Explanation:
1 kJ = 238.85 cal and
1 cal = 0.004187 kJ
so it will be 78.9×238.85 = <em><u>1</u></em><em><u>8</u></em><em><u>,</u></em><em><u>8</u></em><em><u>4</u></em><em><u>4</u></em><em> </em>calories
78.9 Kilojoules (kJ) = 18,844 Calories (IT) (cal)
Answer:
-24.76 kJ/g; -601.8 kJ/mol
Explanation:
There are two heat flows in this experiment.
Heat from reaction + heat absorbed by calorimeter = 0
q1 + q2 = 0
mΔH + CΔT = 0
Data:
m = 0.1375 g
C = 3024 J/°C
ΔT = 1.126 °C
Calculations:
0.1375ΔH + 3024 × 1.126 = 0
0.1375ΔH + 3405 = 0
0.1375ΔH = -3405
ΔH = -24 760 J/g = -24.76 kJ/g
ΔH = -24.76 kJ/g ×24.30 g/mol = -601.8 kJ/mol
To let you know what you are working with. if you have a wrong name for a graph then you will get the wrong answer
Take 23mols and multiply it by the atomic mass of Li:
23mol * 6.94g/mol = 159.62g