Question

The following information is given for mercury at 1 atm: Boiling point = 356.6°C, Heat of vaporization = 70.7 cal/g, Melting point = -38.9°C, Heat of fusion = 2.78 cal/g. How many calories of energy are needed to vaporize a 39.4 g sample of liquid mercury at its normal boiling point of 356.6°C?

A) 2778.58 cal
B) 3549.48 cal
C) 4167.94 cal
D) 4823.83 cal

Answer 1

The amount of energy needed to vaporize a 39.4 g sample of liquid mercury at its boiling point of 356.6°C is 2778.58 calories.

Step-by-step explanation:

To calculate the amount of energy needed to vaporize a sample of liquid mercury at its normal boiling point, we need to use the equation Q = m * Hvap, where Q is the amount of energy, m is the mass of the sample, and Hvap is the heat of vaporization.

In this case, the mass of the sample is 39.4 g and the heat of vaporization is given as 70.7 cal/g.

Substituting the values into the equation, we have Q = 39.4 g * 70.7 cal/g = 2778.58 cal.

Therefore, the answer is A) 2778.58 cal.