h₂o has a $delta h_{vap}=40.7$ kj/mol. what is the quantity of heat that is released when 27.9 g of h₂o…

h₂o has a $delta h_{vap}=40.7$ kj/mol. what is the quantity of heat that is released when 27.9 g of h₂o condenses? use $q = ndelta h$.\n60.00 kj\n61.05 kj\n63.09 kj\n68.60 kj

h₂o has a $delta h_{vap}=40.7$ kj/mol. what is the quantity of heat that is released when 27.9 g of h₂o condenses? use $q = ndelta h$.\n60.00 kj\n61.05 kj\n63.09 kj\n68.60 kj

Answer

Explanation:

Step1: Calculate moles of water

The molar mass of $H_2O$ is $M=(2\times1 + 16)=18\ g/mol$. The number of moles $n$ of $H_2O$ is $n=\frac{m}{M}$, where $m = 27.9\ g$. So $n=\frac{27.9\ g}{18\ g/mol}=1.55\ mol$.

Step2: Calculate heat released

We use the formula $q = n\Delta H$. Given $\Delta H_{vap}=40.7\ kJ/mol$, and since condensation is the reverse of vaporization, the heat released $q$ has the same magnitude as for vaporization. Substituting $n = 1.55\ mol$ and $\Delta H=40.7\ kJ/mol$ into the formula, we get $q=1.55\ mol\times40.7\ kJ/mol = 63.085\ kJ\approx63.09\ kJ$.

Answer:

63.09 kJ