text:
What is the OH in alcohol?
What is the $\ce{OH}$ functional group in alcohol (ethanol)? My teacher told me it isn't the same as the $\ce{OH-}$ ion or hydroxide.
But it does donate one electron to the respective carbon chain and it's very tempting to say it's hydroxide as it fits perfectly into the role. What exactly is it?

Hydroxyl group, not hydroxide, though I'm not the one that gave -1.

You could call ethyl alcohol "ethyl hydroxide" analogously to "sodium hydroxide". But it is a poor analogy, since (unlike NaOH) ethyl alcohol is not an ionic compound, does not dissociate in water to form ethyl cation and hydroxide anion, and does not raise the pH when dissolved in water.

Many people confuse between the three forms of the OH group.
When the OH has a negative charge, it is called hydroxide, and it forms ionic compounds with cations. An example is sodium hydroxide, which contains discrete Na+ ions and OH- ions. When dissolved in water, it separates into Na+ ions and OH- ions (solvation not mentioned for simplicity)
When the OH forms a covalent bond (different from the ionic bond above) with any other group, it is called a hydroxy group. An example is ethanol, H3C-CH2-OH, the oxygen in the OH forming a covalent bond with the carbon in the ethyl group. When you dissolve ethanol in water, it does not separate into C2H5+ and OH- ions like the NaOH would.
The final form of the OH is called a hydroxyl radical. Imagine removing an electron from a negatively charged hydroxide ion to make it charge-neutral. In this form it is very, very reactive, as it lacks an electron to achieve the stable form of hydroxide (like how a halogen behaves).

Well, I don’t understand what you are trying to ask.
But the $\ce{OH}$ you are bothered about can be thought exactly as that of $\ce{OH}$ of water ($\ce{H2O}$)
$\ce{C2H5-OH}$ in the same way as $\ce{H-OH}$

As other answers point out, the hydroxyl group in an alcohol is covalently bonded to carbon and thus is not the same as a hydroxide ion.
But don't assume that the hydroxyl group lacks any basic character.  It can still act as a Lewis base towards strong acids like $\ce{HCl}$, facilitating nucleophilic substitution reactions between alcohols and these strong acids.  The hydroxyl group in an alcohol can also act as a Lewis base towards the $\ce{-COOH}$ function in a carboxylic acid, enabling the formation of esters from such acids.  You should encounter these properties when you study nucleophilic reactions, including ester formation, in organic chemistry.

Alcohols contain a hydroxyl functional group. This is different from the hydroxide ion, $\ce{OH-}$.
By way of example, the formula for the alcohol present in alcoholic drinks, ethanol or ethyl alcohol, is $\ce{C2H5OH}$. If you want to emphasize the hydroxyl functional group, you could also write $\ce{C2H5-OH}$. To symbolize any alcohol, you could use $\ce{R-OH}$. As you can see, this is different from $\ce{OH-}$ because there is no formal charge on the hydroxyl group and it is connected to the remainder of the molecule with a covalent bond.
In terms of chemical reactions, you can find similarities in the reactivity of water and alcohols. Many reactions of water and of alcohols involve proton (hydrogen ion, $\ce{H+}$) transfer. Water can lose a proton (to make hydroxide, $\ce{OH-}$), and  alcohols can as well (to make an alkoxide, $\ce{R-O-}$). Both can also gain a proton (to make hydronium $\ce{H3O+}$ or protonated alcohol, $\ce{R-OH2+}$, respectively). So the alcohol is analogous to water, and the alkoxide is analogous to hydroxides in some ways.
This is illustrated below (top, alcohol; bottom, water; left, deprotonated; center, as is; right, protonated):


text:
Does acetylene give a positive Tollens' test? If not, why is this so?
One of the uses of the Tollens' reagent is to confirm the presence of a terminal triple bond in a compound. Does acetylene give a positive Tollens' test? If not, why is this so?

http://chemistry.stackexchange.com/questions/33854/reaction-of-acetylene-with-tollens-reagent

Tollen's reagent (ammoniacal silver nitrate) reacts with terminal acetylenes to form the silver acetylide which precipitates out of solution
$$\ce{RC#CH + AgNO3 + NH4OH -> RC#CAg v + NH4NO3 + H2O}$$
In the case of acetylene itself, since both ends of the molecule are (if you will) terminal acetylenes, a bis-silver acetylide salt is formed and precipitates out of solution (see page 3, item 6b in a PDF from www.sakshieducation.com/ or via the Internet Archive)
$$\ce{HC#CH + 2AgNO3 + 2NH4OH -> AgC#CAg v + 2NH4NO3 + 2H2O}$$
Tollen's reagent is not very stable, therefore it needs to be freshly prepared before use.  The reagent has a number of other uses in chemistry (test for aldehydes and α-hydroxy ketones), anatomical staining, and silver mirroring.  This Wikipedia article provides a nice, concise overview.

So this means that alkynes don't reduce the Tollen's reagent, as the aldehydes do, right?

@ReeshabhRanjan Yes, that's correct.

Actually, you can skip a couple of steps and react silver nitrate solution directly with acetylene to get silver acetylide.  Also, be careful, silver acetylide is a sensitive explosive.

Terminal hydrogens in acetylene are acidic and are easily removed to form acetylide ions, and in this case, elementary silver will be displaced from the ammoniacal silver nitrate and will form silver mirror on the surface of the reacting vessel.

The article linked in the first answer specifically rules out formation of silver mirror, the bis silver acetylide that precipitates is white.

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text:
Rock salt structure: chloride lattice or sodium lattice?
Source
From this diagram of the rock salt structure ($\ce{NaCl}$) we see that both the chloride and sodium ions have the same environment. That is to say, they each have the same number of neighbours at every distance, except that the charges are opposite.
However, I've always been taught to consider the chloride ions as forming a face-centred cubic (fcc) [or cubic close packed (ccp)] lattice, with sodium ions filling octahedral holes.
Is there any reason why we can't think of it the other way round, with sodium ions forming the fcc lattice and chloride ions filling octahedral holes?
Is it perhaps because sodium ions are smaller than chloride ions, or something else?

Unit cells are just a way of describing a given crystal. You can certainly draw $\ce{NaCl}$ the other way, with $\ce{Na+}$ in the corners and it's still correct. There are a few ways of thinking about this structure:

as two interpenetrating FCC lattices
as a $\ce{Cl-}$ lattice with $\ce{Na+}$ in the octahedral holes
as a $\ce{Na+}$ lattice with $\ce{Cl-}$ in the octahedral holes

They're all equivalent. The figure above actually shows both types of octahedral holes. Some might prefer to visualize the structure as the second case since, as you say, the chloride ions are bigger, but going the other way isn't wrong, nor does it describe anything different.

This is a more important question than one would think and not adequately answered before.  Yes, there is an equivalence so you would wonder why we define one as the lattice and the other as in holes.  However this is an important idea of solid state chemistry.
The important thing is generally (GENERALLY) anions are bigger than cations.  Also more often the same species.  In oxides, this is especially the case.  So you can think of the anions as a close packed lattice (FCC or HCP) in the sense of balls with closest packing.  and the cations in certain holes of the lattice.  This is the same in NaCl where chloride is bigger than sodium.
It's also important
Of course this is chemistry, so everything depends.  But if you want a useful initial organizing idea for MANY structures, think of close packed anions with cations in certain holes (perovskite, spinel, corundum, etc.).
So yes, there IS a reason why we think of the chlorides as the base lattice, not the sodium ions and you are right on that the reason is size.

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text:
Why is ionization potential considered a periodic property?
Why is ionization potential considered a periodic property???
A periodic property is a one which appears at regular property
But we see that every element has some kind of ionization potential

The ionization energy is described as the energy required to remove an electron from an atom in the gaseous phase. Apart from some exceptions, it is possible to identify a general trend in ionization energies within the periodic table: moving from left to right in a period the ionization energy increases, whereas it decreases from top to bottom in a group.


text:
Is there a method for finding the net-ionic equation?
I truly don't understand the method for finding the net-ionic equation for a reaction. It is not clear to me how you should decide what the products are, especially when there are multiple possibilities.
Example:

Excess potassium cyanide solution is added to aluminum bromide solution.

The first thing that came to mind for the molecular equation is:
$\ce{3KCN\+AlBr3 \to 3KBr\+Al(CN)3}$
But apparently the correct molecular equation is:
$\ce{6KCN\+AlBr3 \to 3KBr\+Al(CN)6}$
I see no good reason why the one I thought of is incorrect. And to get the net ionic equation, the spectator ions are $\ce{K+}$ and $\ce{Br-}$. But why should I assume the solution is aqueous?
Another example:

Chlorine gas is bubbled through dilute sodium hydroxide.

Well its easy enough to get the reactants:
$\ce{Cl2\+NaOH \to ?}$
What first comes to mind is $\ce{Cl2\+2NaOH \to 2NaCl\+Cl2O\+H2}$
But the correct molecular equation is $\ce{Cl2\+2NaOH \to 2NaOCl\+NaCl\+H2O}$
What should be the thought process for arriving at the correct answer here. I am truly lost.
Another example:

Lithium oxide reacts with sulfur dioxide.

I recognize this as a synthesis reaction. However how can I predict if the product is lithium sulfite or lithium sulfate?

The 'correct' first equation does not balance. Could you provide us with the source of the first problem. I am not familiar with it.

You are right it does not balance. But the molecular equation can't be correct because the ionic equation is $6CN^-+Al^{+3} \to Al(CN)_6^{-3}$. This is from a worksheet on net ionic equations (but its not homework).

I looked it up but can find no references for Al(CN)6 - it could be that your worksheet has a mistake on it.

I have the impression is "how do I know which products to expect?". If I am right here, there is unfortunately no universal answer. Effectively, you'll have to do the experiment and analyse the products. There are a few guidelines and concepts, but this would get far to broad here. Alternatively, you'll have to look it up in a book or chemical encyclopedia.

But apparently the correct molecular equation is $\ce{6KCN\+AlBr3 \to 3KBr\+Al(CN)6}$

That's not correct, what happenend to the other 3 potassium atoms?
It could form $\ce{Al2(CN)6}$ but you can't exclude $\ce{Al(CN)3}$ without further information beyond balancing of equations.
There are two main steps to balancing equation:

Balancing of each type of atom
Balancing number of electrons

If you've done both of those correctly, you've done all you can with balancing alone.

$\ce{Cl2\+2NaOH \to 2NaCl\+Cl2O\+H2}$

Above is clearly wrong because number of oxygen atoms doesn't balance

$\ce{Cl2\+2NaOH \to 2NaOCl\+NaCl\+H2O}$

Above is clearly wrong because number of oxygen atoms doesn't balance

This does not answer my question. I am asking how we know what are the correct products when there are multiple possibilities, as in the examples I gave.

You haven't given us an example where there are two equations that are balanced that you need to choose from.  When there is such an example, you need experimental data.


text:
Heat given off from an electrochemical cell compared to mixing reactants
My confusion regarding the difference between the heat given off in a reaction that takes place in an electrochemical cell compared to in a beaker/appropriate container when the reactants are simple mixed, arose from the following problem question:
For the reaction:
\begin{align}
\ce{HgCl2~(s) + H2~(g,1~bar) &-> 2Hg~(l) + 2H+~(aq) + 2Cl- }\\
\Delta G^\circ &=-51.64~\mathrm{kJmol^{-1}},\\
\Delta S^\circ &=-61.6~\mathrm{JK^{-1}mol^{-1}},\\
\Delta H^\circ &=-70~\mathrm{kJmol^{-1}}.\\
\end{align}
Part (a) of the question then asks: "Determine how much heat is given off at $298~\mathrm{K}$ per mole of $\ce{HgCl2}$ if the cell is operated reversibly"
The correct approach is to use $T\,\mathrm{d}S=\mathrm{d}q$ giving a value of: $18.36~\mathrm{kJmol^{-1}}$
Then part (b) asks: "Now determine the heat given off if the reaction is brought about by simply mixing the reactants"
I realised that the electrical work done in the cell would now be given off as heat so I just added the value for the Gibbs free energy to the value calculated in (a) and got: $70~\mathrm{kJmol^{-1}}$. This answer is correct but my tutor noted that this value is the value for the enthalpy change which would have been a shortcut to the answer. I didn't spot this and I don't understand why this is the case so wouldn't spot it again, should another question like this come up. Please can you clear up why this value is the enthalpy change.
I have some issues with the above working (it is all correct but I don't fully understand it).
Firstly, in part (a) I had to use the Clausius inequality (except it was an equality since the process was carried out reversibly) to find $\mathrm{d}q$. However, why must I use this equation because $\Delta H=q_p$ - the process was carried out at a constant pressure?
Secondly, if the heat given off in part (b) is $70~\mathrm{kJmol^{-1}}$ doesn't this violate the Clausius inequality? I must be wrong but it seems that this must be the case since $\mathrm{d}S \geq \frac{\mathrm{d}Q}{T}$ and using the value for the entropy change ($\Delta S^\circ=-61.6~\mathrm{JK^{-1}mol^{-1}}$) and the new calculated value for the heat given off ($70~\mathrm{kJmol^{-1}}$) then dividing this by $298~\mathrm{K}$ gives a value larger than the value for the entropy change but the Clausius inequality says that the change in entropy is greater than or equal to $\mathrm{d}q/T$. What is going on there?

I think the problem here is related to what each of these terms means, in a fundamental sense.
$\mathrm{d}S$ is the entropy of the system
$\frac{-\mathrm{d}q}{T}$ is the entropy of the surroundings - the heat change of the surroundings divided by the temperature.
$\mathrm{d}H$ is the enthalpy of the system - it is equal to $\mathrm{d}q$ only when the internal energy change occurs at constant pressure, and when there is no form of work being done besides PV work. In an electrochemical cell, moving the electrons is work, and so we can't leave it out of the total energy balance.
In other words, you can't just use $\mathrm{d}q = \Delta H$ for the 1st question because $\mathrm{d}q_p \neq\Delta H$ when you are doing any work besides PV work.
Instead you are using the Clausius inequality, which is just another way of stating the second law: the entropy of the universe must increase for any process. In this case, the universe is the system plus the surroundings, and so:
$$\Delta S - \frac{\mathrm{d}q}{T} \geq 0$$
Solving that for dq gives you the heat absorbed by the surroundings, which is equal to temperature times the entropy change of the system.
For a process that is only (possibly) doing PV work, at constant pressure, $\Delta H = \mathrm{d}q_p$, and so we can just use that. This would be an example of using the 1st law at constant pressure, when there is no other type of work besides PV.
In your case, you used:
\begin{align}
\Delta G &= \Delta H - T\,\mathrm{d}S\\
\Delta H &= \Delta G + T\,\mathrm{d}S\\
\end{align}
This works because the equation for $\Delta G$ is derived directly from the Clausius inequality at constant pressure. In other words, you are using the 2nd law to derive the same result.
Edit
To answer your last questions, and questions from the comments:

why is $T\,\mathrm{d}S$ less than $\mathrm{d}q$ in my example? I don't really get this bit

It's not - but you are dropping the signs and that makes it hard to see. If you leave the signs in:
\begin{align}
T\,\mathrm{d}S &\geq \mathrm{d}q\\
(298~\mathrm{K})\left( -61.6~\mathrm{\frac{J}{K\cdot mol}}\right) &\geq -70~\mathrm{\frac{kJ}{mol}}\\
-18~\mathrm{kJ} &\geq -70~\mathrm{kJ}\\
\end{align}
Which is true, although it is easier to see if you divide by -1:
$$18~\mathrm{kJ} \leq 70~\mathrm{kJ}$$

The entropy divided by the temperature gives $234.9~\mathrm{JK^{−1}mol^{−1}}$ which is not the value for entropy as it should be according to the Clausius inequality given that the change in entropy is equal to the heat given off in part (b). Why is this so?

Here I assume you meant "enthalpy divided by temperature." The second reaction is not carried out in a reversible manner, so we have no guarantee that $\mathrm{d}S = \mathrm{d}q/T$. All we know is that it will be greater than or equal to $\mathrm{d}q/T$, and if you plug in the numbers, it is.
Remember that all of this is built into the $\Delta G$ equation as well, which is why we use it when doing "regular" chemical reactions at constant T & P.

For a closed system. the change in enthalpy at constant pressure equals heat. See here https://physics.stackexchange.com/questions/446187/why-isnt-dh-q-sum-i-mu-idn-i-constant-pressure?rq=1.

For first part since system is operated reversibly at constant temperature:
$$\newcommand{\d}{{\rm d}}
\d S=\frac{\d q_{rev}}{T}\implies \Delta S=\frac{Q_{rev}}{T}$$
This is what you have done.
For second part, Yes Gibbs energy is the maximum (in reversible process) non-expansion/additional work (here the electrical work)
$$\Delta G^\circ=-\nu FE^\circ=w_{add,max}$$
This value would be same as $\Delta H^\circ$ because from rearrangement of definition of $$\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$$$$\Delta H^\circ=\underbrace{\Delta G^\circ}_{\text{non expansion work}}+\underbrace{T\Delta S^\circ}_{\text{work extracted as heat}}$$
Also it can be viewed in another way, since $\Delta H^\circ$ is "the energy supplied as heat at constant pressure".Another thing to keep in mind is $\d H=q_p$ only under constant pressure and no additional work, both to be satisfied.
And for the Classius Inequality, there's some calculation error by you:
$$\frac{Q}{T}=\frac{-70000}{298}=-234.8\le-61.6(\Delta S^\circ)$$


text:
Electrolysis used to cause a Chemical Reaction?
Electrolysis is used to separate compounds like Water into Hydrogen and Oxygen.
Can it be used to induce a synthesis reaction rather than splitting reactants apart? For example, could you use electrolysis to recombine the Hydrogen and Oxygen?

There is a new question on Kolbe's Synthesis. Where electrolysis is used to form alkanes from carboxylic acid salts.

You wouldn't need to supply electric energy to get oxygen and hydrogen to recombine into water, because the free energy change for that reaction is negative. In other words, the products have a lower potential energy than the reactants, and so the reaction will proceed "downhill" on its own. You would need to overcome the very small activation energy, which you could do with electricity or with a flame or spark. But once it got started, you would not need to supply additional energy, and could in fact generate electricity from the reaction.
This is the basis of how hydrogen fuel cells work - hydrogen and oxygen react in a device that is set up to extract the energy that is produced in the form of an electric current.
An easy way to think about it is that if you need to supply energy to make a reaction happen, then the reverse reaction will produce the same amount of energy (if we ignore losses due to inefficiency) or less (if we don't). The reverse is true as well.
On a side note - electrolysis specifically refers to splitting a compound. The "-lysis" suffix means "break apart," and "electro-" prefix means "with electricty." In your case, you are thinking of "electrochemistry" - which just means "using electricity with chemistry." Electrolysis is one thing you can do with electrochemistry.


text:
How do I know with which oxidation numbers I can form a certain kind of compound?
Take, for istance, phosphorus; it has eight different oxidation numbers, but it can only form two different anhydrides, phosphoric (when the oxidation number is 5) and phosphorous (when it is 3). How do I understand which ones I can use to get anhydrides?
 Or take chlorine, it also has eight oxidation numbers but if I wanted to know the corresponding hydracid, HCl, how'd I know which number to use?
Please forgive me if the question is trivial and written in a poor english, but I'm having trouble with both chemistry and English language classes at present!

Truth be told, in general case you cannot know this beforehand. Some general rules do exist, but everyday chemistry mostly works with special cases, that do not follow rules very well. And oxidation number is a derived characteristic anyway.

Inorganic anhydrides are rather simply called oxides and during hydrolysis usually oxidation numbers don't change (however in case of ClO2 and ClO3 disproportionation occurs).

You need to be familiar with some of the compounds earlier, otherwise you can't tell if it exists. Remember facts come first before explanation. It also depends usually on the atom/ion to which it is bonded. for example since Fluorine is most electronegative it can form $\ce{XeF6,XeF4}$,etc. but $\ce{XeH3}$ doesn't form also oxygen forms large oxidation states as it is electronegative too and there is also more energy released as many bonds are formed (ususlly pi-bonds.)


text:
The Combining of Carbon Dioxide and Hydrogen
How does this work exactly? Does it need some sort of kick in order to come together, or does it come together when put in a pressurized chamber?

I believe you are referring to the Sabatier process. $$\ce{CO2 + 4H2 -> CH4 + 2H2O}$$
The reaction uses a $\ce{Ni}$ catalyst and is done at $300-400\,^{\circ}{\rm C}$ and high pressures. The reaction is exothermic, $\Delta H = −165.0~\mathrm{kJ\,mol}^{-1}$, but requires initial heating in order to start the reaction off due to the activation energy.
Read more on Wikipedia.


text:
ph & pOH - Autoionization of Water
So I was learning about pH and pOH and it made sense. I looked at strong, weak acids & bases and learned that they dissociate completely in water. So then I moved on to water and learned about the autoionization of $\ce{H2O}$.
Here is the process:
$$\ce{H2O + H2O -> H3O+ +OH-}$$
So then I learned about the equilibrium constant and then I figured this:
$$\ce{{[H3O+][OH-]} = 1E -14}$$

What does this mean?

So after that I was doing some basic pH and pOH calculations withe strong acids and bases.
Here is one of them:
Find the pOH of a $\ce{1E-6M} \ce{HCL}$
So here are the steps:
$$\ce{-log[1E-6M] = 6}$$
Ok so the solution had the pH of 6
At first I was like that's nice and I proceeded.
$$\ce{6+pOH=14}$$
$$\ce{pOH=8}$$
So then I was like wait a minute where did the $\ce{OH-}$ come from if it dissociates like this:
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
So then I thought about it and this is what I came up with:

Since there is some $\ce{OH-}$ then there must be some water more water that autoionized into OH and that is the only place where $\ce{OH-}$ could've come from. So that means that even though $\ce{HCl}$ completely dissociated there must be more water than $\ce{HCl}$, because referring to the original equation there is $\ce{1M}$ quantities of $\ce{H2O}$ and $\ce{HCl}$ and so in the $molarity$ of the $\ce{HCl}$ given that means there was more water and less $\ce{HCl}$ and the $\ce{OH-}$ could've only come from the autoionization of that excess water.

So that is what I thought could be the only possible explanation of the $\ce{OH-}$ in a strong acid. And that could only come from $\ce{H2O}$ .
So am I right, and If I am wrong then please tell me where the $\ce{OH-}$ came from.

Why is the $\ce{pH}$ scale only to $14$? Because if I took a $\ce{2M}$ $\ce{HCl}$ and then I get a super acid in which the $\ce{H+}$ ions is a lot more than the $\ce{H2O}$ available to absorb them. So why is it only to $14$? Does it have to relate to the original autoionization of $\ce{H2O}$?

And If I totally grasped this wrong then can you please accurately tell me what $\ce{pH}$ & $\ce{pOH}$ and what autoionization of $\ce{H2O}$ is for a beginner and still make sense.

I think you have it all right. The OH- present must come from the autolysis of the water present, which is far in excess of the HCl. If you figure the pOH of pure water would be 7, then it makes sense to think that adding 1E-06 M (a small concentration) of HCl would lower the basicity (raise the pOH from 7 to 8).
The reason a scale of 0-14 is used for pH is that when you get to either extreme end of the scale (close to 1M in either acid or base), the sheer concentration of ions present, relative to water, is a game changer and the system no longer behaves in a nice linear fashion. So the scale does not work well outside the 0-14 range. Conceptually, you can extend the pH scale down below zero using concentrated mineral acids or strong acids like fluorosulfonic acid, and empirically measure their effectiveness in protonating a very weak base as a proxy for pH.


text:
How do the strong acids dissociate?
In writing net ionic equations involving strong acids, one needs to know how strong acids dissociate. For example, the dissociation of $\ce{H2SO4}$:
\begin{align}\ce{
H2SO4 &-> 2H+ + SO4^{2-}\\
H_SO4 &-> H+ + HSO4^{-}\\
}\end{align}
The reason I ask is because apparently the molecular equation
$$\ce{H2SO4 + 2KOH -> 2H2O + K2SO4}$$
has the net-ionic form $$\ce{H+ + OH- -> H2O}$$
It would be helpful if you could describe in general how I should write the dissociation of the strong acids in the net ionic equation.

This is a little tricky. I would argue that only the first proton dissociates completely and should be considered "strong" - using the rules for writing net ionic equations, where we only dissociate strong electrolytes, this would meant that we would write:
$\ce{H2SO4 + 2KOH -> 2H2O + K2SO4}$
$\ce{H+ + HSO4- + 2K+ + 2OH- -> 2H2O + 2K+ + SO_4^{2-}}$ - here only the first proton dissociates completely
Canceling out spectator ions gives:
$\ce{H+ + HSO4- + 2OH- ->2H2O + SO_4^{2-}}$
Your teacher (or whoever is giving you that answer) is dissociating the second proton as well, to give:
$\ce{2H+ + SO_4^{2-} + 2OH- ->2H2O + SO_4^{2-}}$
Which becomes:
$\ce{H+ + OH- ->H2O}$
There are a few reasons why they might do this, but I would argue that the 1st approach (which only dissociates the first proton) is best because it more accurately reflects what is happening, in the sense that $\ce{HSO4-}$ is a weak electrolyte. This will become important later when you study acid-base equilibria.
In general, you dissociate protons from acids as if they were regular metal cations - just "pull them off" and don't forget the positive charge. For polyprotic acids, the first proton is strongly acidic, or dissociates completely, but the other protons are not. There might be exceptions to this rule but I can't think of any off the top of my head.

Is the presence of excess $\ce{OH-}$ ions maybe forcing the $\ce{HSO4-}$ ions to dissociate as well? Kinda like how a weak acid behaves like a strong one in a basic medium.

@Papul - yes, that is probably the reason, but it would only be true for an excess of KOH, so leaving bisulfate out is not, in my opinion, the best answer.


text:
Potassium thiocyanate and iron(II) acetate
I am trying to find the net-ionic equation for the reaction of potassium thiocyanate and iron(II) acetate.
It seems reasonable to me that the molecular equation is
$$\ce{2KSCN + Fe(CH3COO)2 -> 2CH3COOK + Fe(SCN)2}$$
which would give the net-ionic form
$$\ce{SCN- + Fe^2+ -> Fe(SCN)2}$$
My chemistry teacher says the correct form is
$$\ce{4SCN- + Fe^2+ -> Fe(SCN)4^2-}$$
First I don't know what molecular equation would give rise to this. Second, why in the world is my guess, which is simpler, wrong and this one correct?

In the molecular equation, you have written cyanide (CN-) on the left and thiocyanate on the right.  The next equation you have written, has one SCN- on the left and two on the right so it is not balanced.  The last equation is balanced, but it does not represent the molecular equation because the product is not the same.

As best I can tell, your answer is correct. I can't find any reference for iron(II) thiocyanate in the form your teacher says is correct, and several references that say $\ce{Fe(SCN)2}$ is the correct form.
For precipitation reactions, the charge will never change. You should always assume that the charges in the product will balance. There might be some cases where the product has an unexpected formula that you can't predict based on charges alone, but I can't think of any off the top of my head, and you certainly shouldn't be expected to know them at this stage.
edit:
This is the best reference I could find, it's a google books link to the CRC handbook

I guess there is a difference here whether you talk about the complex (which can be a negative anion with more than two thiocyante coordinating it), and the stoichiometry of a solid, which will be di-thyociano per definition for the charge equilibrium (but it is not necessarily the number of ligands coordinating the metal).

@Greg - could be, but this type of question is usually asked in general chem I, so I don't know how the student would be expected to know that, let alone predict it from the reactants.

The textbook may include reference for given complexes, and teacher would like to test that knowledge. So the short answer is that you don't predict it, you learn it. While I don't necessarily agree with this practice, I have seen it in several occasion (including high-school level texts..).

@Greg - while we are making conjectures, consider also the possibility that the teacher is just wrong. Since I can't find any references to that complex, I think that possibility is more likely.

Well, another trouble is that this is NOT a precipitation reaction. Fe(SCN)$_2$ is water-soluble, so a genchem student who didn't know about complex formation would have to conclude that all of the ions were spectators and that there was no reaction.

@Fred - that is a good point. Without knowing the solubility, you wouldn't be able to predict the products. And if it is soluble, then it would be "no reaction."

@thomij When you have a chance, can you add one of the references so future readers will know where to look?

@jonsca - I didn't see any I would consider authoritative before, which is why I left them out. But, some more searching turned up a reference in the CRC handbook, so I added it.

If you add an excess of cyanide ions to a iron(II) solution you could conceivably get precipitate of $\ce{Fe(SCN)2}$ which would then dissolve to form a complex with four thiocyanate ions. However, I can't find any references to such a complex on the internet.
The equations would be: $$\ce{Fe^{2+}_{(aq)} + 2SCN-_{(aq)} -> [Fe(SCN)2]_{(s)}}$$ $$\ce{[Fe(SCN)2]_{(s)} + 2SCN^{-}_{(aq)} -> [Fe(SCN)4]^{2-}_{(aq)}}$$


text:
How to determine the empirical formula for terephthalic acid from its combustion products?

Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticisers. It contains only $\ce{C}$, $\ce{H}$, and $\ce{O}$. In one experiment, $0.4953~\mathrm{g}$ of the acid was burnt in excess oxygen. The gases produced were passed through $\ce{CaCl2}$ and $\ce{NaOH}$. It was found that the mass of the $\ce{CaCl2}$ increased by $0.1613~\mathrm{g}$ and the mass of the $\ce{NaOH}$ increased by $1.0495~\mathrm{g}$. What is the empirical formula of terephthalic acid?

We have:
$$\ce{C_{$x$}H_{$y$}O_{$z$} + O2 -> CO2 + H2O ->[\ce{CaCl2 + NaOH}] CaCl2 + NaOH}$$
The anhydrous $\ce{CaCl2}$ absorbs the $\ce{H2O}$ produced, and the $\ce{CO2}$ is absorbed from $\ce{NaOH}$? So the products are $\ce{CaCl2. 2H2O}$ and $\ce{Na2CO3}$
I may have just figured the empirical formula out, and was wondering what the final answer is so I can compare and see where I went right/wrong. I realised that instead of $\ce{Na2CO3}$, $\ce{NaHCO3}$ would be produced.
Working out:
Via conservation of mass, the $\ce{CO2}$ and $\ce{H2O}$ must also be $0.4953~\mathrm{g}$.
Therefore: conservation of mass (mass of $\ce{CO2 + H2O}$) - increase in mass of $\ce{CaCl2 + NaOH}$
$0.4953 - (0.1613 + 1.0495) = 0.7155~\mathrm{g}$ <--- mass of $\ce{CaCl2}$ and $\ce{NaOH}$
Mass of $\ce{CaCl2.2H2O} = 0.7155 + 0.1613 = 0.8768~\mathrm{g}$
Mass of $\ce{NaHCO3} = 0.7155 + 1.0495 = 1.765~\mathrm{g}$
     C:               H:              O
 1.765x(12/84):     0.8768x(4/147.1):

 (m) 0.252142857:     0.023842284:   0.4953-(0.252142857+0.023842284)= 0.219314859

 (n) m/12       :     m/1         :    m/16

 0.021011905 : 0.023842284     :  0.013707179

       (Divide by lowest value)

( 1.53      :     1.74        :    1 ) x4

  6         :       7         : 4

Therefore $\ce{C6H7O4}$

You are on the right track, but if you google terephthalic acid, you will know that your formula is not correct.
One mistake you are making is using the final mass of the $\ce{CaCl2}$ and $\ce{NaOH}$ and not the change in mass. For example, let's take the $\ce{CaCl2}$:
First, we don't know what the initial mass of $\ce{CaCl2}$ is, so we cannot calculate the final mass the way you are trying to do. We are only given the mass of the terephthalic acid (0.4953 g). The calcium chloride and sodium hydroxide are not part of this mass.
However, we know that the calcium chloride increased in mass by 0.1613 g, presumably by absorbing water (you are correct that the calcium chloride is absorbing water). So, the mass of water produced by the combustion reaction is 0.1613 g.
Similarly, we only know the change in mass of the sodium hydroxide (1.0495 g), which occurred by absorbing carbon dioxide. So the mass of carbon dioxide is 1.4095 g.
The second mistake is to assume that all of the mass of the carbon dioxide and water produced must come from the original mass of the terephthalic acid. Not so! We have excess oxygen added contributing mass to the reactant side, and this oxygen is also in the products.
$$\ce{C}_x\ce{ H}_y\ce{ O}_z\ce{ +}\left(\frac{4x+2y-z}{4}\right)\ce{ O2 -> }x\ce{CO2 +}\frac{y}{2}\ce{ H2O}$$
We need to do more maths to figure out how much of the leftover mass in the original sample is oxygen. Here is the beginning of the setup, now that you have some sense of the mass of water and carbon dioxide produced:

Mass of water produced $\ce{->}$ moles of water produced $\ce{->}$ moles of hydrogen in the water produced (and in the original sample!) $\ce{->}$ grams of hydrogen (in sample)
Mass of carbon dioxide produced $\ce{->}$ mass of carbon (in sample)
Mass of oxygen in sample (by subtraction) $\ce{->}$ moles of oxygen in sample

You are doing everything correctly after this.

I am afraid I could not follow your reasoning and unfortunately can not pinpoint you to your error therefore. The result you obtain, does not resemble the empirical formula of terephthalic acid. As a general tip: These problems are of entirely mathematical nature, so it is important to use proper equations and keep the units at all times. I am going to tackle this problem from beginning to end.

Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticisers. It contains only $\ce{C}$, $\ce{H}$, and $\ce{O}$. In one experiment, $0.4953~\mathrm{g}$ of the acid was burnt in excess oxygen. The gases produced were passed through $\ce{CaCl2}$ and $\ce{NaOH}$. It was found that the mass of the $\ce{CaCl2}$ increased by $0.1613~\mathrm{g}$ and the mass of the $\ce{NaOH}$ increased by $1.0495~\mathrm{g}$. What is the empirical formula of terephthalic acid?

First of all it is important to formulate the correct reaction equation:
$$\ce{C_{$x$}H_{$y$}O_{$z$} + $\frac{\left(2x+\frac{y}{2}-z\right)}{2}$\,O2 -> $x$\,CO2 + $\frac{y}{2}$\,H2O}$$
The information about calcium chloride and sodium hydroxide is only a way to tell you the mass of the product. It is completely irrelevant to the question itself. However, the correct equation would then be:
$$\ce{C_{$x$}H_{$y$}O_{$z$} + $\frac{\left(2x+\frac{y}{2}-z\right)}{2}$\,O2 ->[\ce{CaCl2, NaOH}] $x$\,NaHCO3 + $\frac{y}{2n}$\,CaCl2.$n$\,H2O}$$
First of all, we need to gather all the information we are given in the question or that we know otherwise:
\begin{align}
m(\ce{C_{$x$}H_{$y$}O_{$z$}}) &= 0.4953~\mathrm{g} &M(\ce{C_{$x$}H_{$y$}O_{$z$}}) &= ?\\
m(\ce{H2O}) &= 0.1613~\mathrm{g} & M(\ce{H2O}) &= 18~\mathrm{\frac{g}{mol}}\\
m(\ce{CO2}) &= 1.0495~\mathrm{g} & M(\ce{CO2}) &= 44~\mathrm{\frac{g}{mol}}\\
\end{align}
We can go ahead and calculate the number of moles of carbon dioxide and water produced. We therefore also know the relative number of moles of carbon and hydrogen in terephthalic acid. The basic formula that needs to be applied here is $$n=\frac{m}{M}.$$
How many moles of carbon are in carbon dioxide, what is the mathematical relationship? What is it for hydrogen and water?

 $n(\ce{C}) : n(\ce{CO2}) = 1$
 $n(\ce{H}) : n(\ce{H2O}) = 2$

How many moles of carbon dioxide and water have been produced? How does that relate to carbon and hydrogen?

 $n(\ce{CO2})=\frac{m(\ce{CO2})}{M(\ce{CO2})}= 0.0239~\mathrm{mol} \therefore n(\ce{C})= 0.0239~\mathrm{mol}$
 $n(\ce{H2O})=\frac{m(\ce{H2O})}{M(\ce{H2O})}= 0.0090~\mathrm{mol} \therefore n(\ce{H})= 0.0180~\mathrm{mol}$

Calculate the mass of hydrogen and carbon in the original sample with
$$m=n\cdot M$$ and use it to find the mass of oxygen in the sample.

 $m(\ce{C}) = 0.2868~\mathrm{g}$
 $m(\ce{H}) = 0.0180~\mathrm{g}$
 $m(\ce{O}) = m(\ce{C_{$x$}H_{$y$}O_{$z$}}) - m(\ce{C}) - m(\ce{H}) = 0.1906~\mathrm{g}$

How many moles of oxygen are present in the original sample?

 $n(\ce{O}) = 0.119~\mathrm{mol}$

Now you only have to relate the number of moles for each element to integer ratios and you will find the empirical formula.

 $x:y:z = n(\ce{C}) : n(\ce{H}) : n(\ce{O}) = 0.0239~\mathrm{mol} : 0.0180~\mathrm{mol} : 0.119~\mathrm{mol}$
 $x:y:z = 4:3:2$
 Therefore the empirical formula is $\ce{C4H3O2}$, which is half of the sum formula $\ce{C8H6O4}$, see wikipedia for more information.


text:
Is this the right conclusion for a thought experiment?
In this thought experiment, let’s consider the surrounding to be air that is composed of entirely water vapor (no other species like $\ce{O2}$ and $\ce{N2}$ are present). The surrounding is infinitely large. The system in question is a pan containing a thin layer of liquid water. The liquid water is brought to its boiling point by uniform heating. This pan is open (not covered in any way) and is exposed to the surrounding water vapor.
By definition, boiling occurs when the vapor pressure of the liquid is equal to the external pressure. So in this case, at boiling point, the vapor pressure of the liquid $\ce{H2O}$ is equal to the external pressure of gaseous $\ce{H2O}$. Thus, the system is in a dynamic equilibrium, and the rate of vaporization is equal to the rate of condensation.
This implies that the amount of liquid water will not change over time. In particular, it will not all “boil off” as steam. Is this the right conclusion? It certainly seems a little absurd.

How fast is the water heated?  If it is heated quickly, the vapor pressure of the water in the pan could reach a point above the pressure of the vapor, a nonequilibrium situation.  How large is the volume of water compared to the volume of the system?  If it is significant, the system could boil for some time before equilibrium is reestablished.  It seems that you are asking whether or not boiling is possible in a liquid that is in equilibrium with its vapor.

i think the statement "(no other species like O2 and N2 are present)" will create side problems

@brinnb Jon Custer below seems to indicate the rate of heating does not come into play. That's not what I was asking - I believe boiling is possible in a liquid that is in equilibrium with its vapor; this can be easily achieved in a closed container.

@ADG Could you explain what side problems this will cause?

Thank you all for your answers and comments - some of the concepts discussed are new to me, so I will need to take some time to learn them before selecting an answer. Thanks again!

This conclusion assumes that the enthalpy of the water remains entirely constant, which is not the case. Given that the water is being uniformly heated, but no end to the heating was specified, the enthalpy of the water will continue to increase, and thus force more and more of the liquid water into vapor form. Additionally, as you specified a very large surrounding, any water that enters the gaseous form would not largely impact the air pressure of the environment, so there would be few things acting to maintain some kind of equilibrium here.

You have a closed system, consisting of a certain volume and containing a certain amount of water.  At a constant temperature it will find some equilibrium consisting of water vapor, liquid water, and ice (depending on temperature -- I'll ignore the ice for now).  As you heat the pan, you are heating the entire system and thus changing the balance of water vapor to liquid water. Where that liquid water actually resides is a problem for the water to figure out - it does not have to reside long term in the pan.  Eventually, as you keep heating, you will get to a temperature of the system where the liquid will be thermodynamically unstable, and the enclosure will contain water vapor only.
Remember, for thermodynamic equilibrium, how fast you heat/cool does not enter into it - those are kinetic questions.

Jon, thank you for your answer. However, I wanted to convey a surrounding that is large compared to the pan. So what happens in the pan would have no bearing on the surrounding. I'll edit the question to make this clear.

If the water vapor you start out with is at 1 atm then that means it has to be at 100 C. It also means that the walls of the box have to be either extremely well insulated or externally heated. Otherwise, condensation will occur on the walls and the gas will cool down. Assuming that the box and pan are somehow held at 100 C then the system will be in equilibrium and nothing will happen (i.e. no boiling). However, if the pan heats up beyond 100 C or the box walls cool off then you will see boiling. In the former case, the water vapor pressure will increase to maintain equilibrium. In the latter case, the water will boil just enough to compensate for the vapor lost by condensation.

Thanks, but a box is not what I had in mind - I wanted to convey a very large atmosphere that is too big to be affected by what happens in the pan.

I think the argument I gave mostly holds regardless of the box size. If the box is very large, the water vapor still has to be at 100 C if it is at 1 atm. If you want the water in the pan to boil then you have to heat it above 100 C. That heat will be transferred to the gas phase, heating it only infinitesimally since the volume is so large. If there is any heat loss from the enclosure then the water vapor that was added from the pan will eventually condense back out as the temperature decreases back to 100 C


text:
Why is it not possible to determine the rate constant this way?
The following equation can be used to determine the rate:
$Rate = \frac{-\Delta [A]}{\Delta t}$
This equation can also be used:
$Rate = k[A]$
By simple substitution, $\frac{-\Delta [A]}{\Delta t} = k[A]$ should be true. Therefore, $k$ can be represented as:
$\frac{-\Delta [A]}{[A]\Delta t}$
However, calculating the rate constant for a given reaction is obviously not that simple. Why does this equation not work then, if the mathematics of it check out? I have tried using this equation for some problems, and as I expected, they do not yield the correct answer as opposed to using the integrated rate equation.

Maybe it has something to do with the fact that you cannot measure $\Delta [A]$ effectively. So we integrate it so that we can put the easily measurable non $\Delta $ values.

The average rate $\Delta[A]/\Delta t$ over time interval $\Delta t$ is only approximately equal to the true instantaneous rate $\rm d[A]/dt$ at time $t$. The difference between the two can be significant if the instantaneous rate changes over the chosen time interval.
Here's an analogy: suppose you're driving your car on the highway at a steady 55 mph. Your average and instantaneous rates of travel will be much the same even over a long time interval. On the other hand, if you're driving through a city with a lot of starts and stops at intersections, your instantaneous rate of travel won't match your average rate of travel unless you're looking at averages over very short time intervals.
To see how far off we'll be if we don't use a small time interval, let's compare the average and instantaneous rates for your reaction with time intervals of 0.1, 0.5, and 1.0 s:

The larger the time interval, the larger the difference between $\Delta[A]/\Delta t$ and $\rm d[A]/dt$ and the larger the error in the rate constants estimated using your method.
So why not just use very small time intervals? Well, there's a limit to how small you can make $\Delta t$ if you're computing $\Delta[A]/\Delta t$ from experimental measurements. You're taking differences between concentrations that are almost the same size. Any error in experimental measurements of $\rm [A]$ may well swamp your rate estimates as you go to smaller and smaller time intervals.

I see, so does this mean that -Δ[A]/Δt is the average rate over a period of time, and k[A] is the rate at an exact point in time?

Yes. That's right.

That rate constant '$\kappa$' is for "instantaneous rate of reaction"(irr) which you wish to calculate from "average rate of reaction"(arr)
$$\frac{d[A]}{dt}=r_{irr}=\lim_{\Delta t\to0 }r_{arr}=\lim_{\Delta t\to0 }\frac{\Delta [A]}{\Delta t}$$
So if the time duration is very small you would get a better approximation.


text:
Why does a liquids height in a Barometer depend linearly on pressure?
Here is a derivation I found relating the pressure P, gravity g, height h, and density $\rho$, of a liquid in a barometer
$$
P=\frac{mg}{A}\\
V=Ah \\
P = \frac{mgh}{V}\\
\rho=\frac{m}{V}\\
P = \rho g h
$$
However, this derivation seems misguided, since $P = \frac{F}{A}$, and P is the pressure of the gas, not of the liquid, as the derivation later relates it to.
Furthermore, it doesn't make sense to me that the height of the liquid and the pressure are necessarily linearly related. Wouldn't this depend on the compressibility of the liquid? Or rather, it seems like depending upon the properties of the liquid, it may be very difficult to change the height past a certain point, for example. In this sense the added energy from increasing the pressure goes into bonds or is stored in electrical interactions.

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].

In a barometer at equilibrium, the pressure of the mercury equals the pressure of the air (at the air/mercury interface).
The compressibility of mercury in principle does make it non-linear.  The density would be 3.8 parts per million higher at the air interface than at the vacuum interface.
Certainly there are other greater sources of error: the vacuum isn't really a vacuum but contains mercury vapor, density varies with temperature, vapor pressure varies with temperature, the size of the tube varies with temperature, there are surface tension effects.
Just like any experimental technique, you need to evaluate sources of error and report a measurement with appropriate uncertainty.

Why does a liquids height in a Barometer depend linearly on pressure?

Since: $$P=\frac FA=\frac {mg}{A}=\frac {\rho Vg}A=\rho g h$$

However, this derivation seems misguided, since P=F/A, and P is the pressure of the gas, not of the liquid, as the derivation later relates it to.

No at equilibrium the pressure of both will be same.

Furthermore, it doesn't make sense to me that the height of the liquid and the pressure are necessarily linearly related. Wouldn't this depend on the compressibility of the liquid?

Since we are assuming uniform density of liquid the mass at any place of same volume will be same, so if we add some water mass adds, so force adds.

Or rather, it seems like depending upon the properties of the liquid, it may be very difficult to change the height past a certain point, for example. In this sense the added energy from increasing the pressure goes into bonds or is stored in electrical interactions.

Such a situation never arrives, as far as theoretical concerns are made, since we assume water as an in compressible fluid.


text:
What's the reason for discrepancy between DFT calculated image and NC-AFM-acquired one in this article?
Looking at the images in this article, I've noticed that the molecule electron density images showed as calculated via DFT look very symmetric, while those obtained by NC-AFM appear somewhat distorted. Namely, the ring at the left (the one with $\ce{OH}$ group) looks elongated, while the one at right is, on the contrary, compressed in horizontal direction, and the $\ce{OH}$ group seems to be tilted to the $\ce N$ atom. Also, the calculated image seems much more smeared, while the measured one gives some highly localized density "rods" for the bonds.
I've made a gif-combination of these two images so that the difference was easier to see:

Here the B is DFT-calculated version, while F is one measured by NC-AFM.
I can suppose one or several of the following might be among the reasons for this:

The substrate of $\ce{Cu(111)}$ wasn't taken into account when calculating(?), while in the measurement it appears to distort the molecule
DFT, being an approximate method, gives too simplified results
Smear vs thin rods difference might because in the DFT case the picture is of electron density while in AFM case the image shows AFM frequency shifts, which is not necessarily the same.
The probe of the microscope distorts the molecule

What are the true reasons for the differences?

This is not really my field, but I've certainly followed it. I'd guess some combination of all of your reasons, plus one more.

I think usually the substrate is considered, but imperfections and defects in the surface may not be properly considered.
Perhaps, although I think this is probably less important than other causes.
Exactly. DFT-computed electron density is not exactly what the AFM measures.
The probe will certainly have different interactions with the molecule, and could be subtly asymmetric.

I'd guess the most likely culprit is different.
In AFM (and all scanned-probe methods) the substrate is rarely perpendicular to the probe. Some minor image correction is performed to "line-fit" the substrate.
Consider if you take a photo of a page on your desk. The page is probably slightly rotated in the plane of the desk, but the camera is almost never exactly parallel to the desk. This results in "keystoning".
The AFM software corrects for this effect, and users can fit a bit after data-collection, but it can cause subtle distortions.

1) Most definitely. Adsorbed molecules often are very distorted, up to chemical reaction with support, especially when reactive surface is used. Despite common misconception, even noble metal surfaces are quite reactive.
2) DFT, while being indeed inexact (as any calclation with limited size of basis set), catches major features of electronic density very well.
3) AFM stands for atomic force microscopy, which is not connected to electronic density directly. Now, naphthalene molecule features large $\pi$-system, which is very polarizable. Since nonvalent interatomic forces beyond electrostatic interactions usually increases with polarizability of interacting particles, I would expect $\pi$-systems  to be highlighted on the NC-AFM images.
4) I'm not an expert on atomic-resolved microscopy, but I never read about this possibility in the articles I worked with, though the works I usually works with used STM. and SEM mostly.


text:
London dispersion forces and distance relationship
Why is the London dispersion force inversely related to distance but not to the second power but the sixth power? Why is there not an inverse square relationship as there is for Coulombic forces?
Is there something non-Coulombic about London dispersion forces?

There is a related question over at Physics.SE with some interesting discussion.

One somewhat simplified way of looking at it is that the London dispersion forces are a dipole-dipole interactions.  The interaction between two dipoles depends upon the relative orientation of the two dipoles.  Some positions are attractive and some are replusive, but averaged over all the positions (and considering that lower energy positions are favored by Boltzmann statistics), the interaction is attractive with potential energy having an 1/$r^6$ dependence.
See for example Stephan Franzen's lecture
and especially the corporate Intermolecular Van Der Waals page
as well as Fritz London's The General Theory of Molecular Forces Trans. Faraday Soc., 1937, vol. 33, pages 8-26.
for more on the 1/$r^6$ dependence of dipoles.
However, a proper quantum mechanical treatment shows that London dispersion interaction is not always proportional to 1/$r^6$
For example, between two hydrogen atoms in 1s states, the potential energy is proportional to 1/$r^6$ but
between a hydrogen atom in a 1s state and a hydrogen atom in a 2p state, energy is proportional to 1/$r^3$
See: Complement $C_{XI}$ of Quantum Mechanics vol. 2 by Cohen-Tannoudji et al.
Also, in 1948 Verwey and Overbeek demonstrated experimentally that the London dispersion interaction is even weaker than 1/$r^6$ at long distance (say hundreds of Angstroms or more).
Casimir and Polder soon thereafter explained with quantum electrodynamics (QED) that the dependence should be 1/$r^7$ at relatively long distances.
For a nice historical overview and QED prespective see Some QED vacuum effects: van der waals forces in The Quantum Vacuum by Peter W. Milonni.

@DavePhD: "but averaged over all the positions (and considering that lower energy positions are favored by Boltzmann statistics), the interaction is attractive" why then might one of my profs ask questions starting with "Based on vdW repulsions, there are how many different bond angles in ...)?

@Dissenter I think he/she is referring to the "12" part of the Lennard Jones 6-12 potential, which has nothing to do with dipole-dipole interaction, it is due to the Pauli exclusion principle (like electron degeneracy pressure).  So once the molecules are close enough that electron wavefunctions overlap, the 1/r^6 attractive transitions to 1/r^12 repulsive

London dispersion forces (or van der Waals) can be rationalized (classically) as a relationship between induced dipole moments.
Some electrostatic (Coulombic) relationships:

Two Point Charges: $V = \frac{q_1 q_2}{4\pi\epsilon_0 r}$ (i.e., $r^{-1}$)
Point Charge and Permanent Dipole: $V = \frac{\mu_1 q_2}{4\pi\epsilon_0 r^2}$ (i.e., $r^{-2}$)
Two Permanent Dipoles: $V = \frac{\mu_1 \mu_2}{4\pi\epsilon_0 r^3}(1 - 3 \cos^2 \theta)$ (i.e., $r^{-3}$)

The last assumes that the polar molecules are stationary with angle $\theta$ between the dipole moment vectors. If the molecules rotate then it falls off to $1/r^6$.
In general, the interaction of a $2^n$-pole (dipole, quadrupole, octupole, etc.) with $2^m$-pole is $V \propto \frac{1}{r^{n+m-1}}$.
So why the $1/r^6$ for dispersion? Here's a quick justification. Let's take the system of a permanent dipole that's creating a induced dipole moment in another molecule:
$\mu_2^* = \alpha_2 E$
where $\alpha_2$ is the polarizability in the second molecule and E is the electric field generated by the polar molecule:
$E = \mu \sqrt{1 + 3\cos^2 \theta} / 4\pi\epsilon_0r^3$
The potential energy is given by:
$V = \alpha_2 E^2 / 2$
$V \propto r^{-6}$
Since there are multiple dispersion interactions (two dipole moments averaged over all rotations, permanent dipole-induced dipole, hydrogen bonding, and induced dipole-induced dipole moment) the general potential energy is classically approximated as:
$V = \frac{C}{r^6}$

what is "$\theta$" here? the third result seems wrong.

That's the general result regardless of the angle theta between two dipole vectors.

As to the difference in constants with induced dipoles, I checked these with Atkins' textbook before posting. IIRC, it depends on the different ways of defining polarizability. I'll check with a colleague who breathes electrostatics.

@GeoffHutchison For permanent dipole- induced dipole $E = \mu\sqrt {1 + 3 \cos^2 \theta}/4\pi\epsilon_0 r^3$ and $V=-\alpha E^2/2$ Then according to London 1937 (explaining Debye's theory), considering all orientations, the $\cos^2 \theta$ term is 1/3.

@DavePhD if you see the still in my answer and see the surroundings, the illustration 27 says  $E=kp/r^3 . \sqrt{(1+3\cos^2\theta)}$

@ADG, yes that is the equation for the electric field of a permanent dipole

@DavePhD Thanks, it's been a while since I haven't taught electrostatics in a while.

@DavePhD I agree with your point (and it's a simpler explanation), but I'm wondering where the 1/4 difference comes in.

@GeoffHutchison one factor of 2 comes from an induced dipole having half the potential energy of a permanent dipole: http://physics.stackexchange.com/questions/38343/why-is-there-a-factor-of-1-2-in-the-interaction-energy-of-an-induced-dipole-with?s=1|1.4295

and I don't know how you got the factor of 2 in numerator of your original V equation, but maybe that explains the other factor of 2

@GeoffHutchinson and DavePhD I know this is a bit late but I remember working these out ages ago, and I think the term comes from the equipartition of the averaging of orientations. I may be mistaken and it might be something else but I'll dig through my old notes and have a look :)

So that the $ V \propto f(\theta)$ and then $\langle V\rangle \propto \langle \bar f\rangle $ where $\bar f=p\cdot f(\theta)$ and $p$ is a conformational weighting factor. You then put the expression $\langle \bar f\rangle $ in a Taylor expansion of $e^{-x}$ and you get something like $\langle f-\frac 1{r^3}f^2\rangle $ then use equipartition theorem on the $\langle f^2\rangle$ term, which is where the 2/3 comes in.

The final step is something along the lines of $\langle V\rangle =-\frac 1{r^3}\langle \bar f\rangle$, which gives you the desired expression. (also note above that the $\langle f\rangle =0$.


text:
Can we make a rusting battery
Since in rusting (oxidation) of Iron, transfer of 4 electrons takes place is it possible to use this reaction under catalytic conditions to create a simple electric cell, even if it just gave a millivolt of potential.
$\ce {Fe + H2O -> FeOH3 + H2}$
Considering that we have complete control over the surroundings, like volume expansion, disposing hydrogen and other such factors.

Yeah in theory is possible, but the efficiency would be se low that it will be practicaly useless :x

@Babounet, I know the efficiency is less but I require it to perform a very small task, that's why. and if we catalyse the reaction it can give good results.

Sure! A nickel-iron battery uses the oxidation of iron at the negative plate to produce about half the voltage. Though invented by Jungner, it's often called an Edison battery. See http://en.wikipedia.org/wiki/Nickel%E2%80%93iron_battery.
There are very low power semiconductor devices which could be powered by an iron battery. You could make three or four small cells from iron nails and a more electronegative element, e.g. lead, to power a digital watch; even without optimization to remove layers of hydrogen and rust, it should operate for months.
For some other low power ideas, see https://www.enocean.com/en/energy-harvesting/.


text:
Looking for a chemical to calibrate a temperature sensor
The best method for calibration would be a bath filled with a chemical at its melting point with a solid/liquid composition.
The constraints being that it must be hydrophobic and relatively non-corrosive and non-toxic with a melting point between (say) 20 °C and 60 °C. Needless to say, but I’ll say it anyway, the chemical must have a very precise melting point. Any ideas?

The standard point recommended in that range by the CIPM's ITS-90 calibration standard is the melting point of gallium at 29.7646 °C. Gallium is not known to be particularly toxic, but it can embrittle metals so is ideally used/stored in glass or plastic containers. It's used as a standard because it is relatively easy to make very pure and is stable in storage. Also, by picking an ITS-90 standard, you can ensure your thermometer is calibrated to the same standard as many other high precision thermometers. N.B. ITS-90 specifies specifically the melting point of gallium, not the freezing point, unlike some of the other calibration standards.

In the old days, you'd look it up in the CRC organic chemical tables, but now you can go to a site such as http://www.colby.edu/chemistry/cmp/cmp.html and enter the desired m.p. and constituents. Here is the result for a search for 50C (+-5C) m.p. without O, N and S so they're likely hydrophobic.
P-DICHLOROBENZENE
PENTAMETHYLBENZENE
1-METHYL-4-PHENYLBENZENE
1,2,3-TRICHLOROBENZENE
1-ETHYL-4-PHENYLBENZENE
1,2-DIPHENYLETHANE
1,1-DIPHENYL-1-PROPENE
CYCLOTETRADECANE
2-BROMONAPHTHALENE
1,2,3,4-TETRACHLOROBENZENE
1,2,3,5-TETRACHLOROBENZENE
P-CHLOROIODOBENZENE
Taking P-DICHLOROBENZENE as an example, you would find it melts at 53.5 C and is mildly toxic, but is used in mothballs, so is a regularly available consumer item not requiring HM permitting to buy. See http://www.epa.gov/airtoxics/hlthef/dich-ben.html and http://en.wikipedia.org/wiki/1,4-Dichlorobenzene for more information.

I like both answers so far and have had some fun looking up possible compounds to use in the colby DB.  Menthol seems to qualify and would probably smell better than dichlorobenzene, although it is slightly less hydrophobic.  p-cresol and 1-methylcyclopentanol also seem like pretty good candidates if you can get a hold of them.


text:
Unusual color from leaking battery. What happened?
I was removing old batteries from a camera the other day. One of the batteries had leaked something very bright blue. I have seen batteries leak before and typically the colors look a bit like a corroded penny - pale green to white. This time it was a beautiful bright blue; what was it? I initially thought this could have been the metal making the connection was corroding as opposed to the color of the contents of the battery. However, I am not so sure about that.

Note: I believe they were double or triple A batteries and not LiION.

It is, as you suspected, likely a copper corrosion product. Copper compounds are usually blue or green, and are used as pigments. Malachite, http://en.wikipedia.org/wiki/Malachite, is one of the likely corrosion products of an alkaline battery and copper terminals.
BTW, you may need to rinse the battery compartment and use a file to remove the corrosion from the contacts to make connection.


text:
Why does reduced mass help when talking about two body problems?
Firstly I would like to say that despite the fact that you may be thinking that this is a physics question and should probably be asked on that part of the website, I came across reduced mass for the first time in Chemistry - physical and inorganic - and it is used a lot in Chemistry so I thought it suitable for this part of the site.
With that said, I will now elaborate on my question. I hear people say that reduced mass ,$\mu$, is used for two body problems, but what sort of problems? Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?
I can follow the derivation (which I shall go through shortly) in a mathematical sense but I can seem to be able to grasp the ideas behind why the steps were taken.
Derivation:
Step [1] - From Newton's laws of motion: $$F_{12}=-F_{21}$$ It follows that: $$m_1a_1=-m_2a_2$$
Step [2] - Simple rearrangement of the result from step 1 to make $a_1$ the subject (it seems a reasonable thing to do but I don't know why it's done). $$a_1=-\frac{m_2}{m_1}a_2$$
Step [3] - The relative acceleration, $a_{rel}$, can be found by subtracting the acceleration of one body from the other (I get how this is done and get that it would give the relative acceleration BUT I don't know why you would need the relative acceleration). $$a_{rel}=a_1-a_2=(1+\frac{m_1}{m_2})a_1=\frac{m_2+m_1}{m_1m_2}m_1a_1=\frac{m_2+m_1}{m_1m_2}F_{12}$$
Step [4] - According to Wikipedia the result from the last step can be used thus (I have no idea what's going on in this step):
$$\frac{m_2+m_1}{m_1m_2}F_{12}=\frac{F_{12}}{\mu}$$ Thus: $$\mu=\frac{m_1m_2}{m_2+m_1}$$
Again, I can do the maths but my problem here is conceptual I believe. I have included the derivation in my question because I believe that it's possible that if you were able to shed some light on why particular steps were taken that this may help.

Reduced mass is purely physical concept, and since you are interested in the concept itself and not its application in some area of chemistry, the question should be asked on Physics.SE.

I am very much interested in its applications. As I mentioned in the second paragraph. I will make an edit to make this clearer

but what sort of problems?

Where two bodies move in the influence of their mutual forces only(that act in pairs and net force on system is zero, so that COM(Centre Of Mass) remains at rest), no external forces!
Actually:
"When two bodies are moving in translational,rotational or vibrational motion only under their mutual interaction, then for simplified analysis, we can consider one body at rest and analyze the motion of other body wrt first body and change the inertial mass of moving body to a new value called reduced mass given as."

Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?

It is such that we are treating the motional parameters of one fixed(i.e. at rest and at rest forever(until something weird happens)). Suppose you are travelling in a train, when you see an oppositely coming train, it seems faster since when you observe, in your reference frame(i.e., wrt you) the speed of the train becomes $\vec v_{other-train}-\vec v_{your-train}$ Since speed of your train has opposite sign as the speed of other train, it rather becomes: $v_1-(-v_2)=v_1+v_2$ and thus speeds add up and it looks fast. Similarly in case of moving in same direction, it appears slow as speeds subtract.
This often helps because sometimes problems are solved easily since one equation is always reduced.
Vibrational Example
suppose two bodies are connected by spring and vibrating, then both are constantly moving, If they are following SHM (Simple Harmonic Motion), then to calculate time period we need to account both of them, but it could be done easily if we treat one at rest and look at the other body while sitting on the body we wish to have at rest. That turns out to be:
$$T=2\pi\sqrt{\frac{\mu}k}=2\pi\sqrt{\frac{m_1m_2}{k(m_1+m_2)}}$$
If we were doing it the traditional way:
$$k_1=kl/l_1=kl(m_1+m_2)/(m_2l)\\k_1=k(m_1+m_2)/m_2\\T=2\pi\sqrt{m_1/k_1}=2\pi\sqrt{m_1m_2/(k(m_1+m_2))}$$
Translational example
Try to fin the separation velocity between two planets at a distance $r$ when their distance reduces to $r/2$ using both methods.
Rotational Motion Example
Try to find the total kinetic energy of two particles at a distance l rotating about their COM due to their mutual gravitational attraction without any external force at angular speed $\omega$ by both methods.

You're transforming a problem in terms of two particles with individual masses and accelerations into a problem with a single fictitious particle. That simplifies the problem and lets you separate the relative motions and interactions of the particles from overall translation of their center of mass.
In step 3, you need the relative acceleration $a_{rel}$ because that's the acceleration the single fictitious particle has.
In step 4, you're calculating the mass $\mu$ of the fictitious particle.
Before, you had $F_{12} = m_1 a_1$ and $F_{21} = m_2 a_2$ (two forces, two masses, two accelerations). Now you've got $F_{12} = \mu \ a_{rel}$ (one force, one mass, one acceleration). The price is that we've thrown away information about the overall translation of the system through space; we're only looking at the relative motions of the two particles.
The coordinates of the fictitious particle are relative or internal coordinates $$x = x_2-x_1,\quad y = y_2-y_1, \quad z = z_2-z_1$$
This is useful because the potential energy $V$ of the interacting particles depends only on these relative coordinates. Making use of reduced masses and relative velocities and coordinates cleanly separates internal motion from translation of the system as a whole. The classical energy of the two body system will be
$$E = \left[\frac{1}{2\mu}(p_x^2+p_y^2+p_z^2)+V(x,y,z)\right] + \left[\frac{1}{2M}(p_X^2+p_Y^2+p_Z^2)\right]$$
where $p_i$ means momentum for motion along the $i$-th coordinate. The first term in brackets is the energy of our fictitious particle with mass $\mu$. The second term is the energy of a second fictitious particle with total mass $M=m_1+m_2$ and coordinates  $X$, $Y$, and $Z$ located at the center of mass of the system (e. g. $X = (m_1x_1 + m_2x_2)/M$).
In chemistry, we don't care about the second term. For example, in an atom, we'll assume that the nucleus is clamped in place so we can just look at interactions within the atom.


text:
What does CH4 form when dissolved in aqueous solution?
So I understand $\ce{CH4}$ has a low solubility, but when it does dissolve, what does it form? E.g. $\ce{HCl_{(g)} + H2O -> H3O+_{(aq)} + Cl^{-}_{(aq)}}$. So what does $\ce{CH4 + H2O}$ form?

Why do you think it would react with water? Generally gases with low solubility in water don't react with it - methane is very, very weak acid and base - it's like opposite of HCl

It's not a chemical reaction per se, other than minor hydrogen bonding. It's more like games with ball bearings that fit into depressions when tilted just right.

Methane dissolved in water is simply methane. It does not react with the water under normal conditions.

There will of course be hydrogen bonding.

Hydrogen bonding is unlikely to be significant as the C-H bond in methane is virtually apolar. There's very little H-bonding even for the ammonium ion in water.

I think you are confusing dissociation with dissolution. Bonds do not need to be broken in order to dissolve a substance. The reason so many covalent compounds do disassociate in water is that water is a highly polar solvent and therefore preferentially solvates ions (provides the driving force for the disassociation).
Methane is pratically insoluble in water, but it will dissolve at low enough concentration to give a solution of methane in water, with only van der Waals forces between the methane and water molecules.

Solvent-solvent interactions must be disrupted in the process of dissolution.

Yes. The solubility of methane in water is 22.7 mg per litre; there are compounds which are actually insoluble for that reason.

A clathrate compound is like a cage. It relies on lattice spacing much more than molecular attraction to stick together; see http://en.wikipedia.org/wiki/Clathrate_compound. Methane-water ice clathrate is comparatively stable almost to the melting point of ice, and there are huge amounts under the oceans and in Arctic permafrost;see http://en.wikipedia.org/wiki/Methane_clathrate.

I think you should make some refercence to the question - clathrates aren't typical solution.

clathrates are not solutions of methane in water

@Mithoron Methane clatrate is a terminal point, it is an idealised structure of a solution of methane in water at highest possible concentration, and it is X-Rayable, so a direct evidence of structure is available. Besides, OP have not clarified directly if he is talking about liquid or solid water =) (OK, it's a bit of stretch, but still).

Yeah, so I say "typical" :D


text:
Can a network covalent solid be amorphous?
All the definitions I'm seeing give examples of crystalline solids like diamond or boron nitride. But I'm wondering if it also includes amorphous macromolecules like vulcanized rubber. I'm thinking "network" probably == "ordered array" but not sure if that's correct.

The fluid behavior of glass lends it to be loosely defined as a amorphous solid as well, many geologists would agree as a similar phenomenon is measurable with mountains and large carbon bodies called the Deborah effect. Perhaps more solids behave as fluids over time than originally thought

The most obvious example is amorphous carbon. As mentioned in a comment, another good example is glass (although, whether it's truly "solid" past the glass transition temperature).
Really, a "network" simply means "solid with covalent bonding" and of course it's easy to imagine many possible amorphous network solids.
Vulcanized rubber might qualify too, but it would depend on whether there were many non-interlinked macromolecules. Usually macromolecules would be considered molecular solids due to the high number of non-bonded packing interactions. If the amount of cross-linking is high, classifying as a network solid makes sense.

The best example I have is glass.


text:
Is there an isotope effect for diffusion of an ion through a crystal lattice?
Consider the diffusion of natural abundance $\ce{Li^{+}}$ and $\ce{{}^{6}Li^{+}}$ through a crystalline matrix, say a spinel such as $\ce{LiMn2O4}$.  The two lithium ions will have the same ionic radius and electric charge.  They will differ only in atomic weight.  Is there any reason to believe that the lighter isotope, $\ce{{}^{6}Li^{+}}$, will diffuse through the crystal any faster than the natural abundance isotope?  When I consider the forces acting upon the ions in the crystal, I come up with coloumbic, dispersion and hard shell repulsive forces.  I cannot see how any of those would change by simply changing the mass of the ion.  Am I missing anything?

Consider diffusion to be an attempt frequency times a probability of making the jump over the barrier.  The barrier likely doesn't change (as you note the forces do not). But the attempt frequency will since the mass of the atom rattling around in a potential minimum does vary.  Also, given you can separate isotopes through liquid or gas diffusion, what is special about solid diffusion to prevent it from happening?

This makes sense. I agree that there should be nothing special about isotope separation in a solid matrix versus liquid or gas...except maybe the magnitude of the separation. Perhaps what I really wanted to ask but didn't realize at the time is how one might go about quantifying the difference in diffusion rates for the two isotopes on a theoretical basis. Probably a very difficult problem though

Focus on the attempt frequency: the atom is to first order a simple harmonic oscillator in the potential well - so it will vary as 1/sqrt(mass)...

One paper to get you started might be N.L. Peterson et al., J. Phys C, Vol 6, 2020-2039 (1973) on "Isotope effect for silver diffusion in AgBr and AgCl"

$F = ma$, so a less massive isotope should move faster, depending on the forces moving it.

Thanks, I guess I was wondering if a situation similar to a terminal drift velocity would be setup in a crystal in the same way it happens in a liquid.

Or, is there any reason to believe that any of the forces acting on the ion might decrease when lithium-6 is involved?


text:
Why is the rate of disappearance negative?
Rate of disappearance is given as $-\frac{\Delta [A]}{\Delta t}$ where $\ce{A}$ is a reactant. However, using this formula, the rate of disappearance cannot be negative.
$\Delta [A]$ will be negative, as $[A]$ will be lower at a later time, since it is being used up in the reaction. Then, $[A]_{\text{final}} - [A]_{\text{initial}}$ will be negative. Therefore, the numerator in $-\frac{\Delta [A]}{\Delta t}$ will be negative.
$\Delta t$ will be positive because final time minus initial time will be positive.
This means that $-\frac{\Delta [A]}{\Delta t}$ will evaluate to $(-)\frac{(-)}{(+)} = (-) \cdot (-) =(+)$
However, we still write the rate of disappearance as a negative number. Also, if you think about it, a negative rate of disappearance is essentially a positive rate of appearance. The reactants disappear at a positive rate, so why isn't the rate of disappearance positive?

Reaction rates are generally by convention given based on the formation of the product, and thus reaction rates are positive. So, for the reaction:
$$\ce{A->B}$$
$$\text{Rate} = \frac{\Delta[\ce{B}]}{\Delta t}$$
To ensure that you get a positive reaction rate, the rate of disappearance of reactant has a negative sign:
$$\text{Rate} = -\frac{\Delta[\ce{A}]}{\Delta t}=\frac{\Delta[\ce{B}]}{\Delta t}$$

When you say "rate of disappearance" you're announcing that the concentration is going down. If you wrote a negative number for the rate of disappearance, then, it's a double negative---you'd be saying that the concentration would be going up!
As you've noticed, keeping track of the signs when talking about rates of reaction is inconvenient. It would be much simpler if we defined a single number for the rate of reaction, regardless of whether we were looking at reactants or products.
We can do this by
a) flipping the sign on rates for reactants, so that the rate of reaction will always be a positive number, and b) scaling all rates by their stoichiometric coefficients.
For example, if you have a balanced equation for the reaction $$a \mathrm{A} +  b \mathrm{B} \rightarrow c \mathrm{C} + d \mathrm{D}$$ the rate of the reaction $r$ is defined
$$ r = -\frac{1}{a}\frac{\mathrm{d[A]}}{\mathrm{d}t} =  -\frac{1}{b}\frac{\mathrm{d[B]}}{\mathrm{d}t} = \frac{1}{c}\frac{\mathrm{d[C]}}{\mathrm{d}t} = \frac{1}{d}\frac{\mathrm{d[D]}}{\mathrm{d}t}$$
This lets us compute the rate of reaction from whatever concentration change is easiest to measure.


text:
how to determine range potential of Pb in cyclic voltammetry?
standart potential of $\ce{Pb}$ is -0.13 V. how we determine an optimum potential of $\ce{Pb}$ sample? and how we choose the range potential?

Welcome to chemistry.se! This seems to be a homework question according to our policy, please share your thoughts and attempts towards the solution. For formatting help visit the [help] and for more information take the [tour].|| I admit, that I do not have any insight into cyclic voltammetry, so I can only give you a warm welcome :D

Could you clarify your question a bit more please?  It looks like you want to use a Pb working electrode for a CV experiment, yes?  And, what kind of solution will you be using for the experiment?  What is the goal of the experiment?  More info would be helpful.


text:
What causes the colour when conducting flame tests on solid salts?
The normal technique for flame tests is to dip a clean nichrome or platinum wire into a solution of the relevant salt, and observe the resulting flame colour when inserting the wire into a non-luminous Bunsen flame.
But what is the species undergoing the electron transition that emits the light of characteristic frequency?
One explanation suggests vaporisation and then atomisation of the sample, with the light emission occurring from the excited atom, rather than ion. Another suggests that it’s the ions undergoing the electronic transitions. The Wikipedia article is somewhat vague on the question.
Now, I was doing some flame tests for a student, and in the absence of an available nichrome or platinum wire I simply took a pinch of the solid salts ($\ce{NaCl}$ and $\ce{KCl}$ respectively), and threw it into a non-luminous Bunsen flame.
The characteristic yellow and lilac flame colours were instantly and repeatably observed.
My question is this:
Given that the outer regions of a Bunsen flame are at less than $1000\ \mathrm{^\circ C}$, and that the boiling points of $\ce{NaCl}$ and $\ce{KCl}$ are $1413\ \mathrm{^\circ C}$ and $1420\ \mathrm{^\circ C}$ respectively (according to Wikipedia), it seems unlikely that the instantly generated characteristic colours originate from individual ions or individual atoms.
So what is the species exhibiting the electron transitions? Is it the metal ions still within the crystal lattice?
Edit: I should emphasise that the flame colour was observed in the outer regions of the flame, not the super-hot tip of the inner cone. So $1000\ \mathrm{^\circ C}$, not $1500\ \mathrm{^\circ C}$ by any stretch of the imagination. And I don’t at all buy the notion that the colours were simply the glow of hot solid particles emitting black body radiation. The colours were the characteristic sodium yellow and potassium lilac, instantly recognisable by anyone that’s familiar with simple tests like these.

Just because something doesn't boil doesn't mean that many atoms or ions from it won't get into the flame. If the emission is intense (like the bright yellow lines in the spectrum of sodium vapour) then you don't need many ions to be in the vapour phase to see it. For example, Fluorescent tubes are driven by emissions from mercury vapour and start working well at room temperature, >300°C below the metal's boiling point.

There is a very detailed discussion of what cause the emission lines in flames with $\ce{NaCl}$ in The Origin of the Spectra (ironically by P. Darwin Foote et al.) 1922, starting around page 165.
They are saying that for $\ce{NaCl}$ most of the transitions are from atomic $\ce{Na}$, the $\ce{NaCl}$ having vaporized, and Na+ then captured an electron, and a small fraction of the transitions are from the act itself of Na+ capturing the electron.
They calculate that a Na atom emits 2000 D-line quanta per second.
They also calculate the fraction of Na atoms with electrons in the $3p$ state is $\ce{10^{-11}}$ at 1000K, rising to $\ce{10^{-7}}$ at 1500K. (Note: the book says "$2p$", but $2p$ in Fowler notation is $3p$ in modern notation)
Also, they say the vapor pressure of $\ce{NaCl}$ in a Bunsen flame is on the order of $\ce{10^{-6} \ mmHg}$.
So the color is from vaporized $\ce{NaCl}$ (vapor phase $\ce{NaCl}$ being molecular), that has dissociated and formed atomic $\ce{Na}$.
Also in The Absorption of Light by Flames Containing Sodium Phys. Rev. 38, 699 (emphasis added):

The dissociation probably occurs in two or more steps. Possibly the
molecules break up into sodium and chlorine ions which then lose their
charges becoming uncharged atoms. There may also be one or more
intermediate steps in the recombination.  But certainly the sodium
giving the light which we are studying consists of uncharged atoms, as
is evident from its spectrum. Evidence that there are comparatively
few charged sodium atoms in the flame is given by a study of the
conductivity of flames. Wilson, (p. 85) for example, estimates that
only 1.6 percent of the sodium in a fame is ionized even when the
concentration is small.
It also seems reasonable to assume that at the
edge of the flame there is more complete dissociation of the salt than
in the center, and that this is the cause of the difference between
the concentration ratios obtained with light from the center and from
the edges. A fairly definite proof that there is more dissociation at
the edges is given by the appearances of a fame into which copper
chloride is being sprayed. Such a fame gives almost no green color
except at the edges, where it is a bright green. This is the more
noticeable if there is a small amount of sodium impurity in the copper
chloride. The center of the flame is then yellow with a distinct border
of green.
The greater dissociation at the surface is no doubt in some way due to the greater amount of air there, but whether it is caused by a higher temperature
or by a difference in chemical action can not at present be stated.

Bunsen flame is up to 1500 C hot depending on the region.

it seems unlikely that the instantly generated characteristic colours originate from individual ions or individual atoms.

It is, however, more or less generally accepted. More precisely, in some flames the color is derived from emission of excited atoms, while in other it is molecules. alkali metal colors are believed to be derived from atomic emission, while copper bluer, strontium red and barium green are emitted by monochloride molecules (well, barium green and strontium red can be, less efficiently, produced by other molecules, but professional pyrotechnic compositions are usually manufactured with chlorine donors).
The mechanism of excitation, however, is a bit less clear, but it is sometimes believed to be chemical and not thermal in nature.
Solid particles in flame usually produce black body radiation, which may be, depending on temperature, red, orange, yellow and whitish (at extremely high temperatures, well above any that a flame can produce it becomes bluish)

Thanks for the reply.  However, I'm troubled by "more or less generally accepted" and "believed to be".  See my edit - I'm throwing a sprinkling of solid salt into the part of the flame that's well below the boiling point.  Are free sodium and potassium atoms really credible as the source of the flame colours?

@ChrisA Yes. Emission lines discoverable through flame test are ridiculously intensive, and, given that there is no much energy released in visible light by non-sooting flames, it is quite easy to color the flame. Non-zero vapor pressure may exist even over solid, say ice slowly evaporates (sic!) in very dry air even set below freezing point. However, since emission lines to be active in flame test must be ridiculously intensive, there is only a few elements active in flame test.

@ChrisA As for excitation mechanism, it is plainly too hard to give a definitive proof, since there is a lot of flames. However, visible light quants have energies 1.6+ eV, which corresponds to 15 000 + Kelvin.


text:
What is the major side product of the reaction between 2-bromo-2-methylpropane and hydroxide ion?

I know that one of the products will be 2-methylpropan-2-ol, arising via a $\mathrm{S_N1}$ reaction. But what else is formed?
Specifically, I am looking for a side product that is gaseous.

On addition of hydroxide to t-butyl bromide there are two competing reactions, substitution and elimination. Depending on the conditions one of these reactions will predominate but both will always be present. If you are taking the $S_{N}1$ as the main reaction then the elimination ($E1$) side product will be 2-methylpropene, which is a gas at room temperatures and pressures.

I'd wager that the E1 predominates here, anyway.


text:
Why might my limelight experiment have failed and how can I improve it?
I was teaching the lime cycle to some children yesterday. I had a limestone chip (about 10g) held at the tip of a blue Bunsen flame for several minutes. It decomposed and disintegrated but only glowed very slightly, and was mostly orange with a bit of white at the tip.
It certainly gave off white light, but nowhere near the quantities required to light a theatre stage, which is what this stuff is famous for, not by a few orders of magnitude.
I tried sitting it on gauze and bringing in a second Bunsen flame from the side, but this did not improve the effect. I recall seeing quite a bright light when this was shown to me as a child.
Any theories on what went wrong? Any gotchas for doing this kind of experiment?

Perhaps your flame wasn't hot enough. Real limelights used oxyhydrogen torches, not wimpy natural gas flames.
Or perhaps your limestone chip wasn't just $\ce{CaCO3}$. If it was dolomitic limestone, there was a lot of $\ce{MgCO3}$, too, and $\ce{MgO}$ doesn't do the trick like $\ce{CaO}$ does.


text:
How is the process of creating PVC fabrics harmful for the environment?
I often hear that the production process for PVC fabrics (such as those used in Gucci and Versace handbags) is very harmful to the environment. What part of the process is harmful and can this be prevented?

Welcome to chemistry.se! If you have questions about this site, you might find answers in the [help]. Take the [tour] to get started.

I wouldn't think that the process of producing PVC fabrics is that much of a problem, for reasons:

PVC is produced by experts that have taken all the necessary precautions for all possible reductions of safety risks it may cause in the lab or in mass scale production.
The process of synthesizing PVC is not a secret. If it was too harmful (rather than its profitability), then companies would have been banned from producing such a thing.

Phew, I thought there was something serious about PVC fabrics.

There is! The problem is not the process, but VCM and phthalates.
Issue 1: VCM
With a chemical formula of $\ce{C2H3Cl}$, a boiling point of $-13.9°C$ and an approximate molecular weight of $62.5 \frac{g}{mol}$, VCM is the raw material for PVC. According to here, the death of four workers had been approved due to VCM in 1974. (Though wikipedia says it was three workers dead) It's reported that in 1975, an annual 5-billion-pounds rate of production of VCM was confirmed.
The industry had the worst ways of producing VCM, and has suffered for decades until better ways of producing it have come around. According to here, a traditional way of producing VCM was the reaction of acetylene with Hydrochloric acid, as following: $$\ce{C2H2} + \ce{HCl} ~->~ \ce{C2H3Cl}$$
It was in the presence of Mercury(II) Chloride as a catalyst! Nothing has been done to reverse the harmful effects done to environment because of this very reaction.
In the modern process, "polymerization" is known to be the most dangerous stage and the peak of the overexposure.
VCM is believed to be carcinogen. Severe symptoms have been reported due to the overexposure of VCM; such as: Hepatotoxicity (liver damage and tumor growth caused by chemicals and more precisely, angiosarcoma).
Issue 2: phthalates
VCM is the fundamental raw material for the synthesis of all kinds of PVC; rigid or soft. But Plasticizers are being added to rigid PVC to "soften it up"; so it would be applicable for the future use as "PVC fabrics". Phthalates are common plasticizers in the production of soft PVCs.

They are divided into two distinct groups, with very different applications, toxicological properties, and classification, based on the number of carbon atoms in their alcohol chain.

Phthalates aren't covalently bonded to the matrix of PVC. Therefore, the lighter ones can leach relatively very easy, usually due to heat or strong solvents.
The most common phthalate is diethylhexyl phthalate(DEHP). Studies show that Americans are exposed to DEHP. DEHP is believed to cause health complications in a variety of ages.
Oh, wait: Issue 3: PVC itself! But when it degrades
PVC is known for its strong linear polymer; but when it degrades (usually due to careless use or disposal) it degrades! Degradation is excessive reduction in a plastic's average molar weight. PVC will "microcrack" when it's degraded, into macroparticles that are a big threat to the environment. These particles "soak up" POPs and afterwards are usually ingested by living things, making it a big deal of our everyday plastic pollution exposure. (as wikipedia puts it)

What should I do now?! Panic?

Considering the advancement of technology and our anxiety to preserve our health we humans aren't sitting around to watch ourselves getting sick! I don't mean the hard-to-come-by sites that say overexposure to PVC isn't carcinogenic, but real efforts into solving the problem. News has been spread that REACH attempts to authorize DEHP and two other phthalates and ban their use in specific materials. DINP is another plasticizer, approved by the EU risk assesment, that will eventually replace the former plasticizers, hopefully.
Scientific researches for over ten years have proven that the amount of exposure to the chemicals mentioned is far from dangerous in current conditions and uses. So sit back relaxed; it's true that these dangers exist, but they're not serious threats to a healthy person, at least not yet.

If incinerated under inappropriate conditions, PVC produces highly toxic dioxines (PCDD/PCDF).


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TPE fabric vs PVC fabrics?
I am currently in charge of purchacing fabrics from asia. My personal policy is to try and stay as enviromentally freindly as possible. Your answer to the following question would be of great help and would prevent me from buying the wrong product in massive quantities.
What is a TPE fabric and how does it compare to PVC fabrics?

By compare, do you mean a comparison of their environmental effects?

TPE in TPE fabric basically refers to a mixture of polymers, whereas PVC is one polymer. But that mustn't have been what you mean, must it?

Hi, yes sorry I meant with regards to environmental effects. Thank you!

This question is impossible to answer without knowing which TPE you mean - TPE is a mixture of thermoplastic elastomers, which is a huge class of compounds.
PVC is polyvinylchloride, which is a specific compound. I found an MSDS here:
http://www.fishersci.com/ecomm/servlet/msdsproxy?productName=AC183320010&productDescription=POLY
My advice is to contact the supplier for each fabric you are considering and to ask them for the MSDS. In that you will see what the acute and long-term exposure effects are for people. "Environmental effects" is a very broad subject, and you are not very likely to find research for any proprietary blends, particulary if they don't come from the US or EU. However, sometimes MSDS list non-proprietary compounds, and you can then check the EPA website for those names to see if there are any reports on broad environmental effects.
If you are also interested in the environmental impact of the entire production chain (which you should be, since end-use is usually the least significant part of that), then you should also ask the suppliers about their manufacturing process, whether they have any "green" certifications, etc. You can look up the patents for each production process (most of the time they will be patented) to see what their raw materials are, and then check the MSDS and EPA reports for those. It is also a good idea to see if you can tour their production facilities, once you know what the potentially dangerous precursors are and how they should be handled. It is easy to temporarily hide bad working conditions, but it is harder to fake treatment and containment facilities. Of course, if there isn't good regulatory oversight, there is a very good chance that the producer will take shortcuts to reduce costs, so keep in mind that what they tell you isn't necessarily going to be the truth.


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Why are d-orbitals required/used for hypervalent molecules (where the central atom has expanded its octet)
Period 2 elements like carbon, oxygen and nitrogen cannot expand their octet because they cannot make use of d orbitals in that energy level (they don't exist). However, period 3 elements like sulfur and chlorine can because they can utilize the d orbitals with the same principal quantum number. However, the the 4s orbitals are lower in energy than the 3d orbitals (from the order of filling of electrons) so why can't they be used/ why aren't they? Moreover, why can't period 2 elements use the 3s orbital?

Interesting, I upvote it then vote to close it! It can be answered by that Q, Rob.

In reality there's no such thing as expanded octet see http://en.wikipedia.org/wiki/Orbital_hybridisation