Can I add the kj/mol and kj/gram together to get final answer. It seems the webpage implies it works by the working out??? Or its wrong and i need to convert the kj/gram to mols ?
33.30 grams of ice at 0.00 °C has heat added to it until steam at 150.0 °C results. Calculate the total energy expended. (Hint: melt, raise, boil, raise
Working out
Melt:
q = (33.30 g / 18.0 g mol¯1) (6.02 kJ / mol)
2) Raise in temperature as a liquid:
q = (33.30 g) (100.0 °C) (4.184 J/g °C)
3) Boil:
q = (33.30 g / 18.0 g mol¯1) (40.7 kJ / mol)
4) Raise in temperature as a gas:
q = (33.30 g) (50.0 °C) (2.02 J/g °C)
5) Add 'em up:
103.7 kJ becomes 104 kJ when rounded to three sig figs
(the joules have been converted to kj/gram
33.30 grams of ice at 0.00 °C has heat added to it until steam at 150.0 °C results. Calculate the total energy expended. (Hint: melt, raise, boil, raise
Working out
Melt:
q = (33.30 g / 18.0 g mol¯1) (6.02 kJ / mol)
2) Raise in temperature as a liquid:
q = (33.30 g) (100.0 °C) (4.184 J/g °C)
3) Boil:
q = (33.30 g / 18.0 g mol¯1) (40.7 kJ / mol)
4) Raise in temperature as a gas:
q = (33.30 g) (50.0 °C) (2.02 J/g °C)
5) Add 'em up:
103.7 kJ becomes 104 kJ when rounded to three sig figs
(the joules have been converted to kj/gram
