Answer:
look below
Step-by-step explanation:
To determine the amount of energy released by the condensation of 246 grams of gold vapor, you will need to use the heat of vaporization of gold. The heat of vaporization is the amount of heat energy required to vaporize a substance at a specific temperature.
The heat of vaporization of gold at its boiling point (2970 degrees Celsius) is approximately 327 kJ/mol. Since the atomic weight of gold is 196.966 g/mol, the heat of vaporization for 246 grams of gold would be:
Heat of vaporization = (327 kJ/mol) * (246 g / 196.966 g/mol) = 402.7 kJ
This is the amount of energy that would be released if 246 grams of gold vapor were condensed at its boiling point. If the condensation occurs at a different temperature, the actual energy released may be different.
It's worth noting that the heat of vaporization is an intensive property, which means that it is independent of the amount of substance being vaporized. In other words, the heat of vaporization for 246 grams of gold vapor is the same as the heat of vaporization for any other amount of gold vapor.