text:
How does hot tub "shock" work?
Hot tub shock is potassium peroxymonosulfate. When added to a hot tub, it creates bubbles. It is said to oxidize organic debris in the water. What is the reaction?

Could you elaborate a bit on the observed effects of a hot tub shock?

It bubbles when added to the hot tub. I misstated the chemical--it is potassium peroxymonosulfate.

There are several properties you can infer from the compound's name.

The compound is ionic, which the structure cation [...]-ate informs us. In this case, we're dealing with the potassium salt of peroxymonosulfuric acid. This should therefore dissolve pretty well in water.
The compound will be an oxidizer, in the conditions present in your bathtub. peroxy means the compound contains an $\ce{O-O}$ bond, which breaks up homolytically (one electron per oxygen) to give radicals (quite reactive species, electronically unhappy).

As per both hints, the compound is expected to be pretty effective at "eating" through organic deposits, by oxidizing them, rendering some soluble. It should also decrease pH, but not so drastically as it is already the conjugate base of sulfuric acid (which can let go of two protons).
About the bubbling: it most probably is $\ce{CO2}$ coming from the oxidation of the organic matter.


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Why does polonium form acidic oxides?
All metals form basic oxides. Metallic oxides are basic in character.
Polonium is a metal, so it should form basic oxides but it forms acidic oxides. Why are its oxides acidic in character?

Polonium is actually a metalloid, and not strictly speaking classified as a metal…

All metals form basic oxides.

That's true. I cannot remember metal, that does not form basic oxide

Metallic oxides are basic in character.

Not necessary. Polonium is not the only one metal, forming acidic oxides. Mn, Cr and some others do it as well.
Oxide is considered acidic, when it forms hydroxide, that is acidic (well, it is simplification, but let's stick with it for a while). Hydroxide is acidic, when it can lose proton, giving stable anion. Acidity of hydroxide and metallicity of element formed it does not connect any way. Acidity depends of ability of remaining part of molecule to stabilize negative charge, which usually correlate with $\ce{O/OH}$ ratio and electronegativity of central atom. So, $\ce{H6TeO6}$ is quite weak acid and $\ce{H2CrO4}$ or $\ce{HMnO4}$ are quite quite strong acids.

Isn't O^2- extremely unstable? I mean I can understand why O would want 2 more electrons but it doesn't really want to be ionic and for example Mg^2+ O^2- would be an ionic oxide(an oxide since O is more electronegative than Mg and ionic since the 2 have opposite charges) and oxygen wants to form covalent bonds like it does in Iron Oxide, not ionic bonds like a chloride would so ionic bonds to oxygen anions or cations are extremely unstable

@caters It isn't related to the topic, but in general there is no purely ionic or covalent compounds. Some oxides, like $\ce{MgO}$, have significant ionic bonding, but just as well there is no chloride that does not have significant covalent bonding. In addition, there is so known lattice energy, an energy, freed as result of bonding of ions into lattice. It is quite big and may be enough to stabilize some unstable ions.

MgCl2 has ionic bonding, not covalent. Same with NaCl. These are all ionic chlorides.

@caters CsCl, most ionic of all chlorides, has only 75% of ionic bonding and 25% of covalent bonding.

MgCl2 has significant covalent character, as can be seen from it's layer structure (acc. to JD Lee)... no time to elaborate on this, you can search yourself maybe

Another way of looking at this is to understand that the structure is always
$$\ce{M-O-H}$$
where $\ce{M}$ is the metal atom.  If the $\ce{O-H}$ bond is weaker than the $\ce{M-O}$ bond, the material acts as
an acid.  Otherwise it acts as a base.
What determines the relative strength is a bit complex.  Suffice it to say that it depends on oxidation state of the $\ce{M}$ atom as well as the constituents in the rest of the molecule and the environment all play a role.
Some materials are amphoteric.  That is they can split either way.  A typical
example is aluminum hydroxide, $\ce{Al(OH)_3}$, which has three $\ce{O-H}$ bonds.
It is usually a base, but under the right conditions it breaks the other way and
is a weak acid called aluminic acid.  The formula for that is usually written as $\ce{H3AlO_3}$, but the structure is the same.

not all metal oxides are this way. For example iron oxides have this structure: FeOx where x is the number of oxygen atoms. No hydrogens here. and yes oxides do tend to be basic When there are hydrogens in the oxide. Otherwise they tend to be acidic(which is how come CO2 acidifies water instead of making it more basic)

While virtually all metals from a basic oxide or amphoteric (reacts with both bases and acids) oxide like aluminium and both Tin(II) and Tin(IV), some also form acidic oxides like Chromium(VI) oxide, Iron(II) oxide.  With oxides its better to describe they according to their reactivity with acids and bases as you are not trying to convert them to the hydrate.
Its not incredible for Polonium to form only an acid oxide.

It does form basi oxide: https://en.wikipedia.org/wiki/Polonium_monoxide

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Can caffeine be separated from coffee?
Is there a way to extract caffeine from coffee completely?

In addition to Richard Terrett's nice answer, this little article from my former neighbouring research institute gives some historical background information on decaffeination with supercritical carbon dioxide.

Regarding "completely":
The short answer is no.
But for all intents and purposes, yes.
By submitting coffee to several small volume extractions, one can drastically reduce the concentration of caffeine present in the grains. This method of extraction (multiple small volume aliquots) is more effective than a single large volume extraction. The link I shared provides a good discussion.

Yes, this is how decaffeinated coffee is made. If I recall correctly, this is done commercially by dissolving the caffeine in dichloromethane (DCM), or more recently in supercritical carbon dioxide. This switch is probably over concerns regarding chronic industrial exposure to DCM, which is mildly toxic.
From personal experience, I have extracted caffeine from instant coffee using a separatory funnel with a system of water and dichloromethane. Purification by sublimation finished the job.

And how did the coffee taste afterwards? :3

@Aesin I'm assuming there would be loss of other organic compounds in the process, thus potentially changing the taste so one can discriminate between caffeine-free and the real deal. I've never tasted it either!

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Understanding the rate of change in titration curves
As I am looking at titration curves, a few things stand out. For now, I have a two part question:

When titrating an acid with a base, for instance, the pH rises more or less abruptly around the equivalence point. Does this have to do with the logarithmic nature of pH? In other words, is this caused by the fact that the number of particles involved in the jump from e.g. a pH of 2 to one of 3 is several orders of magnitude than the number of particles involved in the jump from 6 to 7?
When a weak acid is titrated, the rise to the equivalence point is a lot more shallow. Does this happen because of Le Chatelier's principle? That is to say, as the base is titrating the acid, some of the acid that has not dissociated will dissociate in an attempt to restore equilibrium. This should retard the titration to some degree. Is that the case?

Hi Sadiq. Consider splitting this into multiple questions, so they can better be answered. Also, please use a more precise and clear question title. The current one is too "literary" for the technical nature of the question.

OK just a moment

I have changed the title but if the question needs to be split, I'll allow a moderator to do that—I kind of think that someone who will be able to answer one part will be able to answer the other as they are closely related

Hi Sadiq, it's okay to split your own question up (just edit out the second part and paste it into a new question).  I agree with @CHM that splitting it would probably be to your benefit, but it's your choice.

I just don't want to spam up the board, you know?

This answer from a member of Chemical Forums who seems to have been quite useful to other posters in the past given his or her "karma" would appear to confirm that the shape of titration curves owes in part to their logarithmic nature:
http://www.chemicalforums.com/index.php?topic=6140.0


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How do I balance the combustion reaction of butene?
I have butene being burned in a fuel rich engine and I can't seem to balance this reaction:
$$\ce{C4H8 + $a$(O2 + 3.76 N2)
-> $b$ CO2 + $c$ H2O + $d$ C4H8 + $e$ (3.76 N2)}$$
When I try write equations for the elements, I get 5 equations and 4 unknowns which can't be solved. Here are my equations:
\begin{align}
\text{Carbon}   &:& 4     &= b  + 4d  \\
\text{Oxygen}   &:& 2a    &= 2b + c   \\
\text{Nitrogen} &:& 3.76a &= 3.76e    &  a &= e \\
\text{Hydrogen} &:& 2a    &= 2c + 8d  \\
\end{align}
I'm not sure if this makes a different but it says that water in the combustion products is removed before the dry analysis of the exhaust. The dry analysis has a composition of $14.95\% \ce{CO2}$, $0.75\% \ce{C4H8}$, $0\% \ce{CO}$, $0\% \ce{H2}$, $0\% \ce{O2}$, and balance $\ce{N2}$ (I don't know what that means).

re last sentence: it means you've got a fifth equation for your five unknowns. By the way, at the end of your reaction equation, should the 3(...) actually be e(...)?

as EnergyNumbers pointed out DRY analysis gives you have a fifth equation: $b/d = 14.95/0.75 = 19.933$. Solving those will give you $a = e = 5.994$, $b=3.331$, $c=5.326$ and $d=0.167$.

See this video for a complete description of a matrix-based approach. It starts by showing how to set up the matrix. It then takes each step of reducing the matrix and extracting coefficients.

@RichApodaca your link describe a generic algorithm to solve the problem, which is fine, but probably not very helpful to the specific question as here.

The balanced equation for the combustion of butene is
$$\ce{C4H8 + 6O2 -> 4CO2 + 4H2O}.$$
Nitrogen is included only to indicate that it is present in the medium with some specific ratio to the oxygen amount – well technically it is included to make question look complicated.
$\ce{C4H8}$ in the products side is included to indicate that not all of it can be combusted but some escapes. And it really is arbitrary, depends on the the engineering of combustion room, etc. – which is why you have 4 equations, 5 unknowns: d is free to vary.  But you are given the dry analysis to fix its value.
Looking at the equations given, you have
\begin{align}
  b &= c = 4(1-d)\\
  a &= e = b + c/2 = 6(1-d) = 3b/2
\end{align}
And dry analysis says ratio of $\ce{CO2}$:$\ce{C4H8}$ is $14.95:0.75 = 19.93 (b/d)$ after the combustion. And this is the last equation you need.

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The Evans auxiliary: rotational freedom of the benzyl group - stereoselectivity still guaranteed?
I created the L-Phenylalanine based Evans auxiliary bearing a benzyl group with help of a molecular model set and got into some questions. The structural formula will do:

As I suppose a low rotational barrier, the phenyl group can be aligned off (a) and towards (b) the alpha position of the N-propionyl group. I suppose maximum sterical repulsion for attacking electrophiles for (b).
The phenylalanine and valine based auxiliaries (Bn or i-Pr) are the most commonly involved types of Evans auxiliaries. But why do I get good stereoselectivities for this Bn auxiliary? There are no corresponding carbon-carbon rotational freedoms in the valine species.

You have to imagine the nucleophile attacking from "behind" your screen. The phenyl does have some freedom to move, but it's sitting in front of your screen and thus blocks access from the front, not the back. Assuming the Evans auxiliary is planar, of course.

The comment left by CHM is very helpful.  You need to consider what this molecule actually looks like in 3D.  That may help your understanding greatly, but to be honest there is considerably more going on here than you would expect.
A couple pieces of information would be very helpful for clarifying this question.  The first is what type of a metal are you using for this synthesis step.  The second is what is the outcome.  Believe it or not, based on those two pieces of information the mechanisms for this reaction can be explained entirely differently.
Since the answer basically breaks down into two possibilities I will attempt to explain both.  One of the most helpful articles I came across in my search was entitled "Transition State Models for Probing Stereoinduction in Evans Chiral Auxiliary-Based Asymmetric Aldol Reactions" J. Am. Chem. Soc., 2010, 132 (35), pp 12319–12330.
In this article they go into depth about how the electronic and steric effects balance out to produce the almost complete stereocontrol throughout the course of this reaction.  The scheme below is what I find the most helpful:

Note how the selectivity is varied based on whether the auxiliary is chelating to the titanium or not.  This is why the metal lewis acid used in these reactions is important.
One final note not shown here is the rational behind the non chelating setup.  The opposing dipoles on "y" and aldehyde oxygen in the Zimmerman-Traxler model help to hold the six member ring in place while the aldol reaction occurs.
The scheme shown is incredibly useful for visualizing this transformation and I would like to compliment the authors for their efforts in this publication.
Hope this helps!


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Options for long term storage of helium?
Specifically, if I wanted to purchase a large amounts of helium, what are my options for purchasing helium in a state that is suitable for long term storage?
It's my understanding that if I were to buy helium in bulk it would be shipped cold, as a liquid.  Keeping helium in liquid form for 10-20+ years would involve a lot of serious refrigeration.
Are there other reasonable options for bulk, long term storage?
This may not seem like much of a "chemistry" question but since this is used in chemical sciences, I assume this is a need that some may be aware and may have a good answer since this is an elemental material that some may use on a very common basis.

Steel baloon and helium in gaseous form may be an option, if you have enough space for it. Definitely, it will take much space, but it is still an option. Another option is cryogenic storage, but to my knowledge it is quite complex for liquid helium and requires regular income of liquid nitrogen.

Just out of sheer curiosity, what is the helium for?

Richard Terrett: Retirement. ;)

You should read this page (http://www.linde-kryotechnik.ch/1259/1260/1308/1310.asp) and the pdfs at the bottom. How big is your underground laboratory?

@RLH - I had my suspicions. Personally, I would go for a less... volatile investment. ba-dum-tish

Well, I wouldn't invest a significant portion of my worth in Helium but there have been doom-and-gloom reports for quite a while regarding the coming scarcity of helium. I've tried to do a bit of research but other than pushing for legislation to change the market, no one is doing anything about the problem. I would imagine the major barrier to stockpiling is probably the long term cost of storing enough Helium for 20+ years. If you have to refrigerate it for that long, it's going to cost a fortune. If you can store it in strong, highly compressed tanks as a gas, then it might be worth it.

You want buy creeper helium, and after some years get a lot of money selling on higher price? :) The money inflation will eat all profit. Liquid helium is hard to obtain and quite extremely impossible to storage long-term. http://www.janis.com/4KCCR_Home_KeySupplier.aspx But this helium is used to make high magnetic field. Just for storing it will be huge energy waste. I totally agree with pat - using simple high pressure tanks/cylinders is the best option.

It's 2019.  I never found a way to store helium for the long-term.  I wish I had.

In my experience, for small quantities in laboratories helium usually comes compressed in cylinders. Except possibly in the case that you require extremely high purity, I don't see why helium couldn't be stored for long periods in this form, it would be relatively safe and would avoid the need for refrigeration.
I don't know what sort of quantities you are thinking about, but it seems that you can get big cylinders. For example the largest cyclinder size available from BOC gases, a British company, is 194.7 $m^3$ filled to a pressure of 300 $bar$ (source: BOC website).

Compressed helium stored in cylinders is what I would suggest too. If you have a large quantity, buy a natural geologic gas storage formation…

helium will diffuse through steel over time.
there is no way known to man to keep it indefinitely unless its cooled.
nitrogen and argon though can be stored in bottle over a decade or two.
maybe a ref or two might be in order
Measurement of Helium
Diffusion in Metals
https://www.energy.gov/sites/prod/files/2019/06/f63/Measurement-of-Helium-Diffusion-in-Metals.pdf
DIFFUSION AND AGGLOMERATION
OF HELIUM IN STAINLESS STEEL
https://inis.iaea.org/collection/NCLCollectionStore/_Public/38/111/38111384.pdf
So the simple truth is every balloon is one less person with an MRI.
It is mined from the planet and it is not being replaced at the rate its being taken at.
Try not to use helium if you can guys please.

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review

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Boiling point of vodka
What is the boiling point of vodka?
I have found the boiling point of ethanol, ~173 (degrees) F.
However, I am unable to find the boiling point of vodka. I found some information on it, but I do not trust the source.

This question is fairly imprecise.  It is sensible to discuss the boiling points of water/ethanol mixtures versus composition, but said composition will differ from between the vapor and liquid phases, with enrichment in the lower boiling component.  The presence of the azeotrope mentioned in Spiessbuerger's answer complicates the matter slightly, but that azeotropic composition is so close to pure ethanol, this forms a decent quick guess.  Generally these curves are not linear for any distillation.

Vodka is a solution of water and ethanol.

Water boils at 100 °C
Ethanol boils at 78.37 °C

A mixture of both liquids will start boiling close to the boiling point of ethanol (otherwise distillation wouldn't work).
The lowest boiling temperature is a mixture of:

95.63 mass-% ethanol
4.37 mass-% water

which boils at 78.2 °C.
Read more about this effect at: Azeotrope There is also a picture of a phase diagram there.

Wait, vodka is quoted at 40% ethanol V/V (so mole fraction about 1/5) and not azeotropic! It should be about 83 °C.
Graph of vapour-liquid equilibrium for mixture of ethanol and water

Well, typical vodka has 40%, but one could find even 70% vodka ;) and tobias47n9e didn't claim that it's azeotropic.


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Shape of C₂H₂ molecule
What is the shape of the $\ce{C2H2}$ molecule?
The fact that is has a triple covalent bond between the 2 carbon atoms and 2 other bonding pairs between the hydrogen and carbon atoms leads me to believe that it could be linear shape.
However, I'm not sure if the triple bond counts as a single electron bonding pair or 3. If it counts as 3 bonding pairs then the molecule would have 5 overall, and therefore be a trigonal bipyramidal shape.
So is it octohedral or linear? Pretty sure it's linear

According to VSEPR theory, each carbon has 2 electron "pairs".  These electron pairs are not what you would typically think of as electron pairs because VSEPR theory considers double bonds and triple bonds to still be a single "pair".
So considering each carbon atom, the shape is $AX_2E_0$ ($A$ is the carbon atom, $X$ is a bonding pair of electrons, and $E$ is a non-bonding pair (lone pair) of electrons).
$AX_2E_0$ corresponds directly to a linear molecule.
To directly answer your question, when trying to determine the shape of a molecule, consider double and triple bonds as only a single electron bonding pair.

I am not an expert in VSEPR theory, but is it commonly used to explain the geometry of molecules like acetylene? I only know it for cases where you have one center atom surrounded by different substituents, and all examples on the linked Wikipedia page seem to be like that.


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What is the electron dot structure of an alkyne?
I am trying to determine the electron dot structure of an alkyne like $\ce{C2H2}$.
I know that hydrocarbons with at least one carbon-carbon triple bond are called alkynes, and their general formula is $\ce{C_{n}H_{2n-2}}$, where $n$ is number of carbon atoms.
So what will the electron dot structure of $\ce{C2H2}$ look like?

Have you drawn electron dot structures for alkenes? Recall that pairs of dots represent covalent bonds. As you know that an alkyne has a carbon-carbon triple bond, this should indicate the correct number of dots between the carbons.

Carbon needs four electrons to fill its octet.  In a $\ce{C-H}$ covalent bond carbon shares one electron with hydrogen.  A triple consists of one σ covalent bond and two π covalent bonds.  The carbons share two electrons in each of these bonds; in total six electrons are shared between them.  You can draw this as a dot structure:
$$\ce{H:C:::C:H}$$


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Chemical reactions with a room-scale cooling effect
Are there chemical reactions that could cool down an average sized room by a noticeable amount (say 5 °C)?
I would like to investigate if it is possible to have a situation where I can mix 2 reagents at room temperature and pressure and in open air then they should react and become colder than room temperature without evaporation of some type, with an eye to making a noticeable drop in the room temperature.

As I remember, $NH_4Cl$ dissolves in water with cooling of liquid. Not very much, but seen clearly. There are much reactions, that consume energy, but they are usually either slow or requires high temperature.

The reaction between ammonium thiocyanate and barium hydroxide octahydrate is endothermic. It absorbs heat from the surroundings.
http://www.youtube.com/watch?v=MyAzjSdc3Fc

One way to do it is, as user41631 says, with an endothermic reaction.
Another is to use phase-change materials, from which you remove the thermal insulation, to allow the phase-change to take place. Now, all materials change their phases, at some combinations of temperature and pressure. The phrase "phase-change material" is used not to classify materials, but to identify the paricular use of a material - that is, the material is being used because one of its phase changes happens at a temperature and pressure that makes it useful for a specific process.
Such materials are being looked at for the temperature regulation of buildings.
Let's take a material with a melting point of, say, 18°C, and a high latent heat of fusion. Freeze it, insulate it, and take it into a room that's at 22°C. While the material melts, it will absorb heat from the air, cooling the room.
This principle has been applied in building material design, to provide smoothing of temperature variations. See, for example, Schossig et al's "Micro-encapsulated phase-change materials integrated into construction materials", [DOI], from which the following graph is taken, showing the smoothing of temperatures over a week in August , when a phase-change material (PCM) is used (T_Wall_PCM) - this is compared to an identical room without phase-change materials (T_Wall_REF) - as you can see, the PCM reduces the amount of time that the room spends above 26°C, providing more thermal comfort for people in the room.

Salt + ice takes the temperature down from 0 °C ice to −19 °C salt water, and can be used to make more ice (or ice cream!). So you have an ice-generating engine powered by salt and water, and as much cooling as you might desire (down to −19 °C). You could retrieve the salt later by evaporation. This is obvious, but I’m not aware of any device that does this, so maybe there are obstacles in practice.
Now, I want to know how to get down to −80 °C so I can make dry ice.

May be alcohol with salt?

I think that in practice you would need too much water and salt to make it worth the effort. Also dumping the mixture of melted ice (water) and salt somewhere to make room for more solid ice and salt would be problematic.


text:
Order of elements in a formula?

Possible Duplicate:
Why is methane's molecular formula conventionally “CH4”, while water is “H20” (among others)?

If I have a formula containing 4 hydrogen atoms and 1 carbon
Should it be written $\small\ce{CH4}$ or $\small\ce{H4C}$?
In general, how do I know the order to put elements in, in a chemical formula?

Duplicate of this , and comprehensively answered here

There's something called Hill System. This is the system of writing chemical formulas.
In this system the carbon atoms are first, then hydrogen atoms and then other in alphabetical order. When the formula contains no carbon or hydrogen atoms, then all the elements are sorted alphabetically.
source: http://en.wikipedia.org/wiki/Hill_system
So it should be: $\ce{CH4}$
However, the Hill system is not the only method of ordering elements in a chemical formula. We can instead choose to list the elements based on their electronegativity. ManishEarth explains this in his answer to this question.

Where can i see the entire hill system, I would like to memorize it :)

Here's something you should check: http://en.wikipedia.org/wiki/Hill_system

So its alphabetical?

like Clorine before sulpher?

Yes, but there are some exceptions

@bluepawel Could you give a brief brief summary of the Wikipedia article and incorporate that and the link into your answer?  Comments can get deleted, etc., but if you put the info with the answer it will stay there.  Thanks.


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Is there a way to calculate how fast water will cool?
Is there a way to calculate how fast water will cool if you know the water's starting temperature and temperature of the environment that you put it in, such as a freezer?

Does the container shape matter much to your solution? Cylinder, box?  Resting on a surface?  Embedded in a medium?  in-ground (like a swimming pool?)

Check out Fourier's law. The rate of heat flux (transfer of thermal energy per unit time per unit surface area)  is proportional to the difference in temperature:

Since a change in temperature is directly proportional to the change in energy, you can form this into a fairly simple first-order differential equation. If we take one side (air in a room), at a temperature we'll call $T_{0}$ as a cold reservoir, we can say:
$$
\frac{d(cT_{1})}{dt}=c\frac{dT_{1}}{dt}=-kA\frac{T_{1}-T_{0}}{w}
$$
Where $T_{1}$ is the temperature of the water, $c$ is the water's specific heat, $A$ is the surface area, $k$ is the thermal conductivity of the material between them (i.e. styrofoam cup, say it has a styrofoam lid too so it surrounds the water completely), and $w$ is the thickness of the container.
You will notice that this equation leads to an exponential decay in the temperature difference.
Simply plug in the appropriate constants and solve this equation for any given $\Delta T$ and you can find how long it takes to get that close (it will never quite reach zero).
I'm a bit rusty, so please forgive any notational errors

http://www.physicsclassroom.com/class/thermalP/Lesson-1/Rates-of-Heat-Transfer - much more detailed explanation of this formula.


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How are crystals formed from a gas?
In crystallization, crystals are formed from liquids by cooling.
How can crystals be formed from a gas by cooling?

It's called deposition or desublimation. It's for example how snow is created in clouds or how frost forms on leaf. When leaf is very cold, then vapour surrounding leaf looses so much energy that it changes into solid state.


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How does Phosphorus Pentachloride bond?
The thing I am having trouble with: why does $\small\ce{PCl4}$ form a +1 ion and $\small\ce{Cl}$ a -1 ion?

There is no ionic bonding that occurs, the phosphorus is able to make a hybrid orbital out of its $\small3s$ and $\small3p$ electrons, along with an empty $\small d$ orbital to form 5 $sp^3d$ molecular orbitals, of equal energy, which yields 5 spots for bonding electrons.
This is a known exception to the octet rule.

(graciously borrowed from here)

Thanks, my information came from a 1957 Pauling lecture, so it was bound to be a little off.

@Alyosha No problem, I had to refresh my own memory.

Also- from the link you supplied: 'The 3-level electrons now rearrange (hybridise) themselves to give 5 hybrid orbitals, all of equal energy. ' How do they actually do this? Is the one electron from 3s pulled towards the 5th Chlorine very strongly so to make it a lower potential energy to 'jump out' towards it?

@Alyosha Evidently, the axial (Z) hybrid orbitals are the ones with the most d character (see here for example).  I couldn't tell you the quantum mechanical basis for it, though I think energetically they are close.

@Alyosha However, don't lose heart, there are many others around who do know, so stay tuned. :)

The concept of spd hybrid orbitals should be completely removed from the textbooks, esp. for non-transition metal complexes. Hybridisation is a mathematical concept, that allows neither energy gain nor loss. The so-called hypervalancy is also a concept, that should be treated very critically. A mathematical explanation can be found with more-centre bonds. This concept can also explain the geometry of the molecule: $\ce{Cl\bond{-}PCl3\bond{-}Cl}$.

phosphorus has 5 valence electrons.each chlorine has 7 valence electrons.phosphorus give 1 electron to each chlorine and hence all the 6 attain 8 electrons in the outer shell thus forming an ionic bond.

All the six? All the what six?

They mean all the six atoms end up with a closed valence shell. pappu is suggesting $\ce{PCl5}$ is an ionic substance composed of $\ce{P^{5+}}$ and $\ce{Cl^{-}}$ ions, which is a very poor description of the bonding. Discrete +5 ions are essentially impossible, because their resulting high charge density would force the molecules/counterions around them to donate electrons to the cation and either decrease their positive charge by electrons transfer or at least establish a covalent bond, making a polyatomic cation.


text:
Is there a way to detect bleach residue on a surface?
I was wondering if there is a way to detect for presence or residue of bleach using household items.
I am curious about either chlorine-based or peroxide-based. It can be any sort of reaction that shows the presence of it. Well, as long as the reaction doesn't produce a lot of heat or is somehow toxic.

What immediately comes to mind is something rather obvious, don't use ammonia on any surface that could possibly be contaminated with bleach.
A search revealed a video showing red cabbage juice used as an indicator.  I do not believe the limit of detection would be anything to write home about.  However, this got me thinking about a homemade pH paper.  This would allow you to dampen the surface with a very small amount of neutral water, apply the paper, and possibly get an accurate reading on whether something acidic or basic is present on the spot you tested.
Fortunately, someone else has already put together a procedure on how to make these strips, found here.  With another source showing the expected colors using this indicator.  Since bleach aka sodium hypochlorite is a base, I would expect the color of the test strips to be in the blue to green range.
This could also be a cool home science project.  The strips will be more effective at detection for lower concentrations of bleach than a liquid solution of the indicator. (pH is concentration dependent).
Good luck!
P.S. I linked many articles to fill in some relevant information but I was limited to 2 links.  If you have questions please feel free to ask!


text:
Proper IUPAC name of a substituted cyclohexenol
Here's the molecule:

So far I've tried: 6-methylcyclohex-2-en-ol and 4-methylcyclohexen-3-ol but both are wrong. I'm rather confused as to where I should start: the alcohol or the double bond?

I think you made a counting mistake. It is 6-methylcyclohex-2-en-1-ol

You are right. For those who have the software, many drawing packages have structure to IUPAC name modules (I use ChemDoodle). This is a quick way to verify name rules from structures.

The functional group with highest precedence is the alcohol, so it will be the suffix and bears number 1. The main chain is cyclohexene, and you have two ways of numbering it:

5-methylcyclohex-2-en-1-ol
2-methylcyclohex-5-en-1-ol

However, IUPAC rules state that you must give double bonds at the lowest possible numbering, even before considering prefix groups (rule 6.1 to 6.4 of this wikipedia page). So, it has to be 5-methylcyclohex-2-en-1-ol.

Sorry but as pointed out above in the answer by "jens", this (otherwise great) answer has a typo.


text:
Which species dissociate completely in ionic equations?
I am tasked with writing a molecular equation, an overall ionic equation, and a net ionic equation for the reaction between an aqueous solution of phosphoric acid, $\ce{H3PO4{(aq)}}$ and an aqueous solution of sodium hydroxide. The products are sodium phosphate and water.
I have already figured out the balanced molecular equation of the following:
$$\ce{H3PO4 {(aq)} + 3 NaOH {(aq)} -> Na3PO4 {(aq)} + 3 H2O {(l)}}$$
However, I am uncertain how to write the proper ionic equations. I know that the ions are separated, but how do you format this properly when writing with multiple molecules. If this is based on the disassociation of the chemicals, how do you determine which ions disassociate?
I have arrived at:
$$\ce{3H+ {(aq)} + PO4^3- {(aq)} + 3 Na+ {(aq)} + 3 OH- -> 3 Na+ {(aq)} + 3 H2O
{(l)} + PO4^3- {(aq)}}$$
However, I still do not know why this is the best answer. Why does the $\ce{NaOH}$ show separated in the equation while the $\ce{PO4}$ does not? How would you know which ones separate by simply looking at the problem?

Okay, I think I might somewhat understand it now.  The difference is whether something is a strong acid or a weak acid.  A strong acid will disassociate completely, therefore, it will be split apart completely within the equation.  Other items, including weak acids, will not be split.  Please let me know if my reasoning is off on this.

The second equation you’ve given appears to be the right overall ionic equation. Phosphate is a weak acid, $\ce{NaOH}$ is a strong base. We expect NaOH to dissociate at any pH, where the speciation of phosphate ($\ce{H3PO4}$, $\ce{H2PO4-}$, $\ce{HPO4^2-}$, and $\ce{PO4^3-}$) depends on pH.
A “net ionic equation” doesn’t include spectator ions. Spectator ions exist on both sides of the arrow and can be “canceled like” (think as in algebra). For any acid base reaction, it’s really easy to remember!
$$\ce{H+ {(aq)} + OH- {(aq)} <=> H2O {(l)}}$$
Lastly don’t forget the state: $\ce{OH {(aq)}-}$. I was taught to right them as subscripts, but I’ve seen then as regular text too: $\ce{OH- (aq)}$.

If phosphate is a weak acid, it should not disassociate completely, right?  So, the H3PO4 should stay together I would think.  If that is the case, then should I really be breaking down the ionic equation as follows: $3HPO_4(aq)+3Na^+(aq)+3OH^-(aq) \longrightarrow 3Na^+(aq) + 3H_2O(l)+PO_4^{3-}(aq)$?

Firstly what you have here is an acid base problem. The way you can tell when you have an acid base problem is when your aqueous reactants produce water.
Secondly you know that anything with an (aq) subscript are ions in solution. AKA they are telling you which molecules are disassociated in water.


text:
What are the chemical reactions behind fire?
I've always wondered what the chemistry behind fire is. What are the basic chemical reactions behind a simple wood fire, and how do they manifest into this phenomenon?

Hmm, your question is a bit broad, I've slightly narrowed the scope of it. Could you please elaborate on what you want, and what you already know?

I'm afraid you've completely changed my question. My question is very simple, like we have a composition for every thing(Water H2O, Salt NaCl etc) What about fire? May be I am not framing my question well. Thanks :)

Point is, fire isn't really composed of anything. It's not a substance in the sense you're looking for. Flying soot (carbon) is the closest you can get to a "substance". Fire is a chemical reaction.

Even though fire is one of the Greek classical elements, it is the only one that is not matter in our current understanding. What we experience as fire is the energy (in the form of light and heat) given off by the exothermic combustion of (usually) organic materials, like wood. The large amounts of thermal energy released by combustion often cause the gases in the fire to incandesce, that is, some of the kinetic energy produced by the fire is converted into electromagnetic radiation that we see (visible) and feel (infrared).
The chemical reaction for combustion is pretty simple. Take, for example, the combustion of n-octane $\ce{C8H18}$, which produces more than 5 MJ of energy per mole during combustion (from webbook.nist.gov).
$$\ce{2C8H18 +25O2->16CO2 +18H2O} \space\space \Delta_cH^o=-5430 \text{ kJ/mol} $$
The combustion of wood is more complex. The majority of the organic mass of dry dead wood is lignin and cellulose. Lignin is a highly-cross-linked copolymer of p-coumaryl alcohol, $\ce{C9H10O2}$, coniferyl alcohol, $\ce{C10H12O3}$, and sinapyl alcohol, $\ce{C11H14O4}$. Cellulose is a linear polymer of glucose, with formula $\ce{(C6H10O5)}_n$.
Different species will have different ratios of lignin to cellulose, different cross-link densities in the lignin, and different ratios of coumaryl to coniferyl to sinapyl alcohols. The formula for lignin could be expressed then as $\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z$.
The equation for the combustion reaction for cellulose is:
$$\ce{(C6H10O5)}_n +6n\ce{O2->}+6n\ce{CO2}+5n\ce{H2O} $$
Since lignin is more complex, its combustion equation is more complex:
$$2[\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z]+(21x+23y+25z)\ce{O2}\\ \ce{->}(18x+20y+22z)\ce{CO2}+(10x+12y+14z)\ce{H2O}$$
And assuming the variable composition of wood as a ratio (A:B) of cellulose to lignin, the overall combustion reaction becomes the following monstrosity:
$$A\ce{(C6H10O5)}_n +2B[\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z]+[6An+B(21x+23y+25z)]\ce{O2}\\ \ce{->}[6An+B (18x+20y+22z)]\ce{CO2}+[5An+B(10x+12y+14z)]\ce{H2O}$$
Given the number of variables ($A, B, n, x, y, z$), the $\Delta_cH^o$ for this reaction is difficult to determine (but not impossible, assuming we know some thermochemical reference data). Whatever its value, it is exothermic enough to induce incandescence.

Can you add the mechanisms of the above reactions to your answer?

@ApoorvPotnis Bwahaha xD You really dunno what you asked for! Full mechanism even for burning octane probably isn't known, and even if it was, it would be longer then this post :D

Ben gave the net reactions. All the intermediates would be dozens and dozens of chemical species.

The chemical reactions of a "typical fire" using wood as the inflammable matter are extremely complex and not known to their full extend. The wood isn't actually burning but gasified (mostly superficial) components of the wood, which yield a plasma in an environment that provides sufficient excitation energy. Accelerants are often used to this end.
The situation is aggravated by the many radical species which are produced in chain reactions and the temperature gradient of the flame which favors different species.
The form of fire depends on many parameters, gravity included...
A search for "estimating the number of chemical species in a wood fire" brings up interesting literature which may serve you as a starting point, in addition to Ben Norris's answer.

Ben's answer covers it well. There is another aspect which affects the products of fire; that is the temperature of the fire, which varies according to the phase of the fire, among other things.
I found the following list in some educational materials for fire fighters:
Temperature               Reaction
3920F                       Production of water vapor, carbon dioxide, formic and    (2000C)                       acetic acids
3920-5360F                  Less water vapor - some carbon monoxide - still primarily an
(2000-2800C)               endothermic reaction (absorbing heat)
5360-9320F                  Exothermic reaction (giving off heat) with flammable vapors
(2800-5000C)               and particulates; some secondary reaction from charcoal formed
Over 9320F                  Residue primarily charcoal with notable catalytic action
(5000C)

Fire is fascinating. As already noted in answers above, fire is a mixture of matter and energy. Fire consists of fuel, oxygen (or another oxidizer) and heat interacting in a chain reaction yielding more heat and oxidized byproducts.


text:
What's the name of this electrolysis device?
I've seen it in a couple of videos, it's used to electrolyze water in 2 different containers at the same time to make a comparison and thus is very handy.
What is it called (besides just a hydrolizer)?

It's a precipitator, also known as TDS indicator.

Based on this article, in which the authors describe a system that can do 25 cells simultaneously, the term 'Parallel Electrolyser' would seem appropriate.


text:
Materials used in metallic glass
I've been looking over the lists of materials used together to make metallic glass.
Is it purely coincidental that if you round up/down any decimals from the atomic weight of elements used to construct metallic glass, you're always very close to +-1 (if not at) a prime number for the atomic weight of each element?

Yes, it’s coincidental. There is no expectation that prime (or near-prime) atomic weight means anything for the element itself. In fact, your affirmation is not even true: on the list given by Wikipedia, the first two examples are:

$\ce{Au_75 Si_25}$, your assertion is true
“an alloy of 77.5% palladium, 6% copper, and 16.5% silicon”, and your assertion isn’t met for copper

Let's go further. For a given element, how likely is it that its atomic weight, rounded, is within ±1 of a prime number? Well, atomic weights are relatively low integers, and prime numbers are plenty numerous among small integers, so it is relatively frequent. The answer can be computed with Mathematica:
weights = {#, Round@ElementData[#, "AtomicWeight"]} & /@ ElementData[All];
f[i_] := PrimeQ[i] || PrimeQ[i - 1] || PrimeQ[i + 1]
res = {f[#[[2]]], #[[1]]} & /@ weights
Select[res, ! #[[1]] &][[All, 2]]

74 out of the first 118 elements turn out to satisfy your “near-prime atomic weight” condition, which is 63%. This seems high, but it's not: if you do the same check, not only for atomic weights, but for all integers below 200, 61% of them are near-prime:
Total@Boole@Select[f /@ Range[1, 200], # &]

Thank you for your time.  I have learned two things.  First it's just coincidental.  Second, and more importantly, I have a long way to go in my own fact checking.  Thank you for identifying my discrepancy.
The third one is that an unbelievably high number of elements are near-prime in their atomic weight (statistically speaking).  That's kind of cool.

@Everett it's not such much that many atomic weights are “near-prime”, that many small integers are near-prime. I’ve updated the end of my answer to clarify that…


text:
Specific heat capacity and the periodic table
Can someone explain me how can I use the periodic table in order to find the specific heat capacities of some elements? (for example, of Aluminium )
The question I have encountered is as follows:

Given the specific heat capacities of the following liquids, use the periodic table in order to find the specific heat capacities of the metals $Al, Fe, Cu, Au $ .
$ H_2 O - 4.18 J  /g \cdot C $, $ C_2 H_5 OH - 2.46 $ , $CCl_4 - 0.861 $ , $CCl_2 F_2 - 0.598 $ .

Either some context is missing, or the question as pasted can be nominated for “worst homework of the year” awards…

Either you have a periodic table that features the specific heat capacity of elements in their standard state at STP (standard temperature and pressure), in which case you merely have to look it up, or you don't, and there's nothing you can do.
Specific heat capacity of materials, even elemental solids, are physical properties resulting from complex behavior of the systems, and cannot be determined by simple laws or equations. They have to be determined experimentally, or by first principles calculations.

PS: water, methanol, carbon tetrachloride and dichlorodifluoromethane have got nothing to do with it.

Well...I'll ask our lecturer about this... Thanks anyway !


text:
Using Dalton's law of partial pressures

A container of volume $\pu{4.0 L}$ contains nitrogen at $\pu{25 °C}$ and $\pu{803 kPa}.$ This container is connected by a valve to another container of volume $\pu{10.0 L}$ that contains argon at $\pu{25 °C}$ and $\pu{47.2 kPa}.$ We now open the valve so that the two gases are mixing.
Calculate the partial pressure of each gas after the mixing and the total pressure of the mixture.

Using the Dalton's law of partial pressures, isn't the partial pressure of each gas exactly $\pu{803 kPa}$ and $\pu{47.2 kPa}$ and the total pressure $\pu{803 kPa} + \pu{47.2 kPa}?$
What am I missing here?

Dalton's law of partial pressures means that the total pressure in the final volume $V=V_1+V_2$ is equal to the sum of the partial pressures P(N2) + P(Ar) in this volume. However, the partial pressure of N2 in the total volume $V$ is not the same as the pressure of N2 in its initial volume $V_1$. You have to account for the expansion. This probably means you also need to use the ideal gas law for that.

Is that pressure applied on gas or applied by gas?


text:
Breakdown of valine
What are the chemical reactions responsible for the breakdown of valine in the body?
I know oxidative deamination is part of the process but how is valine further broken down from an keto acid to isobutyrate?

From Wikipedia’s article on propionic acidemia:

[Enzyme propionyl-CoA carboxylase] is required for the normal breakdown of the essential amino acids valine, isoleucine, threonine, and methionine, as well as certain odd-chained fatty-acids.

and

In healthy individuals, the enzyme propionyl CoA carboxylase converts propionyl CoA to methylmalonyl CoA. This is one step in the process of converting certain amino acids and fats into sugar for energy. Individuals with PA cannot perform this conversion because the enzyme propionyl CoA carboxylase is nonfunctional. The essential amino acids isoleucine, valine, threonine, and methionine and odd-chain fatty acids are simply converted to propionyl CoA, before the process stops, leading to a buildup of propionyl CoA.

You can find some more or less detailed descriptions of valine’s degradation pathways by Googling:


text:
Mechanism of oxidation by dichlorotriazenetrione
How does dichlorotriazenetrione supply chlorine to a swimming pool and how does it oxidize organic material?

I am only trying to help when I say that a good rule of thumb when asking questions is to make sure it is not a mere one sentence in length. Otherwise, it shows lack of research. Anyone trying to answer has no idea where you are coming from, why you want to know, and what background of knowledge you already have to make the progression to "learnt" much more efficient. [Google](dichloro-s-triazinetrione water purification)

Dichlorotriazenetrione, also known as dichloroisocyanuric acid, is slowly hydrolyzed in water to isocyanuric acid and hypochlorous acid, which serves as the oxidation agent. The N-Cl bond is easily hydrolyzed because of its relative weakness, which is also reflected in the stability of nitrogen halides. The slow hydrolysis releases a low but constant amount of oxidizers. $\ce{HOCl}$ itself oxidizes organic compounds, being reduced to $\ce{HCl}$ in the process.

In aqueous solution, $\ce{HOCl}$ is in equilibrium with its anhydride $\ce{Cl2O}$ and elemental chlorine, which also serve as oxidizers. Furthermore, it slowly decomposes to $\ce{HCl}$ and oxygen, so that a mixture of different oxidation agents is present in its solution.
$$\ce{HCl + HOCl \rightleftharpoons Cl2 + H2O}$$
$$\ce{2HOCl \rightleftharpoons Cl2O + H2O}$$
$$\ce{2HOCl \rightarrow 2HCl + O2}$$


text:
If a sugar is nonreducing, does that mean it doesn't ionize in water?
I'm thinking of sucrose, which I learned is nonreducing, what I take to mean it doesn't accept/take electrons from the surrounding solution. Does this also mean that it doesn't ionize and thus has a Van't Hoff factor ( i ) of 1?

I believe the confusion lies with the term Reducing sugar. From my perspective sucrose is not an ionic compound, it is covalently bonded. I believe that this should indicate to you that it does not "ionize" or the term that I was taught "dissociate" since it has nothing to dissociate from.
To quote wikipedia:

"The van 't Hoff factor is the ratio between the actual concentration
  of particles produced when the substance is dissolved, and the
  concentration of a substance as calculated from its mass."

Since it does not "dissociate," I would say that the answer is no, it does not "ionize," and therefore has a Van't Hoff factor of 1 when mixed with pure water.
Like anything with electrons, they can be lost. This takes place in our bodies through more complicated reactions than I understand completely. I believe however, according to wikipedia, that sucrose has a Glycosidic bond which when broken, will separate sucrose into smaller sugars, if this takes place in your solution then I believe it will have a Van't Hoff factor > 1.

There's more than one way to skin a cat accept an electron. Actually, in the case of a reducing sugar, it donates an electron. When it does so, it is oxidised. But that doesn't mean that it ionizes in water!
When you keep a reducing sugar in the presence of an oxidising agent, the electron is transferred over to the agent, reducing it. The sugar may form an ion, or it may form a carboxylic acid (depending on the medium). The electron is fuelling the reduction half reaction of the oxidising agent we have used. If we do get a carboxylic acid, there is a chance for it to ionize(again, depends upon the medium) and we may get a Van 't Hoff factor $>1$ {*}. But remember that this is in the presence of an oxidising agent. If we just dunk a sugar in water, it won't be able to reduce anything, and will be chemically inert, giving a Van 't Hoff factor of 1.
*Also, the reaction with the oxidising agent may provide additional ions, increasing the Van 't Hoff factor for the oxidising agent


text:
Which C-C bond would break first, the one in ethane, or 2,2-dimethylpropane?
I have drawn the figures and determined intuitively that the larger molecule's $\ce{C-C}$ would break first. I made this assumption that ethane has the stronger $\ce{C-C}$ bond and I am correct, however I do not understand the underlying reasons why.
In both molecules we have single bonds, but in ethane we have more space for the hydrogen atoms to fill. This was my initial reasoning, that the larger molecule is less compact and will have longer $\ce{C-C}$ bonds. Am I correct in my reasoning, or does it have something to do with the radical stabilities? My book is saying "radical stability increases along the series from primary to secondary to tertiary; consequently, the energy required to create them decreases."
In ethane we would have primary carbons because they bond to only one other carbon. In 2,2-dimethylpropane the central carbon is bonded to four other carbons, making it quaternary. Is this the underlying reason, and if so then what is the reason for this trend?

It does not have to do with steric effects as you suggested. Rather, the reason behind this trend is the inductive effect. Figuratively speaking, carbon side-chains attached to an central atom "push" electron density towards this atom, while hydrogen atoms don't. If you break a C-C bond and produce a carbon radical, it will be stabilized if electron density is donated towards the site of the unpaired electron, which is the case in the tert-butyl radical formed from dimethylpropane, but not in the methyl radical from ethane.


text:
Why are smoke odors persistent in fabrics?
Obviously smoke is very complex, but apparently phenol compounds are the main components of wood smoke that produce the characteristic odor.  (One of more significant of these compounds seems to be syringol.)  The human nose seems to be pretty sensitive to these compounds.
Smoke seems to be very easily absorbed by various fabric-like materials, which then release the compounds back into the air over long periods of time, making the odor quite persistent.  I'd like to know what is going on with these compounds and the fabric-like materials to cause this behavior.

Why is the smell of smoke so persistent, while many other strong smells are not?
In fact this is a tricky question, because you can't make such a comparison. According to IUPAC Gold book Smoke is:

An aerosol originating from combustion, thermal decomposition or
thermal evaporation. Its particles may be solid (magnesium oxide
smoke) or liquid (tobacco smoke).

So it is a fluid colloidal system in which the "the dispersed phase is a solid, a liquid or a mixture of both and the continuous phase is a gas" others strong smells that you smell during your every day life, are mainly already gasses. The interaction areosol-material an a gas-material is completely different.

Interaction between fabric and gasses
We can decompose the interaction in two different phenomena:

gas trapped inside the little voids inside your fabric
gas adsorbed in the fibers surface

The first is governed by Fick law of diffusion this process reach the equilibrium very quickly e.g. if you enter in a room with a gas after a couple of minutes the concentration of gas inside the voids of your fabric will be the same as the outside concentration.
The second is more hard to determine, however is more slower because the adsoprtion occur if there are free sites where the molecule can be placed, and this is a competitive process between all the others gasses present in larger quantities in the atmosphere, that already occupy these sites.
However if you stay a day in a room with a gas is possible the some gas molecule will be ad-sorbed in your fabric. This interaction is more persistent, however I think that the amount of substance is quite negligible.
So interaction between gasses and fabric is not very persistent.

Interaction between aerosol and fabric
This is completely different because colloidal suspensions are unstable. Aerosol are high energy system that tend to lower their energy forming bigger particles or bonds with other materials. So the deposition of these particles over the surface of others materials will be most of the times a spontaneous process. The high surface of the fabrics indeed make it easier.
So what happen when a particulates/aerosols settles on the fabric? This is not  gas trapped inside the little voids of your fabric or gas adsorbed in the fibers surface is a mixture of liquids and solids: many big (relatively!) "clusters" of molecules spread all over the fabric, from the tiny pores to the fiber's ravines. What matter here is the vapor pressure of the substance  you got this equilibrium (for a liquid):
$$\ce{Molecule_{(l)} <=> Molecule_{(g)} }$$ and the analogous with $_{s}$ instead of "l" for the sublimation. For both of these equilibria only the molecules in the liquid-gas or solid-gas interface can evaporate\sublimate, so the cluster surface is very important!

The case of Syringol

In the case of Syringol you have a solid that melt at 50–57 °C. So the sublimation is the only path that allows you to smell it. For most solids the vapor pressure is very low so the equilibrium will keep going to the right for a long long time, releasing syringol, until the source end.

@Michiel thank you very much, would be very interesting perform some quantitative calculation or a real experiment...This question is really interesting I think that absorption of gases in this sort of surface would be very hard to simulate.

My theory is that it's not the molecules that are mentioned that cling to the fabric but much larger (yet still tiny) particles of soot that cling to the fabric releasing the smelly molecules. this provides an explanation for how other smelly molecules do not linger on fabrics. In addition to that Syringol, as mentioned, is a solid at room temperature contributing to its clinging to the cloth.
This could prove for a interesting experiment, unfortunately i do not have a lab available at this time

This is just a guess, but the surface area of 'fabric' is very large.  The smaller
the threads and the more strands there are, the larger the surface area.
Many organic materials are adsorbed (i.e. stuck to the surface of) onto organic materials.  The larger the surface area, the more will be adsorbed.  Since the adsorbtion "bond" is weak, molecules will break loose from the surface over time.
Don't forget, there can be a lot of molecules in even a little bit of smoke.

The main problem I see with this "surface area" argument is that many, perhaps a majority of, other pungent compounds have much lower persistence with fabrics.  Obviously, though, surface area is a big contributor to the effect.

...there can be a lot of molecules... No, there are a lot of molecules...

Combined with the high surface area argument of @Paul J. Gans, I could conceive that since compounds like syringol have several polar and hydrogen bonding groups, they will cling much more readily to typical fabrics such as polyesters. Gasoline and natural gases, for example, have pretty distinctive smells but in my experience tend not to linger in fabrics the way smoke does and they are distinctly non-polar.

This is almost all conjecture. Well as you mentioned many smoke molecules are phenolic. Pehnols combined with other materials are often used to create laminates for fabrics, an example is phenol-aldehyde fabric resins.
Also I believe many types of partial hydrocarbon burnings create aldehydes. Both are able to dissolve in water, and if there is any moisture in the couch it could cling to the water, when the water evaporates from one area you could have some kind of phenol aldehyde remnants sticking around. This would play  well into the surface area argument as well since water likes high surface area, and materials soluble in water moisture would then accumulate readily on high surface area objects.


text:
Should I use specific heat to calculate how much fuel is required?
Question:

A backpacker wants to carry enough fuel to heat $\pu{3.0 kg}$ of water from $\pu{20 ^\circ C}$ to $\pu{100.0 ^\circ C}$. If the fuel he carries produces $\pu{36 kJ}$ of heat per gram when it burns, how much fuel should he carry? (For the sake of simplicity, assume that the transfer of heat is $100\%$ efficient.)

My hunch is that it involves the specific heat equation $Q_\mathrm p = m  c_\mathrm p  \Delta T$ where $Q$ is heat, $m$ is mass, $c_\mathrm p$ is the specific heat capacity (in this case $\pu{4.184 J/(g ^\circ C)}$) under constant pressure and $\Delta T$, but twice I calculated the wrong answer, and would like a more intuitive understanding of the problem.

3.0 grams or kg of water? (I suspect the difficulty is there or going from kJ to J)

$$\Delta H = mc_p\Delta T$$
Where $\Delta H$ is either energy absorbed or released at constant pressure.
Assuming this specific problem refers to $3\ \mathrm{kg}$ of water:
$$
\Delta H = \left(3000\ \mathrm g\right)\left(4.184\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\right)\left(80\ \mathrm K\right)
$$
That means that the energy required is: $1\,004\,160$ joules.
Now, simply divide those 1 004 160 joules by the 36 000 joules your fuel releases per gram.
$$
m_\text{fuel}= \frac{1\,004\,160\ \mathrm J}{36\,000\ \mathrm{J\cdot g^{-1}}} = 27.89\ \mathrm g
$$


text:
How to calculate the solubility of CaF2 at different temperatures?
At $\pu{25^\circ C}$ I used the free energy table for $G$ for the reaction and got $K_\mathrm{sp}= 10^{-9.02}$.
I now need to find out what the solubility is at $10$, $20$ and $\pu{30^\circ C}$. Could you also show what equation you used and how to work out the solubility by one of these other temperatures?

Could you elaborate on how you might approach the problem first, rather than asking for a solution, I think it will be a lot more likely to receive an answer that way.

It depends what you have access to… but if we suppose that you have access to decent thermodynamic tables, here's a hint:
You know how solubility is linked to dissolution reaction (Gibbs) free energy. Now, how does the free energy depend on temperature? In other words, what is $\displaystyle\frac{\partial G}{\partial T}$?

I do have access to thermodynamic tables. I tried using the van 't Hoff equation, because there is the relationship between temperature and the K solubility product.
However, I am not 100% sure that I am using the right equation and I don't have any way to check if my answers are correct.
The answers I get from using the van 't Hoff equation are a little bit counterintuitive.
For 10° Celsius, I get a solubility product of Ksp= 10^−9.02
For 20° Celsius, I get a solubility product of Ksp= 10^−9.03
For 30° Celsius, I get a solubility product of Ksp= 10^−9.00
This just seems incorrect to me.

I think you misread the results.  The 20 C reading is the smallest.  Assuming that Erin used a constant $\Delta H$, that's not possible.  I'd guess that there is a math error involved.


text:
How does simulation software work at a fundamental level?
There are lots of software packages like Schrödinger Maestro or Argus Lab, which are used for simulations of docking.
I have always been fascinated as to how such software works. What is the fundamental principle working "under the hood" of this kind of sofware?
Let's say I wanted to make a simple piece of software which, when I input the atoms and atom number (e.g., C1H4) can be used for finding different energy levels (or other properties) in a molecule such as methane ($\small\ce{CH4}$).
Where does one have to start in attempting to make such a package, given a considerable knowledge of programming? What theory is applicable?

Try to analize one of existing open-source packages, like mpqc.

Well, that's a rather diffuse question. From what I have seen, the software packages you mention do Molecular Mechanics (MM), which is a treatment of molecules with classical physics. For each atom in a given molecule, the forces acting upon it due to chemical bonding with other atoms is calculated using a set of empirically determined parameters (a force field), and then an estimate of the potential energy is provided. But given that this question is tagged with "Computational Chemistry", I also want to mention quantum chemistry, i.e. the application of quantum mechanics to atoms and molecules. This field is both far more interesting and far more difficult than the classical MM treatment.
A first shot would be the Wikipedia article about Computational Chemistry which already points you to other relevant topics, i.e. the article about Molecular Mechanics, which is quite concise about the topic (although I don't know how exhaustive it is in this regard, since I don't deal with MM in my research). For quantum chemistry, the necessary knowledge is more spread-out; relevant Wikipedia articles would be Eigenproblems, the Schrödinger equation and Ab initio methods. I don't know any literature about MM, but if you want to look deeper into quantum chemistry, I would recommend the following two books (and lots and lots of time and patience for reading them):

Attila Szabo, Neil Ostlund: "Modern Quantum Chemistry". Provides a good mathematical introduction and thorough treatments of the Hartree-Fock and Moller-Plesset methods. It's cheap, also.
Ira Levine: "Quantum Chemistry". Covers a wider range of methods, but is in my opinion a bit more difficult to consume than Szabo/Ostlund.


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Why is tin(II) fluoride more effective in turning apatite into fluorapatite?
In more powerful toothpastes, like Crest Pro-Health, stannous fluoride is used in place of sodium fluoride.
According to Wikipedia:

Stannous fluoride converts the calcium mineral apatite into fluorapatite, which makes tooth enamel more resistant to bacteria generated acid attacks. In toothpastes containing calcium minerals, sodium fluoride becomes ineffective over time, while stannous fluoride remains effective in strengthening tooth enamel

Is the stannous fluoride more effective in this reaction simply due to having two fluoride ions per mole, or is there something more complicated in the mechanism of the reaction in which the tin spectator ion is beneficial?

I took a look at the paper that Wiki links to, but the abstract doesn't tell anything about SnF2 and in my opinion just adds to the confusion. Do you know anyone who could provide you with the full article?

@Antimon I will try to get a hold of it in the next couple of days.  Thanks for checking it out anyway.  I get the feeling the actual results may be in another (cited) study in the paper.

Found an answer to this question in a book called 'Pediatric Dentistry'.
Apparently, when Stannous Fluoride reacts with dental enamel (hydroxyapatite), both the fluoride and the tin react to form Stannous Trifluorophosphate $(\ce{Sn3F3PO4})$ -- This is highly resistant to acid attack.
At high concentrations of stannous fluoride (I don't know what counts as "high", the book doesn't say), apparently the following compound is formed as well: Trifluorostannate -- This is also highly acid resistant.
In addition, Calcium Fluoride is also produced, which then further reacts with hydroxyapatite forming Fluorhydroxyapatite (which is also rather acid resistant, although it seems that the stannous trifluorophosphate is more resistant.)
Sodium fluoride only produces Fluorhydroxyapatite as its reaction.
It appears that Stannous Fluoride is a superior agent because it produces 1 or 2 additional compounds (that become part of the surface enamel) that are more acid resistant than just the one that Sodium Fluoride produces. These other compounds (which become a part of the surface of the tooth) also have an anti-microbial effect, thus inhibiting the growth of the very bacteria which are directly responsible for producing the majority of the acid found in dental decay.

$\ce{SnF2}$ appears to have a predominantly covalent structure while $\ce{NaF}$ is ionic. So I think $\ce{SnF2}$ might release fluoride ions slower (more reluctantly) than $\ce{NaF}$.
Since toothpaste contains $\ce{Ca}$ containing minerals as well (as mentioned in the Wikipedia page), highly insoluble $\ce{CaF2}$ is formed easily when more fluoride ions are present. This is bad since $\ce{CaF2}$ will be inactive.
So since $\ce{SnF2}$ doesn't release fluoride ions as much (unless it comes into contact with water – it undergoes hydrolysis fast. Again this is good since more fluoride will be available only while you are brushing your teeth.), it has a longer shelf life than $\ce{NaF}$ I guess.
This also means that it doesn't matter what fluoride you use as long as its covalent. However I suspect tin is used because it happens to be one of the covalent fluorides relatively harmless to us (see this).
This is all just a guess so take it with a grain of salt... I don't have many sources to back it up (except Wikipedia of course!). Do point out any mistakes.
EDIT
As Georg has mentioned in the comments, toothpastes have a significant amount of water. Apparently toothpastes containing stannous fluoride contain a strong complexing agent which prevents oxidation of the $\ce{Sn^2+}$ ion (I don't have access to the paper but I read the abstract here).
Also the abstract mentions that $\ce{Sn^2+}$ has a antibacterial effect (most probably due to it being a heavy metal ion) and this is another reason why it is used (in contrast to $\ce{Na+}$ which has almost no antibacterial effects).

So You think toothpastes are free of water? Think about stability of SnII with respect to oxidation and write a new, hoprfully less nonsensy answer.

@Georg I agree that I didn't consider that toothpastes have a significant amount of water. I have already mentioned that it was just a guess. There is nothing more to be said...

I have edited in why $SnF_2$ is stable even though there is water in the toothpaste.

I'll accept this for now, but if you run into any sources supporting your position, please feel free to incorporate them in.  And, thanks!


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What is the exact definition of Salinity?
I am a bit unclear on the definition of salinity.  I have always thought of it as simply total dissolved ions.
Wikipedia seems to agree "saltiness or dissolved salt content" but many other sources seem to suggest that only certain ions contribute.
So is there a list of specific anions and cations that contribute to salinity and is there an underlying logic involved in their selection or is it simply arbitrary? Is the definition dependent on the solute in question?

The term 'salinity' is measure of 'saltiness', particularly used in reference to sea water.
The exact scientific definition has evolved over time to incorporate our ever growing understanding of physics and chemistry.
In 1889 the International Council for the Exploration of the Sea (ICES) used a measure of salinity based on the total dissolved solids of salt (g/kg) found in seawater after evaporation and drying at 480$^{o}$C.
By taking samples of ocean water from around the world and at different depths, oceanographers have determined that whilst the concentration of salts varies, the relative ionic composition remains essentially constant. The average salinity of the world's ocean water is usually stated as 35 g/kg, with the relative ionic composition of this "average sea water" made up as follows:
The major cations present in 1 kg of sea water are Na$^{+}$ (10.7653 g), Mg$^{2+}$ (1.2942 g), Ca$^{2+}$ (0.4117 g), K$^{+}$ (0.3991 g) and Sr$^{2+}$ (0.0079 g).
The major anions present in 1kg of seawater are Cl$^{-}$ (19.3534 g), SO$_{4}$$^{2-}$ (2.7124 g), HCO$_{3}$$^{-}$ (0.1178 g), CO$_{3}$$^{2-}$ (0.0122 g), Br$^{-}$ (0.0674 g), F$^{-}$ (0.0013 g), B(OH)$_{4}$$^{-}$ (0.0066 g).
The process of determining salinity from evaporation & drying can be quite difficult, particularly on board ships.
In 1963, the ICES adopted a new definition based on "chlorinity" measurement (using titration). The "average sea water" with ionic composition given above was determined to have chlorinity of 19.374%. By measuring the relative chlorinity of a sample of sea water, the salinity could be calculated using the formula S = 1.80655 Cl.
In 1978, with the advent of modern electronics, the United Nations Educational, Scientific and Cultural Organization (UNESCO), International Council for the Exploration of the Sea (ICES), Scientific Committee on Oceanic Research (SCOR) and International Association for the Physical Sciences of the Oceans (IAPSO) adopted a method based on electrical conductivity measurements compensated for temperature. The Practical Salinity Scale of 1978 (PSS78) defines S (a dimensionless quantity) from conductivity and temperature measurements, in the range 2 $\leq$ S $\leq$ 42, by the following formula:
S = 0.0080 - 0.1692$K_{15}$$^{1/2}$ + 25.3851$K_{15}$ + 14.0941$K_{15}$$^{3/2}$ - 7.0261$K_{15}$$^{2}$ + 2.7081$K_{15}$$^{5/2}$
Where $K_{15}$ is the ratio of conductivity of the sea water sample to that of potassium chloride (KCl) solution of mass fraction 32.4356x10$^{-3}$, at 15$^{o}$C and atmospheric pressure.
A formula is also provided for compensation of the conductivity measurement at different temperatures (see http://unesdoc.unesco.org/images/0004/000479/047932eb.pdf)
In 2010, the Intergovernmental Oceanographic Commission released "the international thermodynamic equation of seawater (TEOS-10)" in which it recommended a return to the use of "Absolute Salinity" to take into account variations in the composition of sea water which can occur for example in regions where rivers run into the ocean or where some other regional variability results in "anomalous" sea water composition.
The rational for this approach is based on understanding the effect each chemical component of seawater has on its thermodynamic properties such as density, pressure, temperature and Gibbs function. Various measures of salinity are defined and their relationships with thermodynamic properties described.
The Added-Mass Salinity, which incorporates the Reference Composition Salinity of "standard sea water" (the Practical Salinity Scale) with an adjustment for the Absolute Salinity Anomaly (mass fraction of material which must be added to standard seawater to achieve the concentration levels in the sample) is sometimes difficult to determine in-situ.
Modern techniques for determining Absolute Salinity involve taking various other measurements such as conductivity, temperature, pressure and density by incorporating measurements from sensitive electronic and optical instruments with data-driven computer algorithms.
The Density Salinity, which is the salinity argument in the thermodynamic expression for density which gives the sample's actual measured density, can be used where accurate measurements of density are available, such as a vibrating-tube densimeter or optical refractometer. When combined with Practical Salinity, temperature and pressure measurements, this can provide an accurate and easily determined measure of the amount of dissolved material in seawater (within a certain range).
Composition anomalies can be estimated by computer algorithms which take into account longitude, latitude and depth data. The accuracy of such algorithms is dependent upon the collection of accurate data from field measurements and laboratories across the world and is therefore an evolving process.
(For more details, see: http://www.teos-10.org/).

Thank you theo, I had to read that over several time for clarity but I think I get it.  The distinction between Practical Salinity and Absolute Salinity was particularly illuminating.


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

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

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

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

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


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Looking for a safe exothermic reaction
Is there an exothermic reaction which produces 1 to 2 minutes of heat at around 175 degrees Fahrenheit, no toxic gasses and is considered to be more or less safe?

Safe is a matter of scale. The reactions mentioned in the other answers are all "safe" on small scales, say less than 100 mL or 100 g. What are you planning on heating? How big is it?

Crystallisation of supersaturated sodium acetate is a good one, as seen in my answer here.
Another is the reaction of water with calcium oxide, which is commonly seen in self-heating coffee and soup cans. This reaction can get very hot, though.

Simplest thing that comes to my mind is dissolving sodium or potassium hydroxide in water, and/or neutralizing it. Both steps are exothermic, and also rather safe (given that you don't just blindly throw things together). Should become hot enough, I'd guess.

Use a mix of ammonium chloride, sodium nitrite and acetic acid. It produces water, nitrogen and heat.

Could you elaborate? Maybe add reaction equation

"Ammonium chloride" and "sodium nitrite" are actually aqueous solutions of those salts. The reaction is $$NH_4Cl + NaNO_2 --> N_2 + 2H_2O + NaCl$$.


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What is the chemistry behind this "petrol from air" technology?
A recent news report in the UK claimed a breakthrough in making a petrol equivalent from carbon dioxide and water:

A small British company has produced the first "petrol from air" using a revolutionary technology that promises to solve the energy crisis as well as helping to curb global warming by removing carbon dioxide from the atmosphere.

It sounds like the news reports are over-hyping it a little (and I'm not sure it is that new), so what is the real chemistry behind it? What are the key reactions and how much input energy is required?

Higher rep users might want to think of useful tags for this one. I would have liked something about industrial processes and maybe catalysis but can't yet create tags.

The energy required is certainly more than the amount you'll get out by burning the result, you can't cheat the laws of thermodynamics.

The gist is: water + electrolysis -> hydrogen; hydrogen + CO2 from air -> methanol; methanol + gasoline fuel reactor -> petrol

@EnergyNumbers that would make a great answer if you just summarised the chemistry of the key steps (eg what mechanism, conditions and catalysts covert hydrogen and CO2 into methanol).

@matt_black unfortunately that's pretty much all I know. AIUI they were planning to go from CO2 to CO with a reverse water-gas shift, and then add Hydrogen in a Fischer Tropsch to make the hydrocarbons. But I've seen no tech lit from them, and I guess they're playing their tech cards very close to their chests.

@matt_black Read about Gas to Liquid process in Wiki. Most of the path shown here it used there too.

Step one: trap $\ce{CO2}$. This can be done using specially selected amines, like triethanolamine. Aminosilicones (a common chemical used in hair conditioner), can also do this.
Step two: release trapped $\ce{CO2}$. In case of triethanolamine you'll have to boil mixture.
Step three: mix $\ce{CO2}$ with hydrogen (hydrogen is produced by electrolysis of water). Volume proportion should be somewhere 1:3.
Step four: threat the mixture with proper catalyst. This step may be divided into two stages: reverse water-shift reaction  and some variant of Fischer–Tropsch process with some type of alcohol synthesis as very interesting candidate. There are some multi-functional catalysts that can perform both stages at once.
Another variation on step 4 is the Sabatier process, which involves generating methane from $\ce{CO2}$ and hydrogen.  Typically the catalyst is Rainy nickel (powder).
The required energy in form of electrical current is probably from 1.5 to 3 times of heat of combustion of resulting fuel, which is in turn at least twice more that can be converted into actual work again.

While your estimate of energy efficiency sounds reasonable to me, I'd highly appreciate pointers to  literature where such estimates are given - that would help tremendously in discussions...

@cbeleites well... internal combustion engines have energy conversion efficiency 30-50% and the process is essentially fuel combustion reversed, so at the very least all energy released in combustion must be provided at the time of process. However, some energy is lost during $H_2$+CO_2 -> CH_x + H_2O$ part to shift chemical equilibrium. 30% loss is actually a point-blank shot, but process with 10% losses is actually a very good one and we have at least three steps here. Upper estimate have no actual foundation under it except unbelief that something with ridiculous losses may be used.

I agree with the internal combustion engine efficiency; but I was wondering at the synthesis part. Anyways, thanks. I'm not so sure about not having ridiculous losses, though (given what I found here http://chemistry.stackexchange.com/a/1361/160). I had a quick look at coal liquefaction (which is not the same, but possibly related in difficulty, and has a rather long industrial history): http://www.aspo2012.at/wp-content/uploads/2012/06/H%C3%B6%C3%B6k_aspo2012.pdf. So: 60-70% theoretically possible, existing plant reaches 20-28%. Ouch...


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What would happen to a Water/Lard Mix under pressure and high heat?
If water is mixed with lard oil and heated (creating some super-critical water), how does this affect the volatility of the mixture in comparison if it is a pure mix of either water, or lard oil?
What happens when a mixture of lard oil and water is compressed and then heated to over 800 degrees Fahrenheit almost instantaneously (up to 1200 degrees Fahrenheit)? If it is in a well sealed copper container, would it be volatile enough to burst through the container?
Note: I'm no scientist or chemist, so use of laymen's terms, if possible, would be appreciated

I'm not sure what aspect you are asking about exactly… but all supercritical liquids are completely miscible with each other, so you would get one supercritical phase with the mixture of the two… supposing the oil doesn't decompose.
However, the smoke point of lard is 182°C while the critical temperature of water is 374°C, so what most probably happens is that your lard will burn, breaking down to glycerol, acrolein and free fatty acids.

I edited my question in hopes of clarifying for you... Thank you for your response. I am wanting to know, if it is contained in a soft metal, like copper if it would be volatile enough to burst through the container... And if, the supercritical water and the lard oil combination is more volatile than if it was a pure sample of either liquid. Hope that helps... Thanks again!!

@TirzahVernon this latter question is a matter also of pressure, requiring dimensional figures for the fluid volume and container volume.  If the container is not a primitive (e.g. cylinder) then be sure to indicate.  You may want to ask this question in physics.SE


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What sort of molecules or polymers are used in OLED displays?
OLED displays have been emerging as competitors for LCD displays for some time and are now common in mobile phones (see comparison of a Google Nexus 1 and an iPhone here)
The displays use small molecules or polymers that emit light when a current is passed through them (like a semiconductor LED). But what is the structure of the emitters?

The Wikipedia article you link to lists a number of popular emitters, give their structures, and link to more comprehensive articles about them. I'm not sure what else you want. Is their something deficient about the Wikipedia article(s)? Help us help you.

@BenNorris The wikipedia article doesn't provide much detail. I was hoping that someone who knows the current state of the industry would provide some more recent examples. There are several companies with OLED tech and I don't think they all use the same chemicals. I think a good review would make a good answer for this site.

A typical organic light emitting diode (OLED) structure consisting of a glass substrate, a indium titanium oxide (ITO) anode, and an organic bilayer diffused into a metallic cathode, is given below:

The organic bilayer consists of:

a hole transport layer (HTL), typically of N, N'-diphenyl-N, N'-bis(3-methylphenyl) 1, 1'-biphenyl-4, 4' diamine (TPD) or N,N′-di(naphthalene-1-yl)-N,N′- diphenylbenzidine (NPB)
an emitting/electron transport layer (EML/ETL), typically of tris(8- hydroxyquinoline) aluminum (Alq3)

Source: http://espace.library.uq.edu.au/eserv.php?pid=UQ:8598&dsID=SMS.pdf


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How to fluorinate polyethylene?
I seek to put a layer of fluorine on a rotationally molded cross-linked polyethylene (XPE) plastic gasoline tank. I sell tanks and they will be more eco-friendly if I can accomplish this.
I am an electrical engineer and my last experience with chemistry was Chem 200 lecture and lab at SDSU, a one semester intensive "all of chemistry for engineers" kind of course. I asked a chemistry professor at my alma mater about this, who said:

Xenon difluoride is a powerful fluorinating agent, and is soluble
  (according to Wikipedia) in anhydrous methyl cyanide and several
  fluorinated solvents.  Offhand, this looks to me like it would do the
  job, but it's not a friendly compound (toxic, possibly corrosive).

The concept is already in place all around us: any time you buy a plastic jug with harsh chemicals in it, the jug has been fluorinated. So far, the majority of my learning is confirmation that yes, indeed, fluorinating polyethylene (PE) surfaces with $\ce{XeF2}$ leads to decreased gas permeability.
I've done a bit of Googling and bought a couple of great textbooks, Fluoropolymers 1; Synthesis and Fluoropolymers 2; Properties. Fluoropolymers 1 seems to have great stuff in a couple sections. Perusing Fluoropolymers 2, I haven't seen anything jump out at me yet.
There are two methods I have read about:

Involves an oven, steady high temperature, and plasma discharge.
"Surface Fluorination of Polymers Using Xenon Difluoride", or Gaseous $\ce{XeF2}$ and PE are placed in a "stainless steel reactor" and left alone.

Seeing the options, the 2nd seems the way to go, and this seems to be the epic equation concerning my quest quoted from Fluoropolymers 1 page 231, section 15.3:

The comparative accessibility and desirable properties of $\ce{XeF2}$ enable
  its use for the synthesis of fluoroorganic compounds under “mild”
  conditions. As a fluorinating agent $\ce{XeF2}$ can be used to effect
  electrophilic fluorination of various organic compounds with good
  yield and it is a useful reagent for generating radical cations.
  ... In many cases XeF2 is an attractive alternative to the
  conventional Balz–Schiemann reaction for the preparation of
  fluorinated aromatic compounds, e.g., fluorobenzenes:
$\ce{RC6H6 ->C[XeF2] Rc6H5F + Xe + HF ->C[XeF2] RC6H4F2 }$
Xenon difluoride reacts with carbon–carbon single, double, and triple bonds
  giving addition fluoroorganic compounds, e.g.,
$\ce{R-CH=CH2 ->C[XeF2] RCH2-CHF2 + Xe}$
and
$\ce{RCH2-CH3 ->C[XeF2] RCHF-CH3 + Xe + HF ->C[XeF2] RCF2-CH3 + Xe + HF}$

And this quote from page 236 gives me great hope:

Another argument in favor of the $\ce{XeF2}$ method is the evidence that since 1995 it
  has been used by a manufacturing company that had no previous experience with
  direct fluorination or fluorine at all.

... for I'm in the same position now!

How do I actually perform a fluorination surface treatment my XPE products?
Where / how do I go about getting $\ce{XeF2}$? And if it is soluble in a "fluorinated solvent" like my prof said, is there a way I can flush this through the interior of my tanks in a controlled manner that would accomplish some degree of fluorination?

Edited to add 10-22-12: I called a few companies today. Everyone is pointing me to pay Fluoro-Seal to do this. They are the only company providing this service in the US. I'll need to ship my products to them, wait a couple weeks, and pay $15~45 each tank. I'd still like some kind of bath or wash to do this. This is why no-one making replacement tanks has them emissions legal... all this triples my cost.

Edited to add 12-30-13: Fluoro-Seal has given up on fluorinating rotationally molded PE. I'm pushing a plastics manufacturer to develop their existing acetel product to work with rotational molding. Their solution for injection and blow molding is in production and impervious to fuels (they have parts in nearly all automotive fuel systems in the US right now). So there's hope for a solution; however it seems impossible to fluorinate PE.

FWIW, I've been honing my Googlefu and finding articles by RJ Lagow from the 90s. Hopefully, somewhere he'll describe the physical environment to accomplish this, although my favorite would be a fluid I can "wash" it with.

Update: This doctor's course notes were a boon for me: http://crab.rutgers.edu/~alroche/fluorocarbons.html

"Offhand, this looks to me like it would do the job, but it's not a friendly compound (toxic, possibly corrosive)" - Find me a good fluorinator that isn't toxic or corrosive. The mechanism that makes fluorine gas, and fluorides of the groups on the right-hand side of the periodic table, so useful in reaction chemistry is the same mechanism that will kill you if you get a whiff of it, and that will start a roaring fire with materials you'd bet the rent on being nonflammable (like concrete, sand, asbestos, ceramics).

You send it out to a commercial fluorinator, fluorine in nitrogen gas.  Plastic engine oil jugs are typically surface fluorinated as a solvent barrier.  If you go for acetal, remember that gasoline is 10% ethanol (courtesy of your idiot corrupt government) and other stuff, including detergent.  Solvent attack.  A gas tank must pass collision safety testing - be complaint rather than shatter on impact.

@UncleAl Fluoro-Seal is the only company performing gaseous fluorinating of fuel tanks and they have determined they cannot process rotomolded PE tanks.

They will not be as environmentally friendly as you think because you'll just be contaminating the environment with more perfluorinated organics.

There is a simpler way to get fluorinated polyethylene.
Polyethylene is polymerized ethylene $\ce{CH2=CH2}$. As part of the polymerization, the double bond is no longer there. Polyethylene is essentially a very long saturated hydrocarbon:  $\ce{(-CH2-CH2 -)}_n.$

Another polymer already exists that replaces all the hydrogen atoms with fluorine atoms: poly(tetrafluoroethylene) (PTFE), made from tetrafluoroethylene $\ce{CF2=CF2}$. PTFE is a long saturated ferfluorocarbon: $\ce{(-CF2-CF2 -)}_n.$

The most well known trademarked brand of PTFE is Teflon. Get your coated with or made out of Teflon/PTFE. It will be easier, cheaper, and safer in the long run. If XeF2 will fluorinate polyethylene (which is reasonably inert), imagine what it could do to yyou.

Hi, thanks for your answer! Unfortunately, ptfe as an originating powder is 12x more costly. Next, teflon absorbs water, so the molder has to buy small batches, which makes it again 50% more costly. Last, teflon applications in the field as made by the OEM are absorbing gasoline molecules into the matrix, resulting in odd expansion (very large "bubbles"). PTFE isn't the source... If we could do multiple layer types, it could be considered in a layer, but that industry is 10x more costly than rotational molding.

PTFE is very hydrophobic. It should not absorb water.

I'm going off of what my vendor says. I don't doubt them when they got mad about losing a quarter ton of powder.

Marking this as answered. Everyone is saying use nitrogen gas with 10% fluorine, which is "commercially available". There isn't an aqueous solution to my problem because of the difficulty getting the hydrogen off the surface of the polyethylene solid.

BTW, just realized I had mistaken your suggestion of PTFE (teflon) with Nylon, PA6. I'll ask them about teflon.

@ChrisK Oh, yeah I can imagine nylon as both hydrophilic and more expensive.

There are commercial companies that fluorinate the surface of polyethylene.  Its not clear whether they use fluorine, carbon tetrafluoride and plasma, or xenon tetrafluoride. Given the lower cost and ease of manufacturability of polyethylene compared to PTFE and the extreme hazards in casually using xenon tetrafluoride, I would suspect that utilizing a commercial service would be an optimal solution.


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What is the meaning of '(en)'?
What is the meaning of $\ce{(en)}$ on this two coordination compounds:
$\ce{[Co(en)_2(NO_2)_2]Cl}$

$\ce{[Cd(en)_2(CN)_2]}$
Is it a type of compound? If yes, what is the meaning?

It's an ethylenediamine ligand.
Edit - elaborating as per your request: Unfortunately, you don't provide any information as to how much you already know about complex chemistry, so I'm going back to the roots here. If you already know what a complex is, skip this paragraph. As you probably know, atoms and molecules may bind to each other, forming new compounds. We all know covalent bonds in which two atoms move close together and mutually share their electrons, and ionic bonds wherein one of the partaking partners is more egocentric in terms of electrons, "stealing" it from the other but staying in the vicinity nontheless. What these types have in common is that each atom or molecule provides an electron for the formation of the bond. However, it may also happen that both electrons come fron one partner alone; an electron-rich molecule or ion approaches another atom or ion, "donating" his electrons without really giving them up. This is the situation in coordination chemistry: A "central atom" or ion -- usually, but not necessarily a transition metal -- is surrounded by a varying number of "ligands" which donate electron pairs to the central atom, all while these electron pairs remain "located" at their respective ligands (very dangerous expression, I know ...). The compound that is formed is called a complex.
So ligands are atoms or molecules which can provide lone pairs for coordination bonds, i.e. electrons in non-bonding, doubly filled orbitals. In the above example of ethylenediamine, the binding centers of the molecule are the nitrogen atoms. The en ligand is an example of a multidentate (or chelate) ligand which may form more than one coordinating bond per molecule, while the $\ce{NO2^-}$ and $\ce{CN-}$ are monodentate ligands. The chloride anions in your first compound does not bind to the copper ion, but rather stay outside its "coordination sphere" to balance the positive charge of the cobalt complex (which you might then interpret as an ionic bonding between the complex cation and the chlorine anion).
Note that there are, of course, many quirks to this. Complexes with multiple central atoms, different theories about bonding, bridging or radical ligands, charge-transfer complexes, etc. This is a very wide and important field of chemistry, after all; nothing you could exhaustively explain in one or two paragraphs.

Could you elaborate? Maybe explain a bit about what a ligand is etc. Your answer is really too small in its current form. Make it interesting :)

Oh, I thought it would be just about the nomenclature :) okay, editing ...

Ah, muuch better :) This sort of answer lets people learn more without having to look elsewhere.. Makes Stack Exchange a valuable internet resource.

$\ce{(en)}$ serves as a ligand that binds to the metal ($\ce{Co}$) in the complex dinitro bis(ethylenediammine) cobalt (II) chloride.

en is $\ce{H2NCH2CH2NH2}$, a bidentate ligand coordinating through its nitrogen lone pairs, one pair/nitrogen.  Compare with oxalate.