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

Respuesta :

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:

3440 kJ

Explanation:

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