the heat of vaporization for water is 40.7 kj/mol. how much heat energy must 150.0 g of water absorb to boil…

the heat of vaporization for water is 40.7 kj/mol. how much heat energy must 150.0 g of water absorb to boil away completely? use the formula $q = ndelta h$.\n339.2 kj\n381.6 kj\n610.5 kj\n6,105 kj\ndone

the heat of vaporization for water is 40.7 kj/mol. how much heat energy must 150.0 g of water absorb to boil away completely? use the formula $q = ndelta h$.\n339.2 kj\n381.6 kj\n610.5 kj\n6,105 kj\ndone

Answer

Explanation:

Step1: Calculate moles of water

The molar mass of water ($H_2O$) is $M=(2\times1 + 16)=18\ g/mol$. Given mass $m = 150.0\ g$. The number of moles $n=\frac{m}{M}=\frac{150.0\ g}{18\ g/mol}\approx8.333\ mol$.

Step2: Calculate heat energy

We use the formula $q = n\Delta H$. Given $\Delta H=40.7\ kJ/mol$ and $n = 8.333\ mol$. So $q=8.333\ mol\times40.7\ kJ/mol = 339.2\ kJ$.

Answer:

339.2 kJ