Question

How much energy is required to vaporize 2 kg of gold? use the table below and this equation: q = mlvapor?

Answer 1

If the gold is present in the liquid state, you only have to determine the latent heat of vaporization, or lvap. The empirical data for gold is 330 kJ/mol.

Q = mlvap
Q = (2 kg)(1 kmol/197 kg)(1,000 mol/1 kmol)
Q = 10.15 kJ

It needs an energy of 10.15 kilojoules.

Answer 2

Answer:

3440 kJ

Explanation:

The latent heat of vaporization equation represents how much energy is needed for a substance to change from a liquid to a gas.

Latent heat of vaporization equation:  $Q = mL_{vapor$

m = the mass of the substance

The latent heat of vaporization for gold is 1720 kJ/kg.

So, if you plug in the values into the equation:

Q = (2kg)(1720 kJ/kg)

You get: 3440 kJ