Asked 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
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
Answers
Answered by
DrBob222
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) = ??
q/15.3 = J/gram. For 1 mole,
J/gram x (molar mass) = ??
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.