text:
How can I separate the vinegar from a 5% white vinegar to make a more concentrated solution.
I was under the impression that if I put a semi full bottle of vinegar in the freezer the water would freeze but the vinegar would not until a lower temp.  Therefore I could remove the bottle after the water freezes and pour off the vinegar. Apparently the water and vinegar doesn't separate nicely before freezing so this didn't work so well.
I've seen boiling suggested but I'm afraid I'll lose some of the vinegar in the process.
What is the simplest way to concentrate the vinegar at home?

I know it says that it is for poison ivy rash, but a product like http://www.bayercare.com/domeboro.cfm is aluminum acetate, which would provide you with a stronger "vinegar" that comes in powdered form.

Is there a reason you don't want to buy acetic acid and dilute to the desired concentration?

@jonsca Sorry if I wasn't clear.  I'm using it as an herbicide for poison ivy, not for a rash but the suggestion is appreciated.

@bobthechemist where can I purchase acetic acid? What is acetic acid derived from?

@hortstu No, it's okay, I was trying to clarify that while the label is for ivy rash (just by coincidence), that I understood you were using it against the plant itself. It's still going to give you a source of more concentrated acetate.

@hortstu vinegar is approximately 5% acetic acid.  Enter "buy acetic acid" into a search engine and you'll likely come up with everyone's favorite book-seller as a viable source for acetic acid (!) or look for it in an industrial supply store.

@hortstu Over here (Germany) they sell 25% aqueous acetic acid under the trivil name "Essigessenz" (= the essence of vinegar) in supermarkets.

!! Warning !! Glacial acid acid is nasty. It will acetylate your skin. Took all the skin off my left hand....

@MaxW all substances of a certain acidity are corrosive yes. Thank you.

It would help if the question indicated the desired concentration.

If you have access to a still you can distil the vinegar solution, however this can only go so far. You may find this section of the wikipedia article on azeotropy edifying - to summarise, adding ethyl acetate to a vinegar solution generates a water:ethyl acetate azetrope with a depressed boiling point that makes distillation more effective, however this may be beyond the pale for a home experiment and would obviously render the vinegar non-food-safe.

Thanks, I'm not so concerned about getting 100% pure vinegar or having it remain edible.  I'm just trying to make a more effective yet safe herbicide for some poison ivy at the back of the yard and in the cracks in the sidewalk.    I could improvise a still, but is it the vinegar that gets left behind?  It's not volatile at all?  I was concerned about carmelizing the vinegar if I tried to boil off the water.

Yes,you are right. What you are talking about is a Fractional Freezing. With acetic acid at low concentration you will have ice and a solution of acetic acid more and more concentrated  until you reach the eutectic point. The max concentration you will have is about about 58%. If your freezer go to -18°C you can reach only 38%. Take a look to this article.

If I had a freezer that went to -18C wouldn't everything be frozen at that point? Is -18 the magic number that separates acetic acid from water?

You have to do it in many steps e.g. you freeze it at -4°C, eliminate ice so the new solution is more concentrated and freeze at a lower point, you can do another step to -18°C and so on until you reach the eutectic at -26.77°C. At that point you have ice and acetic acid at 59%, not the best thing for Iced tea! If i have more time I will improve the answer..

The acetic acid has melting point of 16°C, while water has 0°C. Therefore, when placed into a freezer, the acetic acid will freeze first, not the water. So once the solution freezes, separate the liquid (it contains more water than acetic acid) and keep the ice (which has more concentrated acetic acid).

Do you have a source for this?  Acetic acid in solution has different freezing points based on concentration.  At least that's the impression I'm under but I could be very wrong.

Hi Hortstu. I don't have a source for that and I could be indeed wrong (although it seemed "natural" to be like that at the time when I wrote). I also read somewhere that there are different freezing points based on concentration. One could freeze vinegar in the freezer and watch it until a portion of it freezes. Then separate the liquid from ice and then perform titration on them - probably one will be more concentrated than the other.

A more direct way to isolate the acetic acid would be chemically. It requires baking soda (easy to get) and sodium bisulfate (usually found as "spa pH reducer"), and a distillation setup. The idea here is that you can get most of the water out, without also boiling away the vinegar, by first converting it to a salt. Then reform the acid from the salt by mixing it with an acidic salt.

Add baking soda (sodium bicarbonate) to the vinegar until it stops bubbling, or the pH hits neutral. This will take a while as there will be loads of foaming.
Boil the resulting sodium acetate solution until it starts to precipitate. If you want something closer to glacial acetic acid, you'll want to try to get to anhydrous, but "mostly anhydrous" is probably fine. Pour out in a glass dish and let dry, in a 250 C/500 F oven would remove more water.
Mix the acetate very well with 3.5x the weight of sodium bisulfate and put the mix into the distilling flask. Distill over the acetic acid. You'll end up with a brick of sodium sulfate in the boiling flask, takes a long while to get out, but you should also get very concentrated (50% to 90+%) acetic acid in the receiving flask, depending on the level of anhydrousness of the reactants.

More detailed procedures can be found on some other science forums, sometimes split up between 'creating sodium acetate from vinegar', and 'creating acetic acid from sodium acetate and sodium bisulfate'. Here's a video for the latter step.

Ive tired this via distillation and afterwards it left a black substance so at the very least you need to distill it first to remove any solutes in the viniger.

Hi Josh, welcome to Chemistry! Take the tour to get familiar with this site. Could you edit your answer for spelling and grammar? It would make the answer better for future users.


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How would one go about making a cheap freeze spray?
I'm interested in making a liquid freeze spray, for a fun science project. My only experience with this comes in spraying inverting canned air dusters, which spray a -60F mist/liquid (liquid if you spray while the can is inverted). I'd like to make a spray that I can store and use. And if it can't be stored more than a few hours, then hopefully something easy to make on the spot.
My chemistry understanding is high-school level basic, so I do not have a working knowledge of chemistry. On one end of the spectrum, I know there is liquid nitrogen, but it's expensive and complex to store correctly (high-pressure canisters).
I found this list of cooling bath configurations. Could someone point out which configuration can be stored long term (months), stored cheaply (e.g. a plastic or thin metal thermos; not a heavy high-pressure fire-extinguisher container), and be a liquid when "sprayed" (assuming keeping conditions in our pressure and room temperature)? I'd like a spray temperature of less than -60F, so ice and water won't cut it.
I'm thinking of going with dry ice and > 90% concentration ethanol. I have two questions regarding this approach and a random question:

If you have a solution of half dry ice and half 99% ethanol, how can you keep it at a low temperature? Won't the solution slowly return to room temperature? My canned air dusters always spray out at the -60F; I'd like something like that if possible.
Can this solution be stored in a cheap plastic or thin metal container? Will there be some pressure buildup from the dry ice that will eventually crack or explode the container? Assuming a 10 oz container with a solution of half dry ice and half > 90% ethanol.
How are canned air dusters always able to spray their liquid solution (if inverted while sprayed) at -60F, even if they are stored for months at room temperature? Is it because the contents of the can are under pressure and that this pressure makes the liquid cold?

It's worth noting, for anyone else looking for this sort of thing and not set on making it themselves, that sprays designed to rapidly cool small things are commercially available for electronics and plumbing work. For example: http://www.maplin.co.uk/p/maplin-freezer-spray-400ml-n68an

SAFETY - Putting dry ice in a sealed container is DANGEROUS.  As the dry ice warms it sublimes to gas. The gas will build up pressure inside the container. If the container cannot hold the pressure it will EXPLODE.

The reason the 'canned air' dusters and other aerosol-based things like anti-perspirant cans produce a cryogenic effect is from the decompression involved. They are stored at room temperature, but the can is pressurised.
There are two effects that happen when you spray out some aerosol.
Firstly, by the ideal gas law, $PV=nRT$, when you release the pressure (i.e. lower $P$), the temperature ($T$) also drops. (The increase in volume $V$ mostly compensates for this, though.)
Secondly, the contents of the can are stored under enough pressure to keep at least one of the parts of the mixture in a liquid state. When you release the pressure by spraying it out, the liquid boils and evaporates, and takes energy in the form of heat to do so. You'll notice that the contents of a canned air duster, if you hold them upside-down when you spray, include more than just air -- there's also a liquid in there which is designed to be mostly inert to electronics and to quickly evaporate just as described above when the pressure inside the can drops, maintaining the pressure needed to produce a gas jet. If you spray for an extended period of time, you'll normally notice a boiling sound coming from inside the can, and the can getting cold, as that liquid evaporates. (As a safety note, these substances are often flammable, so don't use these air dusters near naked flames or ignition sources unless you want fire.)
Most people don't have the equipment needed to produce this sort of thing at home.
None of the cooling baths in that list are designed to be left as-is for months at a time -- they're designed to be made up as and when needed, with occasional addition of liquid nitrogen or dry ice to keep them cool. While you could produce a cold pressurised spray on the spot using dry ice and ethanol, there's a high probability that any sort of robust switched-release mechanism could make the vessel you're using explode under the pressure of sublimating carbon dioxide. Also, unless you relied on a pressure source other than the carbon dioxide itself, your spray probably wouldn't have time to get very cold.
So, I don't recommend it.

Instead of a cooling bath which can't be left as-is for months at a time, can I use the same concept of the aerosol spray: find a compound such that it can be pressurized into a liquid at room temperature, and when sprayed, will change state (hopefully to a gas), expand, and therefore cool significantly?

Yes, but given that you'd most likely be pressurising flammable gases, possibly using improvised equipment, I'd think really carefully about whether you want to be trying that, and how much skin you plan to have in later life. Having done that, your best bet is probably dimethyl ether.

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Why do the reactions of Magnesium in steam and cold water differ?
"Magnesium has a very slight reaction with cold water, but burns in steam.
A very clean coil of magnesium dropped into cold water eventually gets covered in small bubbles of hydrogen which float it to the surface. Magnesium hydroxide is formed as a very thin layer on the magnesium and this tends to stop the reaction.
Magnesium burns in steam with its typical white flame to produce white magnesium oxide and hydrogen." - From Chemguide.
May I know why the reactions of Magnesium in steam and cold water differ?

Magnesium is a very active metal- highest activity of all engineering metals. Your question gives you part of the answer. In cold water it forms a decent protective layer. Given greater activation energy- simple as hot water, that layers is simply damaged the oxidation of the magnesium proceeds unimpeded at a greater rate. Once its in steam the oxidation is extremely rapid.

So why is magnesium forming the oxide on one case and the hydroxide in the other?  Look up the decomposition temperature of the hydroxide and see how that fits in.

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Calculating the pH of a solution of Ca(OH)2
A solution prepared by dissolving $2.8 g$ of lime, $\ce{CaO}$ in enough water to make $1.00 l$ of lime water ($\ce{Ca(OH)_{2(aq)}}$). If solubility of $\ce{Ca(OH)2}$ in water is $1.48 g$. The $p\ce H$ of the solution obtained will be:
[$\log 2 = 0.3$, Atomic masses are $\ce Ca$ = $40$ , $\ce O$ = $16$, $\ce H$ = $1$]
[Note : $\ce {Ca(OH)2}$ is considered to be a weak base because  of its low solubility but it is 100% dissociated]
(A) $12.3$
(B) $12.6$
(C) $1.3$
(D) $13$
My attempt:As $2.8g$ of $\ce {CaO}$ dissolves in $1l$ water the molarity of the solution is 0.05M. So,the concentration of OH- ions in the solution is $0.1M$.So $p\ce {OH}$ is $1$ and hence $p\ce H$ is $13$. But, the answer is given to be $12.6$.What am I doing wrong?

If solubility of Ca(OH)2 in water is 1.48g are you missing L? the solubility is g/L?

You did not account for the fact that $\ce{Ca(OH)2}$ is partly precipitated. Firstly, to find out the moles of $\ce{Ca(OH)2}$ precipitated:-
The number of moles in solution is $\frac{1.48}{74}M$ which is $0.02M$. The total moles precipitated is just the moles of $\ce{CaO}$ minus $0.02M$ since the initial number of moles of
$\ce{Ca(OH)2}$ is equal to the moles of $\ce{CaO}$ taken and hence the amount precipitated turns out to be $0.03M$. Nevertheless, this is not required. With $0.02M$ $\ce{Ca(OH)2}$ in solution, the $\ce{OH-}$ concentration turns out to be $0.04M$ which, assuming no contribution of $\ce{OH-}$ from water and complete ionization, gives a $p\ce H$of $12.6$.


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Find the number of copper spheres in a box

A box contains a mixture of small copper spheres and small lead spheres. The total volume of both metals is measured by the displacement of water to be $\pu{420 cm^3}$ and the total mass is $\pu{5.4 kg}$. What percentage of the spheres are copper?

I can't get the right answer. I did this:
Given:
$m_{\text{total}} = 5.4\;\mathrm{kg} =5400\;\mathrm{g}$
$V=420\;\mathrm{cm^3}$
$\rho_{\ce{Cu}} = 8.96\;\mathrm{g} ~\mathrm{cm^-3}$
$V\times{\mathrm{\rho_{\ce{Cu}}}} = 3763.2\;\mathrm{g}$
$\frac{m_{\ce{Cu}}}{m_{\text{total}}}=69.9\%$
Can someone tell me how to approach this problem?

umm... the problem is underdefined. Are the spheres the same volume? The same mass? If it's the latter, you don't need the densities. It's possible that the box contains just two "small" spheres of the right size.

hint: $v_c + v_l = 420 cm^3; v_c*\rho_c + v_l*\rho_l = 5.4 kg$ - two equations, two unknowns, two table values.

Here is your formulation :
$m_{\ce{Cu}} + m_{\ce{Pb}} = m_{\text{tot}}$
$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}V_{\ce{Pb}} = m_{\text{tot}}$
$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(V_{\text{tot}}-V_{\ce{Cu}}) = m_{\text{tot}}$
$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(420-V_{\ce{Cu}}) = 5400$
$V_{\ce{Cu}}(\rho_{\ce{Cu}}-\rho_{\ce{Pb}}) + 420\rho_{\ce{Pb}}= 5400$
Put values for $\rho_{\mathrm{Cu}}$ and $\rho_{\mathrm{Pb}}$ in the equation, and rest is simple arithmetic.


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What is the reaction mechanism for the hydrolysis of Amifostine?
I have to do research on a drug called Amifostine: 2-(3-aminopropylamino)ethylsulfanyl
phosphonic acid.

I'm supposed to say how the organic reaction works inside the body.  I understand that it's hydrolyzed by an enzyme into its active form; 2-((aminopropyl)amino)ethanethiol.

I don't get what happens to the phosphonic acid functional group. Does the water break apart the $\ce{S-P}$ bond because it is a weak bond? And what does the acid group become? $\ce{H3PO4}$?

You are expelling a thiol rather than an alcohol via alkaline phosphatase-catalyzed hydrolysis.  One can also view it as a free electron pair on "X" attacking the phosphorus to pop out its =O to an anion, then the charge poking back in to expel thiolate.

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Can we extract KI from Lugol solution?
I want to extract the $\ce{KI}$ component of the Lugol solution, is there a method to do it? If not, how can I make a $\ce{KI}$ solution?

As you know you have to remove the others species like $\ce{I2}$ and $\ce{I3-}$. $\ce{I3-}$ I'm quite sure it exists only in solution so simply evaporating the water from Lugol 's solution you should eliminate it. To remove completely $\ce{I2}$ you have to do some others steps.
$\ce{KI}$ is produced industrially by treating iodine with $\ce{KOH}$.(see Wikipedia for reference).
$$ \ce{3I2 + 6KOH -> 5KI + KIO3 + 3H2O}$$
So I think the simple way is to treat Lugol's solution with $\ce{KOH}$ make an ice bath to precipitate $\ce{KIO3}$, filter the solution to remove it and than evaporate water.
You can also try to remove iodine by means of active carbon I know this method is used to purify water see ref and ref.
You can try evaporating the Lugol's solution and re-crystalize it and than eventually  try different solvents to solubilize one of the compounds and leave the other as precipitate but if think is quite difficult to find a better solvent than  water (as you know iodine is very insoluble in water instead $\ce{KI}$ is very soluble).
If you want to make the $\ce{KI}$ from your self here there are two Home experience:

extract potassium iodide from salt
make $\ce{KI}$ from $\ce{I2}$ and $\ce{KOH}$

A very clear exposition, can I use other thing than $KOH$ to treat Lugol's solution?

I don't know I'm sorry! Maybe you can do something with $K_2CO_3$ and/or $KHSO_3$... Of curse if you use active carbon you don't need it...


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Can 1 mole of H2 react partially with one mole of O2?
I am currently studying stoichiometry (high school level) and we came across the following equation in school:
$\ce{2H2 + 1O2 -> 2H2O}$
which is clear to me. We learned in school that reactions in chemistry only happen if there are enough atoms/molecules to react (so every molecule/atom can react), therefore the following reaction won't do anything:
$\ce{1H2 + 1O2 -> nothing happens}$
So now my first question: Why does the following not happen:
$\ce{1H2 + 1O2 -> 1H2O + 1O}$
And, as a follow up question, why does the following not work?
$\ce{H2 + O2 -> H2O + 1/2O2 (Left) }$
What do I do wrong? (maybe I simply misunderstood our teacher)
Are there any explanations for this? I'm especially interested in the answer of my example with $1$ mole of $\ce{H2}$ and $1$ mole of $\ce{O2}$. According to our teacher nothing will happen at all. I would assume that at least all $\ce{H2}$ molecules react with half of the oxygen molecules to $\ce{H2O}$ - leaving half a mole of $\ce{O2}$ left over.
(otherwise this would mean that all molecules "know" each other so they are "aware" that not all molecules/atoms can react)
Can somebody help me? Or point out what I am doing wrong?

You may also want to have a look at this question for this reaction. Your intuition is quite right.

As you have said, you are studying stoichiometry at High School Level. From this I can guess, that you have probably not studied the concept of Limiting Reagent yet.
What is Limiting Reagent?

In a chemical reaction, the limiting reagent, also known as the "limiting reactant", is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent since the reaction cannot proceed further without it. The other reagents may be present in excess of the quantities required to react with the limiting reagent.

As per your question, $\ce{H2}$ is limiting reagent. Hence it will get totally consumed when reaction happens and 1/2 Moles of $\ce{O2}$ will be left.

1 mole of H2 and 1 mole of O2 won't make 1 mole of H2O and 1/2 mole O2.

1 mole of $\ce{H2}$ and 1 mole of $\ce{O2}$ will make 1 mole of $\ce{H2O}$ and 1/2 mole $\ce{O2}$ will be left.
$\ce{H2 + O2 -> H2O + 1/2O2 (Left) }$
And here is an advice for you. If your teacher is saying that this will not happen at all, tell him to be sure that reactant in reaction are single molecules, not 1 mole.
Read this to be more clear on concept of limiting reagent.

Wow! Thanks for the great answer :) I'll definitely share your answer with my class mates!

I've said it before, but your edits are always appreciated! (I was trying to figure out a way to accept that one anyway, but I figured it would be good to point out the rationale, so please don't take it personally!)


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Does Zinc become black & dull after reacting with dil. HCl?
In the reaction where $\ce{Zn}$ reacts with $\ce{HCl}$ to give off Hydrogen gas & Zinc Chloride solution, does the zinc turn a dull black color after a while? If so, why?

Could you provide a little context? Are you asking because you tried this or saw it done and the zinc turned black? Are you trying to find a way to make zinc turn black?

Under which conditions does it happen? Only when you dip a zinc sheet into hydrochloric acid or also when you hold the sheet above to only react it with the hydrogen chloride in the vapour phase?

Look at it under a microscope.  If the surface is microporous, light goes into the reflecting cavities, bounces around, and is absorbed before it can escape.  Wilkinson Sword sold bolted stacks of double-edged razor blades as low power laser beam dumps,  Each sharp edge was mirror bright with a slight taper toward the edge.  Light bounced in and could not escape.  The sharp sides of the stack were astoundingly black.  375 blades x .004" = 1.5" square target.
Or, you have impure zinc and crud is accumulating at the dissolving surface.  Alloy 8111 aluminum foil contains about 2% or so of strengthening dispersed intermetallics.  Dissolving aluminum foil in acid or base leaves black residues.

The $\ce{Zn}$ cation polarises the $\ce{Cl}$ anions to a good extent and reduces the energy gap between certain energy levels. Therefore the electrons in $\ce{ZnCl2}$ are able to jump this gap by absorbing certain radiation, when this radiation is absorbed all the radiations except these are reflected and can be seen as complimentary color which in this case happens to be black.

It doesn't become pitch black, but it does become dull gray/dark gray. The reason is that even if metal surface is mirror bright, it is usually composed of multiple crystals with slightly different composition and orientations. Because of this differences, different faces of the microcrystals are etched with different speed, forming a complicated surface shape.
For the same reason, fine metal powders usually have dark coloration.

I've been doing this as art for a while and when I let the muriatic acid dry over night on the zinc it not only turns perfectly black but becomes  amazingly hard to sand down or buff back for aesthetic purposes.The image not only illustrates the deeper darker parts but near the center of the image is a dark area that is dark and not deep and I used a muriatic wash to try to get it off. It even resisted that. So my thought was that it was some type of zinc oxide that had occurred and was harder than the zinc.
Curious about everything .
Charlie W.


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Group Theory and Chemistry for Undergraduate Thesis
I am advising a student on her senior math thesis.  She would like to involve chemistry somehow (her other major).  I know no chemistry.  I have read in several places that representation theory features prominently in some areas of chemistry.  When I try to read anything online I am not sure how much chemistry one needs to assume.  So, if my student is in her senior year of a chemistry major, are there any places where she can start to look for a project in group theory and chemistry?

Dover has a good book in this area, but I am not enough of an expert to compare it to other references http://www.amazon.com/David-M.-Bishop/e/B001IQW5FI/ref=sr_ntt_srch_lnk_1?qid=1379379257&sr=8-1

I think in every Physical Chemistry book there is a chapter about group theory I've study it from Atkins

And every Inorganic Chemistry text should have multiple chapters applying group theory and symmetry to the structure and spectroscopy of organometallic compounds.

The first 4 chapters of Chemical Applications of Group Theory by F A Cotton* should be enough for your student. However there are numerous books available.
Again LibreText Chemistry (formerly known as UC Davis ChemWiki, link via the Internet Archive) has a good documentation on group theory.
* The Internet Archive holds both the first edition by 1962, as well as the third by 1989. Both are  available as .epub and .pdf for a 14 day loan.

Several answers and comments have suggested general texts that have good foundation information; however if a student is writing a thesis, these sources will be insufficient since they are geared towards the teaching of group theory rather than the application of group theory to generate new knowledge.  Even at an undergraduate level, there should be an attempt to generate new knowledge in a thesis.  To do this, the student would need to get her nose in the chemical literature, and I assume any student in an ACS accredited undergraduate program has experience with literature searches.  My very brief survey of group theory reviews came up with the following potentially interesting areas:

understanding the structure of natural and synthetic perovskites, which have a number of technological applications. Journal of Physics: Condensed Matter (2013), 25(17), 175902
Stereochemistry of medicinal compounds:  Helvetica Chimica Acta (2013), 96(1), 4-30.
the role of group theory in furthering our understanding of superconductivity:  Comptes Rendus Physique (2011), 12(5-6), 532-541.
Group theory and biomolecular conformations: Journal of Physics: Condensed Matter (2010), 22(32), 323103/1-323103/21.
A somewhat esoteric area of nuclear point groups:  Acta Physica Polonica B (2013), 44(3), 305-318.

Almost certainly, a senior has encountered the importance of symmetry in Inorganic Chemistry and IR spectroscopy, and these areas may be more accessible to the student.  In the end, however, good questions and topics for a senior thesis are going to require getting ones nose in the literature.

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text:
Do the names of some elements have anything to do with planets?
There are elements like neptunium and plutonium in the periodic table. Did their discovery have anything to do with Neptune and Pluto? Or are they randomly assigned with such names?

Neptunium (93) and plutonium (94) were so named because they followed uranium (92) on the periodic table.
Uranium was named after Uranus. German chemist Martin Heinrich Klaproth discovered uranium in 1789, eight years after William Herschel discovered Uranus. At the time, uranium was the densest element known and Uranus was the farthest planet from the sun known.

"At the time, uranium was the densest element known and Uranus was the farthest planet from the sun known." Is that a correlation or a causation? I mean, was uranium named after Uranus because they were densest and farthest (respectively) or because Uranus had been recently discovered?

Uranus is far from the densest planet  (actually, Earth is so), so I assume Uranium was named because of its being discovered shortly after the planet. Elements around it in the Periodic Table are rarer and had to wait longer for discovery.

Asked and answered, but this is a good place to include other astronomically named elements:

cerium named after the asteroid Ceres (or the Roman goddess of agriculture)
helium from the Greek helios or "Sun"
mercury OK, named after the Roman god, but so was the planet
palladium named after the asteroid Pallas, discovered the year that Pd was first isolated
selenium from the Greek "selene" for "Moon"
tellurium from the Greek "tellus" for "Earth"
krypton Sorry comic fans, the name predates Superman

Naming information is from here and a very nice element etymology website.  Now that you know something, you should probably take a mentally stimulating diversion.


text:
Calculate vapor pressure from volume
For a chemistry lab, I'm required to calculate the vapor pressure of some organic solvents... methanol, ethanol, and propanol, in addition to an "unknown" sample of some solvent. In the lab, we were instructed to heat these solvents in a sealed syringe along with 5 mL of air. While they were being heated, I recorded the volume and temperature of the air and solvent in the syringe. Now, from that data, I need to calculate the vapor pressure of these substances at the temperatures at which I recorded the data. How can I do this?

Hi Zane, didn't you first do a “dry run” with only air and no solvent in the syringe? (here is one possible experimental protocol for such measurement).

Nope. We weren't instructed to do so

You should treat everything as ideal gas. Then you calculate the volume change from the air itself, using ideal gas law and pressure being constant (atmospheric, maybe). Calculate also the number of moles of air in the syringe, it will come handy soon.
The remaining change in volume is then attributed to the solvent, again using ideal gas law, you put in the given volume, temperature, pressure and obtain number of moles, which turned into vapor.
Having this, you know number of moles of solvent and number of moles of air, so again assuming everything ideal (Raoult's law), you know the partial pressures (i.e. vapor pressure).
For added fancines, you can take the decrease of the liquid volume into account.
Side note: While using ideal gas and ideal mixture of ideal gases approximation, all quantities are nicely additive, so don't be afraid.


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What is advantage of creating elements that disappear in a flash?
More than dozen of heavy elements of periodic table do not occur naturally but are prepared artificially at nuclear reactors.. These elements are said to disappear in a flash..
So, why invest lot of money for the discovery of such elements? What are the advantages of such elements which disappear in a flash?

Researchers are looking for the Island of Stability, a predicted group of heavy nuclei with halflives in the minutes or even days.
While we have some understanding that certain nuclear configurations and compositions are more stable, we are currently unable to reliably to a priori predict the stability, the half-life, or even the decay mechanisms of a nuclide. Thus, when a new element or isotope is produced, no one knows how long it will be around. Every time a new nuclide is produced, observations of its decay help refine the models of nuclear stability and decay.
The Island of Stability is believed to be centered around 120 protons and 180 neutrons. Element 118 $\ce{^194Uuo}$ has been produced, and while its most stable isotope has a half-life of only 0.89 ms, it is very close to the island of stability and there may be a heavier, more stable isotope of $\ce{Uuo}$ out there.
As in many areas of pure science, the benefits to society (or even other fields) may not be evident except for those few who are actually doing the work. The outcome of pure science research is knowledge and information. Consider, for example, that research on cathode rays led to the development of the television decades later.
The scientists studying cathode rays likely could not even envision their work being applied to the projection of moving pictures (especially those transferred over the airwaves as radio signals), as such technology did not exist or was very new at the time.


text:
Which silver halide is the most soluble?
Halides are group 7 elements. They generally have a low melting and boiling point. They are highly reactive. Silver chloride and silver bromide are not soluble in water. So they will be precipitates. However, I do know that they have varying solubilities. So, they will form different amounts of precipitates. I want to know why they have different solubilities and which is the most soluble halide.

The solubilities of silver halides decreases down the periodic table:

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

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


text:
How is it that the equilibrium constant does not depend on the mechanism?
For a reaction of the form
$$\ce{aA + bB <=> cC + dD}$$
the equilibrium constant is
$$K_c=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$$
regardless of the mechanism of the reaction. Why is this the case?
I've seen derivations that use a reaction in one elementary step to demonstrate this, but this obviously doesn't work in general. The sample reaction used was $\ce{N2O4 <=> 2NO2}$.

You should indeed be able to write every equilibrium reaction in this simple form. I guess that the reason for this might lie in the principle of microscopic reversibility (shameless self-plug: see this answer of mine). But for non-elementary reactions, i.e. those whose reaction mechanism consists of more than one step, you get a more complicated equilibrium constant that is the product of the equilibrium constants of all the intermediate reaction steps (if the reaction mechanism has no branches). An example: Consider the reaction
\begin{equation}
  \ce{H2(g) + I2(g) <=>> 2HI(g)} \ .
\end{equation}
Its mechanism consists of three reaction steps:

$\ce{ I2 <<=> 2I } \qquad \qquad \qquad \qquad K_{1} = \frac{[\ce{I}]^{2}}{[\ce{I2}]}$
$\ce{I + H2 <<=> H2I} \ \ \quad \qquad \qquad K_{2} = \frac{[\ce{H2I}]}{[\ce{I}] [\ce{H2}]}$
$\ce{H2I + I <=>> 2HI} \quad \qquad \qquad K_{3} = \frac{[\ce{HI}]^{2}}{[\ce{I}] [\ce{H2I}]}$

Now, take a look at the equilibrium constant of the overall reaction
\begin{equation}
  K_{\mathrm{tot}} = \frac{[\ce{HI}]^{2}}{[\ce{I2}] [\ce{H2}]} \ .
\end{equation}
Then make the following substitutions: From the third reaction step's equilibrium equation you get $[\ce{HI}]^{2} = K_{3} [\ce{I}] [\ce{H2I}]$. Substituting this into the equation for $K_{\mathrm{tot}}$
\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] [\ce{H2I}]}{[\ce{I2}] [\ce{H2}]} \ .
\end{equation}
From reaction step 2 you get $[\ce{H2I}] = K_{2} [\ce{I}] [\ce{H2}]$ which then gives
\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} [\ce{I}]^{2}}{[\ce{I2}]} \ .
\end{equation}
And finally, from reaction step 1 you get $[\ce{I}]^{2} = K_{1} [\ce{I2}]$. Substituting this into the equation for $K_{\mathrm{tot}}$ you see that the overall equilibrium constant reduces to the product of the equilibrium constants of the reaction steps
\begin{equation}
  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} K_{1} [\ce{I2}]}{[\ce{I2}]} = K_{1} K_{2} K_{3} \ .
\end{equation}
If you have a reaction that doesn't proceed along a chain of reaction steps but branches out the situation will become more complicated but can be treated in an analogous way as above.

Because intermediates doesn't appear in the overall reaction does this mean that their concentrations are zero at equilibirum? Otherwise can we detect them at equilibrium? Is this experimentaly confirmed?

Philipp's answer is a nice application of Hess's Law to equilibrium. Let's look at why this approach works.
The position of equilibrium (as given by the equilibrium constant $K$) is related to the standard Gibbs free energy change that occurs during the reaction:
$$\Delta_rG^\circ=-RT\ln{K}$$
The Gibbs free energy $G$ is a state function - a property that depends only on the current state of the system, not the pathway by which the system reached that state. The value of $\Delta_rG^\circ$ depends only on the values of $G^\circ$ (or $\Delta_f G^\circ$ if you prefer) of the reactants and the products, which depend only on the identity and amounts of the substances. $\Delta_fG^\circ$ is independent of the way in which any of the substances were prepared.
Likewise, $K$ is a state function. $K$ describes the compositions of the mixture which qualify as meeting the requirements for equilibrium. The value of $K$ is independent of the amounts of any specific substance or the pathway (mechanism) by which equilibrium is reached (see Philipp's answer). The individual equilibrium concentrations, however, in the system are not state functions. $[\ce{A}]_{eq}$ depends on the initial concentrations of all species in the mixture.
The rate of reaction is dependent on the mechanism. Different mechanisms may have different orders, different activation energies, etc. Reaction rate (and consequently the rate constant $k$) are not necessarily state functions.

This answer is the same as the one by Philipp, but instead of using a specific example, it makes the argument for any multi-step reaction where each step is an elementary reaction.
The overall reaction is a sum of elementary steps
The overall reaction happens because elementary steps happen consecutively. Along these multiple steps, some species are reactants or products (i.e. appear in the chemical equation of the overall reaction) and others are intermediates (i.e. made by one elementary reaction and consumed by another).
The overall chemical equation is equal to the sum of the elementary steps, with intermediates appearing both on the reactant and the product side (those can be removed to arrive at the net chemical equation of the overall reaction).
At equilibrium, all elementary steps are at equilibrium
The only way the reactant and product concentrations can be constant (i.e. at equilibrium) is when all the intermediates are at equilibrium as well. You arrive at the equilibrium constant of the overall reaction by multiplying the equilibrium constants of the elementary reactions.
The rate laws of the elementary steps have the stoichiometric coefficients as exponents
This is the definition of an elementary step. The rate law for the forward reaction is the product of the reactant concentrations raised to the power of their coefficients. The rate law for the reverse reaction is the product of the product concentrations raised to the power of their coefficients. (Many of these reactants and products are intermediates with respect to the overall reaction.) At equilibrium, the forward and reverse rates are equal, so product concentrations divided by reactant concentrations (raised to respective coefficients) for the elementary steps will equal the quotient of reverse and forward rate constants, and we will call this equilibrium constant.
The equilibrium constant expression of the overall reaction is the product of the equilibrium constant expression of the elementary steps
Because the equilibrium constant of the overall reaction is equal to the product of the equilibrium constants of all the elementary steps, we can derive the equilibrium constant expression of the overall reaction by writing the product of the individual equilibrium constant expressions.
Just like intermediates cancel out of the net chemical equation because they appear both as reactant and as product, concentrations of intermediates will cancel out of this derived equilibrium expression (they appear both in the denominator - as reactant of one step - and in the numerator - as product of a different step). What remains after canceling are concentrations of reactants and products of the overall reaction, raised to their respective coefficients in the net reaction.
So whatever the coefficients are in the chemical equation of the net overall reaction will appear as exponents in the equilibrium constant expression for that same net overall reaction.
For an example, look at the answer by Philipp.

"At equilibrium, all elementary steps are at equilibrium". Does this mean that when we have a catalyst some quantitity of it is consumed as the steps that catalyst participate are all in equilibrium?

@adosar The catalyst is not so much consumed as it is "busy", e.g. bound to one of the intermediates.

"The rate laws of the elementary steps have the stoichiometric coefficients as exponents." I don't think I agree with you on this. For elementary reactions, the rate laws don't depend on stoichiometric coefficients, but on its molecularities. For example, for the unimolecular reaction methyl isonitrile to acetonitrile: H3CNC -> H3CCN. The rate law for this reaction is k[H3CNC]. We can change the stoichiometric coefficients of this reaction to 100: 100 H3CNC -> 100 H3CCN. But the rate laws will still be k[H3CNC], and not k[H3CNC]^100

@cuong.pq Yes, that is accurate. If someone writes an elementary step, though, I would hope they would use coefficients that correspond to the molecularities.


text:
Why are sigma bonds stronger than pi bonds?
Why are sigma bonds stronger than pi bonds?

Simply, it is due the fact that the extent of overlap in sigma bonds is greater than that in pi bonds.
It is also interesting to note that because of this difference in strength, a double bond is not twice as strong as a single bond (which is sigma) because one of them will be a pi bond.

There aren't any cases that I am aware of that have $\pi$ bonds but no $\sigma$ bonds so that the strength of each could be measured independently of the other to substantiate the assertion in the first place. (Rotational barrier of ethylene notwithstanding.)

The reason behind this is the orientation of the overlapped orbitals. Sigma bonds result from head-on(co-axial) overlapping while pi bonds are outcome of lateral(para-axial) overlapping. Here is a pictorial representation of ethene(sp2 hybridized C atoms) :

The greater the extent of overlapping, the higher the probability of finding the valence  electrons in between the nuclei and hence the bond will be stronger & shorter.
In MOT, this can be explained using Overlap Integral. This is how Atkins depicts it :

In simple terms, after forming a sigma-bond (a pre-requisite for pi-bonds), the two atoms get locked along the inter-nuclear axis. As a result, the orbitals available for pi-bonding can only partially overlap, thus forming a weaker bond.

Bond strength and bond length do not necessarily correlate. With a higher overlap, there is also more inter electron repulsion (derived from the Pauli principle). It is also untrue, that a $\sigma$ bond is a necessity for a $\pi$ bond. The reason why $\pi$ bonds are weaker is the same as why p-orbitals have higher energy than s orbitals.

@Martin : I presented as far my understanding. I appreciate that you point out the anomalies. You can post an answer or suggest an edit(doesn't seem viable due to lots of changes), if you care to !

As stated previously, it is due to the head-on overlap of sigma bonds and the lateral overlap of pi-bonds.
The smaller overlap of pi bonds also explains why double and triple bonds basically exist only for 2nd row elements (C,N,O especially) and not for higher row elements. A C=C bond has a length of 133 pm. A Si-Si bond length is at around 186 pm, therefore the contribution of the pi-integral overlap is almost negligible.
The inability of Silicium  to form strong pi bonds is also one of the answer why life on earth is carbon based and not silicium based, since the richness of organic chemistry is in part due to the ability of carbon to form strong double and triple bonds.

Unsaturated compounds (due to the existence of a double bond) are more reactive than saturated compounds, because the electron density at the internuclear axis is zero in a pi bond (in contrast to the sigma bond).

Pi bonds involve sideways overlap while sigma bond involves head on or axial overlap. Axial overlaps have higher degree of overlapping than sideways overlap.

How does this answer differ from the accepted one, apart from it is shorter and has no explanation?


text:
Why is sulfuric acid used in crystallization
I read that:

Take a cupful of water in a beaker and add a few drops of dilute
  sulfuric acid. Heat the water. When it starts boiling add copper
  sulfate powder slowly while stirring continuously.
  Continue adding copper sulfate powder till no more powder can be
  dissolved. Filter the solution. Allow it to cool. Do not disturb the
  solution when it is cooling. Look at the solution after some time. Can
  you see the crystals of copper sulfate? If not, wait for some more
  time.

So why is sulfuric acid required in this experiment?

The sulfuric acid will dissolve any residual copper metal or any copper oxides that exist as impurities in the copper sulfate, converting those impurities into copper sulfate.
$$\ce{H2SO4(aq) + Cu(s) -> CuSO4(aq) + H2(g)}$$
$$\ce{H2SO4(aq) + CuO(s) -> CuSO4(aq) + H2O(l)}$$
$$\ce{2H2SO4(aq) + Cu2O(s) -> 2CuSO4(aq) + H2O(l) +H2(g)}$$

Even if no impurities are present in your copper sulfate,  your water might be slightly basic. (See this question.) Also water typically has dissolved $\ce{CO2}$, which in neutral solution means some $\ce{HCO3-}$.
By lowering the pH throught he addition of dilute sulfuric acid you ensure that the concentration of $\ce{OH-}$ and $\ce{HCO3-}$ is extremely low. Either of these ions, if present in significant amounts, will cause some coprecipitation (since both  copper hydroxide and copper carbonates are much less  soluble than copper sulfate) and as a result you wouldn't get nice big crystals of pure $\ce{CuSO4\cdot 5H2O}$.

I do think the acid is to get rid of the CO2 species. It isn't a problem of coprecipitation but rather nucleation. Any species of copper carbonate precipitating would form nucleation sites for the copper sulfate. So instead of getting a few large crystals forming slowly you'd get a zillion small crystals.


text:
Enthalpy: when and how does it vary?
I have been looking for a physical explanation for entropy and enthalpy variations. An example would be the air system of a vehicle, where people say we have a strong enthalpy and entropy variations along the exhaust line, which is not the case in the fresh air line. How could that be possible?

Do you have another example for your problem? I'm not sure I really understand what your question is about by considering your given example (although I can't say exactly why - so, another example might help).

Hi @Philipp, I just updated the post and added a picture. This is a valve and the arrows are representing the fluid flow. What I still can't figure out hier regarding enthalpy/entropy, is the fact that there will be a substantial increase in the entropy accross the valve, however, the enthalpy remains substantially constant. Do you have an idea?


text:
The ionization energy of H₂⁺ molecule ion
I am trying to compute the ionization energy of $\ce{H2+}$ molecule ion from the electronic energy spectrum. The question is whether one should use the purely electronic Hamiltonian or the Hamiltonian with the internuclear repulsion term to do that? I illustrated two ways to compute the ionization energy below. Which one is correct?

Notice that, nuclei are not at the equilibrium distance since in my problem I can manage this distance by external forces.

I would like to clarify the matter since from OP's own answer I have a feeling he/she does not quite understand it.
So, first of all, in general the ionisation energy has to be calculated as the difference in total energy and not just the electronic one.
Secondly, for molecules there are two types ionization energies: adiabatic and vertical:

The adiabatic ionisation energy refers to the formation of the
  molecular ion in its ground vibrational state and the vertical
  ionization energy applies to the transition to the molecular ion
  without change in geometry. (From IUPAC Gold Book)

Note that in accordance with the Franck–Condon principle the most probable transition is the vertical one. Consequently, usually, when we say "ionisation energy" we actually mean the "vertical ionisation energy".
Finally, if the total energy is understood in the clamped nuclei approximation, i.e. as the electronic energy $E_{\mathrm{e}}$ together with the nuclear repulsion energy $V_{\mathrm{nn}}$, then for the vertical ionisation energy one could indeed use the difference in electronic energy rather than in total one simply because the nuclear repulsion energy stays the same. Mathematically, $V'_{\mathrm{nn}} = V_{\mathrm{nn}}$ since geometry is not changing, and thus,
$$
  E_{\mathrm{i}}
  =
  U' - U
  =
  (E'_{\mathrm{e}} + V'_{\mathrm{nn}})
  -
  (E_{\mathrm{e}} + V_{\mathrm{nn}})
  =
  E'_{\mathrm{e}} - E_{\mathrm{e}} \, ,
$$
where primed quantities refers to the resulting molecular specie and non-primed are that of the original molecular specie.
So, yes, for the vertical ionisation energy in the clamped nuclei approximation one can use the difference in electronic energy. And one can do so regardless of what are the molecular species in question. For the specific case of the vertical ionisation of $\ce{H2+}$ the electronic energy of the resulting $\ce{H2^2+}$ specie is obviously zero, so we specifically get what OP suggested
$$
  E_{\mathrm{i}}
  =
  E'_{\mathrm{e}} - E_{\mathrm{e}}
  =
  0 - E_{\mathrm{e}}
  =
  - E_{\mathrm{e}} \, .
$$
However, this is just a trivial simplification of a general formula for the vertical ionisation energy for the case of a zero-electron resulting specie.

Finally, I have found the answer. The correct way is the second one, presented in the right-hand side figure. Thus, the ionization energy is the difference between zero level and pure electronic energy. One could compute the ionization energy from the potential curve as well, but, in this case, the ionization energy is a difference between the potential energy of $\ce{H2^+}$ and $\ce{H2^{2+}}$. The potential curve for $\ce{H2^{2+}}$ just the Coulomb repulsive potential and, thus, we subtract the repulsion potential from the potential energy curve that leads us to the pure electronic energy.

Both ways are correct (in the clamped nuclei approximation). The first one is strictly according to the definition, the second one is the trivial consequence of considering the vertical rather than an adiabatic transition as well as the resulting specie being the zero-electron one.


text:
Nuclear fusion reactions violate law of conservation of mass, but it has not yet been taken into account. Why?
According to the law of conservation of mass, during any physical or chemical changes, the total mass of the products is equal to the total mass of the reactants. But in nuclear fusion reaction if considered an example, energy that sun emits in its core is due to collision of hydrogen nuclei and formation of helium nuclei. Here conservation of mass is not obeyed as certain part of mass is converted into energy. So, law of conservation of mass is violated here. Then law of conservation must be redefined as: during any physical or chemical change, the total mass of the products is equal to the total mass of reactants provided mass has not undergone conversion into energy.

If you consider the mass-energy equivalence that Einstein discoverd there is no contradiction: http://en.wikipedia.org/wiki/Mass-energy_equivalence

This law apilicable only chemical reactions. If gases are products ,reactants shuld be take in closed system.

Since nuclear physics is physical the first sentence should read; "According to the law of conservation of mass, during any mechanical or chemical changes,..."

The mass loss/gain in chemical reactions is extremely small, but it exists according to relativity. For example, the complete detonation of $100\ \mathrm{g}$ of TNT produces approximately $99.99999999\ \mathrm{g}$ of chemical products, with the remaining $0.00000001\  g$ of the initial mass transformed into energy. In chemical reactions, the mass difference between all the reactants and all the products is on the order of 0.1 parts per billion or less, so mass may seem conserved, but it is not. Compare this to nuclear fusion, where the mass difference can reach the order of 1%.

take out a piece of paper, weigh it, burn it, and weigh the ash of the paper, draw a conclusion.

Nuclear reactions appear to violate both the Laws of Conservation of Mass and Energy because mass is converted into energy or vice versa.
However, the concept of mass-energy equivalence that emerges as a consequence of Einstein's General Theory of Relativity makes the Laws of Conservation of Mass and Energy special limiting cases of the Law of Conservation of Mass-Energy. Since mass is now a form of energy, we can convert back and forth as long as there is no net loss between the two.
The equivalence of mass and energy comes from Einstein's famous $E=mc^2$, where $c$ is the speed of light (in a vacuum).

The Law of Conservation of Mass is no longer valid in its original formulation under the mass-energy equivalence given by $E=mc^2$ - not only is it not violated, it no longer even applies.

... However, Conservation of Mass is still taught in elementary chemistry, because chemical reactions do not result in the changing of atoms from one element to another (or even one isotope to another) as nuclear reactions do. As such, CoM and CoE hold for chemical reactions, and the laws remain useful teaching tools in this capacity.

@KeithS I'd say that in mass is actually transformed in energy in chemical reactions (and vice-versa). However, the amount of mass and energy involved is so tiny that it can be neglected for most of practical purposes and CoM and CoE are still very useful in elementary chemistry. It's the same as situation in physics: Newton's laws don't hold, but they are still very useful when we can neglect relativistic effects.

Since this thread seems to have been bumped by @Community a better answer is in order. Physicists don't typically make a distinction between rest mass $m_0$ and moving mass $m$ but rather consider mass to be the invariant of the energy-momentum $4$-vector: $mc^2=\sqrt{E^2-\boldsymbol{p}^2c^2}$ where $m$ is the invariant mass of the system, $E$ its total energy and $\boldsymbol p$ its total momentum. $\boldsymbol p=\sum\boldsymbol p_i$, the vector sum of the momenta of the system's constituents. For a massless particle, $\boldsymbol p_i=E_i\hat{\boldsymbol T}_i/c$, where $\hat{\boldsymbol T}_i$ is the unit vector in the direction of propagation of the particle, while for a particle with mass $m_i$ its momentum is $\boldsymbol p_i=\gamma_im_i\boldsymbol v_i$ where $\boldsymbol v_i$ is the particle's velocity and $\gamma_i=1/\sqrt{1-\boldsymbol v_i^2/c^2}$ is its Lorentz factor.
The mass is invariant under Lorentz transformations which is to say that's it's the same in all inertial frames, so even when moving we consider the mass of the system to be the same. As an extreme example of a reaction that seems to convert mass into energy, consider $e^+e^-$ annihilation. In the rest frame of the $e^+e^-$ pair, $\boldsymbol p_+=\boldsymbol p_-=\boldsymbol0$ so $E_+=E_-=m_ec^2\approx0.511MeV$. So the total energy is $E=E_++E_-=2m_ec^2$ and the total momentum is $\boldsymbol p=\boldsymbol p_++\boldsymbol p_-=\boldsymbol0$ and so the invariant mass of the system works out to $m=2m_e$.
After annihilation, we get a pair of photons and from momentum conservation we get that $\boldsymbol p_1=-\boldsymbol p_2$ and energy conservation tells us that $E_1=|\boldsymbol p_1|c=E_2=|\boldsymbol p_2|c=m_ec^2$ but what happened to the mass? Doesn't a photon have zero mass? A photon does in fact have zero mass, but a system of two photons has nonzero mass unless the photons are moving in the same direction. Take the $x$-direction to be the direction of travel of the first photon and let the other photon travel in the $xy$-plane with non-negative $y$-velocity. Then
$$\begin{align}\boldsymbol p^2&=p_x^2+p_y^2=(p_{1x}+p_{2x})^2+p_{2y}^2=\left(\frac{E_1}c+\frac{E_2}c\cos\theta\right)^2+\left(\frac{E_2}c\sin\theta\right)^2\\
&=\frac{E_1^2}{c^2}+\frac{E_2^2}{c^2}+2\frac{E_1E_2}{c^2}\cos\theta\end{align}$$
where $\theta$ is the angle between the velocities of the to photons. So for the system of two photons,
$$m^2c^4=E^2-\boldsymbol p^2c^2=\left(E_1+E_2\right)^2-E_1^2-E_2^2-2E_1E_2\cos\theta=4E_1E_2\sin^2\frac{\theta}2$$
So the mass of the system of two photons is
$$m=2\frac{\sqrt{E_1E_2}}{c^2}\sin\frac{\theta}2$$
In our example, $E_1=E_2=m_ec^2$ and $\theta=180°$ so we have $m=2m_e$, so the mass didn't change nor did the energy in the reaction of $e^+e^-$ annihilation. This is emphasized the the Wikipedia section cited earlier: mass and energy don't get interconverted but rather are separately conserved.
If this seems like sort of a theoretical definition of mass, where do you think most of the mass in your body is concentrated? I have had students who, after a lifetime of body shaming answer, "In my butt?", but the right answer is of course: "In your atomic nuclei!" and the quarks that make up your atomic nuclei are moving at a healthy fraction of the speed of light, so if we didn't use the relativistic $4$-vector definition of mass to compute your mass we wouldn't come up with a number that was all that close to your mass as measured by a balance or your resistance to acceleration.
After atomic nuclei or molecules react, the mass of the reactants is identical to the mass of the products, and the same holds for their momentum and energy. A well-insulated reaction vessel would have the same mass after reaction as before, it's only later as heat of reaction leaks out, carrying with it energy and also mass that the mass changes, immeasurably for a chemical reaction but significantly for a nuclear reaction. The mass may be almost unrecoverable if it's carried away by neutrinos in a beta decay or by gravitational waves in collisions of compact objects, but it's still out there in the universe someplace.
The masses of the individual components of the system in general won't be the same for reactants as they were for products, but masses aren't what add, it's the components of the energy-momentum $4$-vector that add like $4$-vectors and from the $4$-vector we can determine the mass of the system.

By considering that mass and energy are the two faces of the same coin then the conservation is valid according to Einstein concept.

I dont really think it violates the concepts of mass and energy conservation as dE=dmc2 and the speed of alpha beta and gamma rays are almost as equivalent as the speed of light hence taking in consideration that einstein stated that energy can be converted from mass if the body is in speed of light or nearly the speed of light ,the statement is correct.
.To be short and non-lengthy i can state that the basic concepts of mass conservation and energy convervation are violated but  the einstein's theory of relativity proves the above statement.

Could you reformat this to not be two run-on sentences? It's almost unreadable as-is.


text:
Will compressed hydrogen float in air?
Imagine a container of fixed volume is taken and hydrogen gas is filled in it, as a result the container floats in air. But if in the same container is filled with so much hydrogen that the hydrogen gas molecules had to compress to fit in the container, will the container still float in air ?

Yes. A simple example is an ordinary balloon, which places the gas inside it under pressure (because the tension of the rubber surface adds to the air pressure from outside the balloon), yet the balloon still floats readily.
What you're probably talking about is a theoretical container that can hold highly-compressed hydrogen, like from a supply cylinder, yet has negligible weight. At what point, then, would the hydrogen weigh more than air and the container would settle to the ground?
To answer that, we'll look at the ideal gas law: $PV = nRT$, or equivalently, $P = \dfrac{nRT}{V}$. For our purposes, we'll ignore the temperature increase of the compressed gas by saying this container, while having negligible weight, also has an efficient cooling system so the temperature delta is negligible.
At STP (273*K, 100kPa), one liter of volume filled with hydrogen gas contains:
$$100 = \dfrac{n(8.3145)(273)}{1}$$
$$100 = 2269.8482n$$
$$n = .0441\text{ mol}$$
With hydrogen having a molar weight of 1.00794 g/mol (and $\ce{H2}$ gas having double that), the density in g/L of hydrogen gas at STP is .08890 g/L.
Borrowing from ashu's answer, since his math and mine are within 1% error, the density of air at STP is 1.225g/L (1 m3 = 1000 L). So, in order to have the same mass of gas in the same volume, and thus the same density, you would need 13.78 times the amount of gas than you'd have in the same volume at STP (pretty much 1.225 mol). Because, given a unit volume, double the gas equals double the pressure, the resulting pressure of a container of negligible mass holding enough compressed hydrogen to not float in air is 1377.953kPa.
In SAE units, that's actually not much in relative terms; only 199.855 psi (call it 200 so it would actually sink in air and not be neutrally buoyant). That's the water pressure at a depth of about 460 feet (140.72m), which human freedivers have matched using dive sleds.

Kudos for the real-life comparison.

Here is a short answer to your question!
The compression will result in increament of density of hydrogen in the container.
Hydrogen will float in the air until its density becomes equal to air through the compression.
Hydrogen's density (0.08988 g/L) is  very much less than that of air's (1.225 kg/m3).
As soon as the density of hydrogen becomes more than that of air, it will settle down.
See this to know how does gas density increases due to compression.


text:
How did the first amino acids appear on Earth?
Is there proof or a theory that shows that the first amino acid syntheses on Earth were influenced by a change in heat, gravity, radiation, or something similar?

Welcome to Chemistry.SE. Since there is quite a large(!) gap between the presence of amino acids and the development of life I just want to be sure: You want the answers to your question to concentrate on the first amino acids on earth and not on the first life, right?

the question is concentrated in the first amino acids on earth what happened any theories or proves?
What happened to atoms to merge into this way

Are you looking for something like the Miller-Urey experiment: http://en.wikipedia.org/wiki/Miller%E2%80%93Urey_experiment

An earlier question maybe a duplicate.

first life??...

The universe cooled after the Big Bang. There was certainly an epoch when the universe’s overall temperature decreased from 200 °C to −40 °C throughout a very large volume. There is your test tube (more like a Florence flask) for abiogenesis and then the panspermia reservoir.
Before you suspect the preceding is clever, how and when (what ambient vacuum temperature) did elements heavier than boron arise?

Though this answer doesn't directly address the asker's concerns, I have to say you got me. I had never considered the possibility that life in the Universe could have arisen while the entire thing was around what we consider ambient temperature, and the idea hit me like a brick for a few seconds. Alas, the Universe cooled down far too quickly.

Redshift z = 140, 10.1 million years old, 111°C radiation temp. A fraction of zero metallicity hypergiant stars might have gone supernova to supply heavier elements, then chemistry. Redshift z = 110, 14.5 million years old, 29°C radiation temp. Complex organics from radiation chemistry with asteroids as substrate. 5-10 million years to evolve even unicellular life buried in asteroids' liquid water inclusions is inadequate. (Somebody post it in physics. It happened in the multiverse!  Once.)


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Migratory aptitude in Baeyer-Villiger reaction
In short, I do not understand how or why the migratory aptitude is as listed in every textbook I've read.
Firstly, they talk about a positive charge being built up that is stabilised by the movement of the most substituted group.
I cannot fathom such a build up in my mechanisms.
Please help me understand this in a mechanistic fashion.

The migratory aptitude list for the Baeyer-Villager oxidation is as it is because that is how the molecules undergoing the reaction behave. When we are learning about new reactions from a textbook, we often read about the mechanism and theoretical explanations of chemo-, regio-, and stereoselectivity. It is easy to forget that the experimental observations came first.
The migratory aptitude list $(\ce{H>R3C>R2HC}\approx\ce{Ph(>H)>RCH2>Me})$ is the result of many experiments. Nor is it a linear list as is usually presented. $\ce{H}$ migrates more readily than tertiary groups, which migrate more readily than secondary groups, which migrate more readily than phenyl groups (which migrate more readily than $\ce{H}$).
The mechanism and theoretical explanations came later. The partially complete mechanism from the Wikipedia page is shown below. The key step (center) is a concerted collapse of the tetrahedral intermediate with a [1,2]-sigmatropic shift of $\ce{R}^1$ and ejection of the carboxylate leaving group. Note that this mechanism shows none of the proton transfer steps.

Since the whole thing is supposedly concerted, I can understand your objection to the build-up of $\delta^+$ on $\ce{R}^1$. However, with the exceptions of the anomalous behavior of $\ce{H}$ and $\ce{Ph}$, the migratory aptitude list is the same as the carbocation stability list. Thus, researchers looked for a link between the two phenomena and a way to relate one to the other. The link may not be real, but the trends are there. To a student learning organic chemistry, it is much friendlier at first to apply an existing trend in reactivity to a new reaction (with an exception), than to learn a whole new trend that is unrelated but cosmetically similar.

How is $\ce{H}$ present twice in the list?

That is mysterious, isn't it.

I guess, if the intermediate shown above is protonated to any degree (conceivable under acidic conditions), that makes the $\delta^+$ more obvious. Without the protonations, the way I like to think of it is that the leaving group ($\ce{RCO2-}$) bears a negative charge, so the remainder of the molecule must have a positive charge spread out over it to compensate.

@BenNorris Can you please give the reason behind this mystery? I don't seem to understand this circular train of thought here. "The migratory aptitude list (H>R3C>R2HC≈Ph(>H)>RCH2>Me)' Why is the H present twice here? And how's that even mathematically possible?

Sorry for pinging you again, but, did you happen to have a closer look at my previous comment? @BenNorris

@GaurangTandon - I may need to amend the answer. The source of this circular order is my undergraduate textbook, but like many textbooks no references are given. I have not been able to find references that MCPBA or other peroxyacids will oxidize benzaldehyde to phenyl formate and not benzoic acid.

Which textbook is that? Please let us know.

@GaurangTandon - I think the order presented is completely experimental. Since in the reaction, you can only compare the relative migratory aptitude of any two groups only at a time. It might have happened like this - with tert-butyl aldehyde, H migrated (in the experiment). With tert-butyl phenyl ketone, tert-butyl migrated in the experiment. With benzaldehyde, phenyl migrated in the experiment. Individually, the three experiments indicate the relative orders as - H > tertiary, tertiary > Ph, Ph > H, and when folks attempted to combine these pairs, we run into trouble.

(Purely my speculation, of course, as an attempt to explain the seemingly impossible order).


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Food safe basic polyelectrolytes?
Pectin is an example of a food safe polyelectrolyte. It acts as an acid because it is a proton donor and develops negative charges along the polymer.
Can you think of any examples of food safe polyelectrolytes that act as bases and develop positive charges along the polymer?

I voted to close this as "Too Broad," but I also upvoted it, because I think you may be trying to ask something like "Is there any reason there can't be a food safe polyelectrolyte?" and I think that's a good question.

Humans (mammals?) like to eat stuff that is neutral or somewhat acidic. Basic stuff tastes more or less disgusting. Exception proving the rule are artisan indigenous specialities like the notorious fermented fish that some Scandinavians reputedly eat.

Gum arabic is negatively charged at pH above 2,3 and I believe there might be a synergy there. All depends on what application are you thinking of, or is it just an academic question?

Pectin is a heteropolysaccharide rich in galacturonic acid:

The carboxylic acid in galacturonic acid gives pectin the ionizable character. Pectin can react with bases to form carboxylate anions along the polysaccharide chain.
There exists a sugar-derivative and polysaccharide with opposite behavior - they react with acids to produce cations along the polysaccharide chain. The modified sugar is glucosamine, and the polysaccharides are chitin and chitosan(http://en.wikipedia.org/wiki/Chitosan). The amine ($\ce{NH2}$) group on glucosamine is basic and will react with acids to become cationic.

Glucosamine is used as a dietary supplement, and chitin and chitosan are found in various foods, like fungi and shellfish.


text:
Molecular chirality and optical rotation
Why does having molecular chirality result in optical rotation? The dissymetry or chirality of molecules translates to the rotation of plane polarized light, the magnitude and direction depending on the concentration and the nature of the substance. But why does molecular chirality cause the rotation of plane polarized light.
If this rotation is due to the assymetry around the bond, the different molecules are oriented in all possible directions(randomly) and hence the light will fall on them in all different orientations, and should ultimately be undeflected since for every orientation of the molecule, we can reverse the orientation such that the light appears to be falling on the molecule from a direction other than the one for our original molecule.
Moreover, why does inverting the configuration invariably reverses the direction of rotation of light?

It seems to me that you get very close to the answer. Citing you: "since for every orientation of the molecule, we can reverse the orientation such that the light appears to be falling on the molecule from a direction other than the one for our original molecule." In order to "reverse" the orientation, you would need to have a position that is exactly the mirror image of the opposite situation. And this is exactly what cannot happen for a solution of chiral molecules.

since for every orientation of the molecule, we can reverse the orientation such that the light appears to be falling on the molecule from a direction other than the one for our original molecule.

This is false.
Let's take 2-butanol. For this stereoisomer, light is turning clockwise when viewed from the right side (I'm not sure of this, but we can assume).

Here, the disks are there to clarify the direction from which we are viewing the entire setup. Here, the "face" of the disks is to the right so we are viewing it from the right.
By the principle of reversibility of light, if light enters from the other side, we get the reverse path:

Rotate this diagram (by $180^\circ$ around an axis perpendicular to your screen):

Note that the disks are being viewed from the other side (the left) now. But we can flip them:

Take a close look at this. This is the same molecule as the first, just oriented backwards. And looking from the right, it still gives a clockwise twist to the angle.
So while rays of light incident on different orientations of a molecule may differ in the exact value of the angle of rotation, rays of light incident on a reversed (not stereoisomeric, just spatially rotated) molecule have the same direction and value of the angle of rotation and do not cancel.
[8]:

Thanks for a great explanation. This explains a lot. On a side note, why does a molecule rotate he light which falls on it? Is it due to diffraction effects?

@SatwikPasani Good questions; see http://physics.stackexchange.com/questions/15503/how-do-optically-active-compounds-rotate-plane-polarized-light


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Thermodynamic vs. Kinetic products
With regards to complexes, I read that thermodynamic products are favored over kinetic products, which are labile. Why wouldn't the first product formed be the most prominent?

The difference between Thermodynamic and Kinetic products is the difference in activation energy and the difference in (Gibbs)-Free-Energy.
When the activation energy is low, the reaction proceeds fast, while the product might not have as low a Free-Energy as an other product (with higher activation energy). When the reaction conditions allow however, the thermodynamically more stable product (with lower Free-Energy) will be formed. Even though the kinetic product first forms, most complexes are form in equilibrium reactions. Therefore, the kinetic product reverts to starting materials and then, slowly forms the thermodynamic product. This thermo dynamic product, though it is also in equilibrium, will not revert to starting materials as easily as the kinetic product will, due to the lower Free-Energy.
To conclude: First, the kinetically favored product will form. Then, when the reaction conditions allow it, the equilibrium shifts towards the thermodynamic product.
Note that the equilibrium is essential for the thermodynamically favored product to form.


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What determines the density of an element?
I used to be under the (wrong) assumption that the density of an element correlates with it's atomic number $\mathrm{Z}$, I thought that since having more protons meant the atom weighed more; but of course that's wrong.
So why does make osmium $\mathrm{Os}$, with an atomic number of $\mathrm{Z}_{\mathrm{Os}}=76$ more dense than say uranium $\mathrm{U}$, with an atomic number of $\mathrm{Z}_{\mathrm{U}}=92$?

I think you would have more luck getting an answer on physics.SE. But don't re-ask it there if you indeed want it on that site, just flag your own post for migration

It's a fine question on Chemistry SE, borderline physics/chemistry. Our current policy is to let it fly here (where it already gather two answers); if you want more physics-oriented answers, asked an amended version of the question on Physics.

There are a couple of factors here that you need to consider.
First of all, increasing Z does mean that there are more particles in the nucleus; however, you're also adding electrons to the systems as well. Electrons dictate the size (i.e. volumen) of the atom, so the size of the atom could increase with increasing Z, which would mean that fewer atoms could fit into the same volume. To use your example, the covalent radius  of Os is 144 pm; the covalent radius of U is 196 pm. So just from this property alone, you would expect more Os than U atoms to pack into the same volume.
There is another factor to consider as well: the crystal system of the solid. At standard temperature and pressure, atoms for different elements in the solid state pack into different arrangements, which are called crystal systems. Some of these are much more efficient than others. For example, there are three variants of the cubic crystal system: primative, body-centered, and face-centered. From Wikipedia:

Assuming one atom per lattice point, in a primitive cubic lattice with
  cube side length a, the sphere radius would be a⁄2 and the atomic
  packing factor turns out to be about 0.524 (which is quite low).
  Similarly, in a bcc lattice, the atomic packing factor is 0.680, and
  in fcc it is 0.740. The fcc value is the highest theoretically
  possible value for any lattice, although there are other lattices
  which also achieve the same value, such as hexagonal close packed and
  one version of tetrahedral bcc.

Now, depending on which crystal system the atoms in the solid take, you could get more or less atoms per unit volume. Unfortunately, determining which crystal system is favored for different elements is not straightforward, and requires some knowledge of the bonding between the atoms and other factors.

Since the density of matter is the mass per unit volume, I think that, from the chemical point of view, lengths of chemical bonds matter. If we consider same element forming different substances, the substance with shorter bond length should be denser. The term "density of element" seems confusing for me. The density depends on the state of matter (solids, liquids, gas) and conditions including temperature etc.

While neucleons (protons and neutrons) are responsible for the mass of the atom, electrons are responsible for its size. The neucleons are in the tiny nucleus of the atom. The electrons take up a vast area around this necleus. If you keep on increasing the subatomic particles (electrons, protons, etc), defintely the atomic mass increased, but the volume (size) of the atom does not necessarily. Why?
Well, you know electrons are kind of orbitting around the neucleus (this simple model well suffices for this explanation) in orbits: s,p,d,f, etc... Each orbit means a shell which determines how far those electrons in a particular shell. A shell can have several fixed number of electrons, so when you keep on adding subatomic particles the mass necessarily increases, while the volume is necessarily not. What happens if a particular shell is already filled with its maximum?? Then the new electrons go to the next higher level/orbit which means a larger volume. If you think in this line of thought, you will be able to understand the densitiy does not linearly goes with the atomic mass, but there is a logic still.


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How do we represent a molecule in text form?
How do we represent a molecule in text form?
A molecule, as a computer scientist I see it as a connected graph of atoms. A connected graph means that beginning at any atom, there is a path to any other atom in the graph.
The interesting question is, how do we represent these as text?
Specific questions are:
1) Is $\ce{H2O}$ different than $\ce{OH2}$? That is, does order matter, or what does order mean?
2) What does this mean $\ce{NH4NO3}$? Why do we write N two times? Is this the same thing $\ce{NNH4O3}$?

http://openbabel.org/docs/dev/FileFormats/Overview.html#file-formats

The representation of a molecule in text form actually has a name: the chemical formula (pl. formulae).
The formula we use for everyday purposes is an empirical formula. That article should give you more information, but in a nutshell, it is a simple way of identifying a compound by the ratio of its constituent elements. Since ionic compounds and other compounds such as macromolecules can have huge absolute numbers of atoms which are technically bonded but in practice are subdivided into their simplest components, empirical formulae are useful for unambiguously identifying most compounds. Note, however, that we run into trouble when talking about isomers or other compounds with the same elemental ratio but different structures.
For these compounds, we must disambiguate by using a structural formula, which is basically a graph of the molecule. Translating this into text can be done many different ways, using a formal system for specifying branches and stereochemistry. The three most widely used formats for chemical database applications are SMILES, PDB, and CML. You can get more information on all these from Open Babel (alternately OpenBabel).
There is also a middle ground, using a formula similar to SMILES, variously called a molecular formula or condensed formula. It implies structure by counting every atom and splitting individual elements apart to show functional groups like hydroxyl ($\ce{-OH}$) or amino ($\ce{-NH2}$) bonded to the main structure.
I think this is what your question is really about. Yes, in a condensed formula, the order of elements and functional groups matters, because you are trying to communicate information about a compound's structure with a minimum of overhead. It wouldn't make sense to say that butyraldehyde is $\ce{C2H4CHOCH3}$ because the formyl group is at the end. Therefore we say butyraldehyde is $\ce{CH3(CH2)2CHO}$ to better represent the compound's structure. This helps distinguish isomers (but not stereoisomers). Even using the Hill system, one must still be careful.
I hope this clears things up for you on how chemists use different formulae in different situations, and how a condensed formula works.

There are many text representations of molecules. SMILES is commonly used but has issues in representing stereochemistry, to the MDL (now Accelrys) CTAB format, to InChi.
Certainly a molecule can be represented as a graph, in which the atom are nodes and the bonds edges.
In terms of the molecular formulae you asked about, for a simple molecule like water, a chemist would recognise H2O and OH2 as the same thing, but things quickly get out of hand when more complex molecules are represented in this fashion, and order becomes important. Any chemist would recognise NH4NO3 as ammonium nitrate, but may fail to immediately see it when represented as N2H4O3. We write the N twice because it makes the context easier to determine - ie that this is a salt, comprised of the ammonium anion and the nitrate cation.
I'm afraid that there are a number of contextual complexities when approaching cheminformatics from the viewpoint of a computer scientist - a knowledge of chemistry is just as important :)

I disagree with the comment that SMILES has issues in representing stereochemistry. By definition, SMILES exactly represents stereochemistry. That doesn't mean all software correctly handles it, but SMILES can specify all types of stereo information.

Agreed, however there exists contextual complexity that SMILES cannot handle.  For example, the drug thalidomide, which exists as a racemic mixture.  How to represent that ?  If drawn 'flat', ie no stereo representation, is the racemate assumed ?  If so, was it meant ?  If so, was it 50:50 ?  60:40 ?  What about chirality brought about by conformational constraint ?  I guess my comment added more complexity than was required by the answer to the OP :)

I think my comment was mainly aimed at indicating your answer needed to be more specific. You can represent stereochemistry in SMILES. I completely agree though that there are vast areas of chemical complexity which are hard to represent computationally, and EE and racemates are good examples. (I'm not sure there exists any format which can represent all of chemistry. Organometallics are a specific personal gripe.)

Correct me If I am wrong and also I am not sure that this answers your full question or not as it only provides an answer to your specific questions.
Here are some rules for writing chemical formulas. Kindly read them first.

1) Is $\ce{H2O}$ different than $\ce{OH2}$? That is, does order matter, or what does order mean?

Literally in these type of molecules, it doesn't matter that you write it as $\ce{H2O}$ or $\ce{OH2}$ because it will have no effect on its structure. These type of molecules follow Hill System

2) What does this mean $\ce{NH4NO3}$? Why do we write N two times? Is this the same thing $\ce{NNH4O3}$?

In this case, $\ce{NH4+}$ (Ammonium) and $\ce{NO3-}$ (Nitrate) are two different ions which makes $\ce{NH4NO3}$.
In these type of compounds, cation is written first and then the anion. In the formation of Ammonium Nitrate, $\ce{NH3}$ reacts with $\ce{HNO3}$ and then cation & anion formation takes place and reaction proceeds.

So basically these two N in ammonium nitrate came from two different molecules. By 'from different molecules' I mean that each N is bonded with different atoms, this is why we need to write it twice.
And I don't think that $\ce{NNH4O3}$ is the same thing. (I am not sure about it so kindly correct me if I am wrong here. and leave a comment with explaination.)
Note: As commented by @F'x, the order of elements in a chemical formula is a complex matter. See this and this, there are some good answers as mentioned by @F'x. I also found them useful.

Regarding your first answer (“cation is written first”): there is no cation in H2O. The order of elements in formula is a complex matter: see there, there and this other excellent answer

thanks for the beautiful knowledge. I also never thought about CH3 and H2O. I didn't knew about the information provided in answer. :) and I have edited my answer now. Kindly review it and edit if anything is wrong.!


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How should elements be colored? And how many bonds does an atom have?
I'm working on my own 3D molecule viewer for my AP Chem class. So far everything's going together nicely. There's a couple of things I'm not sure about though...
Is there a certain rule of thumb or standard for what colors each element inside a molecule should be? For example, Jmol renders oxygen red and hydrogen white:

For my app they are currently grey:

Also, how can I check if there are single bonds, double bonds, etc.? As you can see here, I haven't programmed that yet into my app. Tylenol:

I look up coordinates for each molecule in an NCI/CADD group server online database. This database gives me the number of elements, and the X, Y, and Z coordinates for where I should put each element. For example, water:
3
H2O
O   -0.0000   -0.0589   -0.0000
H   -0.8110    0.4677    0.0000
H    0.8110    0.4677    0.0000

It doesn't tell me anything about where bonds go. More info about .xyz file format here.

Might I ask which language you want are writing your app in? Because if it is C++ maybe you might be interested in Asymptote which is a vector graphics language with 3D support. With it you can 3D PDF documents and furthermore it integrates very well with LaTeX. Maybe you could use it to write you app. See my answer here: http://chemistry.stackexchange.com/a/560/189

I'm writing this in Java. I'm using it as an Android/IOS app for my peers in High School. I was originally using the source code of Jmol. It worked but was very laggy. I decided to write a much lighter program that carries out the same function. I currently get 60+ frames a second. With Jmol I got like 15-20.

Thanks for the info. I wish you good luck with your app. When you're finished I would very much like to try it out if you intend to publish it for everyone.

Related: Why do the atoms in molecular models have the colors they have?

For the coloring part I would suggest you look at the CPK coloring convention.
As for the decision whether a bond is a single or a double bond: You could make this decision based on the bond length between two atoms. But you will need to find typical values for single, double and triple bonds for every combination of elements which seems a bit too much trouble. On the other hand, the depiction of double and triple bonds is used mostly for bonds between carbon, oxygen, sulfur, phosphorus and nitrogen (for other elements like metals the bond order is not readily identifiable, so I would leave those as single bonds). So, the amount of data to be collected might not be that large.
A very naive way to make the decision how many bonds an atom has could be the following (though this might not be very efficient):

calculate the distances between the atoms
for each element set a maximum distance another atom can be away from an atom of the element in question to be considered a bonding partner
for each element set a maximum number of bonds $N_{\mathrm{max}}$ an atom of this element can have, e.g. 4 for carbon, 4 for nitrogen, 3 for oxygen, etc. (for metals this might be a bit difficult to decide)
for each atom in the system order the list of atoms within bonding distance from lowest to highest distance. From this list choose only the first to $N_{\mathrm{max}}$th atoms as bonding partners. Then look whether some of the bonding partners qualify for double or triple bonds - if so you have to kick out some of the bonding members from the end of your bond candidates so that you don't exceed your maximum bond number.

There will be better ways out there to do this but this one worked for me in the process of creating "static" pictures of molecules.
Update:
You asked what size you should use for your atoms. Again, there is no definite answer to this - use whatever looks best for you. Nevertheless, I found that the sizes calculated according to the following formula give rather good looking atom size ratios for the different elements (maybe you have to multiply the resultant values by a constant to give good default sizes):
\begin{equation}
  \text{size} = \frac{1}{2 \sqrt{2}} \cdot \sqrt[6 \,]{\text{atomic number}} \ .
\end{equation}
Furthermore, here is a nice source for some typical bond distances. Other good resources would be the CRC Handbook of Chemistry and Physics or the book by Holleman and Wiberg if you can lay your hands on them. Those might come in handy if you try to adapt my method and look for values for the maximum bond distances. Another advice concerning the maximum bond distances would be that typical values for the bond lengths of bonds containing (transition) metals are between 200 and 300 pm, so a value of 350 or 400 pm might be a good choice for the maximum bond distance for those species.
Concerning the maximum bond number each element can have: here, the typical rules for drawing Lewis structures apply. You can find good resources for this on this site, e.g. here, and in every introductory general chemistry text. But you should be mindful of some special cases. You should not forget the cationic forms of the elements, when thinking about their maximum bond number. For example, carbon usually has 4 bonds, but there are very rare cases where it comes with 5 bonds ($\ce{CH5^{+}}$) although those are practically not important and some not so rare cases where it has 3 bonds (carbenes, carbocations, carbanions), so you might consider a maximum bond number of 5 for carbon, although for nearly all practical purposes 4 might do. Another example is oxygen. It usually has 2 bonds but when it is protonated or in Meerwein's reagent ($\ce{Me3O^{+}BF4^{-}}$) it has 3 bonds. Transition metals are very flexible when it comes to the number of bonds they have - it can be everything from 1 to 6 sometimes even more, so you should choose a high number, e.g. 8, for them.

Is it necessary to include colors for different isotopes? For example: deuterium, an isotope of hydrogen.

I'd say that this is pretty much your choice. You don't need to stick with CPK coloring if you don't like (parts of) it. Some conventions are pretty widely spread, like white for Hydrogen or red for oxygen, but there are a lot of things I would consider "safe to change".

To extend on Philipp's excellent answer, I'd like to offer some advice on judging bond orders. Pekka Pyykkö and Michiko Atsumi published covalent (single-, double-, triple-) bond radii for (almost) the entire periodic table in Chem. Eur. J. 2009, 15, 12770-12779. It is summarised in figure 3. These are obtained from experimental or theoretical data.
The figure itself can be found on Pyykkö's homepage as a pdf-file. This version has been extended to include tetrahedral covalent radii. I include a screenshot for reference.

Another interesting and comprehensive publication for sizes of atoms is the list of Van der Waals radii of elements by S. S. Batsanov published in Inorg. Mat. 2001, 37 (9), 871-885. A mirror of the publisher's pdf can be found here. This is probably also a good starting point for deciding the default size of your atoms. Additionally it might help you identify weak interactions.
One last word on colouring: the mentioned CPK colouring scheme is widely known and many chemists understand it intuitively now, but it is not uncommon that programs specify their own schemes. That mostly depends of the background you would like to use. One of the most common deviations is carbon, which originally used black, but nowadays often is displayed grey because many viewers prefer dark backgrounds. Also frequently changed is the colour for hydrogen when using light backgrounds as white doesn't identify as well then.


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What causes the smell in rotten meat-grease and how can I get rid of it thoroughly?
We've just spilled the can of used meat-grease we collect so that it doesn't go down the drain and it turns out it was rotten. Now the granite counter-top and the wood-laminate floor of the kitchen smell very bad even after repeated cleaning with (1) dish-soap, (2) vinegar and (3) Murphy's Oil Soap as well as an enzymatic product called anti-icky-poo.
Extensive googling suggests that the smelly chemical is either putrescine or cadaverine. Am I right about that? How can I clean my kitchen such that it does not smell so bad?

did you try febreze?  not kidding...

If the stench is caused by putrescine or cadaverine, then you are in luck! Both putrescine and cadaverine are amines ($\ce{RNH2}$), which react with acids to form water-soluble ammonium salts ($\ce{RNH3+}$). Most soaps, surfactants, and other cleaning products are basic to help remove fats and oils (by hydrolysis). Try white vinegar, which contains acetic acid. If that fails, you can buy muriatic acid (a dilute solution of hydrochloric acid), which is used to clean pools. Find it wherever pool supplies are found.
Caution: While both of these solutions are fine for the granite, muriatic acid may not be fine for your floor - test in a small hidden area first. Vinegar should be okay for the floor as long as you don't let it soak.
Putrescine:

Cadaverine:

Thank you so much! The white vinegar did the trick, I just had to leave it soaking there for a few minutes.

I'm not sure about muriatic acid from the pool store, but muriatic acid from a home improvement store is likely to be on par with concentrated hydrochloric acid (about 12 molar, 37%). That concentration is extremely corrosive and not safe to wiping a kitchen.

Muriatic acid is HCl at about 6% m/v.

If you are right, and the smell is coming from an amine, then you may need to clean a lot more than you have.  Even if you think you've done a thorough cleaning, amines have very low odour thresholds.  I can't find specific numbers for putrescine and cadaverine; however, another amine generated in decaying animal protein, trimethylamine, has an odour threshold of 0.00021 ppm.
We can do a back of the envelope calculation here.  Volatile organic amines need to be at a concentration of 1 ppm in order for the vapor pressure to be high enough to generate an odour.  Your odour is "strong" so let's choose a concentration of 10 ppm.  The goal is to use our cleaning product to get the concentration of amines from their current level (10 ppm) to the desired level (0.00021 ppm) so that we don't smell it any longer.  I read on many bottles of commercial cleaning products "kills 99.99% of the germs!" so let's go with that number as the cleaning efficiency.  If I get rid of 99.99% of the amine in one wash, that's:
$$10-10\cdot0.9999=0.001~\text{ppm remaining}$$
That means we need two good cleanings to get the amines to a level where they cannot be detected by our noses.  Note that this is a simplified problem and assumes that the fats don't interfere with the cleaning efficiency and that you are able to clean every surface exposed to the grease with 99.99% efficiency.  These are typically the limiting factors in cleaning.
These types of odours are a serious issue for swine and poultry farmers and one of their approaches to eliminating odours is through proper air flow.  Get some fans going to dilute the smell, as this will make life a bit more manageable and will also help in ultimately eliminating the odour.

I think that the problem is that you have meat-grease that creates a hydrophobic layer inside tiny holes of your granite and wood.
I recommend, with all the necessary precautions (eyeglasses and gloves), a solution of hot caustic soda for your granite. This will saponify the fat so you can remove it.
For the amines, you have to use an acid: HCl and vinegar are good choices but I would try citric acid too. Hot solutions tend to be more efficient.