posted by Stephen .
When 15.3g of NaNO3 was dissolved in water inside a constant pressure
calorimeter, the temperature fell from 25.00 °C to 21.56 °C. If the heat capacity of
the calorimeter (including the water it contains) is 1071 J/°C, What is the enthalpy
change for dissolving 1 mole of NaNO3? NaNO3 → Na+ + NO3- ∆H = ?
what formula do i use to solve this? i know q=C(change in T) but how does the 15.3g of nano3 affect the problem
You determine q for the system and that is q for 15.3 g NaNO3.
q/15.3 = J/gram. For 1 mole,
J/gram x (molar mass) = ??