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Why is water a dipole?
Water ($\ce{H2O}$) is a dipole. The reason why is simply because it is not symmetrical, there are more electrons on the oxygen side than on the hydrogen side, and the electronegativity of oxygen.
But why isn't $\ce{H2O}$ symmetrical like $\ce{CO2}$? Why isn't $\ce{H2O}$ non-polar like $\ce{CO2}$? Does it have anything to do with the orbitals?

Since when isn't water symmetrical? It's C2v, so there are two planes and a C2 rotational symmetrie.

@Julian We are talking about symmetry around the axis that goes through all the atoms in $\ce{CO2}$. Such an axis does not exist in $\ce{H2O}$. What I really meant was $\ce{H2O}$ is a dipole because the atoms form a V-shape.

For clarity, water exhibits a dipole.  Water is a molecule.

Yep, it has to do with the orbitals.
$\ce{CO2}$ is linear, so even though the $\ce{C-O}$ bonds have individual dipole moments, the overall dipole moment is zero as these cancel out (they point in opposite directions, as shown in the diagram below).
On the other hand, $\ce{H2O}$ is "bent", which means that the individual dipole moments of the bond are at an angle to each other. They add up to give a net dipole moment (shown with grey in the diagram).

The colors indicate electron density, red is more dense/blue is less dense. Dipole moment is from low density to high density.
OK, so why do these molecules have differing shapes? This is where orbitals come in. I'll try to explain as much as I can without going into orbitals.
Carbon has an outer shell electronic configuration as $2s^22p^2$. Out of these four electrons, two are used in $\pi$ bonds, and two in $\sigma$ bonds. If you don't know what those are, just look at it like this for now: A set of bonds between two atoms will have one and only one $\sigma$ bond, with the rest of them $\pi$ bonds. So, any single bond is made up of just a $\sigma$ bond, a double bond is made up of one $\sigma$ and one $\pi$, and a triple bond is made up of one $\sigma$ and two $\pi$ bonds. What these types of bonds actually are can be explained if you know what an orbital is.
Now, what VSEPR says is that the geometry of the molecule is only decided by the $\sigma$ bonds and lone pairs on the central atom. You count up the $\sigma$ bonds and lone pairs (lets say they add up to $x$), and decide the geometry based on that. The geometry is the most stable configuration of $x$ hybrid orbitals. In simple terms, if we took $x$ balloons and tied them together, the directions the balloons point in help us correspond to where the bonds and lone pairs lie:

In $\ce{CO2}$, we have two $\sigma$ bonds and two $\pi$ bonds (as each double bond has one of each type). Each bond takes up one electron from carbon, so we have no leftover electrons for forming any lone pair. Since we have two $\sigma$ bonds and 0 lone pairs, $x=2$, giving us the structure given by the first set of balloons, which is linear. And $\ce{CO2}$ is indeed linear:

Now let's take water. The central atom (Oxygen) has a valence configuration of $2s^22p^4$, that is, 6 electrons. In water, since we have two single bonds, we have one $\sigma$ bond each (and no $\pi$ bonds). So we have total two $\sigma$ bonds.
But this leaves us with $6-2=4$ unpaired valence electrons. These form two "lone pairs" (pairs of electrons which do not bond). With two lone pairs and two $\sigma$ bonds, $x=4$. This gives us a tetrahedral structure (third in the balloon diagram). Two of the four points in the tetrahedron are occupied by the lone pairs, and two by bonds:

(Note that the angle 104.5 is not the same as the angle in perfect tetrahedra, 109.25--this is due to the lone pairs repelling each other)
So finally, we have the following "bent" structure for water:

From the structure, as shown above, it is very easy to check if the molecule has a dipole moment.

Fundamentally, dipolar simply means that a region of the molecule has a center of positive charge and a center of negative charge. We can say that water has a dipole moment because it is not a 'balanced' molecule, like $CO_2$ is. I understand what you are really asking though.
I believe it likely has something to do with orbitals. You can look into VSEPR Theory for an explanation of why water has a bent shape. It might have to do with both reasons you mentioned, the electronegative properties, and also that oxygen just has more electrons in general (and therefore needs to put them somewhere that has a low amount of energy).
According to VSEPR, the electron's try to separate themselves out as far as possible, that is one explanation for why the molecule has a bent shape. The electrons orbiting oxygen will inevitably occupy what they call the p orbitals. These extend outward a greater distance, allowing the electrons to occupy a space that is further away from the other electrons, and therefore requires less energy to do so (because electrons repel each other).
This model seems slightly counter-intuitive though, and requires a bit of imagination and understanding of the theory itself, if not a great deal quantum physics. But, like any good model, it tends to predict molecular properties and shapes, and this is why it is taught in schools.

Another great definition of a dipole is simply a molecule whose center of mass does not equal its center of charge density.


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Does "covalent" AgCl (aq) exist and if so, how to quantify its concentration?
I am reading Ionic Equilibrium by James N. Butler.  Using silver(I) chloride in excess aqueous chloride as an example it speaks of several forms:  $\ce{AgCl (s)}$ i.e. the ionic crystalline salt, $\ce{Ag+(aq)}$ and $\ce{Cl^{-}(aq)}$ the aqueous ions from dissociation, and finally $\ce{AgCl(aq)}$, which Butler purports to be "dissolved covalent silver chloride molecules as distinct from solid salt or dissolved ions."
Does this "molecule" of $\ce{AgCl}$ really exist in solution?  If so, is there a way to quantify the concentrations of this covalent species?  Equilibrium solubility product constants seem to give the concentration of the aqueous ions only.

For your second question: If you can determine the quantity of ions present in solution, you should easily determine the other quantity, i.e. the "covalent dissolved" AgCl.

Some amount of $[\ce{AgCl(H2O)}]$ particles should exist in solution from pure thermodynamic point of view. See, $\ce{Ag+}$ ion can coordinate two ligands, and in solution both $\ce{Cl}$ and $\ce{H2O}$ are present. However, it should be really hard to detect trace amounts of this particle, as $\ce{AgCl}$ is almost insoluble. In case one can get concentrated salt solutions, similar particles may exist and hints for their existence may be got from conductivity measurements and osmotic pressure. Sometimes a direct study using NMR is possible, which can give some structural information if used properly (see 2D-NMR methods). As I remember, $\ce{Cr(CH_3COO)2}$ hydrate is slightly soluble in water but undergoes little to no dissociation.

So...It sounds like the AgCl (aq) exists but a near or non-detectible concentration.  So there is no need to account for it in equilibrium calcs?

It depends, there is no way to figure it without problem and numeric data for it. Usually, when particle concentration is very small, you can drop it's concentration from equation where it is part of sum with other much bigger operand, but not from other equations.

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Simulation of a particle and chemical molecule
I am currently studying the basics of simulation and want to try out some experiments. It is a known fact that gold nanoparticles can bind to chemical groups which contains thiol (R–SH group). I want to simulate the binding of a thiol group to a gold nanoparticle. How can I do this? Where do I have to start? Which software can be used for this purpose?

Density Functional Theory is worth investigating if you're looking for a fairly efficient way of investigating these interactions at an electronic level. The term 'nanoparticle' is wonderfully vague as it encompasses two to three orders of size magnitude - the size of the nanoparticle will inform the specific choice of method:
Small (as in, only a couple of atoms) nanoparticles are amenable to computation with atom-centered basis sets in programs such as Gaussian, ADF, ORCA, GAMESS and so on. The latter two may be acquired free for academic use. If your particle/ligand complex is highly symmetric and/or you have supercomputer time you might be able to work with a moderately large particle. I recall a paper abstract where researchers did TDDFT on a 120 atom silver nanoparticle in ADF (here it is!), aided by high symmetry.
Large systems, where it's a) computationally unreasonable to model the whole particle and b) to a first-order approximation the ligand-particle interaction can be treated as occurring on an infinite plane or edge, may be more amenable to periodic DFT using plane-wave basis sets (PW-DFT). There are many codes available, ranging from free to very expensive. Off the top of my head, VASP, SIESTA, Quantum Espresso, Abinit and CASTEP are fairly popular.
There's also an atom-centered periodic franken-version of ADF called ADF-BAND which can model 1D to 3D periodic systems, which is worth mentioning. It is not particularly well documented, however.

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Methane seems to dissolve more readily in cold water. Why?

When you are diluting air in the system and/or properly evacuating it, you must consider water temperature. Methane is completely dissolved in water at 42 degrees Fahrenheit (5.5 C) but can be completely released (as a gas) at 58 degrees F (14.5 C). (source)

This would seem to imply that the enthalpy of solution for methane in water is negative (i.e. the process is exothermic). But, intuitively, it seems that you would have to input energy to force non-polar and polar to mingle.
Can anyone explain why this might be the case?

Did you check on how other gases are dissolved in water?

In that class? Yeah I looked for other specific linear alkanes, like ethane, and alkanes in general. No dice.

Spießbürger asked for "other" gases, not alkanes. Starting from propane or Butane things change a lot: this is about "permanent" gases only. Do what Spießbürger wrote. You will learn something very important.

Dissolution increases in liquid with decrease in temperature, as the

"Spießbürger asked for 'other' gases, not alkanes." Other gases, which may not share important properties with methane. I have an idea of how carbon dioxide dissolves in water, for example. I still haven't got a clue about methane, which is unlike CO2. I was hoping for an answer, not an unhelpful, cryptic comment.

"Dissolution increases in liquid with decrease in temperature, as the" ... what?

The enthalpy of solution going from the ideal gas state to the solution at infinite dilution is exothermic.  However, more importantly, the Gibbs energy of solution is positive meaning that dissolution of methane into water is not a favored process.  Most of this comes from the relatively large negative entropy change associated with the dissolution at room temperature.  This positive Gibbs energy is manifested in the very small solubilities observed and gives rise to the whole concept of something being "hydrophobic."  The fact that any methane dissolves in water is due to equilibrium effects based on a finite positive Gibbs energy.  Since methane is dissolved to a greater extent at lower temperatures must mean that the Gibbs energy is less positive at the lower temperatures.

Wilhelm et al.  "Low Pressure Solubility of Gases in Liquid Water"  Chemical Reviews. 1977. p.233

Never mind, I think I found the answer.
In the gas phase, attractions between solute particles are negligible, while solvent-solute attractions may reduce the energy of solute particles, resulting in an exothermic process that favors lower solvent temperatures.
For the record.


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Percent composition of nitrogen in fertilizer
The question was posed as follows:

Fertilizers supply the essential elements, nitrogen, phosphorus and potassium for plant growth. A bag of fertilizer contains 500g of ammonium sulfate, $ \small \ce{(NH4)2 SO4}$, and 500g of potassium nitrate, $\small \ce{KNO3}$. Calculate the percentage by mass of nitrogen in the bag of fertilizer.

This came in a test of mine and I got the answer wrong, now I did a redo and here's my solution:
(42 * 2)/227  * 100 = 37%

Is this answer correct?

I got 17.542% by mass nitrogen.

An explanation for beginners, a refresher for the rest of us:
To make sure anyone reading knows what molar mass means: The mass of one mole of something, in our case the mass of a mole of nitrogen, or the mass of a mole of ammonium sulfate, etc. A mole is just a way of saying a really big number which is actually $6.022 * 10^{23}$. It is a word we use instead of saying that big number, like when we say a dozen of eggs, but we know it means $12$ eggs. The same concept applies to a mole of eggs (that's a lot of eggs). To calculate the molar mass you need to add the atomic masses of each atom present in the molecule. If it is water, then we must count hydrogen's atomic mass twice (two hydrogen atoms per molecule), oxygen only once, since the formula is $H_2O$. Atomic mass can be found on any respectable periodic table, be sure not to confuse the atomic mass with the atomic number! The atomic mass always represents the weight in grams of a mole of that element. This convention makes our calculations much easier.
Ultimately we need to know how many $mol$ of nitrogen exist in this bag of fertilizer, so we can find out the mass of nitrogen in said bag (atomic mass can be used both ways because it is a ratio of grams per mole). To find that we need to do this process for both of the compounds separately because they have different molar masses. If there are two $chickens$ per $hen$ and we have $4_{mols}$ of $hens$ we should be able to tell that we have $8_{mols}$ of $chickens$. The same concept is used to determine how many $mols$ of nitrogen we have, based on how many nitrogens we expect per molecule of each substance.
Here is how that calculation might look in dimensional analysis in case it surprises you:
$81_{grams} * \frac{1_{mol}}{12_{grams}}$
Notice units of grams cancel each other out, and the remainder is in moles. This is how you would find the number of moles of carbon there exist in a pure sample of 81 grams.
Notice we used the ratio $12$ $\frac{grams}{mol}$ which is carbon's atomic mass.
I highly recommend writing out the entire problem using dimensional analysis so that you can be absolutely sure you have the answer correct for whatever problem you are solving. You will know this is true when all the units cancel correctly.

To solve the problem it may be good to take a step back from the calculator and consider what we know. What is the molar mass of these two substances? It should be the summation of the atomic masses of all the atoms in the molecule. Just like a water balloon's mass is not only the mass of the water, but also the rubber. Similarly the mass of water is not just the mass of the hydrogen, but the oxygen as well.
There is $500_{grams}$ of ammonium sulfate, but if we are trying to figure out the moles based on this information, we will need to do some dimensional analysis. Use the molar mass of ammonium sulfate which has units $grams/mol$ to multiply this ratio we found by the total mass of ammonium sulfate $(500_{grams})$ in such a way that the $_{grams}$ cancel each other, leaving $mol$ units. This is how many moles of that molecule exist in a 500g sample. We need to know this because the relationship of nitrogen per molecule is not a gram to gram relationship. You could of course, go another route and figure out the percent by mass of nitrogen per molecule and your final calculation should be equivalent. You should get about $21$% nitrogen by mass in a molecule of ammonium sulfate, and therefore 21% of the mass of that substance should equal the mass of nitrogen in that substance. It will be different for the other substance in the bag.
Once you know the total moles of nitrogen, convert back into grams of nitrogen, make sure your units cancel. From there you know the mass of nitrogen in the fertilizer bag, and the total mass of the fertilizer bag.
find the answer: what percent mass is the nitrogen atoms compared to the total mass of the fertilizer bag (all the atoms/molecules)?

I'm stuck on finding the amount of moles in nitrogen. As you said, there are 2 moles in ammonium sulfate and 1 in potassium nitrate. Since the formula for finding out the no. of moles = mass/molar mass. The molar mass in ammonium sulfate is 28 and 14 in potassium nitrate, but that would give the value of m.

@Joseph Potts You are forgetting that the 500g is how much mass all of the molecules of ammonium sulfate weigh, your goal is to figure out how many molecules of both substances exist. How much does one mole of ammonium sulfate weigh? How many moles of nitrogen are in a mole of ammonium sulfate? This information for both substances will help you find the way to the right answer.


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Is there a similar characteristic of the lanthanide contraction in the other transition metal groups?
Basically, does the lanthanide contraction also occur with the 4d, 6d groups as well? I am at a lack of understanding in this topic, I am just wondering what specific elements does this jump in a lower atomic radii occur. It seems to happen due to the d orbitals being occupied? I am making that judgement based on this graph.
I am hoping someone can explain, as one orbital seems to make all the difference.

Hope I interpret your question correctly:
A decrease of atomic radii occurs in every period (from left to right). Atomic radii of lower periods are in general larger (Increase from top to bottom). This simple behavior only holds true for elements that only have p-block and s-block electrons.
The first anomaly of this behavior is the d-block contraction. This can be seen in the 4p, 5p, 6p and 7p elements (Elements to the right of Ga, In, Tl). For example the atomic radius increases about 60 % from C to Si. The difference of Si and Ge with the added p-block contraction is just 20 %.
The lanthanoide contraction occurs when the f-block is added. For this reason the 4d and 5d elements are very similar in size. For example Hf (5d, 160 pm) is even smaller than Zr (4d, 159 pm).
Finally the actinoide contraction should have the same effect on 6d elements. But I'm not sure if realiable numbers for the atomic radius exist for those elements.

even if there is no data for actinoids' atomic radii, they can be calculated rather well.


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Why is it called "carbonation"?
Why is it referred to as "carbonation" and we drink "carbonated" beverages when carbonate is $\ce{CO3}$ while $\ce{CO2}$ (carbonite?) is present in carbonation?

Nope. Carbonate is $\ce{CO}_3^\color{red}{2-}$ (an ion), not $\ce{CO3}$. Carbonite is $\ce{CO}_2^\color{red}{2-}$.
Carbonation involves dissolving $\ce{CO2}$ gas in water. It turns out that $\ce{CO2}$ reacts (maybe not the best term to use here*) with water via the following reaction $$\ce{CO2 + H2O -> CO3^{2-} + 2H+}$$
So you have effectively carbonated the soft drink.
When you depressurize the bottle by opening it, the reverse reaction occurs and you get carbon dioxide.
*Such dissociation is normal and integral to dissolution of polar solutes in water, so generally it is considered as part of the dissolution

I've always wondered about carbonite. Has anyone ever produced salts of it? Can you be frozen in it?

@BenNorris: Apparently it's a hypothetical radical, used to explain certain phenomena. Its conjugate acid exists in the gas phase, but not yet known to exist as a solution.  Go figure :S

https://chemistry.stackexchange.com/questions/64416/does-the-carbon-dioxide-dianion-exist


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Is "ice point" the same as "freezing point"?
I am using an old book and they almost use both terms interchangeably sometimes. Do they mean the same thing? Similiarly for "steam point" and "boiling point"

Could you gives us a few sample sentences?

Sounds like it, but I've never heard "ice point" or "steam point" used before.

I've heard the term 'ice point' used and I think this is an interesting question.  Certainly, and ice phase is different from 'freezing' - the latter which is transitional and regionally-dependent.

Ice point is the freezing point of water.
Freezing point is the point where liquid turns to solid.
The difference is that in the case of ice point is substance is always water.
(Note that there is a slight discrepancy of 0.01 K between the triple point of water, melting point, freezing point and/or ice point. And technically, Ice point is the point where water and Ice are in equillibrium.)
See e.g. http://www.thefreedictionary.com/freezing+point, http://www.thefreedictionary.com/ice+point.


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

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

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

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

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

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

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

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

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

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

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


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

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

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

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


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

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

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

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

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

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


text:
Why are equilibrium constants unitless?
I haven’t quite reached the point where I can read a full-fledged text on chemical kinetics and thermodynamics yet, so bear with me, please.
I’m wondering why a value like $K_\text{eq} = \frac{[\ce{NO}]^2[\ce{O2}]}{[\ce{NO2}]^2}$ wouldn't have units of M?

I goofed up the first time I tried to answer this question, erroneously applying dimensional analysis to your equilibrium expression.
It turns out that Silberberg[1] gives a good explanation of why $K_\text{eq}$ is dimensionless, which is often glossed over as the terms of the equilibrium expression are generally taught as concentrations. In actual fact, the terms are ratios of the concentration or activity of each species with a reference concentration (1 $\mathrm{mol\cdot{L^{-1}}}$ for solutions.) For example, a concentration of 2 $\mathrm{mol\cdot{L^{-1}}}$ divided by a reference of 1 $\mathrm{mol\cdot{L^{-1}}}$ yields a ratio of 2, with no units. As each term has no units, so too does $K_\text{eq}$.
[1] Silberberg, M.E.; Chemistry – The Molecular Nature of Matter and Change 3e; 2003, p. 719

I don't quite follow but that's a step in the right direction at least.

This only works if the number of products is equal to the number of reactants. In the example Sadiq gives it is not, and so the equilibrium constant as written has units of $(\text{mol}\cdot\text{L}^{-1})^3/(\text{mol}\cdot\text{L}^{-1})^2 = \text{mol}\cdot\text{L}^{-1}$.

I understand this much, but I what I don't understand is how it can have significance in equilibrium reactions

The stated facts is true for the standard equilibrium constant, not for an equilibrium constant.

This source explains it well. It looks like part of class material, but it clearly explains the dimensionlessness of $K_{eq}$.

The resolution of this apparent paradox is that the above equation, while perfectly satisfactory for everyday use, is not technically correct.  A more correct version is:
$$K_{eq} = \frac{\frac{\lvert B \rvert_{eq}}{\lvert B \rvert_{ss}} \frac{\lvert C \rvert_{eq}}{\lvert C \rvert_{ss}}}{\frac{\lvert A\rvert_{eq}}{\lvert A\rvert_{ss}}}$$
where the "ss" subscripts refer to the concentration of that species in the standard state.  (By this definition, Keq is always unitless.)

It then goes on to state:

Strictly speaking, division by the standard state concentrations is also necessary in every thermodynamics equation in which you take the log of a concentration product, otherwise the units don’t come out right.)  We will NEVER use this "correct" version of the equation in this class (well, never except in one problem on this week’s problem set...), and $K_{eq}$ for a reaction with unequal numbers of reactants and products is ALWAYS given with units,
even in published papers.

[1] http://www.bio.brandeis.edu/classes/archives/biochem102/102-02stand.pdf

The first definition refers to a modification of the  standard equilibrium constant. An generic defined equilibrium constant is more convenient to use and can also have a dimension.

Dimensional analysis is useless. The correct answer is the one already given involving activities, which are dimensionless.
Activities are defined as ratios. For example a pressure activity (there are many kinds) is defined in terms of the ratio of the actual pressure of a gas divided by the reference pressure, often 1 atm or 1 bar.
In the present example, the activity is the ratio of the molality divided by the reference molality of 1 molal. This assumes ideal solutions, which is good enough if the solutions are dilute. If the solution is not ideal one has to correct the molality for nonideality.
Because of these complications, detailed discussions of activities are usually left for a course in physical chemistry.
So while we use square brackets and molarities, we have to understand that we are really dealing with activities.
By the way, the activity of a pure liquid or solid is 1, which is why $[\ce{H2O}]$, for example, is dropped from equilibrium calculations.

Wait a sec. Don't you mean molarity (as opposed to molality)?

The issue of activity vs. concentration (or activity vs. pressure) is subtly different from the issue of dimensional vs. dimensionless.  "Activity" can be related to the chemical potential, which does have dimension of energy/particle or energy/mol.  It is the inherent definition of energy relative to a standard state that leads to $K_{eq}$ being dimensionless.

Note
This answer is incorrect, but because it is a common pitfall I'll leave it to show. It is wrong 1) because the OP meant $\lvert \ce{NO} \rvert ^2$ in the numerator not $\lvert \ce{NO2} \rvert$ as I took it (edited) and 2) because the $K_{eq}$ formula is incorrect. I will post a better answer.
The concise answer: Because the units cancel. Here I'll use $\lvert \rvert$ = concentration = moles/liter = m/L
$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert^2} (?) units$
$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} (?) units$
$K_{eq} = \frac{\lvert NO_2 \rvert m/L \lvert O_2 \rvert m/L}{\lvert {NO}_2 \rvert m/L \lvert {NO}_2 \rvert m/L}$
$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} \frac{m^2/L^2}{m^2/L^2}$
$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} \frac{m^2}{m^2}$
$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} (unitless)$

Unfortunately not so - check the exponent on [NO] in the numerator. Straight dimensional analysis on Sadiq's example results in residual units.

So he meant $\lvert NO \rvert ^2$ and not $\lvert NO^2 \rvert$ (which I took to mean $\lvert NO_2 \rvert$)?

Yeah, I read it like that as well. But the stoichiometries don't work out otherwise.

I've edited the question. Sorry. Yes I meant $[\ce{NO}]^2$. (A reminder: you can use \ce{ } to format molecular formulae easily in LaTeX.)

Thanks for the tip. What's with the left [ being darker than ] in formulas? I switched to \lvert and \rvert just because it was ugly.

What is the difference between NO^2 and NO_2 ?

In chemistry, the subscript $NO_2$ would mean one molecule with one atom of Nitrogen, two of Oxygen. $NO^2$ could mean a molecule with one atom of Nitrogen, one of oxygen with a net charge of +2 (an ion), though it should be written $NO^{2+}$ to be more explicit. Superscripts before the element would be the atomic weights for various isotopes ($^{18}O^{16}O$).

If you were referring to the typo in the original question, it's now been corrected to $[NO]^2$ which is the same as $[NO][NO] = 2[NO]$, unless I'm mistaken.

I think this post should be edited

@tatan I see those same errors on my mobile phone, but on a desktop computer (Chrome 61) it displays correctly. There may be differences in the latex engines, but I'm not familiar enough to do a cross-platform fix. Can anyone else suggest edits?

authors: Христо Христов


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

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

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


text:
Is it possible to get energy from frozen methane without oxygen?
In one of the solar system documentaries it was mentioned there are moons in the solar system with lakes of frozen methane.
Assuming humans or robots or whatever could make it there, would there be a way to extract energy from frozen methane without using oxygen? Otherwise, it would seem that the oxygen would have to be carried there as well.
Is it possible to recycle the oxygen efficiently enough to burn methane again and again?

I would imagine that any oxidizing agent should work. While oxygen has worked on earth, sulfur should do the trick.

... or add a bit of chlorine trifluoride

The planet will have gaseous oxygen if it does have an atmosphere, and if that is the case then with enough energy(like a spark) atomized hydrocarbons could be lit for energy. What is atomizing you say? Well its just breaking a liquid apart into little particles, making it easier for the oxygen to react with said methane.
As for recycling the oxygen from the combustion reaction, it will not be possible to do in an energy efficient way, CO2 is too stable and the bonds are too strong.

"The planet will have gaseous oxygen if it does have an atmosphere." - Do you have a source for this? Any atmosphere will have a little oxygen, but I don't think the mostly methane atmospheres have enough to make it a viable source of energy.

Methane in the atmosphere will be low because it will be in liquid form, boiling pt of methane -164°C boiling point of oxegen -182.95°C. Also oxygen is the 3rd most abundant element in our Galaxy. http://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Universe

Oxygen may be the third most abundant element, but not necessarily in its free form. Silicon is one of the most abundant element in the earth's crust, but not as Si(s).

Is it possible to recycle the oxygen efficiently enough to burn methane again and again?

Don't think so. Sounds a little like perpetual motion. Assuming you're burning the methane into $\mathrm{CO_2}$ and $\mathrm{H_2O}$, you would need a large supply of something that would regenerate the $\mathrm{O_2}$. The water's a much easier target -- you could electrolyze it, for example, but you'd only recover 1/2 the oxygen each cycle.
I can't think of many things to act as chemical $\mathrm{X}$ in:
$\mathrm{CO_2}+\mathrm{X}\rightarrow{}\mathrm{CX}+\mathrm{O_2}$. That would be a killer reaction for the "carbon capture and storage" crowd -- we could just process our carbon emissions into oblivion! Goodbye greenhouse effect! Would sure help terraform Mars!

Nor perpetual motion ideology, think more like trees or algea or what ever process that take sunlight or a planets core thermal energy for recycling

No, unless you bring an oxidizer with you
Considering just the chemistry, any oxidiser would work. Unfortunately there isn't any free oxidiser on such worlds.
There is in fact only one solar system body with known hydrocarbon lakes, that is Saturns moon Titan. And while it does have a lot of oxygen, that is all bound up in water ice. There is no free oxygen (or other oxidiser) in Titans atmosphere.
In general you shouldn't expect any free chemical fuel lying around on a celestial body. Chemical systems tend to move to an equilibrium at their lowest energy state, unless there is some mechanism that pushes them out of that equilibrium, like photosyntesising plants on earth. If there were any free oxygen in Titans atmosphere that would have reacted with the hydrocarbons billions of years ago (mediated by e.g. photodissociation in the atmosphere).
You also can't re-use oxygen or anything else without some other form of energy input. In the theoretical best case the energy that you gain from burning methane is exactly the same energy that you need to put in to extract the oxygen from the combustion result (CO₂ and water) and re-form the oxygen and methane. In practice you need to put in a lot more as such reactions are far from 100% efficient.


text:
Are redox half-reactions "real"?
Are the half-reactions used in analyzing redox reactions a real thing or are they a conceptual shortcut?
Is there a way to, however briefly, run a half-reaction by itself without the corresponding other half?  For example, reduce metal atoms to metal ions without the corresponding oxidation half-reaction?

Electrolysis/Galvanic cell is basically redox reaction split in half-reactions, running on different electrodes. For example, hydrogen fuel cell has two electrodes. On one hydrogen donates electrons to electrode:
$H_2 = 2H^+ + 2e^-$
on another, oxygen drains electrons from electrode
$O_2 + 4 e^- + 4 H^+ = 2 H_2O $
The electrons travel from one electron to another via external electricity consumer. If electrons are forced to travel by external force, the reaction may be reversed. This ideas are utilized in electric chemical batteries, accumulators, fuel cells, industrial electrolyzers and so on.
This way, half-reactions are separated between different electrode. Running half-reaction by itself without other half elsewhere is problematic at best, because of electroneitrality violation. You'll need extreme voltage source and even then it will stop quickly.

When you say "extreme voltage source" -- aren't the voltages involved rather small (1-3 volts)? I understand that if the electrode becomes surrounded by charges that counteract the half-reaction, it will be squelched quickly.  Assuming there was another mechanism to remove the half-reaction products, the half-reaction might proceed indefinitely.

@thinker If half-reaction products will not discharge (wich requires other half-reaction to run), you will simply charge up some capacitor (in electrical engineering meaning). Molar charge of electron is ~96 485 C/mol.  So to charge a capacitor with chemically visible amount of electrons (some millimoles) one need to store about 100 culons. Typical capacitances are, to my knowledge, measured in microfarads, so about several MV are required to run measurable amount of half-reaction without product discharge.


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

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

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

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


text:
Calculating laser wavelength/power to cause emission of light in a gas?
It's been a long, long time since I've had to worry about electron excitation levels and band gaps. Please forgive (and correct!) my terminology and misconceptions.
I've become interested in volumetric displays. In particular I'm looking at a pair of lasers scanning a volume of gas such that the intersection of the lasers would supply enough energy to cause the gas to emit visible light. Since this is the start of my research I would like to investigate a number of gasses in order to determine a number of gasses which might be suitable for this application. I have, however, forgotten how to determine the amount of energy needed to cause an outer shell electron to emit a photon, and then trying to back-calculate the required wavelength(s) and power requirements of the laser emitters.
I'm looking for some information on how to identify the required information for a given gas and how to calculate the energy required to cause visible light emission, and also information on how the wavelength of laser light impacts the ability of the gas to absorb the energy with the intended effect of emitting visible light. Links to introductory texts are always helpful, as are answers which help me get the terminology correct so that I can improve my searches for this information.

Maybe adding some information such as what kind of gas you're planning to use would be helpful. Basically you're looking for fluorescence.

Two-photon excitation using femtosecond lasers that emit in the infrared range is a technique used in living cell imaging via fluorescence microscopy. Advantages of the technique, such as the minimized light scattering (~ 1/$\lambda^4$) using long wavelength for excitation and the fluorescent dyes used are discussed here.
For large volumetric displays, a different technique was/is under development in a joint project of the Japanese National Institute of Advanced Industrial Science and Technology (AIST), the Keio University and an industrial partner.
Here, fluorescence does not play a role and special gases are not needed either.
The images can be displayed in air and are formed - pixel by pixel - by plasma dots generated in the focal point of pulsed infrared lasers with a pulse width of 1 ns and a repitition rate of 100 Hz.


text:
What chemical file formats are compatible with 3D printing?
I would like to print some molecules rather than building them using a kit. What file formats work?

None of the chemical file formats are typically accepted by 3D printing services. The reason is that chemical file formats are fundamentally different from those describing 3D objects.
In the simplest case, a chemical file format describes a list of atoms and their positions (plus possible metadata). An example is the XYZ format, which stores exactly that. Other “chemical objects” which can also be described in some chemical file formats include bonds, protein structure, crystalline structure information (unit cell, symmetry, occupancy of each atomic position, …), etc.
3D file formats, on the other hand, describe a series of objects. In the simplest case (e.g., the STL format or the Wavefront .obj file format), the object itself is not stored, but an unstructured triangulated surface delimiting the object. In those, you can't describe a sphere, but only a tesselation of the sphere.
So, you need a conversion step to create a 3D representation of your chemical structure. Chemical file formats don't typically contain representation information: you need to pick the one you want (CPK, liccorice, protein cartoon, …) and use software to create a 3D scene. VMD, for example, allows export to STL format, which is suitable for 3D printing.


text:
How does the work that won the 2012 Sustainable Chemistry Award contribute to sustainable chemistry?
I'm seeking a lay explanation for how the work of Dr Marc Taillefer that won the 2012 European Sustainable Chemistry Award, contributes to sustainable chemistry.
From the press release, Dr. Taillefer

is being recognised for his seminal contribution to the field of
  homogeneously catalysed coupling reactions leading to C—C, C—N, C—O, C—P bonds. His team at
  the Institut Charles Gerhardt, ICG (Montpellier, France) is investigating for a decade the
  environmentally sustainable conversion of small molecules into more valuable substances
  catalysed by copper and iron molecular complexes. This renaissance of “Ullmann type arylations” is now often used at the academic or industrial level and avoids the use of more expensive catalysts based on palladium.

The objectives  of the award are (to quote from this press release) to:

Recognise individuals or small research groups which make an outstanding contribution to sustainable development by applying green and sustainable chemistry.
Promote innovation in chemistry and chemicals that will deliver clear improvements in the sustainable production and use of chemicals and chemical products.
Demonstrate that chemistry and chemicals can play a central role in delivering society’s needs, while minimizing and solving environmental problems.

His work is about developping new catalysts based on copper and iron, to replace to traditional catalysts based on palladium. Copper and iron are both very common elements in nature, while palladium is considered a high supply risk (see the 2012 British Geological Survey risk list for details).
The new RSC Visual Elements Periodic Table can be used to check this kind of information.


text:
Why are noble gases stable?
Generally, noble gases are very unreactive. Why?
I understand that it has to do with electron shielding prospective electrons that could join the atom or leave and the energies associated with both actions, but why are 5 electrons in the p orbital (to take Fluorine as an example) so much worse in shielding the nuclear charge than 6 (Neon)?
What IS electron shielding, anyway? I understand it in the classical lie-to-children way, but not satisfactorily.

It's all about the energy levels of electrons. Electrons exist in different energy levels and each energy level can only be occupied by a specific number of electrons (google quantum theory).
Example: Neon has an electron configuration of $1s^2 2s^2 1p^6$, which simply means that it has a full $1s$ orbital (the maximum is 2 electrons as denoted by the superscript), and has a full valence shell which is the $2s$ and $1p$ orbitals (which is the valence shell for molecules in row 2).
Now for neon to lose an electron (to bond) it would have to take an electron out of a full orbital ($1p$), which is what makes molecules reactive (a non-full valence shell), and would just take the electron right back.
Think of Ne$^+$ as more electronegative (wanting electrons) than elemental Fluorine, so it will pull an electron from almost anything and go back to its elemental form.
If this answer is not satisfactory then I can explain further, without diving head first into quantum theory.

You're right, it wasn't quite satisfactory, maybe I didn't state it obviously enough in the question, but I understood most of that already. As useful as that explanation is, it contains some tautologies and because of that doesn't explain much of WHY (simply restates the meaning of stability). So if possible, could you wander to the "real" explanation?


text:
Why does 9-fluorenone dissolve in hexane?
I know like dissolves like, but I am still unclear.  9-fluorenone has a polar $\ce{C—O}$ bond.
Is it because 9-fluorenone can only form London forces with other 9-fluorenone molecules and hexane can break these forces?

Firstly, as a general rule, organic compounds with more then 6 carbon have higher London forces then polar forces.  This is a very general rule and doesn't work in all circumstances. Thus, this molecule is not as polar.
Secondly, cyclic compounds have much weaker London forces then molecules that have straight chains. Thus, this makes the London forces not so high(relative).
Thirdly, to have something dissolve in hexane it must have weak intermolecular forces, but not weaker then hexane itself, as hexane only has to suspend the solute molecules in solution and prevent the solute from clumping up and falling to the bottom.
So, quite simply, what happens is that the hexane molecules surround the 9-Fluorenone molecules, which diffuse through the solution.

I am curious about your claim that “cyclic compounds have much weaker London forces then molecules that have straight chains”. I don't believe it is true for aromatic rings, but in any case, do you have a reference or explanation of that statement?

Aromatic rings have an additional attraction due to their permanent quadrupoles. See my answer to another question. I am not sure if I would classify the pi-stacking interaction as a dispersion force since it is neither induced nor transient.

I would argue differently from Mike. 9-fluorenone has one polar $\ce{C=O}$ bond, but it also has two aromatic rings. Dispersion interactions are strong for such aromatic rings, and the point can be made that this factor will drive the solvation of 9-fluorenone in hexane.
This is confirmed by the fact that 9-fluorenone is insoluble in water.


text:
How do non-polar substances dissolve in non-polar solvents?
The case of polar solvents is clear to me - we get an attraction between opposite charges. However, how do non-polar substances dissolve in non-polar solvents? How could it be explained on a molecular level?

The electron density distribution in molecules (including nonpolar ones) is not static. Therefor, as a function of time, the electron density is not uniform. Occasionally, randomly, the electron density in a molecule will shift to produce a spontaneous dipole: part of the molecule now has more electron density ($\delta^-$) and part of the molecule now has less electron density ($\delta^+$). This spontaneous dipole is transient. The electron density will shift back to negate it and then shift again to create a new spontaneous dipole. In a vacuum, this behavior would be a curiosity.
In the presence of other molecules, these spontaneous dipoles have a propagating effect. If molecule A develops a spontaneous dipole, then the electron density in neighboring molecule B will by respond by developing a spontaneous (but opposite) dipole. The $\delta^+$ region of B will be close to the $\delta^-$ region of A. This new dipole in molecule B is an induced dipole. This induced dipole in B will then induce a new induced dipole in another neighboring molecule C. By this time the transient dipole in A is already fading, perhaps to be replaced by a new induced dipole from another molecule.
These random, transient, but continuously propagating dipoles have attractive forces associated with them. These forces are named London dispersion forces after the physicist who proposed them. The magnitude of these forces scales with increasing surface area of the molecule. Thus, larger molecules will overall have stronger London dispersion forces than smaller molecules. Linear rod-like molecules will have stronger forces than spherical molecules (spheres having the smallest surface area to volume ratio of the 3-dimensional solids).
More complex attractive forces also arise from nonpolar molecules. Benzene (and other aromatic molecules) can pi stack, which relies on the strong permanent electric quadrupole. Benzene also has a (weak but) permanent magnetic dipole due to its ring current, so some component of the attraction between benzene molecules may be magnetic and not coulombic. Or, the two attractions may be one and the same, as benzene's electric quadrupole may result from its having a magnetic dipole. As his part of the discussion is devolving into physics, I will end here.

By virtue of the explanations provided by Ben Norris and Colin McFaul, shouldn't polar solutes also dissolve in non-polar solvents? After all, polar solutes can also induce formation of temporary dipoles in the non-polar solvents and entropy also increases?

Certainly more polar and less polar molecules can mix. However, polar molecules tend to exclude non-polar molecules because their interaction is much stronger than a dipole-induced dipole interaction.

Some chemical phenomenon can never be explained. They are results of experiments.

@user1.618 - One of the goals of science is to explain (or at least develop predictive models for) observable phenomena. It is short sighted to suggest that some phenomena can never be explained. Better to say "There are some experimentally observed phenomena that cannot be explained yet."

I think @Ben Norris's answer is great, but I wanted to add one other reason that non-polar molecules will dissolve in non-polar solvents, even in the absence of interactions: entropy. If there are no intermolecular interactions at all (ie, the solvent molecules ignore each other and the solute molecules, and the solute molecules ignore each other and the solvent molecules), then the entropy of mixing will be positive, and mixing will be favored. As a result, species always want to mix; it takes unfavorable conditions to cause non-solubility.

Very true. I forgot about $\Delta S_\text{mix}$.

In other words; equivalent to diffusion of gases.

Great, but how is entropy changes unfavourable when polar solutes are dissolved in non polar solvents and vice versa

The reason non-polar molecules won't dissolve in polar solvents is believed to be due t0 enthalpy. When a non-polar molecule is added to a polar solvent (eg.water) stronger hydrogen  bonds (or permanent dipole bonds) have to be broken to allow the non-polar molecule 'dissolve' (get surrounded by solvent molecules). The bonds which are then formed between the non-polar molecule and the solvent are much weaker than those between the polar solvent and itself. Therefore there is an energy cost to dissolving non-polar molecules in polar solvents that is believed to outweigh the entropy gain.


text:
Is crushing an avocado a chemical or physical change?
Applying pressure can result in chemical changes, does this hold true for an avocado?

It's physical change, but it can increase the chemical reaction rate as surface area of Avocado is increased.

Your title and question are a bit different.

Is crushing an avocado a chemical or physical change?

Physical. Crushing (or any other physical manipulation of) any substance is, by definition, a physical change.  However, those physical changes can affect the conditions for a chemical change, potentially causing a noticeable change in the rate of said reaction. You somewhat acknowledge this in your question, though I'm not sure it's quite intentional:

Applying pressure can result in chemical changes, does this hold true for an avocado?

Under reasonable conditions, I'm fairly sure that nothing in an avocado is reactive enough to undergo a chemical change because of any conditions the physical crushing altered. Except that it will increase the surface area and speed up oxidization.  Given enough of a crushing pressure (e.g. equivalent to the pressure put on coal to create diamond) I'm sure some chemical reaction or another will occur.

I don't think so.  A chemical change implies changing molecular-level structure usually accompanied by transferring electrons.  When you squish an avocado you're only rearranging the fats, carbohydrates, and fiber in it - I don't think a chemical reaction occurs.
If you left it out, and it turned brown (oxidized), that would be a chemical change!

I second Phillip’s answer, which is right in practice, with a small additional detail: it depends on the pressure applied. If you apply pressure by hand, you are not likely to create or destroy chemicals bonds, and the changes will be physical. Applying pressure can change chemical structure, as in the case of solid phase transitions or pressure-induced reactions, but the pressure has to be much larger than that: typically of the order of gigapascals (1 GPa is rougly 10,000 atmospheres).

The actual act of crushing the avocado (applying simple pressure) is a physical change.  While the act of applying pressure may allow oxidation to happen faster, the application of pressure is still physical.  The type of pressure you would apply to an avocado under normal circumstances would be insufficient to result in chemical changes.  Chemical changes result in the breaking or formation of molecular bonds which are simply not possible by normal kitchen appliances.  Pressure and heat are related which could cause the enzymes and proteins within the avocado to denature, but the act of applying the pressure is not directly responsible for any chemical change in this case, and the application would only be a physical change.  If there were chemical changes, these changes would be caused by factors other than the pressure itself.

My 5 cents.
In the beginning it's only physical. However if you crush it a lot of things change. The most obvious is that oxygen can penetrate it much easier than before. There are a lot of substances that can be readily oxidated (f.e. unsaturated compounds). You gain bigger surface so the cells on surface will dry. The nice equilibrium in which this fruit exist is changed. I can't explain it on the level of biochemistry but as every living thing avocado must have some mechanisms that protect it from decomposing. Some of them can't work properly if there is not enough moisture and to much O2.


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

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


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

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

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

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

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

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

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

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

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


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Find empirical formula when individual masses are known
I'd like to illustrate my question with an example:
Imagine nickel reacts with fluoride and you start with $0.766\ \mathrm{g}$ of nickel, an unknown amount of fluoride and you end with $1.261\ \mathrm{g}$ of nickel fluoride. The mass of fluoride in the end must then be $1.261\ \mathrm{g} - 0.766\ \mathrm{g} = 0.495\ \mathrm{g}$.
So, in the end, there are $0.013$ moles of nickel and $0.026$ moles of fluoride. Now the question is, what is the empirical formula for nickel fluoride? ($\small\ce{NiF2}$). I see that for every mole of nickel, you need $2$ moles of fluoride. Then why is the empirical formula $\small\ce{Ni2F4}$ incorrect?

An empirical formula is, according to IUPAC’s Gold Book definition:

Formed by juxtaposition of the atomic symbols with their appropriate subscripts to give the simplest possible formula expressing the composition of a compound

Wikipedia goes even further:

the simplest positive integer ratio of atoms of each element present in a compound

and then

An empirical formula makes no reference to isomerism, structure, or absolute number of atoms.

So, NiF2 is an empirical formula, while Ni2F4 simply isn't. Another example is given by, again, Wikipedia:

For example, formaldehyde, acetic acid and glucose have the same empirical formula, CH2O

Their molecular formulas (which account for the absolute number of atoms in a molecule) differ, though: formaldehyde CH2O, acetic acid C2H4O2, glucose C6H12O6.

And what about H2O2?

H2O2 is a valid molecular formula, but not an empirical formula. The empirical formula is HO.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Why are isomers difficult to separate?
I recently learned that attempts to compare the spectra of two isomers of $\ce{C_3H_3^+}$ were frustrated by the difficulty of separating the two species.
What makes these isomers difficult to separate?

I'm not even sure how one would try to separate them. Could you provide some more detail that would help someone answer the question? What kind of spectroscopy and what kind of separation were discussed?

It would be important to know what kind of "separation" you are talking about -- spectral separation, which would just mean suppression on the spectroscopic response from undesired species, or physical separation, where you actually remove those species from your sample?
The problem is that not only may your sample contain different isomers, but these may actually interconvert through rearrangement reactions after ionization. This is a very common phenomenon in mass spectrometry, for example. If you want to single out the spectral signals of a certain isomer in a bulk sample, the simplest method would be to take the spectra of all unwanted components (from the literature or, preferably, separate measurements) and subtract them from your "mixed" spectrum. You would need information about the composition of your sample, though, or at least some indicator which tells you how much and what you have to subtract. Physically separating isomers may also be possible, e.g. if one of the isomers can be selectively ionized by laser irradiation, then separated from the remaining non-ionized sample. Such techniques are actually somewhat common in physical chemistry (not so much in everyday analytics, I think), although their applicability does depend heavily on the compounds and isomers under investigation. And there certainly are cases where different isomers are just too difficult to separate, or interconvert so readily that one structure virtually don't exist without the other.

Molecules which are made up of exactly the same elements but differ in atomic arrangement are called isomers.
Since the two isomers have the same mass and charge, any seperation techniques which rely on a difference in mass and/or density would be very difficult.
Mass spectrometry is an analytical technique in which components of different mass are separated and detected, typically by ionizing (charging), accelerating through an electric field to a given velocity and finally, 'bending' through a magnetic field. The final position of the is detected as a function of its charge-to-mass ratio.
Any two isomers of $\small\ce{(C3H3)^+}$ have the same charge-to-mass ratio and therefore would be very difficult to separate and detect with mass spectrometry.

So it's just that isomers in general are difficult to separate, not something special about this particular configuration?

Isomers can be identified by techniques other than mass spectrometry. For example, since each monomer would have different electron configurations and bond lengths, it is possible to detecting the energies associated with molecular orbitals using infra-red spectrometry. Each isomer can also be modified and separated by some chemical means by forming ligands within in a complex.

Firstly, the reasons above—identical density and molar mass—are very true.
I also know that at least for some enantiomers they racemize, or readily and spontaneously convert between both forms. This means that even if you start with a pure sample of one form, you will eventually end up with both. A famous example of this is with the banned drug Thalidomide, made in the 50s to treat the morning sickness of pregnant women. The R-form worked fine. Unfortunately, its S-form caused severe birth defects. If someone could figure out how to separate the two without it racemizing (there is some work on similarly structured molecules, I'm sure), they would have done so.


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Which element is being reduced in this reaction?

Which element is reduced in this reaction:
  $$\ce{2KMnO_4 +3NaSO_3 +H_2O ->2MnO_2 +3NaSO_4 +2KOH}$$

I am trying to figure out which elements lose their electrons, but how am I supposed to know when elements are neutral and when they are ionized? I have tried the following, but I don't believe it to be correct
$$\ce{2K^+Mn^+O_4^{-2} +3Na^+SO_3^{-2} +H_2O ->2Mn^{+2}O_2^{-2} +3Na^{+}SO_4{-2} +2K^+OH^-}$$
My reasoning being that:

Oxygen atoms are always doubly anionized
Sulfate and sulfite are also doubly ionized
Sodium and potassium are always singly cationized

Based on these rules I tried to make the bonds neutral, but my assumptions must be incorrect. Any help would be greatly appreciated

Your reasoning is not entirely right. In those compounds, O is always (-II), K and Na always (+I). However, it's not true for S, which can have different oxidation states. So does Mn.
So, in KMnO4, Mn must be (+VII). In MnO2, it is (+IV). Thus, Mn is reduced.
For sulfur: in NaSO3, it must be be (+V). In NaSO4, it must be (+VII). Thus, sulfur is oxidized.
However (big however, in fact): the sulfate anion is not SO4–, but SO42–. Similarly, sulfite is SO32–, not SO3–. So, your equation should be instead:
$$\ce{2KMnO_4 +3 Na_2SO_3 +H_2O -> 2MnO_2 +3Na_2SO_4 +2KOH}$$
which means that sulfur is oxidized from (+IV) to (+VI).


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In orbital hybridisation, what determines the number of orbitals that hybridise to be at the same energy level?
For example, in phosphorus pentachloride five orbitals hybridize, but why not 6, as in the hydrated Al ion? I have read that the reason six orbitals hybridize is that a maximum of six oxygens can stably fit around it. Why is it not the same case for phosphorus?
Secondly, why does water $sp^3$ hybridize even though it has enough lone pairs in its p orbital to bond with 2 oxygens. Surely it would take more energy to pull up the lower $s$ orbital to bond than to allow the $p$'s, already elevated to that energy, to do the bonding?

The difficulty in talking about hybridization lies in the fact that it tends to confuse atomic and molecular orbitals, or (equivalently?) one- and many-electron wave functions.
Mixing atomic orbitals generally only becomes possible when we do a mathematical approach which treats more than one electron at once. That, however, is mathematically rather complex, and the resulting wave functions are molecular orbitals, which are entirely different from atomic orbitals. Molecular orbitals are constructed as combinations of atomic orbitals, which means that it is possible to e.g. add s and p wave functions together to obtain something that looks like sp hybrids. But, and this is important, there are actually many different possibly ways for such combinations, and none of them should really be considered physical reality; and furthermore, this "mixing together" has to follow very strict mathematical rules which in the end affects the entire set of molecular orbitals at once. Describing molecular orbitals in terms of hybrid orbitals (or something similar) is possible if one does it thoroughly, in terms of a full and correct mathematical treatment. Still, even when applying the concept in this way, it is somewhat arbitrary, since it represents only one of many possible solutions (and depending on the problem, not even the best one).
In the simple and qualitative terms often introduced in chemistry (especially organic chemistry), it is a very crude concept which should be regarded with great caution and especially not as something that would be some sort of reality. Hybridization is a mathematical procedure for many-electron wave functions; nothing more, nothing less.
In more technical terms: The role of one- and many-electron wave functions in atoms is crucial here. Atomic orbitals are assumed to be hydrogen-like wave functions; in the one-electron atom, the orbitals are $\ell$-degenerate, so linear combinations of wave functions for different orbitals would mathematically still be solutions to the Schrödinger equation. In many-electron atoms, the degeneracy breaks down, so combining different orbital wave functions would be invalid, since they belong to different eigenvalues. We do, however, employ a many-electron Hamiltonian such as the Fock operator in this case, and unless I am mistaken, the electron-electron interactions lift this restriction. So, by some unitary transformation of the many-electron wave function of a single atom, I guess one might arrive at something like hybrid orbitals. Then again, this does not reflect physical reality.

This is only partly an answer to your question. I think teaching hybridization in depth is a bad pedagogical choice, so I am not the right person to explain the inner workings of this theory.
Don't spend too much time try to get the details of the how's and why's of hybridization…  There are arguments over how useful it is as a concept in theoretical chemistry, and it certainly has severe limitations. For example, the picture derived from hybridization for the water molecule is simply wrong: it predicts two equivalent lone pairs on the oxygen atom, and they are not.
Just think in terms of molecular orbitals!


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Separating NaClO from its aqueous form in bleach
I was wondering, can I separate the $\ce{NaClO}$ from the $\ce{H2O}$ in bleach. The decomposition point for $\ce{NaClO}$ is $374.14\ \mathrm{K}$, so I'm risking decomposing the compound if I do simple distillation.
What should I do?

PS: I don't have a vacuum pump.

Sodium hypochlorite, $\ce{NaClO}$, is unstable and can't be separated from the water in which you usually find it without it decomposing.
That's the short answer.  To learn more about it check Wikipedia for sodium hypochlorite.

I checked Wiki. It says it is hygroscopic, but not necessarily unstable.

Well, you can't buy it (from e.g., Sigma-Aldrich) in solid form, but only in solution. That says something about intrinsic stability…

Sodium hypochlorite is unstable in substance, but calcium hypochlorite, $\ce{Ca(ClO)_2}$, isn't, any chemical supply store won't hesitate to sell it.
There's other chlorine equivalents: chloroamine T, trichlorocyanuric acid, chlorosuccinimide are but a few.


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When considering Van der Waals forces, why do dipoles form?
Imagine two atoms, and only consider the Van der Waals force. The electron cloud will jitter due to its quantum mechanical nature- some of these jitters forming dipoles, some not. However, on average they form dipoles. Why?
This question is equivalent to asking (although it is useful here to ask in another way): why is the potential energy of a dipole lower than a non-dipole?
What determines which electrons are going to move to the middle of the atoms to form a dipole, and which ones move to the outside of an atom (I assume that a dipole is formed like this $e^-..ion...........e^-..ion$ for the centre of positive charge shifted rightwards of the centre of negative charge): given that the jitters are instantaneous, and photons travel only at the speed of light, how are dipoles formed quickly enough before the next jitter?
Am I falling into a rut because I am not considering the wave nature of electrons?

Hi Alyosha! Your two recent questions are heavy with anthropomorphism (atoms or electrons “decide to do X”, etc.). I don't know how to phrase it exactly, and when I manage I'll try to make an answer to these questions, but I believe this view negatively impacts your understanding of these questions.

Thanks for that. Sometimes anthropomorphisms help in that they remove confusion when asking a question quickly by stressing the fact that I want to know WHY rather than be given a superficial explanation, so I suppose I've gotten into bad habits. I've tried to edit my posts as I agree that here they detracted heavily from the questions' posts.

on average they form dipoles.

Not quite. On average they form the nice (electrically neutral) electron orbitals we know and love. However, when they happen to be in a dipole state, there is an attraction between the two atoms, and that is evidently more than enough to overcome the randomly occurring opposing dipole states. That's just mathematics.

why is the potential energy of a dipole lower than a non-dipole

It isn't. But quantum mechanics allows electrons to visit higher energy states, with a probability dependent on the energy difference.

What determines which electrons are going to move to the middle of the atoms to form a dipole, and which ones move to the outside of an atom

Nothing. The  "movement" of electrons is just a random statistical fluctuation in the location of the electrons' centers of charge. Quantum-mechanical events like this are inherently nondeterministic.  And strictly speaking, you can't tell electrons apart anyway. There is no "which electron," at one point in time you have one electron here and another there, the next they may have switched places. Or not, you can never know.

(I assume that a dipole is formed like this $e−..ion...........e−..ion$ for the centre of positive charge shifted rightwards of the centre of negative charge)

Don't think of those as being electrons, just the statistical average of the negative charge distribution.

given that the jitters are instantaneous, and photons travel only at the speed of light, how are dipoles formed quickly enough before the next jitter

First, They're not quite instantaneous (that would have them basically everywhere possible at once), and not quite jitters.
In summary, don't think of electrons as particles with a distinct position moving around, just think of them as waves, with a peak that occasionally forms, and that peak causes the electric arrangement of two nearby atoms to, on average, be mildly attractive.

I still don't see why the particles will attract, because:
1- Given that the average state of the electrons is the neutral and non-attractive AND an attractive dipole is no more energetically favourable than a repulsive opposing dipole, can I conclude that the electron peaks are arranged so the atoms are attracting for as much time as they are repulsing?
2- If (1) is true, then it seems to be that the electron peaks are arranged in an attractive way attract more strongly than the repulsive arrangement repulses.
I suppose all my former misunderstandings boil down to the above.

The atoms will only be repelled when the peaks of both electrons are between the two atoms. Anything else and they will be attracted.

Also- I take it that this http://www.dycem-cc.com/imagedownloads/vdw.gif model of inducing a dipole is an erroneous simplification, then?

''What determines which electrons are going to move to the middle of the atoms to form a dipole,...?''
Electrons are indistinguishable particles, so the question doesn't make sense. Electrons are excitations of the electron field (a quantum version of like wavelets on a lake), and their formation somewhere is just by concentrating the field at some place.
In some molecules such as benzene, some electrons don't even clump at all near particular locations but are strongly delocalized.

I was sloppy with my wording, yes.


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Clogged drain, commercial products are ineffective
I have a clogged drain from my shower. Commercial products like Savo or WC Net totally don't work. Is there any chemical substance which can be bought without permission or which can be made easily to clear the drain?
I live in Czech Republic, it's part of European Union, just for your information about the law system here (can be helpful to decide what's legal).
My first idea is sodium hydroxide, is it powerful enough?

if you decide to go with chemical products, do be careful. I would wear protective eyewear in case the chemicals splash out the drain or boil out or something.

I don't know what the products you mention contain, but you basically have four chemical weapons, and two physical weapons against a clogged drain. Choosing depends on the nature of the clog, which is hard to guess.
Chemistry first:

Acidity: sulfuric acid. ’nough said.
Basicity (causticity): lye, a.k.a. “caustic soda”, a.k.a. sodium hydroxide.
Oxidation: bleach or peroxide.
Enzymes: take time to act, but can be very powerful. This you should find at an eco-friendly store or supermarket aisle.

Now, on to physics:

Hot water: if the pipe is clogged by solidified fats, boiling water can melt them and unclog. Hardly typical for a shower drain, but very often useful for kitchen sink.
Last, but not least: mechanical action. Either a drain snake, or a plunger. Most often, the clog is a localized problem, and by breaking it into smaller parts, you will restore the flow.

For a typical shower drain, hair and other organic matter is most likely the issue, so I would choose drain snake first, then sodium hydroxide, then enzymes.

Not a chemistry answer: but a very successful one, in my experience: get a Ropump. It's a cross between a plunger and a bicycle pump. It makes drain-unclogging a pleasure.
Mechanical action is just so much more successful for this task, than chemical.  Because water is extremely resistant to being compressed, the Ropump's action, in turning water into a liquid drain rod and forcing it down the drain, has a pretty high success rate.
You pull plunger out, while the far end sits in a reservoir of water. This fills the body of the pump with water. You then fit the far end over the blocked drain, and push the plunger in, forcing the water down the drain, and displacing the blockage.
In the end, you don't know what the chemistry is at the blockage. So it's easier to anticipate the unintended consequences of a mechanical, rather than chemical, intervention.

“It makes drain-unclogging a pleasure” — I laughed at that one. Some things should not become a pleasure… or else you'll end up enjoying Saturday afternoons unclogging others’ pipes for fun :)

OK, but that is probably too expensive for 1 time unblocking.

@Salda sorry to be the one to tell you, but this won't be the last blocked drain that you have to deal with. Think of it as an investment for the future. It should be about €20-25

I am able to detach the bend to the outside drain. I then push a hose pipe in as far as possible. This clears a lot of stuff. I have also inserted a water-blaster at both ends. This was most successful but the shower cubicle then needs rinsing.


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Identifying a glass frosting chemical
I have purchased an "unknown" white coloured powder from the market which can frost glass objects. Here are some properties of this powder

White in colour
When mixed with HCL or water, it results in the soluion becoming ice cold, the reaction is endothermic.
When kept for a while (10 minutes) the white powder settles at the bottom of the plastic vessel and the acid floats above it.
When stirred, the entire solution becomes milky
When glass is dipped in this solution for 90 seconds, it becomes frosted/translucent.

Since I am not sure what this chemical is, I cannot determine the safest method to dispose it. The person from whom I brought this told me to put it down the drain, but I wouldnt choose to do that before I know if it will not effect the water table near my locality.
Can someone point out what could this powder be, and the best way to dispose it off?

I have been working with acids for over 20 years. The best solution to get rid of any type of acis is with baking soda. Baking soda will neutralize the acid to an organic matter and then can be discarded down the drain. Once neutralized it will have no effect on the water system

The most common glass-etching powders are solid ammonium bifluoride and solid potassium bifluoride. See for example the Material Safety Data Sheet for “EtchON Acid-Frost Glass Frosting Powder”, or this one. This is not a definite identification, but it fits your description, and those two are the most commonly available such chemicals.
Disposal is a bit tricky. Definitely don't put it down the drain! The first MSDS linked above says:

Ecological Information:
  Noxious for the water inhabitants. Avoid disposal into rivers, sewer system and land penetration.

The second gives a hint at disposing of it:

Disposal Instructions:
  All wastes must be handled in accordance with local, state and federal regulations. Material can be converted to a less hazardous material by weak reducing agents followed by neutralization.

So, contact your local waste management authorities, and ask them how to proceed. Do not attempt it neutralize it by yourself unless (i) you know what you are doing, and (ii) you have the appropriate hardware (fume hood, e.g.).

The first link is from the company I brought it from, makes me wonder why I wasnt provided this information when asked. Looks like its not worth dealing with this chemical, doesnt feel right.

@Rupin: Just to reinforce how little you want to mess with this stuff without good reason: bifluoride, when combined with an acid, becomes hydrogen fluoride. "Aqueous hydrofluoric acid is a contact-poison with the potential for deep, initially painless burns and ensuing tissue death. By interfering with body calcium metabolism, the concentrated acid may also cause systemic toxicity and eventual cardiac arrest and fatality, after contact with as little as 160 cm² (25 square inches) of skin."


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Would the detaching of dendrites from electrodes in a cell, result in the crystal dissolving back into the electrolyte?
Dendrites that form on the electrodes of cells shorten the cell's lifetime, and result in reduced capacity before the end of life.
If the dendrites could be detached from the electrode, would they then dissolve back into the electrolyte, or would they remain as solids suspended in it? How does this effect battery lifetime and capacity?