Balloons and lifting gases Hydrogen is the lightest element, so it's cable of lifting the most weight in out atmosphere (probably not the best terminology there, but you get the picture)
Would hot hydrogen (in the same sense as hot air) be able to lift even more mass?
Would a higher or lower density of hydrogen in a ballon lift more?
If you could have a balloon which had nothing in it (it was a vacuum inside) would that lift more than a hydrogen balloon?
Basically what are the physics to balloons and lifting?
(really not sure what to tag this, so if someone else could that'd be great)
 A: 
Would hot hydrogen (in the same sense as hot air) be able to lift even more mass? 

Yes. Though I suppose the fire danger goes up, and you certainly can't use a propane burner to warm it...

Would a higher or lower density of hydrogen in a ballon lift more? 

Lower density always means higher buoyancy.

If you could have a balloon which had nothing in it (it was a vacuum inside) would that lift more than a hydrogen balloon?

Yes, and this has been proposed in various ways in science fiction literature. The engineering challenge is finding a away to confine the vacuum that is as light as a gas bag so that you don't loose the advantage to extra weight.
In general a volume $V$ of material of density $\rho$ immersed in a fluid of density $\rho_f$ experiences a buoyant force of
$$ F_b = gV\rho_f $$
and a weight of 
$$W = -gV\rho $$
so the available lifting force is 
$$ F_l = gV(\rho_f - \rho) .$$
Where the object is floating at the surface of a liquid the buoyant force is modified to reflect the volume of liquid displaced $F_b = g V_d \rho_f$ where $V_d$ is enough to cover the weight of the floating object.
A: dmckee's answer is a great not-too-technical description of buoyancy. Read that first. But in case you're interested, I thought I would go into some more detail.
The buoyant force on a submerged object (e.g. a balloon submerged in air) is equal to the weight of the displaced fluid,
$$F_b = \rho_f g V$$
as dmckee said. The physical origin of this force is actually the pressure difference between the top and bottom surfaces of the floating object. Pressure in a fluid at a certain height is related to the depth of the fluid above that height by
$$P(z_2) - P(z_1) = \rho_f g (z_2 - z_1)$$
that is, density of fluid times gravitational acceleration times height difference. If you have a rectangular box whose top and bottom surfaces are flat, then it's pretty easy to calculate the buoyant force as the pressure differential times the area of those surfaces,
$$F_b = (\Delta P)(A) = \rho_f g \Delta z A = \rho_f g V$$
For an irregular shape, you'll have to do an integral of some sort. For example, I once wrote a blog post which discusses, in part, deriving the buoyant force (and weight) from the minimization of potential energy, and that method can be easier to apply to irregular objects. (There are also a couple of interesting applications, even if you don't care about the math.)
You can also take into account variations in density (or gravitational acceleration) over the size of the balloon by doing an integral. But, according to the US standard atmosphere model, the density of the atmosphere takes about $20\ \mathrm{km}$ to drop off to near zero, which corresponds to a fraction of a percent change over the height of a typical hot air balloon (a few tens of meters). That fraction of a percent is generally negligible, so you're pretty safe just using a single value for the density.
However, you can't neglect differences in density between vastly different altitudes. Remember that the buoyant force on the balloon is equal to the weight of the amount of fluid displaced. As you go higher, the density of the air drops, which means the balloon displaces a lower mass of air. Therefore, as the balloon rises, the buoyant force drops. Eventually it reaches a height at which the buoyant force exactly balances out the weight of the balloon (and basket), and the balloon levitates at that level. As has been said in the comments, for a controlled balloon, the operator can adjust the level by either heating the gas inside the balloon (thus making it expand and displace more air) or by letting some gas out (thus making the balloon contract and displace less air).
A: A vacuum balloon is a possibility, but I doubt it can provide more lift than a hydrogen balloon: you need a rigid shell to prevent implosion of the vacuum balloon, and my feeling is the shell will be too heavy. However, while vacuum balloons cannot compete on lift, they can have better altitude control: you can bleed air in to lose altitude and pump out air to gain altitude.
Together with my coauthor, I proposed some designs of vacuum balloons made of commercially available materials, such as boron carbide and aluminum honeycombs (US patent application 20070001053 (11/517915), or http://akhmeteli.org/wp-content/uploads/2011/08/vacuum_balloons_cip.pdf ). The main problem is so-called buckling (loss of stability). It is possible, but difficult, so nobody has built a vacuum balloon so far, as far as I know, although the idea of vacuum balloon is centuries old. 
A: Approximate mean atomic mass of air: 29
Approximate mean atomic mass of hydrogen: 2
So with both gases at the same temperature and pressure, you are getting $\frac{27}{29}=93\%$ of the theoretical lifting capacity.
You would have to heat the hydrogen to about 200 C in order to get to 95% efficiency. It would be much better to work on making the surrounding/supporting structure lighter than to heat the hydrogen - the apparatus plus fuel for the heating is likely to weigh much more than you gain in lift.
