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What is the geometrical structure of OF₂?
Is it linear, angular, triangular or tetragonal? Well, since $\ce{O}$ has 2 electron matches of his own, I think it's tetragonal, but in wikipedia it is angular... Why? What about $\ce{CS2}$, is it linear?

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Oxygen difluoride ($\ce{OF2}$) is analogous to water ($\ce{OH2}$) both in terms of structure and hybridization.

$\ce{OF2}$ has 4 ligands around the central oxygen, 2 $\ce{O-F}$ bonds and 2 lone pairs.  Just like in water these 4 ligands assume a roughly tetrahedral geometry around the central oxygen. Again like in water, the $\ce{F-O-F}$ angle is less than the perfect tetrahedral angle (109.5°), in this case being around 103°.
Carbon disulphide ($\ce{CS2}$) is linear.  It is analogous to the 3 carbons in allene, in both compounds the central carbon is $\ce{sp}$ hybridized making the $\ce{X=C=X}$ structure linear.

You're right, if you're talking about the arrangement which take the nonbonding electrons into consideration,
Which is not the case in VSEPR geometry,
Since we have a difference in electronegativity between F and O (F is more electronegative)  ,  so we will have AX2 geometry which is angular.

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The angular structure is actually a distortion produced by the bond pair electrons (between oxygen and fluorine) and lone pair electrons of oxygen since every structure tries to decrease its net potential energy they assume this shape. This doesn't happen in case of Carbon-disulphide because the sulphur atoms each make a double bond with carbon leaving no lone pair or relative high electron density on carbon atom, instead the two sulphur atoms with 2 lone airs each after forming double bonds, repel each other into a configuration of linear to minimise their relative potential energies.


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What are the products of the reaction between fluorine and ammonia?
What are the products of the reaction between fluorine and ammonia? Is it the same as with chlorine and ammonia, which depend upon which reactant is taken in excess?

Fluorine reacts with ammonia to give dinitrogen and hydrogen fluoride as the major product.
The Formation of Dinitrogen Tetrafluoride in the Reaction of Fluorine and Ammonia - J. Am. Chem. Soc. 1959, 81 (23), 6338-6339 - states that dinitrogentetrafluoride accounts for 6% of the yield while the major reaction is:

$\ce{3F2 + 2NH3 → N2 + 6HF}$

The same article quotes in its first paragraph that the result remained the same irrespective of the concentration of the reactants. (In the case of chlorine it matters which reactant is taken in excess)

I wonder if the N2F2 produced is just the product of NF3 degredation...

The reaction is this
$$\ce{5 F2 {(g)} + 2 NH3 {(g)} -> N2F4 {(g)} + 6 HF {(g)}}$$
And it doesn’t depend upon which reactant is taken in excess; I believe it’s because fluorine is just too reactive.

"And it doesn’t depend upon which reactant is taken in excess."  I find that hard to believe.  Can you provide a source?  As worded your statement implies that reactive 1 mol of F2 with 100 mol of ammonia will still form N2F4.

N2F4 is a minor product , major product is always N2

Actually, there are three types of reaction occuring between ammonia and fluorine which are given here. They all depend on reaction conditions and concentration of reactants. All the reactions are dangerous and produce volatile compounds. Reaction should be done under expert supervision. Please do not try this if you new in chemistry field.

$\ce{2NH3 + 3F2 ->[\Delta] 6HF + N2}$ (mentioned)
$$\ce{2NH3 + 6F2 → 6HF + 2NF3}$$

Ammonia react with fluorine to produce hydrogen fluoride and nitrogen(III) fluoride. The reaction takes place at the low temperature.

$$\ce{4NH3 + 3F2 → 3NH4F + NF3}$$

Ammonia react with fluorine to produce ammonium fluoride and nitrogen(III) fluoride. Ammonia should  be aqueous. In this reaction, the catalyst  can be metallic copper. Impurities: tetrafluorohydrazine $\ce{N2F4}$.

When fluorine reacts with an excess of ammonia it forms $\ce{NH4F}$.
When fluorine reacts with a limited supply of ammonia it forms $\ce{HF}$.
Both these reactions yield nitrogen ($\ce{N2}$) gas.

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Build the geometric structure of SnBr2
What is Sn valence in this case? Because  it depends on it if it will be linear or tetragonal.I'm talking about a molecule.

Clarify, if you are talking about a molecule, a water solution or a solid state.

In the gas phase (molecular), SnBr2 is bent with an angle of 95 degrees.
The IUPAC definition of valence is such that there is only one valence for each element.
Valence is independent of the compound being consider.

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If Alkynes can be reduced to Alkenes using Sodium in liquid ammonia, why can't Alkenes be reduced to Alkanes?

Firstly, I would like to ask why the alkyne reacts at all with the solvated electron? Surely the alkyne group is extremely nucleophilic (more so than the alkene). This, to me, seems to suggest that an alkene would be more likely to react in this manner than the alkyne. However, this is clearly not the case; why?

As shown in your diagram, when the initial electron is transferred to the alkyne a radical anion is formed; the transferred electron pairs with an electron from the alkyne to form an anion at one end of the molecule. These two electrons are in an $\ce{sp^2}$ orbital.  If you try to repeat this process with an alkene, you wind up with the paired electrons in an $\ce{sp^3}$ orbital.  An $\ce{sp^2}$ is lower in energy than an $\ce{sp^3}$ orbital since it contains more s character and an s orbital is lower in energy than a p orbital.  The electron solvated in ammonia has enough energy to transfer into a lower energy $\ce{sp^2}$ orbital, but not enough energy to transfer into a higher energy $\ce{sp^3}$ orbital.

But we can do Birch reduction of Benzene, wouldn't in that case too, the electrons pair in sp3 only?

@GurjotSingh But in the case of benzene the intermediate radical anion is further stabilized by conjugation.  Generally simple alkenes won't undergo Birch reduction, but conjugated systems will.  1,3-Cyclohexadiene is reduced to cyclohexene, but 1,4-cyclohexadiene is not.


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Structure of Metallic Hydrogen
I am simply curious what the theoretical arrangement of metallic hydrogen would be.
I found this article which gives a picture, but I don't understand what the labels mean (e.g. is H5 an isotope, some kind of charge, or just denoting the layer?).
I would like to know:
1) If the image is the correct one.
2) What the labels mean if it is correct.

The figure seems OK. The H1 , H2,H3 ,H4,H5,H6 numbers are just a label of simple hydrogen atoms, how they match up to the graphene like rings "graphene-like layers of hydrogen in irregular six-atom rings"

There are two substances here: metallic hydrogen, which has yet to be produced in the lab, and graphene-like hydrogen, which is what you link to. Which one are you asking about?

Just whether the arrangement depicted for metallic hydrogen is accurate to the actual theoretical arrangement. Sometimes people add pictures that are not accurate, but easier to grasp to make their point.

As annav states, the labeling H1, H2 etc is labeling of hydrogen atoms within a graphene-like hexagonal ring structure. Another way of looking at it, is in the image below:

Image source
In an article published after the web article you link to, Aromaticity, closed-shell effects, and metallization of hydrogen. (Abstract only), the authors confirm that

Recent theoretical and experimental studies reveal that compressed molecular hydrogen at 200-350 GPa transforms to layered structures consisting of distorted graphene sheets.

This structure, according to the article, forms a stable structure at those pressures.  As per the comments, this is form graphene-like hydrogen, the abstract goes further, discussing the process that would result in hydrogen metallization:

Metallization occurs upon further compression by destroying the closed shell electronic structure, which is mainly determined by the 1s electrons via lowering of the bonding bands stemming from the unoccupied atomic 2s and 2p orbitals.

This metallization of hydrogen is, according to the article High temperature superconductivity in sulfur and selenium hydrides at high pressure theorised to occur at pressures of 400 GPa or greater.


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Color changes of a solution of acetone and sodium hydroxide
I have put acetone and sodium hydroxide solution (approximately $\pu{1M}~ \ce{NaOH}$) in a beaker and didn't see any colour change for a day (actually, it was transparent, no colour).
After almost 2 days, I saw a dark brown colour (no precipitate, just diffused colour). Now (after 25 days) its colour seems black, but when I take a small amount and dilute it with water, it is brown. What chemical compound has formed that has this colour? What reaction has taken place? (I doubt it is an aldol condensation)

Probably your acetone has polymerized into short chain by a aldol reaction.

Acetone may have self condensed with itself. As time passed it would have got polymerized too.

Under basic condtions, acetone does condense with itself, first to form mesityl oxide, which is colorless, and then to form isophorone, which has a yellow color.
Both mesityl oxide and isophorone are still reactive carbonyl compounds likely to be attacked by deprotonated acetone carbanions.  That leads to bigger and more conjugated condensation products, which leads to darker and darker colors.
I found a paper which looks in some detail at the base-catalyzed polymerization of acetone, albeit with a different base and at high-temperature conditions in the gas phase, but I suspect a similar pathway happens, more slowly, at ambient temperatures in the liquid.  In the liquid, even larger molecules can likely form because they will stay solvated and can keep reacting (as opposed to in the gas phase where big molecules "crash" out of the phase as soot/gunk/etc.)
The color is likely in part from isophorone but much moreso from large polyconjugated molecules derived from isophorone containing many consecutive sp2 hybridized carbons. Such compounds are very often colored due to intense absorption of visible light.

So, you suggest the colour is because of isophorone?

Partially.  The color is like from large polyconjugated molecules containing many consecutive sp2 hybridized carbons. Such compounds are very often colored due to intense absorption of visible light.  I'll update my answer to reflect this.


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How to RAISE the melting point of water?
It is common that adding common salt into water will lower its melting point. But, is there any way to RAISE the melting point of water? Does it lower the boiling point of water as well? Why?

source
One way to raise the melting point of water is to increase pressure beyond about 635 MPa.  By raising pressure you could get the melting point to be even greater than the normal boiling point.
A second way is to lower the pressure, but this can only increase the melting point by 0.01 K.
A third way would be to added enough of a high melting point solute that you pass the eutectic point. In other words, generally when you add a solute to water, the melting point decreases, but only until you add enough solute that the eutectic point is reached, beyond which adding more solute increases the melting point.

source: http://www.phasediagram.dk/binary/CaCl2.PNG
For example, in the diagram, above, A is the eutectic point.  Beyond this point, adding more CaCl2 increases freezing point.  Then beyond 38% CaCl2, melting point exceeds ) degrees C.

Wow. Thank you. That graph helps me a lot :D.
But, TT what is that eutectic point?

The graph only relates to the pressure changes, not eutectic point.  Each solute would have a different eutectic point. For example, 23% NaCl in water is a eutectic point, beyond 23% melting point will start to increase.

http://upload.wikimedia.org/wikipedia/commons/thumb/d/db/WatNaCl.png/1024px-WatNaCl.png
From: Wikipedia
why is there liquid solution beyond 23% salt solution and below 0 degree Celsius?

@Al3 eutectic point is the LOWEST melting point.  I'm saying that adding more solute starts increasing (instead of decreasing) the melting point once you are at the eutectic point.  I'm saying in some cases, if you go far enough past the eutectic point, the system will have amelting point greater than 0 degrees C.

@Al3 I added another figure to the answer.  Hopefully it is more clear now what I mean.

You can do it with methane under pressure (but more moderate pressure than in @DavePhD's answer) (https://en.wikipedia.org/wiki/Methane_clathrate).  Methane is more soluble in the caged clathrate structure than in liquid water, so the former is stabilized over the latter.  For example, at 50 atm (5 MPa) methane pressure the melting point is raised to $+4°C$.
Such clathrates complicate natural gas processing, an issue also identified in the reference cited above.

Just remember 'water' will not freeze at hypereutectic concentration above liquidus point. Below solidus point, you still won't get ice, only precipitate of the salt.

If what you want is for the water to solidify into ice at a higher temperature, it is impossible to do by adding solute to water, or indeed any other substances.

Nope.  The water still freezes as part of the eutectic mixture after the excess solute has precipitated out.  This eutectic freezing occurs at the eutectic point and does not qualify as raising the melting point of water.  You can raise the melting point with a few solutes that are more soluble in the solid phase than the liquid; methane under pressure does that as described in another answer.

Water is a liquid, thus already melted. Your question should either be about raising the melting point of ice or raising the boiling point of water.
To raise the melting point of ice, meaning trying to keep ice as a solid at a temperature as high as possible, reduce impurities.
To raise the boiling point of water, meaning trying to keep water liquid at a temperature as high as possible, increase the pressure, that is, do it in a pressure cooker instead of with a surface exposed to atmospheric pressure.

Your first statement is obnoxiously pedantic and wrong. Yes there is a special word for frozen water, but it's still water. It goes without saying that we are talking about melting ice if we say 'melting water'.

Man you are obnoxious. It must suck to not be able to infer.


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What is the Lewis Structure for CO?
In many sites I see that they take $\ce{C}$ with valence 4, bur our teacher takes it with valence 2. With what valence should I take $\ce{C}$?

Carbon is typically viewed as having a valence of 4.  Its valence can be satisfied by any combination of bonds and lone pairs of electrons.

There is resonance between various Lewis structures as:
$$\ce{:\!\overset{\ominus}{C}#\overset{\oplus}{O}\!: ~<->~ :\!C=O\!:: ~<->~:\!\overset{\oplus}{C}-\overset{\ominus}{O}\!:::}$$
The term "valence" is not used much nowadays, and "oxidation state" is used but has a different meaning.
The IUPAC definition of oxidation state requires the oxygen have a -2 oxidation state except in peroxides.  All the oxidation states must add up to 0 for a neutral molecule.  Therefore the oxidation state of carbon in CO must be +2.
There is an IUPAC definition of valence:

The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.

This imposes a unique valence upon each element, as explained in this article, in which case valence is always 4 for carbon.
There are other definitions of "valence" such as:

The number of hydrogen atoms that can combine with an element in a binary hydride or twice the number of oxygen atoms combining with an element in its oxide or oxides

(Chemistry of the Elements by Greenwood et al.)
According to this definition, carbon would have a valence of 2 in CO.

How many valence electrons are there in these cases? I'm only considering the middle example since I'm not getting the others. In that case C has 4 valence electrons, but you said it has the valence 2. Can you explain me please?

So it's correct the connection C=O, in which C has no unpaired electrons.

C has one lone pair of electrons in CO. Do you see the lone pair in each Lewis structure of my answer to the left of C.

I do, but if it has a lone pair it means that the number of its electrons is bigger than 2 and we said that it only has 2 valce electrons (the C valence is 2 in this case, as we said)

"valence of 2" does not mean "has 2 valence electrons".  Read the two definitions of valence quote in the answer and see that there is no mention of electrons.

I would have interpreted the Greenwood definition differently. I mean there is a can and an or, there's too much fuzzyness about that.


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What is the Lewis structure for HClO?
In many sites, I see that they put $\ce{O}$ in the middle, but our teacher said that we should put $\ce{O}$ near the extremity. Which is the best way?

The issue is that $\ce{HOCl}$ will dissociate, as usually, into $\ce{H+ + OCl-}$, which can only be possiblbe because $\ce{Cl}$ is formally $\ce{Cl+}$ and not the expected $\ce{Cl-}$. If $\ce{O}$ was to be put in an extremity, one would end up with $\ce{OHCl}$, which makes no sense because $\ce{H}$ cannot make two bonds, or $\ce{OClH}$, which makes no sense because $\ce{HCl}$ exists independently.
Hypohalous acids, like $\ce{HOCl}$, are a point of entry into the "weird" chemistry of halides, which at least in Euopean countries I know is not covered until late in college.

On closer examination I cannot agree with your answer. ‘$\ce{OClH}$ […] makes no sense because $\ce{HCl}$ exists independently’ — this is about as valid an argument as ‘$\ce{CO2}$ makes no sense because $\ce{CO}$ exists independently’. $-1$

I completely agree with Gonçalo Justino's answer, but would like to add another shade to it.
In principle there are quite a few possible arrangements thinkable for this molecule - at least in the gas phase. In this thought it simply refers to the fact that they are local minima on the potential energy surface. The most stable of all is the one that has been discussed here before. Hypochlorous acid is also one of the compounds that disrupt the ozone cycle. Because concentrations in the stratosphere are low, we can find most possible arrangements of the three atoms. It should also be noted, that these reactions usually occur via a radical pathway.
Structure of $\ce{HOCl}$
The bonding situation in these "weird" compounds is not obvious at all, classifying it as a purely covalent bound is equally as wrong as calling it ionic.
Interestingly the $\ce{O-H}$ bond is much more covalent than the $\ce{O-Cl}$ bond, when we look at the most stable, i.e. bent, conformation. I am not going into detail about possible linear intermediates, as they are only very short lived species.
The analysis of a DF-BP86/def2-SVP calculation with the quantum theory of atoms in molecules (QTAIM) shows us this somewhat unexpected behaviour.

What we see here is that the bond between the hydrogen and the oxygen is strongly covalent as the electron density is $\rho_\mathrm{BCP}(\ce{O-H})=0.34~\mathrm{e\,a.\!u.^{-3}}$ and the Laplacian has a high negative value, $\nabla^2\rho_\mathrm{BCP}(\ce{O-H})=-2.02$, i.e. much charge accumulation. On the other hand the chlorine oxygen bond has quite some ionic character. The electron density is $\rho_\mathrm{BCP}(\ce{O-Cl})=0.18~\mathrm{e\,a.\!u.^{-3}}$, while the Laplacian is slightly positive, $\nabla^2\rho_\mathrm{BCP}(\ce{O-Cl})=+0.01$, i.e. low charge depletion. This makes this bond about half ionic and half covalent. (An ideal ionic bond would have an electron density close to zero and a much larger positive Laplacian value.)
From this analysis it makes sense, that in the gas phase the molecule more likely dissociates via $$\ce{HOCl ->[h\nu] HO + Cl}.$$
In aqueous solution there are quite a few more equilibria at work and the reaction and stability of this compound is highly pH dependent. In general this compound will decompose whenever it gets the chance.
Structure of $\ce{OClH}$
This structure is about $61.5~\mathrm{kcal\,mol^{-1}}$ higher in energy and therefore quite unstable. But that does not mean it does not exist.
The interesting point here is that both bonds are about the same strength and both can be considered predominantly covalent. The oxygen chlorine bond, $\rho_\mathrm{BCP}(\ce{O-Cl})=0.25~\mathrm{e\,a.\!u.^{-3}}$, has a little higher electron density than the chlorine hydrogen bond, $\rho_\mathrm{BCP}(\ce{Cl-H})=0.21~\mathrm{e\,a.\!u.^{-3}}$. Both have a negative Laplacian, $\nabla^2\rho_\mathrm{BCP}(\ce{O-Cl})=-0.19$, $\nabla^2\rho_\mathrm{BCP}(\ce{Cl-H})=-0.57$, i.e. charge accumulation.

Other possible structures of $\ce{HOCl}$
Any other structure, for example linear or bent $\ce{OHCl}$, are no minima on the potential energy surface. They might exist as excited or transition states, but they are incredibly short lived. One only has to care about these species when dealing with complex gas phase reactions. (There is now way they could be stable in solution.)
Conclusion
While it is theoretically possible to have the oxygen in a terminal position, it is certainly not the ground state. Another aspect to think about is also, how accurately can a Lewis structure represent the molecule. Especially when it comes to small molecules one always has to take it with a grain of salt as most of the theories brake down at this level.
For up to highschool level of chemistry, however, the answer to this question is the one given by so many others before.

If that did not bore you to death, might I suggest additional literature: Kirk A. Peterson, Sergei Skokov and Joel M. Bowman; J. Chem. Phys. 1999, 111, 7446. (mirror)

So when we say that water is the compound with the chemical formula $\ce{H2O}$ would it better to specify also its structure? Shouldn't do this with all compounds? For example why we identify $\ce{CO2}$ as having the structure $\ce{O-C-O}$ and not the structure $\ce{C-O-O}$? There is no reason that a chemical formula must correspond to a single structural formula.

@ado I have no idea, where you are heading with this. It is obvious that a sum formula dues not contain any structural information, a name however might imply a common structural information. This is the case for water. For compounds with the same sum formula, there are usually plenty of structures, think isomers. Obviously you must include this information to know what you are talking about. How your comment applies to this answer, I don't understand.

Putting the oxygen in the middle is definitely the correct way to go.

Lewis structure for HOCl

The properly way to determine the Lewis structure, based on this example, is:

Total valence electrons: $7 + 6 + 1 = 14$
Total electrons needed for octets/doublets: $8\cdot2+2\cdot1 = 18$
Total shared/bonding electrons: $18-14 = 4$ (In other words, there are only two single bonds.)
Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 14 - 4 = 10$ (In other words, the are only 5 pairs of lone electrons (2 for $\ce{O}$ and three for $\ce{Cl}$))

Furhter information on how to draw Lewis structures, please follow the link

The lewis structure of Hypochlorous Acid:

Other answers have shown the chemical correctness well. But from a high school perspective there’s nothing stopping you from drawing the structure differently, is there? Well, let’s look at both possibilities in Lewis structures (lone pairs omitted for clarity):

Consider each atom to abide by the octet rule (hydrogen: dublet rule) and thus the terminal non-hydrogen having three lone pairs and the central atom having two. Which structure is more stable? Consider the following rules:

All-octet structures are more stable than those that expand or fall short of an octet for every atom.
Since these are both all-octet structures this point is moot.
The structure with the least formal charges is more stable.
The right-hand structure wins; it has zero formal charges, while the left has $\pm 1$. The following point is now longer needed.
For structures that cannot avoid formal charges it is more stable to have formal negative charges on electronegative atoms and formal positive ones on electropositive atoms.

Thus, both are possible structures (Martin did nice calculations there) but we can correctly predict the more stable structure according to these rules (again taking Martin’s results as a starting point). Your teacher basically told you to skip this analysis and go straight to what is correct.


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How do we know whether we will get a mono substituted or tri-substituted product during electrophilic aromatic substitution?
For example consider the bromination of phenol we get the tri-substituted product (even though bromine is deactivating so I suppose it should deactivate the ring with each addition) while sulphonation essentially gives the mono-substituted product (even at high temperatures).

The OH group is strongly activating, the bromines are only slightly deactivating.

@ron Okay.Then what about Nitration in phenol? It also gives tri-substituted product. NO2 is strongly deactivating.

A lot of it is experience, you pick it up as you go along.  That the nitration of phenol occurs so readily is a further reflection of just how activating the OH group is.

@ron Take a look at this link http://chemwiki.ucdavis.edu/Textbook_Maps/Organic_Chemistry_Textbook_Maps/Map%3A_Bruice_6ed_"Organic_Chemistry"/16%3A_Reactions_of_Substituted_Benzenes/16.09_The_Synthesis_of_Trisubstituted_Benzenes      It's a good activator not excellent.

I see this, "In the case of the nitrogen and oxygen activating groups displayed in the top row of the previous diagram, electron donation by resonance dominates the inductive effect and these compounds show exceptional reactivity in electrophilic substitution reactions" (emphasis mine).  What do you see in the link?  You need to compare relative rates of reaction.

@ron Hmmm...Okay, you have a point. But then why do we not observe tri substitution in other products ,say sulphonation to phenol even though SO3H group is less deactivating then NO2. Or let's say Friedel crafts alkylation(CH3 being moderately activating) ?

Sulphonation is a little different because it is easily reversible, therefore product distribution is often thermodynamically controlled.  The weakly activating methyl group does undergo tri-nitration, but it requires forcing conditions.

@ron Is the exceptional reactivity of phenol strong enough to counter the effect of the three nitro groups that get substituted one by one ? Since nitro group is the most deactivating molecule for an aromatic ring .

@DelPate  Yes, the hydroxyl group activates the ring enough that 3 successive nitrations can occur.  As I mentioned in an earlier comment, even the weakly activating methyl group is strong enough to allow 3 successive nitrations under forcing conditions.

It is like others have already noted that phenol is very strongly ortho-para directing, while bromine is only slightly deactivating (or sometimes slightly activating). It is so activating that you do not even need a Lewis acidic catalyst or elevated temperature to let the reaction go fast and in good yield. Moreover, if you want the mono substituted product you will actually need to cool the reaction down to < 5 °C and use the rather dangerously flammable solvent $\ce{CS_2}$, yielding 4-bromophenol in 85% yield (source: Vogel). I assume the para product is formed due to the fact that there is no significant difference in electron density between ortho and para carbon atom while ortho is slightly sterically hindered.

Okay, so basically the answer is that we will get a tri-substituted product(if activating) unless- 1)Steric hindrance prohibits it    2)We use reaction conditions(like temperature,solvent) to favour some product. Would that be correct to say?

In aqueous medium Phenol get converted to phenoxide ion which has more +M and it highly activates the ring and hence all the positions get occupied

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text:
How do I calculate the pressure of a known liquid in a sealed container heated above boiling point?
If I have water in a sealed container heated to say 150 degrees, how do I determine the amount of pressure being generated in the container? What about for other liquids? I have searched extensively and cannot figure this out.
I was looking for a formula of sorts, but for example reasons let's say a 5000 ml container with 4500 ml of water in it, with the rest of the space air, heated to 150 degrees Celsius.

First we would need to know how much water per volume in the sealed container.  Also we would need to know if any air is in the container or not.  If we knew those things, then the next step is to consult steam tables (either equilibrium or superheated) to determine all the thermodynamic properties of the steam at the right temperature and bulk density.

Note: for convenience, "gas" refers to any gas at a temperature beyond its boiling point; "vapor" refers to any gas evaporated from its liquid state, also implying that the liquid itself is below its boiling point, such as the 150 degrees water in your example.
Dalton's law of partial pressure would come in handy here. The law states that the partial pressure generated by each type of gas particles sum up to the total pressure in a sealed container. The partial pressure of "gasses" can directly be calculated using the ideal gas law. Of course the partial pressure of "gasses" is 0Pa if there is no gas but only vapor in your container.
Now the partial pressure of the vapor of liquids. In a sealed container, if left to attain dynamic equilibrium (which is the steady state in this case) the partial pressure of the vapor of any liquid would be equal to its equilibrium vapor pressure, which is a function of temperature only. Though not excessively accurate, the Antoine equation can be used to estimate this:
$\log P = A- {B \over C+T}$
where P is the equilibrium vapor pressure, A, B and C are substance-specific constants and T is the thermodynamic temperature of the substance. The three substance-specific constants can be checked up on the internet, or provided by other sources, whereas conducting experiments can be a last resort (as it is a very tedious experiment that you need the whole curve of the function to approximate the constants).
If the liquid in question is a mixture of more than one liquid substance, then you'd need a third law - Raoult's law to determine the partial vapor pressure of each component of the mixture:
$p_i = p_i^\star x_i$
where $p_i$ is the partial vapor pressure of component $i$ of the mixture, $p_i^\star$ is the equilibrium vapor pressure of a pure sample of component $i$(in itself, not in a mixture), and $x_i$ is the mole fraction of $i$ in the liquid.
At dynamic equilibrium, combining Dalton's law of partial pressure and Raoult's law yields the expression for total vapor pressure:
$p_{total} = \sum p_i^\star x_i$

Since steam (water vapor) is the most common working fluid in external-combustion engines, steam tables, as mentioned by Curt F above, are widely used and will show the relationship more accurately than the ideal gas law, $PV=nRT$.
See http://www.wolframalpha.com/examples/SteamTables.html,
http://www.efunda.com/materials/water/steamtable_sat.cfm or http://www.tlv.com/global/TI/calculator/steam-table-pressure.html for an online steam table.

DrMoishe, what about for other liquids such as organics? Water was just an example of convenience. Will the ideal gas law alone allow me calculate the pressures that will build up in a tank if I heat a liquid beyond its boiling point?
I ask because I feel as if the ideal gas law would be come unapplicable when the pressure is so much that more water cannot evaporate. Is there simply not straightforward way/formula to calculate how much pressure would build up in a container at certain temperatures?

There are additional factors not included in the ideal gas equation, such as van der Waals force, http://en.wikipedia.org/wiki/Van_der_Waals_force. In theory, if you know all the forces acting on a substance, you can calculate its pressure/volume/temperature relationship, but in practice the formula is derived empirically. Tables are available for many substances such as CO2 and NH3. J. Willard Gibbs, http://en.wikipedia.org/wiki/Josiah_Willard_Gibbs, was one of the first to explore chemical thermodynamics; you might want to read his work.

Increased gas pressure does not stop the liquid from evaporating.  The ideal gas law works best at lower P and higher T so might be a first approximation. This sounds like you should be considering experimentation.

The volume of the headspace above the liquid (here 500 mL) is not important unless all of your liquid boils while bringing it to the set temperature. If it doesn't, then the system pressure will be on the saturation (or coexistence) curve, which is the curve collecting P/T pairs at which the liquid and vapor coexist (see a phase diagram such as provided by Wolfram, link courtesy of Dr. Pippik). At higher T there might not be a coexistence point, but inspection of the phase diagram reveals that this is not an issue at the desired T, here $\pu{150 ^\circ C}$.
To check if liquid and vapor coexist, find the pressure corresponding to $\pu{150 ^\circ C}$ on the saturation curve. That pressure, according to the Wolfram site, is 450 kPa.
You can now perform a quick check using the ideal gas law to see approximately what amount of liquid water corresponds to 0.5 L of vapor at the saturation pressure. It comes out to 0.064 mol, which is about 1 mL of liquid, or negligible relative to the 4.5 L of liquid in the container.
So you can conclude that the container at that temperature will be at a pressure of 450 kPa, with most of the water remaining a liquid.

There are 3 main factors involved with a liquid in a sealed container: If the container contains no head space and is allegedly strong heating the liquid will cause the pressure to increase. This will continue until there is a phase change or the plastic or tensile limit of the container is reached. If the first the most likely change is that the liquid will become super critical and behave as a gas.  The pressure profile of just the liquid if the external pressure is just sufficient to prevent vaporization [a piston] will follow the liquid-vapor equilibrium curve. The pressure necessary to prevent expansion will follow the liquid-solid equilibrium curves. Above the critical T the gas will obey the appropriate equation of state.[The common atmosphere gases are liquids above their critical T.]
If there is head space and enough liquid to remain until the critical temperature the pressure inside will follow the vapor pressure curve of the liquid[s] with allowances for deviations from Raoult's Law if more than one liquid.
If there is head space and an inert gas the original pressure is atmospheric, a sum of the vapor pressure and the inert gas pressure. As the temperature is raised the inert gas will follow the appropriate gas law and the vapor will follow a modified vapor pressure curve. The vapor pressure is a function of temperature and inert gas pressure, the pressure affects the liquid phase more than the gas phase increasing the vapor pressure.  I do not know the effect of inert gas pressure on the critical parameters.
A vapor is a GAS in equilibrium with its liquid, nothing to do with boiling points etc.


text:
How can n = c/v?
I know this question is probably really rudimentary or I may just be fundamentally misunderstanding something but I would really like to just understand this.
So $n=\pu{1 mole}$, $c=\pu{1 moles/L}$ and $v=\pu{1 L}$
This is how I learned this and it makes sense. However, this way, the equation doesn't make sense. It reads: $$\pu{moles}=\frac{\pu{moles/L}}{\pu{L}}$$
Which simplifies into $$\pu{moles}=\frac{\pu{moles}}{\pu{L}^2}$$
http://www.wolframalpha.com/input/?i=x%3D%28x%2Fl%29%2Fl
Which is obviously wrong. What am I not getting here?

It should be $n=cv$
Then, your first equation
$$\pu{moles}=\frac{\pu{moles/L}}{\pu{L}}$$
becomes$$\pu{moles}=\pu{\frac{moles}{L}}\times\pu{L}=\pu{moles}$$

How are n, c and v then defined? Because if they are defined the way I defined them, then my equation still holds.

Or do you mean to say that the equation is n=cv not n=c/v?

yes this is correct. the equation is n=cv


text:
Carbonate + acid reactions - How to recognize these type of reactions?
So I've got $$\ce{CaCO_3 + HCl \rightarrow CO_2 + H_2 O + CaCl_2}$$
I'm wondering how I can recognize these types of reactions? I know that when I mix an acid and a base I get salt and water but in this reaction there is also some $\ce{CO2}$ which is confusing me. When can I expect this type of salt base reaction and should I expect more like these with other by products other than $\ce{CO2}$?

You say "this type of salt base reaction" , do you mean salt and acid reaction ? Since there is no base in the reaction.

The reaction you showed is actually a salt "exchange" reaction (also called salt metathesis), in which a salt is transformed into another one during the course of the reaction. In your example when calcium carbonate is treated with hydrochloric acid, calcium chloride is formed. At the same time, carbonic acid is generated and it spontaneously decomposes into carbon dioxide and water. The equilibrium is shifted towards the products due to the release of carbon dioxide gas from the reaction mixture, thus driving the reaction to completion.


text:
Biochem Book age and suggestions
I'm taking chem classes, and am interested in learning a some biochem on the side, on my own. I've bee looking for a decent book, though I'm on a bit of a tight budget so not brand shiny new ones upwards of a hundred. Hopefully. Obviously I don't want something ancient just for reduced price, so two questions. How many years back is ok to use as a cutoff for publication date, and any suggestions?

Lehninger's book is really something to consider. We study it for our chemistry olympiad, so it's worth noting, isn't it?

I would suggest the Lehninger Principles of Biochemistry, which is one of the best in my opinion since it covers most of the key aspects in biochemistry.
It is quite expensive to buy as new, but since it is quite common you can maybe borrow it from a library. In this way you can get an impression about the book and eventually decide if it is worth buying it or not. If you are interested in having it I would buy a secondhand one and I would say it does not matter if it is not the last version, the previous one should be good as well.

That's actually one I was looking at myself. The last edition is pretty cheap too. Looking through the preview to get a bit of a look at it now.

I had to read biochemistry by berg. I think the 7th edition is the latest one, but you can't go wrong with the sixth either. This is a good text to get a general overview of biochemistry that you can later apply to research systems in the future.

I took the class biochemistry last year so both books are pretty new.

I know you are asking for a textbook, but if you want to do some casual learning here are a few pop-sci biochemistry-focused books that I recommend giving a try:

Oxygen by Nick Lane.  I like this book best but any of Nick Lane's books would be a good start.  This one focuses on the profound changes oxygen has had on life on Earth, from evolution to biochemistry.
Blondes in Venetian Paintings... by Konrad Bloch.  A series of essays on completely different topics, all recounting various aspects of the Nobel-winning career of a great biochemist.
The Chemicals of Life by Isaac Asimov(!).  Sure this one is very dated but by all accounts it's extremely well written.  (I haven't read this one yet myself but it's on my list.)

Cool, thanks. Can always use something a bit lighter/non text book to read while still sticking with science.


text:
Does fluorescein decompose from long term exposure to light?
I've had about one liter of fluorescein solution setting on the shelf for at least several years and one day noticed that it suddenly lost its characteristic green color. It's now a light yellowish-green solution that seems to have lost the property of fluorescence altogether.
The container is a reagent bottle with a leak-proof cap so I doubt that the solution reacted with anything in the atmosphere other than maybe 100 mL of air that was in the bottle to begin with.
The bottle was exposed to direct and indirect sunlight on a daily basis.
Does it have to do with the decomposition of the fluorescein, and if so, what are the expected decomposition products?

Have you investigated what happens if you add a little NaOH to some of the solution? If the color and fluorescence is restored, it might have been a change in pH, perhaps caused by the very slow reaction with a glass container. Good question! Please post your findings.

@DrMoishePippik I will try that and get back to you. Thanks for the idea.

@DrMoishePippik That was a no-go on the NaOH. Increasing the pH did not restore the fluorescence.

Yes, Fluorescein decomposes due to exposure to light.
See The Conservative Behavior of Fluorescein for quantitative data.


text:
Interaction between soy milk and coffee producing bubbles
I noticed when I poured my soy milk (which I shook as per the directions on the carton) into my coffee, a large number of very fine bubbles rose to the surface. What causes this? Is there a chemical reaction involved, or is it just a physical effect?

I can't see here any reaction. Probably it were just bubbles of air - shaking could mix milk with air a little.

@Mithoron the bubbles were very fine when I did it almost like effervescence


text:
Can a negative pH exist?
I am programming a small model to simulate diffusion of $\ce{H+}$ ions in human tissue. It's quite a simplistic model based on a naive algorithm, and I'm mostly doing it to familiarize with the concept of agent-based models.
My space is an $n\times n$ grid where each position $(x;y)$ hosts $h$ $\ce{H+}$ ions.
I understand that $\ce{pH}$ can be calculated based on $h$ as $-\log(h)$. As I want my initial $\ce{pH}$ to be stable at average human $\ce{pH}$, I'd set $h$ across the matrix to be $10^{7.5}$. (In fact, as I don't want to simulate $10^{7.5}$ protons I decided each particle will represent a bulk of $10^5$ protons and made my equation $-\log(10^5\cdot h)$. )
However, I do not understand the equation, wouldn't it at some point yield a negative $\ce{pH}$ as $-\log(x) < 0$ for $x > 1$?

Yes, pH can be negative.
The official definition of pH is: negative log of H+ activity, not negative log of H+ concentration, however.  See pH Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+] J. Chem. Educ., 2006, 83 (5), p 752.
By either definition pH can be negative in concentrated solutions of strong acids.  For example 5M HCl would have a negative pH.

Like you said the $\ce{pH}$ of blood is approximately $7.5$. So since
$$ \ce{pH} = -\log[\ce{H+}] \\
\text{or}\\
[\ce{H+}] = 10^{-\ce{pH}}$$
this means that the proton or hydrogen ion (same thing) has a concentration of $10^{-7.5}$ not $10^{7.5}$ (notice the negative). If you're dealing with blood you'll never have to worry about negative $\ce{pH}$ values. Like DavePhD stated, negative $\ce{pH}$ values only occur with high concentrations of strong acids.
In short, I think where you're going wrong is with $-\log(10^{7.5})$. It should be $-\log(10^{-7.5})$.
As a side note, you are correct. $\log(x)\geq0$ for $x \geq 1$. So $-\log(x) \leq0$ for $x\geq1$, but when do you ever have proton concentrations above 1 in human biology? Practically never.

http://pubs.acs.org/doi/abs/10.1021/ed083p752?journalCode=jceda8

$\text{pH}$ can definitely be negative.
It is defined as $$\text{pH}=-\log[\text{H}^{+}]$$
Note that if the $[\text{H}^{+}]\geq 1\text{M}$, the $\text{pH}$ will become negative.
Practically, this is only possible when you have an ultrapure solution of a strong acid.
Hope this helps.

http://pubs.acs.org/doi/abs/10.1021/ed083p752?journalCode=jceda8


text:
Can powdered beta-tin be made from raising the temperature of alpha-tin?
When white tin ($\beta$-tin) is cooled to a temperature below $13.2\ \mathrm{^\circ C}$, it creates the allotrope of gray ($\alpha$-tin), a gray, amorphous powder.
My question is that once you have the powdered gray tin, doe just raising the temperature above the point of stability turn it back to white tin, but in a powdered form?
Is there a change in appearance of the gray powder when that happens, or is it mostly unnoticeable?
Does the conversion to gray tin change the melting point?

I'd love to know; this seems like it would make an amazing demonstration.  All I can find is that reheating will convert the tin back into its beta form.  I imagine that would happen below its melting point at 231 C, but I don't know for sure.  As long as it did, you'd wind up with tin powder.  Really cool 20 second video of this, for anyone interested.

Yes, powdered beta tin can be made from powdered alpha tin, down to micrometer size. See, for example, this paper on exactly this topic: Powder Technology 1995, 84 (1), 35–38.

$\alpha$Sn and $\beta$Sn are the two solid allotropes of Sn.  As you not, below 13C the stable phase is $\alpha$Sn, which has a diamond cubic crystal structure (like diamond, Si, and Ge) and is a semi-metal.  Above 13C, the thermodynamically stable phase is $\beta$Sn, a body-centered tetragonal crystal. So, cycling back and forth around 13C varies the thermodynamically stable phase back and forth, with the only question being the kinetics of the transformation.  The video in the comment by @JasonPatterson shows that the transformation indeed takes place reasonably quickly (unlike diamond to graphite for carbon).
As for melting, if you rapidly heated $\alpha$Sn and avoided the phase transformation, you would find that the melting temperature would be lower. Using the SGTE thermodynamic data (A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991)), one finds that the $\alpha$Sn -> liquid phase transition would occur at about 430.7K, ~75K lower than the standard $\beta$Sn -> liquid melting point at 505.06K.

Do you mean that an undercooled melt of tin would actually be produced by rapidly heating αSn > 430 K? I find that a bit hard to swallow. It's clear that the theoretical melting point of the unstable phase would be lower, but do you think this could be observable?

Why would it not be observable? For example, the melting point of amorphous silicon (a distinct thermodynamic phase, not a glass) was measured through pulsed laser irradiation (Phys Rev Lett 52 2360 (1984). So, yes, you would observe a lower melting point, provided you heat rapidly enough to bypass the (relatively slow) solid-solid transition.

Phase changes are fickle. Much like supersaturated solutions.
Add a crystal to a supersaturated solution and it will precipitate rapidly.
Bring a crystal of the thermodynamically favorable phase into contact with a "supersaturated" phase and it may just do this: https://www.youtube.com/watch?v=sXB83Heh3_c
I am not sure how exactly the proceedure in the video was done, so do not take it as evidence that my statement is true.
Also, Diamond is stable well below its phase transition temperature. Surely there are other compounds like that.


text:
Does a change in partial pressure for a gas-solid reaction, changes the equilibrium?
This is my reaction, $$\ce{Cd (g) + 1/2 Te2 (g) -> CdTe (s)}.$$
Considering the solid CdTe, $$K = P_{\ce{Cd}}^\mathrm{Eq}\cdot \left(P_{\ce{Te}}^\mathrm{Eq}\right)^{\frac{1}{2}}$$
I only have elementary knowledge on chemistry. I know that the equilibrium values of pressure are dependent on temperature. My question is - does increasing the partial pressure of one of the elements only changes the equilibrium pressure? If yes, for one or both?

The equilibrium constant will stay the same at the same temperature.  If the pressure of one component of the system in question is changed, equilibrium will be reestablished by either an change in the relative rates of the forward and reverse reactions.
Increasing the partial pressure of one reactant in this system will result in an increase in the rate of the forward reaction.  Pressure of both both reactants will be reduced until equilibrium is reestablished.

Thanks for your answer. I also sort of assumed the same, just wanted to verify with someone who knows these stuff. But the problem is, I am actually doing this experiment in the lab to deposit CdTe from elemental vapor. However, the observation doesn't match the theory. Increasing Te vapor increases the CdTe deposition rate, but increase of Cd vapor reduces the rate. Do you think, some kind of gas  kinetics is involved here?


text:
Is it possible for this Iron-Carbon Compound to exist?
Can this compound be created?

Where the orange is Iron and the black is Carbon.

Short answer: No
Long answer: Assuming you had hands tiny enough to manipulate individual atoms, you could make this molecule. The amount of time it would stay together would be as close to 0 as it gets. To start iron can't accommodate that amount of electrons, and the formal charges of this molecule are extreme to say the least.

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

that molecule is not stable.
There exists
(L)xFe=C(R1)(R2) carben complexes and
(L)xRu=C=C=C=C=Ru(Lx) molecules,
so their might exists also a (L)xRu=C=Ru(L)x or (L)xFe=C=Fe(L)x molecule in theory
but at least the 3 not centered Carben-C-Atom (connected to each Iron atom) have to be stabilized.


text:
Why does Mg2+ have the ionic radii bigger than Ca2+
If they were atoms, $\ce{Mg}$ should have a smaller radii. How does this change?

Can you point to a source that says calcium ion is smaller than magnesium ion?

An exercise book.Also look here: https://answers.yahoo.com/question/index?qid=20071216145936AAgZZnv "To answer your question: (out of Ca2+ and Mg2+ )Mg2+ is larger because it they are cations and Mg is above Ca;"

Source: http://upload.wikimedia.org/wikipedia/commons/thumb/9/9f/Atomic_%26_ionic_radii.svg/551px-Atomic_%26_ionic_radii.svg.png
Ca2+ is bigger than Mg2+

as shown in the above pic Mg2+has more smaller size then Ca2+...
I'm sorry but your question is wrong,
Why Ca2+ has bigger Ionic Radii then Mg2+? because

No of Shells increases down the group

-Property of Periodic table
therefore ->

As you move down a group in the periodic table , additional layers of electrons are being added, which naturally causes the ionic radius to increase as you move down the periodic table.

you can check reference this for more information
Image source : http://searchpp.com/ionic-radii-periodic-table/

Mg is smaller the Ca because it is above it in the periodic table, magnesium being a period above it shows that it has one less electron shell so it's smaller, calcium has another ring of electrons, as to say, that are around it so it's radius would be bigger.
Your question is wrong.


text:
How accurate is this polar mechanism for the Diels-Alder reaction?
So the prof got to Diels-Alder without discussing it in the context of molecular orbital theory. Instead we got a resonance picture of what happens... this is what he told us:
1) 1,3-butadiene has both an electrophilic and a nucleophilic terminal carbon atom because we can draw resonance structures showing dipolar terminal carbon atoms ...
2) For ethene we can similarly draw a dipolar resonance structure.
From there we're told to simply use Coulomb's law - opposite charges attract - to complete the mechanism.

So, how adequate is his electrostatic explanation? Is it adequate - probably ... if one simply wants to learn it for a test. But is it conductive to further understanding of chemistry? I feel rather short-changed here since I'm paying him the big-bucks (tuition is insane) and the book goes into far greater detail ...

ETA: I asked the guy about and grilled him on the electrostatic explanation. For example I raised Martin's excellent point about the fact that the dieneophile cannot be that polarized - otherwise wouldn't it be a liquid instead of a gas due to the strong intermolecular attractions?
He instead went on a tangent on electron-withdrawing groups and how the electrostatic explanation can explain these ... what do you think?

I chuckled. I laughed a little. I cried. I cried more. I cried much more. I laughed myself into insanity.

@Martin what happened?

I read your question... This mechanism is so wrong. There are so many publications about DA. It is impossible to describe it correctly without MO theory. That is like abandoning the last fifty years of research. I really feel sorry for you having to endure these ancient concepts without any meaning.

@Martin is the mechanism wrong or very, very oversimplified?

Those resonance structures are atrocious, that diene simply doesn't have a positive and a negative charge at each end- Just because you can draw curly arrows until it does, doesn't mean that this particular resonance structure has anything to do with reality.

That's what I was thinking, this isn't a traditional electrostatic reaction. The book makes considerable reference to HOMOs and LUMOs which the guy ignored,

Indeed, it wasn't possible to explain these types of reactions (pericyclic) until Woodward and Hoffmann (and yes EJ Corey too) started discussing it in MO terms.

Disclaimer: This is not a full answer. I was just too long for a comment.

I think the suggested mechanism is not entirely wrong, it might be a contributing factor of the whole process, but not a very important one. In the last fifty years, the mechanism has been discussed in almost countless publications. One of the more recent ones by Goldstein et. al. concluded that a concerted pathway is the most economical.[1]
They also included possible diradical pathways and the similar. Of course there is no uniformly applicable mechanism, it is very dependent on the substrates. Many of the reactions can be effectively described by frontier molecular orbital theory, so it is also possible to do so for the simplest Diels-Alder reaction.
There might be many more publications out there discussing the whole mechanism in pretty exhausting detail, but for a start I thought the cited one will do just fine. Just to name some more, see [2] through [4].
The electrostatic interpretation is tempting, as it is simple and it explains the simplest case very well. I believe it explains it too well for the wrong reasons:
 I would not expect the polarisation, i.e. induced dipole, in ethylene and butadiene be strong enough to make this happen at all. Both are colourless gasses at room temperature. If there were more induced dipoles, they would be liquid, which means that these molecules are basically unpolarised.
In a publication by Gian Franco Tantardini and Massimo Simonetta it was stated, that the contribution of the polarised Valence Bond Structure is very little, see [5]. If I read it correctly, the contribution is less than 1%. This alone should suffice to disregard a mechanism purely based on electrostatic considerations. In the same publication it is stated, that ethylene has quite a significant contribution from its polarised structure (about 25%). However, these consideration would in most cases not be able to describe streospecific reaction products, that are obtained from different substrates.
As another aspect, one must not forget, that the ground state of 1,3-butadiene is trans and it requires just a bit of energy to rotated it to the necessary cis formation, which should be about 10 times less present. So what are the odds having a well polarised ethylene hitting a polarised cis-butadiene in the right angle at the right speed. There has to be more to it...

E. Goldstein, Brett Beno, and K. N. Houk; J. Am. Chem. Soc., 1996, 118 (25), 6036–6043.
R.B. Woodward, and Thomas J. Katz; Tetrahedron; 1959, 5 (1), 70-89. Mirror at znu.ac.ir.
Michael J. S. Dewar, Santiago Olivella, and James J. P. Stewart; J. Am. Chem. Soc., 1986, 108 (19),  5771–5779.
J. Sauer; Angew. Chem. Int. Ed., 1967, 6 (1), 16-33.
Gian Franco Tantardini and Massimo Simonetta; Int. J. Quant. Chem., 1978, 14 (S12), 131-141.

I think that a major teaching opportunity was missed.  If MO theory had previously been introduced in the course, then there is absolutely no excuse, here is another topic where it was meant to be applied.  If MO theory had not been previously introduced, then it still remains a teaching failure, here's why...
This was a great opportunity to introduce a new class of reactions.  So many of the reactions discussed in introductory organic chemistry are polar.  Here was a chance to point out that there are also many non-polar reactions. In order to explain these reactions mechanistically, the concept and power of aromaticity could have been reapplied.
The class of concerted reactions (variuos cycloadditions, electrocyclic reactions and sigmatropic rearrangements, etc.) should have been (at least) mentioned.  In addition to presenting the Diels-Alder reaction as a member of this class of reactions, the intriguing example of the sterospecific, thermal ring opening of cis- and trans-3,4-dimethylcyclobutene could have also been presented in passing.  This example almost always elicits a "why does it open like that" response and whets the student's imagination and interest in further learning.

If this wasn't the time to delve into MO theory, the concept of aromatic (4n+2 electrons) and antiaromatic (4n electrons) transition states could have been introduced.  Teaching this approach doesn't take long at all and it makes for a nice tie-in back to benzene aromaticity. This approach neatly explains why the Diels-Alder reaction occurs so readily, but the dimerization of ethylene to produce cyclobutane does not.

Diels-Alder reaction is a cycloaddition not a electrocyclic reaction, right? Don't you mean pericyclic reactions as opposed to ionic reactions (i.e. with ionic intermediates)?

@Jori - "In organic chemistry, an electrocyclic reaction is a type of pericyclic rearrangement where the net result is one pi bond being converted into one sigma bond or vice-versa."

@ron do radical reactions fall under the category of polar or non-polar?

@Jori Thanks, I'll revise the sentence to remove any ambiguity.

@Dissenter  I think that generally, most radical reactions are relatively non-polar.

Pretty big points you raised here, I absolutely agree, kudos. As for the radical reactions, @Dissenter, they have to be mainly orbital driven and hence, polarity should play a minor role.

The mechanism of the butadiene + ethylene Diels-Alder is not wrong as drawn. It shows the concerted movement of electrons from the diene to dienophile resulting in cyclohexene. This correctly describes the bond formation. It is unusual to see the parent Diels-Alder reaction explained through those minor resonance contributors. We often draw mechanisms through minor resonance contributors to explain aspects of reactivity.
However, there are several problems with describing the Diels-Alder reaction this way. First, this tries to force the Diels-Alder reaction into the paradigm of "polar" reactions, i.e. nucleophile-electrophile. Although we learn some reactions of dienes and isolated alkenes as nucleophiles (electrophilic addition of H-X, X-X), there are essentially no reactions of electronically neutral dienes/alkenes as electrophiles. One of the key lessons of the Diels-Alder reaction is that it introduces a class of reactions that are "non-polar" (related question), the pericyclic reactions.
The second problem is how to handle substitution. The Diels-Alder reaction can create four stereocenters, which suggests that upto 16 stereoisomers can result. However, that is not what is observed. There are only specific stereochemistries of products that result from specific stereochemistries of reactants: the Diels-Alder reaction is stereospecific. Without invoking molecular orbitals, explaining this is impossible. One can learn the rules for stereochemical outcomes, but the rationale relies on the interaction of MO's.
The addendum to the question gives some indication of what the instructor was thinking when introducing the Diels-Alder reaction. Although the Diels-Alder reaction is not a traditional nucleophile-electrophile reaction, it does follow frontier molecular orbital theory: one of the component's HOMO interacts with the other component's LUMO. Typically, the dienophile provides the LUMO, which is the traditional role of the electrophile. An alkene conjugated to an electron withdrawing group (such as a carbonyl) is more electrophilic. It's also a more reactive dienophile. So there is a parallel between electrophilicity and dienophile reactivity both explained by having a lower LUMO.
The same is true with respect to the diene component. Typically, the diene provides the HOMO, which is the traditional role of the nucleophile. A diene substituted with an electron donating group (such as a methoxy group) is more reactive toward dienophiles and electrophiles.
So while the Diels-Alder reaction is non-polar, it can have polar character. It turns out that drawing resonance structures of the reactants is very useful for predicting the regiochemistry reactions with polar components. The major product will result from bond formation of the most nucleophilic end of the diene with the most electrophilic end of the dienophile (normal electron demand DA). Those resonance structures are simply substituted versions of what the instructor started with.

This is not a full answer either, but it needs to stand alone from the previous discussions.
You can't draw resonance structures without MO theory.  This discussion is formally impossible.


text:
Can we use peaks in any chromatogram to identify a compound
Does a compound give a specific chromatogram peak, using which we can identify it in a mixture?

Compounds have specific retention times that are used to identify their respective peaks in chromatograms. Usually these times are dependent on many factors (temperature, column type, mobile phase...), and as such some kind of pre-calibration of the machine with respect to compounds in the mixture has to be done. The area of the peak is then indicative of the amount of substance, also subject to calibration (or comparison with reference substances with known concentrations).
For structure elucidation of previously unknown compounds in mixtures other methods such as GC-MS are much better suited.


text:
Why is only bromine used in the degradation reaction?
Why specifically is bromine used in the degradation reaction (Hoffmann bromamide)? Why can't the same be performed using chlorine or iodine?
In other words, what makes bromine a more convenient reagent as compared to the other halogen gases? For instance, during halogenation of alkanes, we don't use fluorine due to the explosive nature of the reaction, and we don't use iodine because of the slow rate. Is the reason somewhat analogous to this?
I thought because of kinetic factors, maybe somewhat similar to how monochlorination and monobromination are much more easier than monoiodination and monofluorination

It isn't.
Under alkaline conditions, the Hofmann rearrangement of primary amides to amines can be performed with chlorine too. But why work with chlorine when bromine is much more convenient.

Choosing bromine over chlorine when possible is an obvious decision for everyone with practical lab experience!
Under standard conditions, bromine is a liquid and can be added using a dropping funnel, while chlorine is a gas, which makes adding stoichiometric amounts much more tedious.

Hypohalites have been used too: in the presence of alkanols, rearranged carbamates are formed. Under acidic conditions, hypervalent iodine compounds have been shown to be effective.


text:
Does the boiling boint of miscible liquids follow a linear relationship?
I want to create a steam bath for a reaction the proceeds well at 105℃.  I cannot use pure (demineralized, anyway) water for this, since it would boil at 100℃ at STP.
As such I figured that maybe I could get the required temperature by mixing water with another liquid (with which water is miscible) which has a higher boiling point (e.g. vinegar at BP around 118.5℃ or vegetable glycerin with BP at 190℃).
Does the boiling point of the mixture of liquids proceed in a linear manner between the two temperatures?  E.G. would a volume of 50% water & 50% VG boil at 145℃?
As it is being done right on the boiling point of the combined liquids, I expect to lose some of the liquid to steam and need to add more. Will the lower temperature liquid boil out first?

Except if molecular composition of your two liquids are similar, the boiling point of the mixture will not proceeds linearly but like this for instance :

With the lower blue curve the boiling curve. In between the two curve you'll then have a mixture of vapor and liquid and only vapor on the upper part the binary diagram.
So you should find out an already drawn binary diagram for your mixture or draw it out by yourself first.

I was looking at a few of the diagrams and they all seemed to follow the same basic shape as seen above.  I'm thinking that if I can draw a grid over one of them and scale my temperatures (along with 105℃) onto the diagram, it should give me a rough idea of the mix required.  Does that sound feasible, or was I seeing a limited subset of them that followed 'nice logical curves'?

The shape will often be similar yes. But I doubt the values will.

This can't be responsive to the question: If the liquids are merely mixed then when the boiling point of the first is hit it is will predominately vaporize.  Yes, some of the second will be evaporating in the process, but essentially at the temperature above lower liquid's boiling point it is leaving the mix en mass.  This is the basis of distillation!  So replenishment would not be proportional to the initial mix, and I don't see how it makes sense to characterize the mix as having a unified boiling point.


text:
Why does brown sugar have a strong smell but white sugar doesn't?
I noticed something while cooking and was wondering what the chemical explanation is for this. White sugar has a smell, but it's different, and much fainter than that of brown sugar.
Brown sugar is unrefined, so does this have something to do with why odorous chemicals are present?
What are these chemicals and why are they present in one, but not the other?

Sugar is made by repeatedly boiling and cooling cane syrup (or sugar beet syrup).  After each cooling, the solution becomes supersaturated with respect to sucrose, causing sucrose to crystallize out of the "mother liquor" (the industrial term for the liquid solution from which crystals form). These crystals of "raw" sugar are heavily processed (washed in syrup, dissolved, treated with carbonate or phosphate minerals, recrystallized, and subject to ion exchange or charcoal color removal) to result in white sugar.
Nowadays, brown sugar is usually made by mixing refined white sugar with blackstrap molasses. In former times, the brown sugar was simply unrefined or unwashed crystal fraction from the 2nd or 3rd crystallization of cane juice. But precise control over the ratio of "contaminating" liquid to white crystalline sugar was difficult, which is why industry moved towards reconstituting brown sugar by mixing molasses and white sugar.
Blackstrap molasses is simply the spent "mother liquor" remaining after the third cycle of boiling and cooling to produce sugar crystals.
The smell of brown sugar thus comes primarily from molasses.  The smell of molasses is very complex.  Here is the abstract of a relevant paper:

The characteristic sweet-aroma components in refinery final molasses were isolated by using a combination of sensory evaluation, column chromatography and gas chromatography, and were identified by infrared spectrometry, mass spectrometry and color reactions. Identified were ethyl n-hexanoate, ethyl phenylacetate, phenylacetic acid, n-butyl benzoate, isopropyl benzoate, l(+)-pantolactone, benzoic acid, o-toluic acid, m-toluic acid, β-phenylpropionic acid, succinic anhydride, maltol, isomaltol, and 2-methyl-5-hydroxy-6-ethyl-γ-pyrone.
The aroma of the mixture of these compounds and vanillin and vanillic acid was similar to that of refinery molasses by sensory evaluation. The results indicate that these compounds are important in producing of sweet molasses aroma.

If I had to bet, I would say that maltol and isomaltol are probably responsible for "most" of the odor of molasses, with the other listed compounds forming important complementary notes.


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Chemiluminescence - Determining Temperature from Luminol Imaging
I'm working on an imaging project and I have no background in chemistry or chemiluminescence for that matter. This is a Computer Science project.
While most of the details for it have been worked out, I'm having difficulty figuring out the scope and possibility of an important requirement.
Given a luminol image, I will develop a heat-map. For now the heat map is built based upon the intensity of the light.
However, given the intensity of the light and with the knowledge that luminol is being used, is it possible to determine the approximate temperature of the point based on its intensity and the intensity of the surrounding region?
I would assume that this chemical would burn differently at various temperature and would have a specific signature of light emitted depending on it its temperature but I'm having difficulty pinpointing whether this is true, possible and backed by research (or science).
The closest parallel I've found is combustion analysis. The heat release rate is calculated by measuring the amount of light radiated from a flame at various wavelengths, since certain chemicals give off radiations at specific wavelengths.
Can anyone confirm this, suggest alternatives or point me int the right direction?

However, given the intensity of the light and with the knowledge that luminol is being used, is it possible to determine the approximate temperature of the point based on its intensity and the intensity of the surrounding region?

Probably not! The energy released from the excited state of the 3-aminophthalate dianion formed in the oxidation of luminol (or rather its dianion) by hydrogen peroxide in alkaline is released in form of light. There's little to no heat released in this chemoluminescent reaction.

There is some research out there, e.g.
http://www.ncbi.nlm.nih.gov/pubmed/3749789, Influence of pH and temperature on luminol-dependent chemiluminescence and
http://pelagiaresearchlibrary.com/der-chemica-sinica/vol4-iss3/DCS-2013-4-3-161-164.pdf, Effect of temperature on chemiluminescence of luminol ethyl amine.
Note that there is also a time dependency, so that increased luminous output is offset by shorter duration.
That's an interesting use of chemiluminescence and implies use of tomography if the sample is three-dimensional.


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Why is SN2 favored greatly over SN2' in this reaction?
Secondary allylic halides can be produced in high yield with the following Mitsunobu-type reaction:[1]

The yield being $94~\%$.
The last step in the mechanism is:

Why is the $\mathrm{S_N2}$ reaction favored so much over the $\mathrm{S_N2'}$ reaction?

Reference: Magid, R. M.; Fruchey, O. S.; Johnson, W. L.; Allen, T. G. Hexachloroacetone/triphenylphosphine: a mild reagent for the regioselective and stereospecific production of allylic chlorides from the alcohols. J. Org. Chem. 1979, 44 (3), 359–363. DOI: 10.1021/jo01317a011.

By "Sn2-dash" do you mean conjugate addition ?

Yes, it's the standard notation, Sn2-prime.

Presumably because $\ce{Cl-}$ is a hard nucleophile and is driven by electrostatic interaction to the $\ce{C-O}$ bond?

@Jori the chloride ion is a "hard" nucleophile?

But chloride doesn't seem so hard to give almost quantitative yield.

While pi-based resonance effects are frequently important, you must always consider inductive effects in every reaction.
Electrostatically, there is a much higher positive charge on the ipso carbon than the beta carbon, because of inductive effects.  The net effect is that there is a lower energy of activation for nucleophile attack at the electrophilic center because the coloumbic forces are more favorable for attack there.  The softer the nucleophile and electrophile become, that less important this effect is, but it is still quite considerable.
Without doing the QM calcs, I would rate the phosphonium electrophile as quite hard.

could you perhaps expand this answer a bit? can you give a more detailed explanation of why this is the case

I no longer have a computational package to run a quick calculation on an analogous molecule to draw a nice picture.  However some degree more elaboration has been added.

The ipso carbon will have a slight positive partial charge, while the beta carbon will most likely have a negative charge. However, the orbital coefficient of the LUMO is larger on the beta carbon, therefore soft and/or large nucleophiles will most likely attack there. After all, chlorine is quite small and probably also not well solvated in these reactions. (I might add some calculations supporting that on Monday, if I have the time.)

Could it be that Coulomb interactions between the chloride and phosphonium put the chloride near the ipso reactive center?

@Marko With molecules tumbling in solution, I think the use of coulombic forces to explain colloision frequency by location on a molecule is a much weaker argument.  However, I can't dismiss the possibility.

The $\mathrm{S_N2'}$ pathway has a strict requirement (similar to $\mathrm{E2}$) for the correct alignment of the orbitals involved in the reaction, in this case, the alkene pi-system and the C-X anti-bonding orbital (in this case C-O anti-bonding orbital). That means that only particular conformation(s) of the electrophile will be reactive.
In the direct $\mathrm{S_N2}$ attack of the electrophile, while certain conformations will be more reactive than others, the C-X is always oriented correctly for reaction to occur.
I suspect that the product distribution reflects the competition between the steric hindrance of the direct $\mathrm{S_N2}$ attack and the conformational requirements of the $\mathrm{S_N2'}$. The reaction given in the question uses a relatively unhindered electrophile and a small nucleophile, which likely explains the preference for the $\mathrm{S_N2}$ product.

It would be splendid if you post an image for the SN2' conformation requirements.


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Preparation of oxalic acid from acetylene
I think that using ozonolysis would be a good idea to prepare oxalic acid from acetylene. Upon ozonolysis of acetylene we would get glyoxal:
$$\ce{O=CH-HC=O}$$
Would I obtain the oxalic acid upon oxidizing glyoxal with hydrogen peroxide?

Ozonolysis of acetylene won't produce oxalic acid.  The triple bond holding the two acetylenic carbons together will be attacked as shown below.  The reaction is complex and depending upon reaction conditions, different products can result (ref_1, ref_2).

Oxalic acid can be produced from acetylene by reacting acetylene with alkaline $\ce{KMnO4}$ (ref_3).
$\ce{H-C#C-H ->C[KMnO4/OH^-]\ HOOC-COOH}$
EDIT: Response to OPs comment
Here is a rough idea of what is going on mechanistically:
1) the permanganate forms a cyclic intermediate
2) that produces a diol
3) since a double bond is still present the reaction can be repeated to yield 2 gem-diols that can equilibrate to an alpha-diketone
4) now the standard alkaline permanganate oxidation of a carbonyl occurs
5) hydroxide can add to a carbonyl and
6) the manganese ion complexes with the carbonyl oxygen
7) the alpha-proton donates its electrons to reform the carbonyl which is now part of a carboxylic acid
8) repeat steps 5-7 on the other carbonyl and you have oxalic acid

Isn't the reaction with hot alkaline $\ce{KMnO4}$ basically oxidative ozonolysis?

I've never heard it used that way. "Ozonolysis" implies a specific reaction involving ozone.

But the two will give different products. So it cannot be the same reaction. What is the mechanism behind the reaction that you have proposed?

Yes, two different reactions, two different mechanisms and two different sets of products.  I've edited my answer to include a basic mechanism.


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Why is zinc more reactive than copper?
In terms of electronegativity, from what I understand electronegativity increases going across the period, so surely this should mean that zinc less readily loses its outer shell electrons than copper? Zinc has a greater nuclear charge but the outer shell electrons are in the same shell, so should the outer shell electrons in zinc not experience a stronger attraction to the nucleus? I have a feeling that this has something to do with which sub shells the electrons are removed from in copper and zinc...

By reactivity, do you mean standard reduction potential?

I am learning about half cells, so it is in this context. Zinc loses its valence electrons more readily than copper, and also when zinc is placed into copper sulfate solution it will displace the copper, so is more reactive...

For an extremely thorough discussion of the findamentals, see this paper: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95(10 (2018) 1801-1810. The Zn and Cu Daniell cell is addressed at great length (way too long for an answer here) and the metal cohesive energy is the source of more than 75% of the energy in the Daniell cell (p. 1805.).

In a metal, Zn isn't a standalone atom, it is a part of a weird large lattice. Unlike a standalone atom, Zn has no valence electrons of its own. Instead it shares the delocalized electron cloud with all the other Zn atoms in the Zn crystalline structure.  There is no particular reason why Zn inside a lattice would be attached stronger to its delocalized electron cloud than Cu.

You have to think about the whole process.  When a metal loses electrons to make a metal ion the following happens:

The metallic bonds holding the metal atoms together are broken.
The metal atom loses the electrons.
The resulting metal ion is hydrated.

In your analysis you are only focusing on step 2.  The enthalpy and entropy of the entire process factor into the reduction potential.

So why do we need to ionize two electrons?  Do we need two electrons from sodium or silver?  Is two electrons enough for aluminum or chromium?

I edited the answer to make it more general for any number pf electrons lost.

Zinc has last sub orbital as $\mathrm{4s^2}$ which it gives willingly to become electronically stable. Also, zinc's pull on its last electronic shells is much less and is hence more willing to combine. Copper has a last sub-orbital as $\mathrm{4s^1}$ which decreases its capacity to form bonds like zinc. Ideally, copper should form more bonds but due to lack of electrons in the last shell, it cannot form many bonds.

Zinc's relatively higher reactivity than copper is not because its highest occupied atomic orbital is $\ce{4s^2}$ while copper's is $\ce{4s^1}$; according to that magnesium should be more reactive than sodium as the highest occupied atomic orbital is $\ce{3s^2}$ while for sodium it is $\ce{3s^1}$. Therefore, this is not the case.
Now, the question that arises is: what is the real reason zinc is more reactive than copper?
The answer is that zinc is able to lose its outer electron more readily than copper. This phenomena occurs because copper metal is able to delocalize its outer electrons more readily than zinc. The metallic bond of copper is therefore stronger than zinc, so more energy is required to break this bond in order for it to react. Since more energy is required for copper to react, we can say that zinc is more reactive than copper.


text:
Using stoichiometry to split a complex mineral
Here is the problem I am stuck with, in particular just the first part of it.

I have calculated the moles $4.492 * 10^{-3} \,\mathrm{mol}$ and I am quizzed with regards to the equation that occurs. First assumption I am making is that the gas released is carbon dioxide.
Is it possible to separate out $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ into simply the $\ce{(Mg_{x}Cu_{1-x})(CO3)}$ part that reacts with the dilute sulfuric acid to produce the carbon dioxide, the moles of which I know?

Please write informative titles. And, homework tag isn't for homework only. It's a symbol you could use whenever you wanted to indicate that you're looking for a thorough explanation, rather than a direct answer.

Is it possible to separate out $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ into simply the $\ce{(Mg_{x}Cu_{1-x})(CO3)}$ part that reacts with the dilute sulfuric acid to produce the carbon dioxide, the moles of which I know?

$4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{CO2}$ implies $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$.
$4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ is 0.8167 g.
Then, considering the atomic weights of C, O and H, subtract the contribution of these elements from 0.8167 g to determine the mass of $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2}$.
Then divide by 2 to get the mass of $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{Mg_{x}Cu_{1-x}}$;
Then use the atomic weights of Mg and Cu to solve for $x$.


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What are disadvantages of having an impurity?
I thought about this while thinking about alloys and why they are better than pure metals, even though they are technically impurities. It got me wondering, are there actually any good real-life applications where having an impure substance is completely disadvantageous?

Impurities are the UNWANTED substances that are present in the substance..
There are Impurities in most of stuff including a lot of chemicals and it's very hard to remove them too
However If impurity exceeds the limit its completely Dangerous..
Like for example :

Surface water (river or lake)
  Surface water can be exposed to acid rain, pesticide runoff, water runoff, and industrial waste.
Groundwater (private wells or public water supplies)
  Groundwater can be contaminated by disease producing pathogens, careless disposal of household wastes, and leaching from landfills and septic tanks.

which can lead to Nausea
Lung irritation,Vomiting, Dizziness, Skin rashes and even Death
For the Chemistry and Chemical Industry impurities might slow down the reaction or may lead to formation of unwanted bi products , Might affect the rate of Crystallization - Most common issue I think
There are lots of articles online out there here are some for you : Impurity in medicine
Impurity in Electrical cells
Impurity in water
Effect on Solder machine
you can google and find more


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Cannizaro Reaction, Hydride Shift Reason?
In Cannizaro reaction when concentration of OH- is less in the rds step hydride shift occurs, though it is very difficult for hydride ion to shift yet the reaction proceeds but what is the driving force here??
My teacher told me that this mechanism was proposed because the hydride shift occurs but there must be some reason as to why the hydride shift?

Related: http://chemistry.stackexchange.com/questions/22023/cannizzaro-reaction-versus-nitroaldol-reaction, http://chemistry.stackexchange.com/questions/21576/please-explain-the-mechanism-of-cannizzaros-reaction

Hydride shift occurs because in the intermediate dianion there is no other suitable leaving group. It is either $\ce{H-}$, $\ce{R-}$ or $\ce{O^{2-}}$. Of those $\ce{H-}$ is the best leaving group. Note though that even under these conditions $\ce{H-}$ will of course not be solvated on its own (it is way too unstable for that). It is directly transferred from the dianion to the aldehyde.
This reaction will go "to the right" since the hydride shift is irreversible. There is simply no mechanism to form a dianion from the alcohol and carboxylic acid.


text:
Calculating vinegar strength
The pKa of vinegar is 4.76. If I know the  pH of the vinegar, what would be the formula for determining the percentage acid base on the relationship between pH, pKa and concentration? Can you show the formula so that I only need to plug in the ph?
I am not a chemistry student and I have not done these kinds of calculations. I am searching for a convenient way for farmers in developing countries to quickly approximate the strength of vinegar. The final test would be done in a lab but I need a way for them to test vinegar other than titration because titration can seem more like witchcraft than science to the uninitiated.

What is the definition of pKa? Ka is the equilibrium constant of which reaction? Try to write something. It is not difficult.

I understand that, "pKa = -log10Ka The lower the pKa value, the stronger the acid."
I am not a chemistry student and I have not done these kinds of calculations. I am searching for a convenient way for farmers in developing countries to quickly approximate the strength of vinegar.
The final test would be done in a lab but I need a way for them to test vinegar other than titration because titration can seem more like witchcraft than science to the uninitiated :-).

If $pK_a = 4.76$, then $K_a = 1.74 \times 10^{-5}$
Then approximating that $K_a$ can be written in terms of concentration rather than the true definition which is in terms of activities:
$K_a = 1.74 \times 10^{-5}M = \frac{[A-][H+]}{[HA]} = \frac{[H+]^2}{c-[H+]} $
$1.74 \times 10^{-5}M = \frac{[H+]^2}{c-[H+]} $
where $c$ is the molar concentration of the vinegar
If you want pH explicitly in the equation, then approximating pH = -log[H+], then:
$1.74 \times 10^{-5}M = \frac{(10^{-pH}M)^2}{c-(10^{-pH}M)} $
So, for example if pH is 3, $10^{-3}$ is 0.001, so
$1.74 \times 10^{-5} = \frac{10^{-6}M}{c-(10^{-3}M)}$
$1.74 \times 10^{-5}c -1.74 \times 10^{-8}M = 10^{-6}M$
$1.74 \times 10^{-5}c = 1.02 \times 10^{-6}M$
$c = 0.059M$

Thanks.
What does the H+, the HA and the M stand for?
If the ph is 3, what would the math look like?

pKa = -log(Ka), the Ka value calculated is not correct

http://chemistry.about.com/od/chemistryglossary/g/Pka-Definition.htm
What is it?

Hope this isn't double posting, but I am not understanding some of the symbols and I am not familiar with the conventions.
What do the M, H+ and HA stand for?
I am trying to come up with a formula where the only variable is the ph which the farmer would get with a ph meter, do the math and calculate the acidity of her vinegar.
The Ka, according to the formula DavePhD gave, is .000174 because the pKa for vinegar is 4.76. The molar mass is 60.05196 g/mol. So assuming the ph is 3, could some write out the "long math" for this? Are other numbers needed to solve this?

I want to solve for concentration or percentage of acetic/ethanoic acid in the solution.

Concentration is c in the final formula.  The units are M which means moles per liter. In the final equation there is only c and pH, and since pH will be known, c is the only unknown.

How do I read the (10−pHM)2/c−(10−pHM)? Am I multiplying the ph by the moles per liter (60...), subtracting it from 10 and squaring that number. Then dividing that by the same (10−pHM)and the solving for c?

M is just there to make the units work out.  M is not a variable.  You raise 10 to the (-pH) power. So for pH 3, 10 to the -3 power is 0.001.

@VinegrMan I added an example with pH=3 to the end of the answer

I believe the definition for the acid dissociation constant already includes all approximations and is therefore correctly defined with concentrations.

@Martin I don't think there is an IUPAC definition, but for example in Levine's Physical Chemistry it is a(H3O+)a(X-)/a(HX)a(H2O).  Then Levine says a(H2O)=1 is an approximation, and considering only the activity coefficients of the charged species is a further approximation.

@Martin definition in terms of activity discussed here: http://pubs.rsc.org/en/content/articlelanding/1998/an/a705491b/unauth#!divAbstract

I think it is this lack of official definition, that may cause confusion here. I thought of it, as it is presented in the wikipedia article. However, for the purpose of most calculations, the difference should be negligible. Thanks for the article, it was quite interesting to read.

I think for 5% acetic acid (typical vinegar) proper consideration of the decrease in water concentration from 55.5M to 52.8M and the activity coefficients of all the species would make a significant difference.

As we know:
$$\rm pH=\frac12(pKa-\log C)\quad\mid\quad C=\frac{\rm weight\;of\;acid}{\rm Mol\;wt.*Volume of sln.}\quad\mid\quad Strength=C*Mol.\;wt.$$
Calculating:
$$\rm S=10^{pKa-2pH}*M$$
So finally:
$$\huge\fbox{$\rm S=10^{4.75^{\#}-2pH}*60.05$}$$
#You can as well take that 4.76
Theoretically It is correct but some deviations are bound to occur in reality. I suggest finding out that in few samples first.


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Solve this mole fraction calculation with incomplete information
I am currently trying to solve a problem that seems quite simple at the beginning but the complexity reveals itself soon thereafter. The problem is the following:

You mix a known mass of water with a known mass of ethanol. Then you fill up the flask to the 50 mL mark with water, but cannot weigh the contents when you're done. What is the molar fraction of ethanol in this mixture?

My attempt at solving this problem is to basically calculate the initial molar fraction, and then stepwise add small virtual amounts of water. Then the new molar fraction is calculated, and somehow the excess volume (which is negative for the water-ethanol mixture) would also be included somewhere here. This is repeated until the final volume is reached.
An other attempt would be to simply neglect the excess volume. While this might be fine at the edges of the molar fractions, it would introduce a grave error towards the middle. Or is the error still negligible?
Are these reasonable attempts or am I completely missing something here?

I don't think you can solve that unless you know additional information like mass of water added or volume of water added, or the density of the final solution.
You could approximate by dividing mass of ethanol by density of ethanol to get volume of ethanol, and saying volume of water added is 50mL-(volume ethanol), but this is only an approximation because volumes are not additive when the two solutions are different.
The error could be estimated from this graph:

source: http://en.wikipedia.org/wiki/Partial_molar_property
or the data from the graph could be used to improve the accuracy of the answer
or use this density table: http://www.handymath.com/cgi-bin/ethanolwater3.cgi?submit=Entry

Do you know how large the error would be if the simple approximation were chosen?

I added a graph to the answer, see if that helps


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What does Pb(NO₃)₂ + KI form?
What does

$\ce{Pb(NO3)2 + KI}$

form?

This is just a metathesis (double replacement) reaction.
We know that nitrates are soluble, and alkali metals are soluble (at least in these general cases).
Let's ignore balancing for now, and do that last.
$\ce{Pb(NO3)2 + KI -> Pb^{+2} + NO3^- + K+ + I-}$
$\ce{Pb^{+2} + NO3- + K+ + I- -> PbI2 + KNO3}$
But the $\ce{KNO3}$ is also soluble, and would dissociate.
$\ce{Pb^{+2} + NO3- + K+ + I- -> PbI2 + K+ + NO3-}$
Evidently, we can cancel out the spectator ions $\ce{K+}$ and $\ce{NO3-}$ and are left with a net ionic equation (which is easily balanced).
$\ce{Pb^{+2} + 2I- -> PbI2}$

But if you're actually just looking for what was formed and not for an equation, it would be $\ce{PbI2}$ and $\ce{KNO3}$.


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Can Liquid CO₂ dissolve plastic
I know that subcritical and supercritical $\ce{CO2}$ are excellent solvents. I've seen some examples of plastics being damaged by the $\ce{CO2}$ being impregnated into the plastic at very high pressures, but in general would liquid $\ce{CO2}$ be able to dissolve plastic?

What kind of plastic?

Good question; since it is used in commercial dry-cleaning, you know it does not attack plastics used for buttons (unless they have to be removed). Anyone in the dry-cleaning business know?

From the abstract of Solubility of Polymers and Copolymers in Supercritical $\ce{CO2}$

[…] data to temperatures of 270 °C and 3000 bar […]
Over the same range of conditions, $\ce{CO2}$ cannot dissolve polyethylene, poly(acrylic acid), poly(methyl methacrylate), poly(ethyl methacrylate), polystyrene, poly(vinyl fluoride), or poly(vinylidene fluoride) […]

also I found this handy webpage as a guide (although supercritical is not an option)http://www.coleparmer.com/Chemical-Resistance

Depending upon the plastic polymer Liquid $\ce{CO2}$ I believe It dissolves plastic. The chance of this are low. Here is why:
$\ce{CO2}$ solvent has many desirable properties; technically speaking
it is a very ‘weak’ solvent: The polarity, dielectric
constant and dipole moment are less than that of
most conventional organic solvents. Consequently,
high-molecular-weight polymers, proteins and polar molecules are only sparingly soluble or even insoluble in $\ce{CO2}$
Scientist have tried to make this more efficient by adding fluorine to the solvent,  Unfortunately, fluorinated surfactants are not a good way to solve the problem, being both expensive and harmful to the environment. This research is ongoing, however.

Supercritical carbon dioxide or liquid carbon dioxide is a relatively poor solvent for polar molecules. Roughly, it is similar to pentane. If a certain plastic is soluble in pentane, probably, it may dissolve in $\ce{CO2}$.
Whenever we work with supercritical $\ce{CO2}$ (for chromatography), we usually pump a small amount of methanol along with it to increase its solvation power, otherwise most of the molecules will not dissolve in pure supercritical $\ce{CO2}$, except "fatty" molecules.


text:
Why do no known atoms have electrons in the g or h subshells?
I'm learning about orbital quantum numbers. While checking several elements on the periodic table I noticed that there aren't any atoms that have electrons in the g or h subshells. Why is this?

Well, the first available sub-shell for "g" electrons would be 5g (i.e., 1s, 2p, 3d, 4f, so 5g). Based on current trends, we'd need row 8 of the periodic table. We just haven't found those elements yet.
Let's look at it this way.

After row 1 (1s electrons only), there are 2 rows of the s and p-block elements before "d" opens up with $\ce{3d^1 4s^2}$
After two rows of s- p- and d-block elements, the "f" block opens up with cerium: $\ce{4f^1 5d^1 6s^2}$

So we'd need two more rows of the f-block (i.e., the lanthanides and actinides) and then complete the row.
Right now, we have found element 118, so if we can synthesize a few more, we can open up the "g" block.
Now, there are elements that have g electrons in excited states, so g and h-orbitals are relevant to some chemistry. But the ground-state elements haven't been discovered yet.

Could you elaborate on elements having g electrons in excited states? Any references? Thanks!

According to Umemoto and Saito,[1] starting with element 126, elements would possess $\mathrm{5g}$ electrons. The calculated ground-state electronic configurations for elements 126–131 are:

element 126: $\ce{[Og] 8s^2 8p^1 6f^4 5g^1}$
element 127: $\ce{[Og] 8s^2 8p^2 6f^3 5g^2}$
element 128: $\ce{[Og] 8s^2 8p^2 6f^3 5g^3}$
element 129: $\ce{[Og] 8s^2 8p^2 6f^3 5g^4}$
element 130: $\ce{[Og] 8s^2 8p^2 6f^3 5g^5}$
element 131: $\ce{[Og] 8s^2 8p^2 6f^3 5g^6}$

[1] Umemoto, K.; Saito, S. Electronic Configurations of Superheavy Elements. J. Phys. Soc. Jpn. 1996, 65 (10), 3175–3179. DOI: 10.1143/JPSJ.65.3175.

Why filling of g-orbitals do not start with element- 121 as $\ce{[Og] 8s^2 5g^1}$ according to order of energy of orbitals

@Apurvium due to relativist effects there is large splitting between the 8p1/2 and 8p3/2 levels and the 8p1/2 is lower energy than 5g, 6f and 7d.  Calculations before 1970 suggested that 7d would fill at 121, but later calculations at higher levels of theory showed 8p https://aip.scitation.org/doi/10.1063/1.1674338 You could ask another question about this for more information.  Maybe someone else can explain better.

Although, through a massive amount of energy, you could excite the electrons into g and h level orbitals no elements have electrons in those orbitals at the ground state. Think about how difficult it would be to have an electron in such large and excited state. You would need more protons than any of the elements we currently know about.
In short, it takes to much energy for any known elements to hold an electron in a nonexcited state in the g or h orbital.

This is incorrect.   The reason there are no 5g atoms in the ground state is due to the instability of large nuclei.  Any element with enough protons to charge-balance 5g electrons would fall to bits due to nucleons being able to overcome the strong force and leave the nucleus by quantum tunnelling.   It would not require energy to "have an electron" in a 5g orbital since its energy would still be lower than a free electron (though not by much).