Answer:
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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 value of the activation energy of an uncatalyzed reaction is greater than that of a catalyzed reaction. As we know, a catalyst provides an alternative path for the reaction to happen at a faster rate. So, for a catalyzed reactio, activation energy is lesser than the original path.
C₀=2 mol/l
c₁=0.400 mol/l
v₁=100.0 ml = 0.1 l
c₁v₁=c₀v₀
v₀=c₁v₁/c₀
v(H₂O)=v₁-v₀
v₀=0.1*0.400/2=0.02 l = 20 ml
v(H₂O)=100 - 20 = 80 ml
It is necessary to mix 20 ml of the feed solution and 80 ml of water.