526 L O2 x 1 mol O2 / 22.4 L = 23.5 mol O2
Answer:
Check explanation
Explanation:
From the question, the parameters given are 64.7g of benzene,C6H6; a starting temperature of 41.9°C and bringing it to 33.2°C.
Molar mass of benzene,C6H6= 78.11236 g/mol.
Things to know: heat capacity of benzene, C6H6= 1.63 J/g.K, the heat of fusion = 9.87 kj/mol.
STEP ONE(1): ENERGY USED IN MELTING BENZENE SOLID.
Using the formula below;
Energy used in melting the solid(in JOULES) = (mass of benzene/molar mass of benzene) × heat of Fusion.
=(64.7 g of C6H6/ 78.11236(g per mol) of C6H6) × 9.87 kJ per mol.
= 8.175 J.
= 0.008175 kJ.
STEP TWO (2): ENERGY OF HEATING THE LIQUID.
It can be calculated from the formula below;
Energy= heat capacity (J/g.K) × mass of benzene× (∆T).
= 1.63 J/g.K × 64.7 × (41.9-33.2).
= 917.5J.
= 0.9175 kJ.
Energy required to boil benzene= Energy required to melt the bezene + energy required for boiling.
= 0.008175+ 0.9175.
= 0.93kJ
Approximately, 1 kJ
The density of the gold is 19.3 grams/cc so each cc weighs 19.3grams. Now we can obtain the volume of gold from the given dimensions ie 4.72x8.21x3.98= 154.23 cc. So for the answer, just multiply the volume or 154.23 x 19.3= 2976.6 grams is the answer.
