Final answer:
To vaporize 2 kg of gold, 3440 kJ of energy is required. This calculation uses the latent heat of vaporization value, which is 1720 kJ/kg for gold.
Step-by-step explanation:
The amount of energy required to vaporize 2 kg of gold can be calculated using the latent heat of vaporization. According to the provided table, the latent heat of vaporization for gold is 1720 kJ/kg. Using the formula Q = mLvapor, where Q represents the energy required, m is the mass of the substance, and Lvapor is the latent heat of vaporization, we find that the energy required to vaporize 2 kg of gold is:
Q = (2 kg) × (1720 kJ/kg) = 3440 kJ.
Therefore, 3440 kJ of energy is needed to vaporize 2 kg of gold.