Answer:
The heat required to vaporize 454 grams of liquid ammonia at its boiling point is 624 kJ.
Explanation:
Let's consider the vaporization of ammonia.
NH₃(l) ⇄ NH₃(g)
The heat of vaporization is 1374 J/g. The heat required to vaporize 454 grams of liquid ammonia at its boiling point is:
[tex]454g.\frac{1374J}{g} =6.24 \times 10^{5} J = 624 kJ[/tex]