Ammonia (nh3) boils at -33∘c; at this temperature it has a density of 0.81 g/cm3. the enthalpy of formation of nh3(g) is -46.2 kj/mol, and the enthalpy of vaporization of nh3(l) is 23.2 kj/mol calculate the enthalpy change when 5 l of liquid nh3 is burned in air to give n2(g) and h2o(g).

-87098.82kJ

Explanation:

Hello,

This process could be attained into two steps due to the specified conditions at assuming that it is at -33°C at the beginning:

ΔΔΔ

Thus:

Δ

Once the ammonia is gaseous, the burning chemical reaction is:

So the enthalpy of reaction for the given amount of ammonia is:

Δ

Finally, the total enthalpy is:

Δ

Best regards.

answer; i believe that the correct answer is (a) potassium(k)