn't possibly be desirable. The body also has ways to eliminate these toxic metabolites assuming you don't overwhelm these systems. Hence why a sufficiently small amount of methanol won't make you blind.
Researchers have measured the amount of these undesirable metabolites in the blood after having people, like me, ingest nearly $\pu{1.8 g}$ of aspartame in one sitting. Then they looked at urine and blood concentrations and they couldn't detect any formic acid in the blood but they could clearly see it in their urine for, up to, $\pu{8 hours}$ after a dose. This evidence suggests that acute methanol poisoning isn't going to happen after a ~$\pu{200 lbs}$ person ingests $\pu{1.8 g}$ of aspartame. Let's stop for a second and put this in terms of something we can imagine.
I don't have any reliable numbers, but let's just say that for a $\pu{12 oz}$ diet pop there are $\pu{200 mg}$ of aspartame. That means for every liter of pop, there are approximately $\pu{564 mg}$ of aspartame. That means $\pu{3 L}$ of pop would get me to $\pu{1.8 g}$ of aspartame. That means I would need to drink more than $\pu{3 L}$ in one sitting to overwhelm the excretion systems. This suggests an acute poisoning isn't very plausible.
There is, however, long-term toxicity to consider as well. This is a much more difficult question to probe. "Sure, there is no acute toxicity, but what happens in the long-term with a low dose?" For the answer to this question, we have to turn to toxicologists who have not reported long-term issues. They are looking too. The general consensus is that there is no long-term harm known for the general population. It is hard to quantitate these things, because to perfectly do the experiments necessary would require controlling a bunch of parameters that human experimentation laws prevent, let alone funding constraints. So, researchers make the best with what they have available and using these tools they cannot find any measurable problems with this particular compound in the long-term for the general population. Their analyses are limited by uncertainties in their measurements and this is just the general nature of science.
Tomorrow the majority of scientists in this area of interest could be proven wrong by a very careful experiment, but over forty years of research hasn't shown this chemical to be dangerous at the doses we're likely to be exposed to for any duration of time. We have to make decisions off of the best evidence we have available. I think it's okay for people to apply some level of skepticism to these things, because in science we're continually correcting our accepted knowledge base. Once in a while the general community is proven wrong about something long thought to be true and everybody benefits when that is the case, but we can't say the general community is wrong without substantial evidence. It is for this reason that we have the FDA to evaluate the best available evidence and via a network of reliable strangers, we can enjoy a very cheap/sweet drink that is low in calories. Frankly, I am more worried about the amount of sugar I take in than I'll ever worry about the amount of aspartame I consume -- we have already good evidence that sugar, with a dose too small to cause acute poisoning, is probably killing hundreds of thousands of people due to long-term exposure.

Aspartame’s sweetening properties were discovered by accident. the G.D. Searle company was trying to develop drugs to treat ulcers in the 1960s when one of the scientists accidentally contaminated his finger and licked it. Knowing how many versions of molecules get tested in drug development, my guess is that an ethyl ester version was made, and probably failed due to improper taste or some other property. Aspartame is already “sparingly soluble” in water; I assume the ethyl ester would be even less soluble, making it less useful for sweetening drinks. However, the amount of aspartame that would have to be dissolved is so small I don’t know if it would make a difference. It's probably taste related.
As for toxicity concerns, it takes a lot of methanol to cause toxicity, Wikipedia claims 10 mL can cause permanent blindness, and 30 mL can be fatal, but the amount of aspartame a human would need to ingest to produce 10 mL worth of methanol would be ridiculous.

I've just done the math, and to get 10mL of MeOH strictly from Aspartame contained in food, you'd have to drink at least 405 soft drinks, or chew 10 390 sticks of gum. In a short time. :)

The most likely reason to use methanol instead of ethanol is cost. [1]  Ethanol is roughly twice as expensive as methanol.
Although both aspartame and the ethanol analog can be produced with reasonably high yield[2], the ethanol analog has not yet been produced commercially, so it would be even more expensive to develop industrial scale production of the ethanol analog.
Finally, to address the toxicity part of the question, aspartame is not at all toxic.  It is most stable at lower pH (~4.3) which is in the range of carbonated beverages.  So methanol is not generally produced while the food is in storage, and the body gets rid of methanol as it is produced.
One more note, aspartame can break down via heating.  This is why food that is intended to be baked is not usually sweetened with aspartame (or it is mixed with saccharine).  Although the fear here is not toxicity so much as cake that isn't sweet!

http://web.anl.gov/PCS/acsfuel/preprint%20archive/Files/41_3_ORLANDO_08-96_0880.pdf
http://www.patentstorm.us/patents/4730076/description.html

I didn't say aspartame itself is toxic. :) I said the concern is because one of its possible breakdown products is methanol.

Yes, that was what I was referring to (methanol vs ethanol).  @J. M. - Sorry if I wasn't clear enough in my answer - the point I was trying to make is that aspartame isn't toxic in use since it rarely breaks down.  You pointed out the low quantity used in food too, which is another mitigating factor.

Speculation, not an answer. Besides cost isn't a big issue with things like aspartame.

Methyl ester was chosen by the aspartame manufacturer for two specific reasons:

The methyl ester molecule binds the Phenylalanine and Aspartic Acid molecules together forming the Aspartame molecule.
The unnatural methanol in the methyl ester provides the artificial sweetness of aspartame.

As an artificial sweetener, aspartame’s job is complete when it has tricked our brain into thinking we have consumed glucose when we have not, its toxic waste (methanol) is then swallowed for our body to get rid of.
Aspartame per se can do us no harm; it does not enter the bloodstream (EFSA 2013)
At a temperature of 86 degrees F in the GI tract, the methyl ester in aspartame converts back to pure methanol releasing its bond with the two amino acids. Aspartame’s three components are then free to enter the bloodstream to metabolise separately (EFSA 2013)
Methanol is by far the most toxic of aspartames components – the other two are amino acids.
To assess the severe metabolic toxicity of methanol as a food in humans is first necessary to establish its ADI. This can then be used to compare this toxicity to the current aspartame ADI.
As has been mentioned in your comments, the blinding dose of methanol in humans is 10 ml (one tablespoon) and the fatal dose is 30 ml (3 tablespoons) a simple calculation can express these in (mg/kg) as follows :

The blinding dose of methanol in a 70 kg adult is (114 mg/kg) and the fatal dose is (343 mg/kg). To calculate the ADI we need an NOAEL, which must be between (0.0 mg/kg to 114 mg/kg). If we use for example 10 % of the blinding dose = (11.4 mg/kg) to establish the ADI, this must be divided by a safety margin of 100; therefore the ADI of methanol in humans is (0.114 mg/kg). The ADI of aspartame (40 mg/kg) contains 10 % w/w of methanol or (4 mg/kg).

Conclusion: The current ADI of aspartame contains 35 times more methanol than is safe for humans.

Your conclusion that 40mg/kg is a bit high for an ADI may be valid, but it takes some effort to consume 40mg/kg. In any case, this is not really the place for propaganda. There are also a number of issues with your science. A. 86F?! Body temperature of humans is 97.8F. B. examination of the aspartame molecule shows a phenylalanine residue in the middle with aspartate at one end and methyl ester at the other, so if anything it's "phenylalanine" holding the molecule together, not "methyl ester" C. Methanol isn´t sweet, it's the whole molecule together which has different properties to its parts.


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What are known examples of drugs that racemize/stereoconvert in vivo, and how are they converted?
It is known that although only the (S)-enantiomer of the infamous sedative thalidomide possesses teratogenic properties, it is not very useful to administer the pure (R)-enantiomer since it is racemized within the body. Are there other known examples of drugs that are stereoconverted in vivo? How does the body perform the conversion?

""only the (S)-enantiomer of the infamous sedative thalidomide possesses teratogenic properties"" How was this fact estabished? You always have both enantiomers in vivo, right?

@Georg probably through study of the mechanism of interaction.

Although I can't think of any drug examples other than thalidomide, here's information on thalidomide's mechanism:
The chiral carbon of thalidomide can tautomerize in basic conditions into an enol, which is achiral.  A reversal back to the ketone results in a mix of (R) and (S) enantiomers.

In the body, this tautomerization is generally catalyzed by basic amino acids.  Specifically, albumin is the main catalyst in humans.
While this is beyond the scope of your question, the reason that only (S)-thalidomide causes birth defects is that it can insert itself into DNA and suppresses certain genes necessary for embryonic development.

Reddy - Chirality in Drug Design and Development
http://www.ncbi.nlm.nih.gov/pubmed/9860497
http://www.ncbi.nlm.nih.gov/pubmed/9499573

According to the relevant article, HSA does catalyze interconversion at pH 7.4, but once other components of blood plasma are added to the mixture, catalysis is inhibited. Don't forget that amines can themselves be chiral, as in this case, but that their interconversion barrier is low. The barrier is greater when the amine is cyclic, as in thalidomide, but it still can spontaneously interconvert.

I added that article to the list.  Certain components do inhibit the reaction, although blood plasma on the whole is also a minor catalyst for the interconversion although not as good a catalyst as HSA.

Drugs racemised in-vivo
The following table shows some of the most common (not an exhaustive list) of drugs racemised in-vivo.
Since the biological targets are necessarily chiral, the molecules formed upon epimerisation (inversion at the centre being racemised) are usually inactive against the desired target (annoying, but ultimately fine), issues only arise when the epimerised molecule has its own biological activity, which is often undesirable (for instance in the case of toxicity of thalidomide).

Pathways of racemisation
The mechanisms by which drugs are racemised is not always well known/studied (i.e. it isn't always possible to pin down the exact thing responsible, however two generalised 'mechanistic pathways' have been proposed: 1
1) formation of a covalent intermediate followed by a stepwise enzyme-catalyzed transformation
2) interaction of two opposing metabolic processes, such as oxidation and reduction
One thing that is generally agreed upon, is that the racemisation is the result of more than simple deprotonation/protonation (such as the way chiral centres adjacent to aldehydes tend to epimerise), since the pH in the body is, in most places, not basic enough to do so.
The interconversion of (R) and (S) ibuprofen is one example where a mechanism has been proposed, and involves multiple intermediates:

1: Curr. Drug Metab. 2004, 5, 517–533.
Apologies for the lack of reference to the table, it was taken from a set of lecture notes, though I suspect the lecturer had previously copied it from the primary literature somewhere

Ibuprofen is another example (have a look at e.g. here).
Ibuprofen is an optically active compound with both S and R-isomers, of which the (S)-isomer is the more biologically active. Although it is possible to isolate the enantiopure compound, Ibuprofen is produced industrially as a racemate. An isomerase (alpha-methylacyl-CoA racemase) converts (R)-ibuprofen to the active (S)-enantiomer.


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What do typical results from inductively coupled plasma mass spectrometry (ICP-MS) look like, and how do I interpret them?
I have a degree in chemical engineering and remember having seen a inductively coupled plasma mass spectrometer (ICP-MS) during the laboratory part of a chemistry course I took while at university, but it is many years ago. Now I would like to know what a typical output from a ICP-MS looks like and how I interpret the results.
There is no picture at the wikipedia page but some frequency domain like plots can be found with a Google search and I understand the Fourier transform and how it is used in number theory.
Are the ICP-MS results anything like the spectrum from the Fourier transform of some signal?

I am not an expert in ICP-MS, but an ICP-MS should be considered an ICP ion source coupled with a mass spectrometer. There are some key issues raised by the coupling between both instruments due to the high temperature of the ICP source as well as due to the fact that one wants to analyse atomic ions and remove any interfering molecular ions (mostly oxides, but other molecules and clusters are possible).
Therefore, to attempt an answer to your question, the standard output of an ICP-MS will be presented as a mass spectrum (intensity versus m/z or intensity versus amu). There are a couple of sample spectra in the brochures published by instrument makers, see for instance the Bruker brochure or a Thermo application note. How this mass spectrum is obtained from the detector will lead to the type of data that is acquired:

For a quadrupole mass filter, as well as for magnetic sector instruments, two modes of operation are possible. Either a single mass is selected, and therefore ont will only get a trace of the abundance of a given mass as a function of time (chromatography or on-line sample analysis). Otherwise the mass filter can be scanned, leading to a mass spectrum.
For time of flight detectors, the mass spectrum is built by the measurement of a flight time for an ion to reach the detector. Therefore the mass spectrum is converted from a time dependent signal to a spectrum.
Finally there have been some attempts to couple an ICP ion source with a Fourier transform mass spectrometer. As far as I know, these are not commercial instruments. In such instruments, the ions oscillate at m/z dependent frequencies, and the recorded image current is converted to a spectrum through the use of Fourier transform.


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The Chemistry behind creating Polylactic Acid (PLA)
I've seen a few videos of persons creating plastics from starch by adding an acid, glycerol, and water, however no explanation of the chemistry.
Does anyone know what $\ce{(C6H10O5)_{n} + C3H8O3 + C2H4O2 + H2O -> (C3H4O2)_{n} + (what)}$?
or does it?
I only have a basic understanding of organic chemistry, so sorry if the equations don't make sense.

Polylactic acid polymers are creating a lot of interest these days because they're biodegradable. On the downside, corn starch is often one of the starting materials...and corn is also used as a food source, so there's some controversy over the wisdom of this approach.
I think I've found a site that takes the approach you're basing your question on...but there are a couple of issues with the equation you've written. So, I'll answer this question in two parts. Plastic from starch is what you saw in the video; PLA from starch is what you asked about.
Plastic from starch. The recipe for the plastics-from-starch uses starch, glycerol, acetic acid (the $\ce{C2H4O2}$ in your equation) and water as a solvent. (They do not add lactic acid, the $\ce{C3H8O3}$ in your equation, but lactic acid can be produced by hydrolysis of the starch.) With the starch-acetic acid-glycerol recipe, you do not actually make a PLA polymer, but a starch polymer that is somewhat like it instead. (Ref)
So, starch is a polymer that looks like this:

In water, it tends to ball up into globular clusters due to hydrogen bonding. By adding acetic acid and heating the starch in water, you "denature" the starch and turn it into "disordered chains" that stretch out. When you let these disordered chains dry, they become entangled and form a flexible film, the plastic. In this recipe, the glycerol is included as a  plasticizer, so it is unchanged chemically in the reaction. The glycerol also helps to keep the starch stretched out by stabilizing it with hydrogen bonding. Plasticizers (as the name would indicate) are additives that increase the plasticity of a plastic and make it more durable or flexible. Changing the amount of glycerol in the recipe can have a big effect on the strength of the plastic film you make. This is the process you're seeing in the videos, I think.
Making PLA from starch. This is going to be a bit of a challenge to explain depending on your understanding of organic chemistry, but hopefully the pictures will help. When you make PLA commercially, you digest starch, usually with a bacteria, to turn it into lactic acid (see picture below for the structure of lactic acid). This lactic acid must be purified to make a high quality product. If you just combine a bunch of lactic acid molecules to make PLA, one water molecule would be lost each time they combine...sort of like this. (This is an acid and alcohol combining, not two acids, but the idea is the same.) The water in the polymerization reaction leads to a poor quality plastic.

So, in commercial synthetic schemes, two purified lactic acid molecules are reacted to form lactide...they come together in a controlled condensation reaction, with the loss of water. (Ref for picture.)

Now you can use the lactide, i.e. "lactic acids with the water removed", to do the polymerization to PLA. You add an initiator to open the lactic acid by breaking the bond between the $\ce{O-C}$ in  $\ce{O-C=O}$ and this gives you 2 lactic acid molecules hooked together...which react with another lactide molecule to add two more lactic acid molecules to the growing chain, and so on. And that's how the PLA forms.
So the reaction sequence is strach-> lactic acid-> lactide + $\ce{H2O}$ -> PLA
Probably more than you wanted to know!

No, actually it's exactly what I wanted to know. How to make PLA, as well as what kind of plastic those start to plastic videos are actually making. It's answers like this that will quickly take chemistry out of beta :)

To all: Here's a nice little pdf about actual PLA, including time and temperature: Synthesis and Characterization of Poly (Lactic Acid) for
Use in Biomedical Field

The link to springerlink.com is broken. I am also not able to find any copy saved on the Wayback Machine.


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Basic chemistry equation with ammonia
A table in the book 'Chemistry NCEA Level 1' has 4 headings, the first of which is 'Volume (mL) ammonia solution provided (conc 2 g L-1)'
Could someone please provide a comprehensive explanation of what the end part in brackets means.
This is for a chemistry newbie, as you may gather from the book title.
(this isn't really homework but I couldn't find a relevant tag. It's just someone studying chemistry for fun.)

Hi Mark, welcome to the site.  I think that this question is too localized, as it would only apply to people using that book in the future.  To answer the question, though, it's saying that the concentration of the ammonia solution is 2 grams per liter, so there are 2g of solute (ammonia) in 1 liter of solution (ammonia + water).

I don't understand why this has been closed. If someone without physics knowledge asked the meaning of e=mc2 would that be closed? If someone didn't know that 3n meant 3 times n would their question be closed? I don't know what 'conc 2 g L-1' means. Does the L-1 part mean L to the power of -1? Why does it not mean 2 times g times L-1?

I understand where you are coming from, but the answer to your first two questions is almost always, yes.  These issues are known as "general reference" questions and can be looked up online.  You are welcome to start a Meta discussion about  why this question was closed, if you'd like.  The "L-1" is an old typewriter notation that did indeed mean "L raised to the -1 power" or 1/L.  If you peek ahead in your text to the section on concentration, it should answer that question and more.

@Mark: The L-1 is just a way to easily note  the unit g/L (grams per liter). Regarding the closing of the question, we allow simple questions here. They just shouldn't be easily Googled. One major point to take into account while asking the question is this: "Will it be useful to future visitors/will it help them learn?". We are not one of those quick help fora, actually, we aren't a forum at all. We aim to make the Internet better by keeping good, conceptual questions that will be useful to others. Simple questions are OK as well, though they shouldn't be easily Googled.

This question serves as a bad example for the homework policy.


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Where can I obtain tables of P-V-T/compression factor data?
Some recent research I'm dealing with has required me to look for tables of experimental pressure/volume/temperature values (equivalently, a table of compressibility/compression factors in terms of any two of pressure, volume, or temperature) for various substances. Thus far, I've managed to obtain compression factor tables for a short list of substances from Perry's Chemical Engineers' Handbook.
I am wondering, however, as to where I should be looking for more extensive tables (e.g. tables with finer sampling), as well as tables for substances that were not included in that handbook. What books/websites/journal articles should I be looking at?
My attempts at searching on the Internet have not been very fruitful; I would appreciate knowing what keywords should I be using with the usual search engines to get better results.

This borders on being one of those infernal make-a-list questions, but in its current form, it seems OK to me :)

@Manish, I know, but I wouldn't be asking this question if it wasn't very troublesome to pull up references for this sort of thing... :( as I said in a comment to the answer, tabulating data like this is hard work.

Yep. I doubt there are many lists like this--so it's fine :)

One place to look for simple fluids is http://webbook.nist.gov/chemistry/fluid/.  But there are only a limited number of compounds treated.
Be warned that the data in the NIST Webbook is smoothed.  Collecting data of this sort is very difficult.  Many different sets of observations are combined and fitted to an appropriate equation.  What is at the URL given above is computed from that equation.
If you are looking for overall compression factors, this will do very nicely.  If you want five figure precision in compression factors, well, this won't do.
Another place to check is the  which is a primary source of raw data.

"Collecting data of this sort is very difficult." - yes, I've had to do a simplified version of that experiment myself, so I understand very well how difficult it was to obtain those data. Speaking of which... you speak of raw data. Those would be what I need (and somehow I think asking for five-figure precision here is a bit much to ask, so I didn't say so in the OP). Would you happen to have pointers on where I can obtain the raw data you speak of? (Corollary: I don't want the smoothed data, as I'll be doing comparisons with some equations of state.)

Try the NIST data first. It is most likely the most accurate data available. Individual experiments almost never totally agree for a large number of reasons, which is why raw data may not be a good thing to use. Raw data will have to be found in the literature. The "Journal of Physical and Chemical Reference Data"
(an American Institute of Physics publication) is a place to start.


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Does ozone (O₃) gas have a color?
Does ozone have a color or is it colorless? I have heard that ozone has a bluish color.
Could someone explain this observation scientifically?

This is pretty Googleable, and I'm closing it as such. If anyone disagrees, please let me know (with a reason)

Well, of course this is googlable, but why closing for "too localized"?

I googled this. I got two answers. I just say it in question. Bluish and colorless. Also a specific explanation not getting. Also I think a question closing because of its Googleable not a good practise and discouraging.Thanks.. :(

@Manishearth see this meta question.. http://meta.stackexchange.com/questions/8724/how-should-we-deal-with-google-questions

@georg best fit. Unfortunately, the "general reference" option is only enabled as a trial on some sites. Still, current consensus seems to be that such questions ought to be discouraged.

Hmm, that meta post is from the earlier days of SO, when rules were much more lax. Issue is, we want good, conceptual questions, not stuff that can be easily Googled. Such stuff is better asked on [chat]. in this case, it did get a conceptual answer, though the question itself still remains not that good. And don't let closes discourage you--if you can improve it, it may be reopened. If you still feel that this shouldn't have been closed, feel free to post on [meta], we can see what the community thinks. (as it is I was unsure about closing this, but another mod agreed)

You may not realize, but Google has picked up this question and answer as the featured link at the top of the page for "colour of ozone".  The advice to google it leads back here. What's more, another chemistry stack exchange answer similarly closed also is in the first page. Also a Wikipedia article on Ozone which says almost nothing on the topic.
Luckily Quora comes second. I've added an answer there for those wondering where they can find the colour of ozone https://www.quora.com/What-is-the-colour-of-ozone/answer/Robert-Walker-5. Others here might like to add answers there as well.

Ozone gas does indeed have a slight blue color. It is dark blue in a liquid form and purple-black in solid form.
The color comes from the fact that it absorbs all wavelengths of light other than those in the blue region. As the phase changes, the molecules get closer together, shifting the area of absorption, changing the color of the material.


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Reaction of soap with water
I've seen it many times... Why does soap become foam-like when it reacts with water (moisture) or when we touch it with wet hands.. What compound actually causes this physical change? And why does this physical change occur?

So you're thinking of solid 'bar' soap?

yeah,..! Why are there any other different species in your mind.>?

Sorry, what I meant is that I guess you're in particular thinking about the physical change of the properties of solid soap when you get water on it.

Yeah man.. I mentioned the word 'physical change' in the question

The simplest kinds of soap are just a fatty acid chain with something more ionic on the end - something like the ubiquitous sodium laureth sulphate:

When it's in bar form, it behaves a bit like a solid block of plastic, because chemically speaking, those long chains of carbon are quite a lot like a common plastic such as polythene. These chains attract each other and repel and exclude water, keeping the block together. The sulphate group on the end is hydrophilic, though - it's chemically stable surrounded by water. This means that when you add water, it infiltrates and teases apart the structure of strands, getting to the ionic ends on each chain and forming a "solvation shell" of waters around each one. There's no real chemical reaction here - as you say, it's a physical process, which is why rubbing the soap helps - you're applying shear to the structure and helping to separate the strands.
Once they're separated from the bulk, the soap forms tiny bubbles called "micelles", which have the chains pointed inwards towards each other and the ionic sulphate groups meeting the water at the outside.

While this question should get a more elaborate answer, mine will at least give you leads.
Soaps act as surfactants, that is, they are amphiphilic (both hydrophobic and hydrophilic) molecules that lower the surface tension of a liquid.
By lowering the surface tension of water, soap bubbles can form, the surface becoming more "elastic" - the molecules at the interface experience less inward/normal force than they would were surfactant not present. On the contrary, adding soap to a pond would prevent water striders from gliding on the surface of the pond.

Thanks. But I didn't ask about surface tension. Its in my 10th grade syllabus. I'm asking 'why is this physical change?' Please read the question and its not necessary for ponds & surface tension

@CrazyBuddy Then your question is ambiguous. The question you ask in the main post and the one you ask in the comment are different. Consider clearing up your question.

One thing that should be made clear is the question you want to ask, and also what you mean by "solid-like".

sorry for that.. I told solid-like 'cause i thought soap as something like colloid due to its appearance

The soap dissolves in the water. Dissolution is not a chemical change because the same chemicals are present, they just interact with each other differently. Like CHM said, soaps act as surfactants and the difference in surface tension improves the survivability of bubbles, thus creating a "foam".

There. Is. No. Chemical. Change. We have all been trying to tell you this. We've been telling you about physical changes and you've been ignoring us.

We're talking about physical changes and you keep going "no I mean the physical change man".


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What is the name of cyclopropane with a double bond?
I was drawing out all the isomers of $\ce{C5H8}$ and came upon two structures of cyclopropane attached with double bonds. The group is neither a methylene nor an ethyl... what is it? What are the names of these two molecules?

 Ethylidenecyclopropane
 Vinylcyclopropane (or Ethenylcyclopropane)

Note that these aren't the IUPAC names, care to add those as well? Also a bit about how the two functional groups have (common) names as ethylidene and vinyl respectively. Thanks :)

Ethylidenecyclopropane is the IUPAC name for that molecule. What else could you call it? See http://www.acdlabs.com/iupac/nomenclature/93/r93_271.htm. The vinyl substituent has been recently accepted by IUPAC (in the last 10 years), but I haven't found supporting proof on the web.

As a side note, IUPAC names are nice and all, but completely worthless if practicing chemists do not use them: 2-propanone, methanoic acid, and 1,2-didehydrobenzene are all significantly more frequently referred to by their common names of acetone, formic acid, and benzyne.

@Manish, as already noted, "ethylidene" is in fact valid IUPAC nomenclature.


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Polymer Slurry used in boring foundation holes?
I watched a video on the construction of Burj Khalifa, Dubai. The construction engineers answered that they're using a special type of protection known as Cathodic protection to protect from the corrosion due to local ground water. They told a compound named "Polymer Slurry" (What's that?). They're using viscous Polymer Slurry as their basic coat to dump the high density & low permeability concrete in order to create a "GOOD FOUNDATION". What is it made of? (Simply any polymer?) And what does it do? Does it actually repel water? Many organic compounds repel water... Then Why this?

"Polymer slurry" just means a slurry of polymer. It could be anything.

After reading the link provided in the question, and doing a little googling, of "polymer slurry" and "foundation", I found something useful.
Polymer slurry seems to be made of one or more hydrophilic superabsorbent polymers, such as polyethylene glycol, polyvinyl alcohol, or carboxymethylcellulose that produce a thick viscous slurry when mixed with water. It does not repel water so much as the opposite. It is so hydrophilic that it absorbs water before the water can reach the concrete or the steel. Then this slurry becomes thicker the more water it absorbs so that the rate of diffusion of water through the slurry decreases. Ultimately, this action creates a hydrogel, which may be quite strong, despite being mostly water, due to extensive hydrogen bond crosslinks.
Other applications of polymer slurry seem to be in the drilling industry to modulate the viscosity of drilling fluids.

The polymer slurry has the purpose of keeping the hole stable, while it is still being worked on.
While building a 50 m deep hole, I think it's common to need to change or replace the tool, when it broke, or the type of ground changed.
Changing the thing on the other end of 50 m steel rod means to pull out the rod, make the change, and let it down again.
But even without any change, the rod needs to be pulled out before filling with concrete, I think.
The slurry does not repell water on the surface like oil would, but it repels ground water from entering the bore hole by having a higher pressure.
The pressure comes from a higher density and same depth below ground.
One useful effect of the polymer could be that it does not soak into dry parts of the hole wall.
Over 50 m of hole wall, a touching rod will ceartainly loosen some sand, but id will not just drop to the end - depending on the slurry and the particle size, it will sink slowly, or even not at all.

From Burj Dubai - Preparing the Foundation:

So, large boreholes were dug 164 feet (50 m) deep and then filled with viscous polymer slurry to maintain the stability of the hole. The slurry is denser than water, but lighter than concrete. The concrete, when poured into the hole, displaces the slurry and hardens to form a foundation pile.


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Are there any nonflammable thermal insulators which can be made at home?
I am trying to build a homemade thermos. It needs to be able to conserve the heat in 125 ml of water for 15 minutes.
However, unlike a normal thermos, it also needs to be able to be heated. What homemade substances can I use that are not flammable, yet provide a relatively high degree of insulation?
Note that my thermos will be heated with a flame from a candle for 25 minutes.

Strictly speaking an insulating flask prevents heat moving in either direction. So heating something in the flask will always be very inefficient (the whole point of the flask is to inhibit the flow of heat). In a vacuum thermos (sometimes called a dewar) this will be virtually impossible. You could design a flask that is mostly insulated but with a gap where you can heat the internal space. But this makes it a worse insulator.

Consider using a ceramic container.  You could adjust the density, thickness, and size to match your insulating requirements.
"home made' is loose, here, since you would need to acquire oven bake-able ceramics, or a similar product.  (unless you have a kiln lying around).


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Calculating entropy: why consider a reversible path?
I am reading up on entropy in a textbook and I got confused by this:
It says that to calculate the entropy for an irreversible process using heat flow, one must imagine the reversible process in which the initial and final states are the same as for the irreversible process. Why is this so and why do we need to think of it in the reversible way?

Short answer: you don't. Just calculate the entropy of the initial and final states and take the difference. If there's a heat bath you have to include the change in entropy of the heat bath as well, but that's fine. I've never understood why some textbooks teach this weird and complicated reversible vs. irreversible path stuff, it's just unnecessary.

There are two parts to this answer.  The first is that the entropy change is defined only for reversible processes.  The second is that the entropy is a state function.  Many, but not all, the functions in thermodynamics (including the internal energy) are state functions.
A state function is one that is independent of path.  That is, if one goes from state $A$ to state $B$, the change in internal energy is independent of how the change takes place.  The same is true for the entropy.
So the entropy change in going from state $A$ to state $B$ is independent of how the change takes place.  But we can calculate the change ONLY for reversible changes.
So we mentally find a reversible path from state $A$ to state $B$ and calculate the entropy change.  Since entropy is a state function, that must also be the change for a non-reversible path.

For a closed system, if you calculate the integral of $dq/T_I$ over all possible process paths between two thermodynamic equilibrium states of a system (where $T_I$ is the temperature at the portion of the system interface (boundary) with its surroundings at which the heat transfer takes place), the integral will differ from path to path.  The maximum value of the integral over all possible process paths will be found to occur when the path is reversible.  Over any reversible path, the system temperature T will be essentially uniform throughout, and the temperature at the interface $T_I$ will be essentially equal to the system temperature T.  All reversible paths will give the same maximum value for the integral.  This maximum value is the change in entropy of the system between the two thermodynamic equilibrium states.  It is a function of state because the maximum value  of the integral is determined only by the initial and final equilibrium states.


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Molecular orbitals of heteronuclear diatomic molecules
For drawing MO diagram for heteronuclear diatomic molecule, how do I know which is lower in energy than the other?
For example, given a heteronuclear diatomic molecule $\ce{NO}$, which one is lower in energy?

The rule as generally stated in textbooks is that “atomic orbitals of the more electronegative element are lower in energy than the corresponding orbitals of the less electronegative element”.
Now, MO diagrams are only simple for elements of the second row of the periodic table ($\ce{Li}$ through $\ce{Ne}$). Involving heavier atoms makes it harder to guess at molecular orbital diagrams, and there is need for quantum chemistry calculations. Thus, the rule becomes: the further to the right your element is, the lower its energy levels are. The reason for this is that the charge of the nucleus increases with its atomic number, and thus lowers the energy levels.


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Bond order for carbonate ion for resonance
I saw in a textbook that for carbonate ion, there are 3 resonance forms and the bond order is 1 and 1/3. So in general, how should we calculate the bond order for resonance structures?
Is there any definite way to do this?

There definitely is an easy way to do this.
One of the easiest examples that comes to mind is benzene.  If we look at the structure:

we see that each of the carbon bonds are either single bonds to the adjacent carbon or double bonds to the carbon in the other direction.  If we draw the other resonance structure these bonds shift by 1 atom to look like the following:

Thus, the overall structure can be obtained by looking at those two structure and averaging the bond order between them.  For benzene one resonance form for each bond will be single and the other double $(2 + 1)/2$. The "$2$" on the bottom of the equation is the number of resonance structure being considered.  Thus, for benzene the bond order is $1.5$.
Applying this same method to the carbonate ion, we have 3 resonance structures with bond orders of 2, 1, 1 when considering the bond between carbon and a single oxygen.  Since there are 3 structures the answer is $(2+1+1)/3 = 1.33333$ or $4/3$.

Indeed. I'll add that this only works for resonance structure that are equivalent by symmetry… since they have the same weight. You cannot, for example, calculate in any simple way a bond order for furan, even though you can easily write its resonance forms: they don't have the same weight.

And given that resonance is a figment invented to make valence bond theory handle delocalization, there is likely no theoretical approach to determine the relative weight of furan's resonance structures. Comparison of experimental bond lengths for related molecules (for example, the cyclopentadienyl anion is symmetric) and for tetrahydrofuran and the the dihydrofuran isomers may get you an estimate. Using MO theory will only get you the total bond order for the molecule, since there are no localized bonds.

Only in simple cases, it is possible to make reliable estimates of the relative weight of resonance structures from the Lewis structure alone. If you want a somewhat quantitative yet simple approach to bond orders in delocalized systems, have a look at the Hückel method. It is a quantum-mechanical theory for determining the energies of $\pi$ orbitals, and it defines a method to determine a measure of the $\pi$ bond order between two atoms. If you want to go beyond that, you will have to look further into quantum chemistry and different schemes for bond localization and population analysis. In many cases there is no one definite answer, but rather, your results will depend heavily on the quantum-chemical calculations and assumptions for the bond-order analysis itself.
(There are probably also ways to tackle the problem from an experimental side, although I'm not all sure about this; I'm thinking about perhaps determining electron distributions through NMR chemical shifts, or bond strengths through vibrational spectroscopy.)

There is a formula for this too... it is = 1×[(no. Of pi bonds b/w resonating atoms of resonating structures)/(no of sigma bonds  b/w resonating atoms of resonating structures)]....for example: in benzene, hydrogen atoms do not participate in resonance so bond order = 1+(3/6) =1.5; in carbonate ion, it is 1+ 1/3=1.33 hope this helps....

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].|| Maybe you could extend your answer a little more, be a little more descriptive. Especially remove uncommon abbreviations, like b/w.


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Product/(s) of the following reaction
I was wondering what the product will be in the following reaction:
$$\ce{H3COOCH3 ->[\ce{H+}]} \:?$$
I was thinking about $\ce{CH3COOH}$  being a product but I do not know the mechanism.

This is a specific case of the general reaction of acid-catalysed ester hydrolysis, which should be discussed exhaustively in any good organic textbook. This reaction is closely related to base-catalysed ester hydrolysis, aka saponification. I think any detailed answer to this question is quite redundant, however hopefully giving you the name of the reaction will point you in the right direction.

My school text book unfortunately does not provide the reaction mechanism. :(

here's a good overview: link

Wrong answer/comment. Without water no hydrolysis will happen.

If you're sure that the initial formula is $\ce{H3COOCH3}$ (and that's a big if), then I disagree with Richard's answer in comments. $\ce{H3COOCH3}$ is dimethyl peroxide. While peroxides are not the most stable of molecules, and decompose rapidly (often by oxidizing other compounds), they do not spontaneously transform in acid conditions.
If the formula were $\ce{H3C–COOCH3}$ (note the added C), then yes, it's be acid-catalysed ester hydrolysis.

Whoa, totally missed that. Well spotted. I was mainly looking at the acetic acid product.


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Why doesn't water play a role in reactions in aqueous reactions?
$$\ce{Ba(NO3)2(aq) + Na2CO3(aq)}$$
In the above reaction both the substances are in water, however water plays no role in the reaction. Why is it that although water is present in the reaction, it does not play a reactant role? To clarify, why isn't the above written as:
$$\ce{Ba(NO3)2(aq) + Na2CO3(aq) + H2O}$$
Why does water act as a bystander and not a reactant?

Water is far from a bystander in aqueous solutions. It has a number of features that greatly facilitate chemical reactions, as well as being an important ligand (the aqua ligand) of transition metal complexes.
Here are some (unbalanced) examples:

Water has a rather large dielectric constant and as such is very effective at shielding charges from one another, facilitating the charge separations inherent in the dissolution of ionic compounds. This is why you can easily dissolve sodium chloride in water, but not in hexane. The nature of this shielding is often articulated as several concentric 'solvation shells', where 'shells' of water molecules cluster around an ion with their dipoles approximately aligned towards or away from the ion.

Here's an image from Wikipedia qualitatively depicting the first solvation shell surrounding a sodium ion:

More generally, the large dipole moment of water allows it to partake in hydrogen bonding, where water will weakly associate with other polar molecules or functional groups. This extends to water forming (short lived, highly fluxional) clusters with itself.
Water is also happy to accept (forming hydronium, $\ce{H3O+}$) and donate (forming hydroxide, $\ce{OH-}$) protons in solution. So much so that it will actually do this with itself, in a process called autoionisation. This process makes water molecules significant in a wide range of acid/base and redox equilibria.

Additionally if the ions were some of the transition metals such as Fe3+, then the water plays a greater role as the iron hydrate complex is acidic.

It does play a role--it's all hidden in the $_\ce{(aq)}$. Whenever an ionic compound is dropped in water, it dissociates into the ions. This dissociation is facilitated by the water, as it solvates the ions.
What happens is that a certain amount of energy is required to dissociate an ionic compound (lattice energy). If a pure compound was dissociated, there would be quite a bit of energy required to do this, and you would end up with something unstable (the molecule would probably just recombine.)
Now, water, being polar, solvates the ions. What happens is that water molecules aggregate around the ion, and the net charge gets effectively dispersed. This leads to a gain in stability (and a decrease in potential energy known as the "hydration energy"). Ionic compounds with enough hydration energy (specifically, when the absolute value of hydration energy>lattice energy) are said to be "soluble", and dissociate readily in water.
So the actual reaction is:
$\ce{Na2CO3 ->[\ce{H2O}] 2Na+_{(aq)} + CO3^{2-}_{(aq)}}$
Where the $\ce{_{(aq)}}$ signifies solvation. There will be a similar one for barium nitrate.
So, when one mentions $\ce{Ba(NO3)2_{(aq)} + Na2CO3_{(aq)}}$, it is actually $\ce{2Na+_{(aq)} + CO3^{2-}_{(aq)} + Ba^{2+}_{(aq)} + 2NO3^{-}_{(aq)}}$.
Now, what happens is that the barium and carbonate ions react to form insoluble barium carbonate. This precipitates out, and we get the following reaction:
$$\ce{\underbrace{Ba(NO3)2_{(aq)} + Na2CO3_{(aq)}}_{Mixture~of~ions} ->[\ce{H2O}] \underbrace{BaCO3}_{Insoluble~solid} v + \underbrace{NaNO3_{(aq)}}_{Mixture~of~ions}}$$
Water isn't reacting here, it is more of a catalyst. Without it, the reaction wouldn't occur (unless you melt the compounds--molten ionic compounds behave similar to their aqueous counterparts).

Thank You. I still don't understand why water isn't reacting. I understand why water is required for the reaction to happen, but why doesn't water itself react?

@FarhadYusufali: Counter-question: Why does it have to react? It sort of does react since it forms an aggregate during the solvation, but there's no need for it to react--so it doesn't. Reactions aren't arbitrary; mixing A and B doesn't mean they'll react.

@Manishearth Hey just wondering idly, on the right hand of the bottom most reaction.  Should barium carbonate be labelled with (s) and sodium nitrate with an (aq) or is it just understood?

@user17753: Well, the down arrow indicates precipitation and usually one omits the (s), though I missed adding (aq) to NaNO3, thanks :)


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What makes a molecule a polyatomic ion?
I checked the Wikipedia page, but it did not seem to provide a very intuitive explanation.
What distinguishes a molecule such that it behaves like a polyatomic ion? I understand that it is a covalent bond, but what distinguishes it from other covalent bonds that allow it to act like one element. Why does it have a charge, unlike many other compounds (ionic and covalent)?

A polyatomic ion is simply a molecule that has gained or lost one or more electrons, thus acquiring a net charge. This can occur due to oxidation or reduction of the molecule, by association of an atomic ion with a neutral molecule or by autoionisation of a neutral species into a anion/cation pair.
A concrete example of the latter two processes is the autoionisation of water, which to a first-order approximation may be considered as $\ce{2H_{2}O <=> H_{3}O^{+} + OH^{-}}$. This is coincidentally an example (one of many) of an aqueous reaction involving covalency with a water molecule, addressing in part one of your other questions.
There also exist polyatomic ions known as zwitterions (from German 'zwitter', meaning 'hermaphrodite' or 'hybrid') which possess localised opposite charges on different atoms, but are neutral overall.

Thanks. Why are polyatonic ions treated as a singular element though? (as in the case of double displacement, single displacement etc)

@FarhadYusufali - Because for study purposes you have been shown a subset of reactions to study that don't involve the covalent bonds within the ions being broken. Instead, I imagine you are looking at reactions involving ionic bonds being broken in solution.


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What causes reactions to react the way they do?
We were taught that there are primarily four kind of reactions, synthesis, combustion, single displacement, and double displacement. When the reactants are given to you, the reaction can be predicted.
For example:
If the reactants are AB + CD, it will be a double displacement (two pairs of compounds)
If the reactants are AB + C, it will be a single displacement (one pair and one compound/element)
If the reactants are A + B, it will be synthesis.
Why do the "format" of the reactants determine the type of reaction? Why can't, for example, AB + CD synthesize to create ABCD? Or AB + C synthesize to create ABC? Why is it assumed that if the reactants are in a specific "format" a specific reaction will occur?

If the reactants are AB + CD, it will be a double displacement (2 pairs of compounds)

An example of this might be two salts that dissociate into their respective ions in a solvent, and then the cation and anions that were not previously paired find their way to each other, if they form a salt that is not easily dissociated in this solvent, we have a new ionic compound (salt) which precipitates out as a visible solid in the mixture container. This might not always be the case, when we write net ionic equations we leave out anything that was not involved in creating the precipitate.
If we have the net ionic equation: $\ce{A^{+}_{(aq)} + B^{-}_{(aq)} -> AB_{(s)}}$ occurring when the solutions mix, then $\ce{AB}_{(s)}$ is said to be the precipitate. Notice that we leave out the other ions C and D because they were observers in the chemical equation, perhaps their attraction to each other is not strong enough to overcome the solvent. We should not always be able to tell whether this will happen or not unless we refer to experiments in which this substance had been seen to precipitate out before, perhaps even measuring its solubility in some experiments, the extent to which it will dissociate can also be calculated from experimentation.

Why can't, for example, AB + CD synthesize to create ABCD?

If you are trying to determine if a covalent bond is compatible, one method is to use the octet rule and see what combinations might occur. For example, you have reactants of silicon and fluorine. What will form? Looking at the periodic table to see the electrons they contribute, draw a lewis structure for fluorine and silicon. Silicon contributes 4 electrons, and 4 fluorine atoms contribute the rest to give all elements full octets. Silicon tetrafluoride seems to be the most stable using this model. This is one method, but often times reactions occur the opposite way too. However, perhaps for some reason it is more likely for the reactants to be present in a given reaction, even after it seems to have stopped on a macroscopic scale (even thought it really has not stopped activity), we need to come up with more things to explain why this behaviour might happen. So as scientists, we devise new models.
This is a less than obvious and bigger topic than you might realise. A general chemistry textbook might have a chapter called chemical kinetics or reaction mechanisms which describe this ten times better than I can. Some things that will help you in understanding some models of reactions and their mechanisms are things like rate laws and how to determine rate constants in the laboratory, or how to determine the activation energy proposed by Svante Arrhenius in 1880's which basically means that two chemicals need to have at least a certain amount of energy (the activation energy) to overcome their repulsive behaviors and other limiting factors (the energy required to actually break apart a molecule's covalent bonds). For a collision of two particles to produce a different particle, there are other things that determine if a new molecule is formed from a collision of two molecules. First, like I said they must overcome repulsions with kinetic energy but they often need to be aligned spatially for a certain molecule to form in some proposed mechanisms. A mechanism is a proposition that an in-between step occurs in the reaction, but it is very unlikely that two hydrogen molecules will smash into a carbon dioxide molecule and create methane and oxygen. We must propose intermediate steps to make it sound more probable. although this would represent the reverse reaction of combusting methane, and will likely never occur because the potential energy of the system would increase, and we know how much the universe hates that kind of behaviour, if not then read on.
Some concepts that will also help you understand "why reactions happen" are some laws like the 2nd law of thermodynamics which states that the entropy of the universe is always increasing. This explains why a reaction in which a reaction of type that you describe $\ce{ABC + D -> ABCD}$ is actually quite possible. The problem is that the reverse reaction also takes place with a higher probability, these molecules might actually break apart more than they combine. This two-way speciality is why this reaction will likely be written as such: $\ce{ABC + D <=> ABCD}$ indicating that it does possibly happen, but can go both ways, and most often goes backwards. The reason for this is that there is more entropy (aka: disorder) when there are more moles of molecules present. In your examples the reactions are unlikely because the reactants ($\ce{ABC}$ and $\ce{D}$) are twice the number of moles than the products ($\ce{ABCD}$). This is a limiting factor for the reaction to commonly take place. It is useful to note that this reaction might occur once in awhile, but it is more often that molecules fluttering about will break apart than create larger and more complex molecules. Indeed, it is more likely that this reaction will be reversed and proposed as a mechanism of a larger process, if at all, where D breaks off and combines with something else, lets say E and ABC still remains along with a very unbreakable compound called ED which has a triple bond that will likely not be broken at the given temperature.
Lets talk more about entropy: You have a snowball you just rolled up in your hand (it is composed of many flakes of ice). Think of each ice particle that makes up the snowball as an atom in a larger molecule (the snowball). Chances are when this snowball starts moving (like molecules tend to do) it will run into something, and when that happens it will become many different smaller flakes and chunks (atoms and smaller molecules). This is because the universe prefers disorder and chaos. If you threw your clothes freshly out of the dryer into your closet, will they miraculously fold themselves neatley and in order? No, because that would make them more orderly and less chaotic, instead they will flail about in the air and land in a disorganized manner, this then makes sense by the phenomenon of the second law of thermodynamics.
Look back on your equations now, why can't $\ce{AB + CD -> ABCD }$ occur? Hopefully this is more obvious to you now. There is more disorder in the reactants than the products. This reaction might still occur, but normally balanced chemical equations represent the net affect that tends to occur. Perhaps occasionally someone will win at the game BINGO, but that does not necessarily mean that the same person will keep on winning at BINGO, more likely it will be someone else in the room, and then someone else, etc. With that BINGO analogy, the person who won at bingo played again, and lost. The same way ABCD might have been created, but then almost immediately was destroyed again.
Chemical engineers might desire to create your molecule $\ce{ABCD}$ and to do this what they might do is add a large amount of $\ce{AB}$ molecules into the reaction chamber. Due to the overwhelming number of these molecules, the reaction is more likely to take place. Not everything in the container will eventually be $\ce{ABCD}$ molecules, but there will be more of them that there was before. These kinds of reactions are actually called equilibriums meaning that the reaction can occur both ways. We write the equilibrium reaction as $\ce{AB + CD <=> ABCD}$ to explicitly show this. The quotient of reactants to products will not be 1:1. Sometimes the rate constant is very large for a reaction implying that it will go to completion, in which case we simply write the reaction in a single direction as you often see with chemical equations.
That is far as I am willing to explain because I am merely repeating snippets of what any chemistry book will explain in a chapter on chemical kinetics, hopefully my repetitive explanation will help you get a better idea of what is actually going on here, but there is still a lot more that you could learn or I will bother to explain.

Food for thought: In the equilibrium reaction $\ce{AB + CD <=> ABCD}$, A chemical engineer decreases the pressure of the reaction chamber, will the reaction shift more in favor of the reactants or the products? These are some questions that chemistry books will ask you after chemical kinetics are discussed. But I do not expect you to be able to answer it yet.


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Why do H₂ and O₂ create H₂O?
Both $\ce{H2}$ or $\ce{O2}$ are stable molecules. However when given energy they heat up and form $\ce{H2O}$. My teacher explained that the increased energy breaks the bonds, allowing new ones to form. However, why don't the old forms reform instead? Why are new bonds created?

Are you familiar with activation energy?  Consider a hole at the top of a large hill.  If a ball or some water was in that hole it would seek the lowest point it could find.  Clearly, the bottom of the hill would be preferable to the bottom of the hole at the top of the hill, however, in order to get there it must first be brought up over the top of the hole.

In this picture, the $H_2$ and $O_2$ are on the right side of the curve.  Just like water in the above analogy, they'd like to move to the lower area, but first must be brought up over the bump.  So some energy must be added to accomplish this, and then they end up in the much lower energy position on the left.
Once over the bump, they stay as $H_2O$ because that is the state they'd like to be in at any time, but were prevented from being in before by the activation energy bump.

Thank you. One question: why is $H_2O$ considered to be the lower part of the hill? In other words, why does $H_2O$ have less potential energy?

The hand waving answer is that the total energy in all the bonds of $2 H_2O$ is more than the total energy in $2 H_2 + 2 O_2$.      http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html

That is not hand waving at all, it is thermodynamics.  Enthalphy gains offset entropy losses.

Adding on to DaleSwanson's correct answer: The ball-and-hill analogy is just saying that a ball will roll down a hill. This is because if the path is available it will find a way to the bottom of the hill due to there being less potential energy in the ball down at the bottom than up at the peak. Think of the activation energy as the little 'push' needed to get the ball moving off the edge of the cliff, once it is nudged off the side it will release almost all the potential energy that was released through kinetic energy when it hits the ground (small amounts of potential energy will also be released in other forms, i.e: friction of the air and ground).

To answer your question, I ask you this: you accidently knock a pen off your desk, it required so little energy it was almost unnoticeable. Would it require more energy for you to reach down and put the pen back on your desk? What position is "more stable" for the pen?

Hydrogen and oxygen are stable, yes, they have full valence shells and are satisfied. Just like a ball sitting on the top of the hill is stable, until you give it a nudge. Now think of the top of the hill as the hydrogen and oxygen molecules. Also, think of the bottom of the hill as the water molecule. Take a look at the balanced chemical equation now:
$\ce{2 H2 + O_2 -> 2H_2_O}$
There is less energy at the bottom of the hill, the same way that there is less energy in the covalent bonds that make up the products. There is more energy stored in the reactants.
It is very possible to calculate the energies of the bonds and compare the energies of the reactants to the products in the balanced chemical equation by looking up heat of formation's for these species, and calculate the change in energy from going to reactants to products. That is, we can calculate how much energy is released when two moles of $\ce{H_2O}$ are created from this reaction.
Similarly we can calculate how much energy a ball will release when falling from the cliff. The release in energy to the surroundings during a chemical reaction is said to be an exothermic process, whereas an endothermic process would take away energy from the surroundings, such as vaporizing water molecules into steam.
Now using the analogy you have learned, would it be reasonable to say that an equally small nudge that caused the ball to fall would be enough to cause the ball to rise up back onto the peak of the hill? You tell me.
A better picture of the path from reactants to products would be different than a smooth 2d hill picture. Sure, the change in energy is not as outrageous as I portray it, and perhaps in some miraculously rare situation some invaders from another country find the ball at the bottom of the hill, they put it in a catapult and launch it at their enemies at the top of the hill. This is extremely rare and very situational. In other words there is almost no chance it will happen. As it turns out, you are thinking very rationally when you ask why the reverse process does not happen! The problem is you have to think on the molecular scale where molecules are colliding into each other. It is very likely that a $H_2$ will collide with the $O_2$ and one side of the oxygen breaks into the bond of the hydrogen and you now have $H_2O + O$. But how would seperate $H_2 + O_2$ form from two $H_2O$ colliding, the answer is that it is just extremely unlikely, it would more likely create $H_2O_2 + H_2$ if at all! There are other steps in-between that would need to account for going backwards in the reaction. You can think of this as a separate path around the hill. There are more ways back up the hill in other words, but they may lead to dead-ends!

Thank you. One question: why is $H_2O$ considered to be the lower part of the hill? In other words, why does $H_2O$ have less potential energy?

@Farhad Yusufali The strength of the bonds in the products are stronger than in the reactants. The heat of formation for water is negative (the system, or chemical process, lost/gave energy to the surroundings). For elemental hydrogen and oxygen gas it is a zero value (kJ/mol). Engineers think of energy change as positive when the rest of the universe gains energy, we think of it as the other way around (positive when the system gains energy).


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Polypropylene 55 gallon drum
Can a polypropylene 55 gallon drum previously filled with chemicals used in the chrome plating process (possibly Nickel Sulfate) ever really be "cleaned out" for re-use? Can polypropylene ever absorb and retain such chemicals, or are these drums practical for this use because they do not?

Polypropylene is hydrocarbon and does not adsorb polar and ionic compounds. So, several treatments with mild hydrochloric acid followed with several treatments with clean water should work. However, I wouldn't store drinking water there without a proven test. For example, leave water there for several weeks and then perform analysis of the water for metal ions. If all concentrations are within safe levels, it should be alright.

Many of the chemicals used in metal plating are extremely poisonous (cyanide), cause severe allergies (Ni), and are carcinogenic (Cr(VI)).
I'd consider it as extremely unprofessional (besides violating many laws and regulations), to re-use anything that has been in contact with these materials for any purpose that requires a 'clean' container, regardless of the theoretical possibility of cleaning everything off.
To use such a drum for food or drinking water would be criminal negligence.
I'd only consider it ok if it was thoroughly cleaned (my myself) and only used by myself as a trashcan in my machine shop outside of my home.
I would even disallow to use it as a trashcan in the machine shop at work. Why risk it?
Even if PP does not absorb ions in principle, there is always the real world:

The guy who does the cleaning gets lazy and thinks you are an idiot requiring him to clean something off he doesn't see and only rinses the drum with water.
The plating bath contained organics that made the ions stick to the drum.
Some stuff precipitated from the plating bath and stuck to the bottom of the drum.
During the cleaning process cleaned and contaminated drums get mixed up.
Some plating solution dried in a crack, or under the rim, or wherever you can't clean in off.
Cleaning is itself a hazardous action that creates a lot of contaminated wastewater.
You don't know where the drums might end up. Of course the risk of using them as a trashcan in the machine shop is minuscule, agreed. But what if a machinist takes them home to his home shop? What if his kids use them to play with? I'd have been delighted as a kid to have a 55-gallon drum as a toy. Maybe the machinist's wife will take them to store food.

There are hundreds of accidents where contaminated containers have been improperly cleaned and re-purposed. Please don't do it, even if it is actually safe.
Even if the drums can be cleaned, imagine anything happening, like some kid being admitted to the ER with a Nickel allergy.
You'd be in a tough position proving that no one got injured from your unprofessionally cleaned drums. The people who used the drums in the first place will have all the paperwork to 'prove' that it wasn't them.
Btw. we had an accident with a metal plating bath that was not properly cleaned and injured a technician who had do do some modifications to it.
It might not hurt anyone, but it's better not to do it.


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Boron-Boron bonds
I am reading up on boron in my notes and there is a certain paragraph that I do not understand

The tendency of $\ce{B}$ to form $\ce{B-B}$ bonds is less pronounced than that of
  carbon, its periodic table neighbor on the right, to form $\ce{C-C}$ bond,
  but it is much greater than that of the nearest element in the group,
  $\ce{Al}$, to form $\ce{Al-Al}$ bonds

Anyone can explained to me why is $\ce{B-B}$ bonds can be formed in the first place?

Why wouldn't they? — In order to ask a good answerable question, please give more information about what you know already, and what you don't understand. Otherwise, we don't know what helps you or not. For example, my direct answer to your question would be “why not? have you looked at the molecular orbital diagram for homonuclear diatomics?”… but I have no way to guess if that would actually help you or not.

I agree with @F'x. A boron-boron bond will form because it is lower in energy than the two atoms separately (definition of a bond in all bonding theories). A better question is why the B-B bond is less common than the C-C bond, but more common than the Al-Al bond.

I am not sure boron-boron bonding is inherently less favored thsn carbon-carbon bonding.  The variety and complexity of known boron-based clusters seems suggestive.  We may not have studied boron-boron bonding widely because there does not seem to be a lot of boron-based life around.

As mentioned in the comments, why not? A boron-boron bond will form as long as there is an increase in stability of the system. Forming a bond requires energy, but there is a lot of energy released once it gets formed. The feasibility of a bond depends which energy is greater.
One example of where a boron-boron bond occurs is in Aluminium diboride / Magnesium diboride

Here, you have graphite-like "sheets" of boron (pink), with aluminium layers between them.
This will probably be quite stable, due to delocalization. The singly-filled aluminium $p$-orbitals will overlap with the empty $p$ orbitals of Boron, and will delocalize across the graphite-like sheet (I cannot confirm this, though). This will lead to a large increase in stability.
A few more compounds with $\ce{B-B}$ bonds are Lanthanum hexaboride and various Yttrium borides. These are probably very stable due to the crystal structure (though delocalization may be a factor as well)

It's even possible to stabilize the B_{2} dimer with carbene ligands; this complex has an isolated B-B triple bond. See http://www.sciencemag.org/content/336/6087/1420.

See this answer for a discussion of magnesium diboride.  The boron layers (and three-dimensional extended networks in other electropositive-metal borides) contain electron-accepting, bonding orbitals that drive ionic metal-boron bonding -- even with an electronegativity difference less than 1.0.


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In sp² hybrid orbital, why does the p orbitals only have 3 sides instead of 4?
Usually two separate p orbitals would have '4 lobes' while in a $\ce{sp^2}$ hybrid those 2 p orbitals would only have '3 lobes'. why?
I'm referring to diagrams such as the one shown below:

First, an $sp^2$ orbital does have two lobes. One is large, and one is tiny. Secondly, $sp^2$ means "a 'mixture' of one $s$ and two $p$ orbitals". Like the following:

So, you plug in three orbitals, and you get out three more orbitals. When drawn in place, you get this:

There are a total of six lobes here, but they are distributed amongst three $sp^2$ orbitals.
The number of lobes need not stay constant--it's the number of orbitals that does, since each orbital corresponds to two electrons, and there's no way for new electrons to be added/removed (yet). What's happening is that the wavefunctions(orbitals) of the electrons are "mixing up" to distribute the electrons in a way that stimulates bonding (one per orbital). The lobes have a significance, but it's much deeper.


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Why can an s orbital only hold two electrons?
According to Pauli’s exclusion principle, an $s$ orbital contains at most two electrons with the opposite spin (up and down). Why can't an $s$ orbital contain a third electron whose state is the linear combination of spin up and down?

Electrons are magnets, they have magnetic fields. Those fields have only two possible orientations, and a single orbital can only be occupied by two electrons if those orientations are mutually opposed.
OK, but why? Why are electrons magnets? Why are there only two possible orientations for the fields? Why can the two electrons in the orbital only exist with opposite directions for the magnetic moments?
Well, there are answers to those questions, but they don't make sense a lot of sense. The magnetic moment of electrons is due to a property called "spin"; the spin value of an electron is 1/2, so it can adopt one of two spin quantum states, +1/2 and -1/2, which correspond to the "up" and "down" orientations of the magnetic moment. Since it has a half-integer spin, an electron is a member of a class of sub-atomic particles called "fermions" which obey rules called the Pauli Exclusion Principle and Fermi-Dirac statistics - one key result of these rules is that no two identical fermions can simultaneously occupy the same quantum state: you cannot have two electrons in the same orbital if they have the same spin orientation, so if one is +1/2, the other must be -1/2, and no more can be added because a third would have to adopt the same quantum state as the one of the first two.
OK, but why? Why does an electron have a spin value of 1/2? Why can't two fermions adopt the same quantum state (when integer spin particles, called bosons, can do so)? Why does the property of "spin" even exist? You need quantum mechanics to even begin to answer those questions -- again, that's more math than anybody is prepared to show you for at least five years -- and even then, the "why" of the Pauli Exclusion Principle is an empirical observation (it has no theoretical basis, it's just an after-the-fact thing that happens) without relativity, but there is a firm theoretical rationale and a reason "why" if one revisits the postulates of quantum mechanics and takes into general relativity into account. But relativistic quantum field theory is more math than any mortal should be forced to endure.
OK, but why? Why is quantum mechanics how the universe behaves? Why is general relativity the way the universe behaves?
You can play the "why" game for a very long time. None of your high school teachers will be able to give you more than one or two repetitions. University professors will be able to go a lot farther. Eventually, you get to questions to which the only possible answer is "nobody knows.... yet". And those are the questions that scientists -- including those university professors -- spend their day trying to answer.
Which is terrific fun. So keep asking "why", even to the answers your teachers give you. And if they can't answer, go ask someone else, or read about it, or go to university. And if they don't know, be a scientist, and figure it out yourself, and tell the rest of us.

Nice answer! By the way, if you are planning to contribute more answers - you can read more about formatting mathematical and chemical expressions here, here, and here. If you are already familiar with $\LaTeX$, then none of it will be new :)

I really loved this answer, in particular the last paragraph. Very nice motivation :) !

Because we can have only one electron per quantum state.  Spin up and spin down are two different states.  A linear combination of the two is not a new independent state.  It is obviously formed from the spin up and spin down states.

-1. First, it does not answer the question, it just paraphrases it in a different way. Secondly, it is just plain wrong. A linear combination of two states is a new state, this new state is obviously linearly dependent with the two states, but it is a different state.

The complete wave function of a two electron system is written using slater determinants. So, for helium it would look like this:
$$\begin{pmatrix}1s(1)\alpha(1) & 1s(1)\beta(1)\\\ 1s(2)\alpha(2) & 1s(2)\beta(2)\end{pmatrix}$$
For lithium, in what you are saying, it would look like this:
$$\begin{pmatrix}1s(1)\alpha(1) & 1s(1)\beta(1) & 1s(1)(\alpha(1)+\beta(1))\\\ 1s(2)\alpha(2) & 1s(2)\beta(2) & 1s(2)(\alpha(2)+\beta(2)) \\\ 1s(3)\alpha(3) & 1s(3)\beta(3) & 1s(3)(\alpha(3)+\beta(3))\end{pmatrix}$$
This obviously is zero, as can be seen from basic principles of matrices and determinants, which is not possible, as the wavefunction obviously cannot be zero.
(Note: 1s(i) refers to the spatial part of the wave function and $ \alpha,\beta $ represents the spin part.)


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Electronegativity of zinc vs copper in galvanic cell
I am reading up on how galvanic cell works and I realised that the flow of electrons is from Zn to Cu. But Zn is more electronegative compared to Cu according to periodic table trends
I read it somewhere that when comparing transition metals, it may be different, but it was not  further elaborated. Can someone explain this to me how Cu is more electronegative than Zn?

There are two common confusions in your question. I hope I can clear them up.

Electrons are not flowing from zinc to copper. Not exactly.

In the Zinc-Copper galvanic cell, the net reaction is:
$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}$$
Zinc is being oxidized:
$$\ce{Zn(s)->Zn}^{2+}(aq) \ce{+2e-} $$
Copper is being reduced:
$$\ce{Cu}^{2+}(aq)\ce{+2e- ->Cu(s)}$$
Electrons, therefore are being transferred from zinc not to copper, but copper (II) ions, which are different than the copper atoms in Cu(s). We need to be comparing something about zinc atoms and copper (II) ions.

Electronegativity has nothing to do with electrochemistry.

Electronegativity is the attraction an atom has for a pair of electrons in a covalent bond.
Electrochemistry is about controlling the electron transfer that accompanies a redox reaction. Usually this involves ionic species, and so electronegativity is not the relevant quantity to compare. In this case, we care about the energy change that occurs when copper and zinc lose electrons (ionization energy) and the stability of the resulting ions in solution. Ionization data mined from this NIST publication.
For zinc:
$$\ce{Zn->Zn+ + e-} \ \ \ \ \ \Delta H= 9.585\times 10^2 \text{  kJ/mol}$$
$$\ce{Zn+->Zn}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.733\times 10^3 \text{  kJ/mol}$$
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{  kJ/mol}$$
For copper:
$$\ce{Cu->Cu+ + e-} \ \ \ \ \ \Delta H= 7.454\times 10^2 \text{  kJ/mol}$$
$$\ce{Cu+->Cu}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.958\times 10^3 \text{  kJ/mol}$$
$$\ce{Cu->Cu}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.703\times 10^3 \text{  kJ/mol}$$

Why is zinc oxidized and copper reduced?

Because doing so leads to a lower energy state. It takes more energy to remove two electrons from copper than to remove two electrons from zinc. Thus, to get a negative energy change (spontaneous), when need to add the zinc reaction as written and the copper reaction in reverse:
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{  kJ/mol}$$
$$\ce{Cu}^{2+} \ce{+ 2e- -> Cu(s)} \ \ \ \ \ \Delta H= -2.703\times 10^3 \text{  kJ/mol}$$
$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ \ \Delta H= -11 \text{  kJ/mol}$$
The estimate above assumes that the solvation energies of both ions are similar and their standard entropies are similar. I could not find a good source of those data.
In electrochemistry, we care more about the potential of the cell, $E_{cell}$, which is the potential for the cell to do electrical work. Potential, $E$, is related to free energy change $\Delta G$ by the following equation, where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant $F=96,485 \frac{\text{J}}{\text{V mol}}$.
$$\Delta G=-nFE$$
For a process to be spontaneous, $\Delta G$ must be negative. $\Delta G$ is negative when $E$ is positive. Thus, in a galvanic cell, electrons will flow in the direction that produces a positive cell potential.
How can we determine which direction that is? First, we could set up the zinc-copper cell, and connect it into a circuit and let it run. In this cell, a piece of copper metal serves as one electrode and a piece of zinc metal serves as the other. Whichever electrode increased in mass is the electrode at which reduction occurred: metals cations are converted to metal atoms, coming out of solution and increasing the mass of the electrode. The other electrode must be the one responsible for oxidation (metal atoms become cations and dissolve away); the mass of this electrode must decrease.
Or, we could do a simpler experiment. Dip a piece of copper metal into a solution of Zn2+. Nothing exciting happens (no reaction). Then, dip a piece of zinc metal into a solution of Cu2+. Zinc reduces the Cu2+ ions and copper metal plates onto the zinc.
Or, we could rely on tables of standard reduction potentials for half cells, measured against the standard hydrogen electrode SHE (for which E is defined to be ZERO). Standard reduction potentials come from Wikipedia, which mines them from a number of legitimate sources.
SHE:
$$\ce{H2(1 atm) + 2e- -> 2H+(1 M)} \ \ \ E^o =0 \text{ V}$$
For zinc:
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=-0.762 \text{ V}$$
For copper:
$$\ce{Cu->Cu}^{2+} \ce{+ 2e-}  \ \ \ E^o =+0.34 \text{ V}$$
To generate a spontaneous reaction, these half cells are combined in such a way so that the overall cell potential is positive:
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=+0.762 \text{ V}+$$
$$\ce{Cu}^{2+} \ce{+ 2e- -> Cu(s)} \ \ \ E^o=+0.34 \text{ V}$$
$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ E^o=+1.10 \text{ V}$$
This analysis only applies to standard conditions. Under nonstandard conditions (for example [Zn2+]>>[Cu2+], then the reverse reaction could be spontaneous.

The answer by Ben Norris is nice but it seems to have an additional confusion.  The difference between reduction potentials of copper and zinc can be calculated based on the empirical ionization data that Ben describes, but this doesn't mean that the difference has nothing to do the electronegativity of the two different elements.  A nice discussion of this issue is provided at the following link.

Could you try to elaborate on this; maybe paraphrasing/summarizing the link here?

Electronegativity is tendency of an atom or a functional group to attract electrons (or electron density) towards itself. Adding an electron to copper is very easy because of its valence orbital which has one free electron, which wants to pair with another electron, whereas in zinc, the valence shell is completely filled, i.e., d-orbital.
Config of Cu: {Ar} 4s2, 3d9
Config of Zn: {Ar} 4s2 3d10


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Reaction of magnesium with the surrounding air
In my lecture notes for inorganic chemistry, it is stated that magnesium has high affinity for oxygen but is rather stable at room temperature in air.
Can anyone explain why this is the case?

In fact we could have the reaction O2 + 2Mg --> 2MgO , so can you have some point missed. Because they can have ionic repulsion, but if I am wrong correct me.

I believe that magnesium exposed to air quickly ends up with a coating of magnesium oxide on its surface.  This oxide coating is quite uniform and does not allow any further oxygen to reach the surface.
Several metals, but NOT iron, have this property of a protective oxide coating.  In iron the coating forms a spongey surface easily penetrated by further oxygen.

Magnesium's oxide layer is not protective- its not tenacious like aluminium oxide on aluminium. Its simply that virtually all metals form an oxide layer almost instantaneous with cold dry air- the activation potential is small because the surface of metals has high energy due to deformed metal bonds to form the surface. After the initial oxide is formed, the reaction always slows down as a greater activation energy is needed for it to continue- both iron and magnesium are still corroding but just its not obvious.  Also don't confuse iron oxide (which would initially FeO) with rust Fe2O3.xH20- with the normal thickness of FeO on iron its just about clear.


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Where did the rest of the elements come from?
There were ninety original elements on the Periodic Table. Over time, new ones were founded and put on. However, they were all either the original elements experimented on. For example, Plutonium was discovered by deuteron bombardment of uranium. Therefore, I would like to know: How were the rest of the elements discovered? It would also be nice if an example could be given. Thanks!

By the "original periodic table" I assume you're talking about the "elements ocurring in measurable quantities in nature".--everything till Uranium sans Technetium and Promethium.
The method of preparation of these elements is pretty much the same -- bombardment. Just the projectile and substrate differ.For example, the (accidental) neutron bombardment of Molybdenum in old cyclotron parts led to the discovery of Technetium.
There are some other sources as well. Analyzing fission products can get you new elements--that's how Promethium was discovered. The process of fission creates smaller nuclei--but there is a shower of neutrons created as well (which can bombard everything in sight). Most of the discoveries have been due to the "shower of neutrinos", because generally all "smaller nuclei" are of already-known elements. Promethium was synthesized by the "formation of smaller nuclei", though.
Since fission occurs in nature (Usually it's just a little bit, though we sometimes have natural fission reactors), we actually have all the elements up till californium existing in minute quantities in nature (source).