A 0.0500-mol sample of a nutrient substance is burned in a bomb calorimeter cont
ID: 998910 • Letter: A
Question
A 0.0500-mol sample of a nutrient substance is burned in a bomb calorimeter containing 2.00 Times 10^2 g H_2 O. If the formula weight of this nutrient substance is 114 g/mol, what is the fuel value (in nutritional Cal) if the temperature of the water increased 5.70 degree C (shown your calculations) One-eighth of a sample of thorium-227 remains undecayed after 54 days. What is the half-life of this thorium isotope? For the equilibrium: 2l(g) rightarrow I_2 (g) the rate law is: rate = k[I]^2 at 23 degree C, k = 7.0 Times 10^9 M^-1 s^-1. What effect will doubling the [I] have on the rate?Explanation / Answer
Heat of combustion of 0.05 mol of substance = mass of water in calorimeter*specific heat of water*temperature rise
Heat of combustion of 0.05 mol of substance= 2.00 *102 g *4.184 J/g0C * 5.70 0C = 4769.76 J
The amount of energy produced by the complete combustion of a material or fuel. Measured in units of energy per amount of material is the fuel value.
Weight of 0.05 mol of substance= 114*0.05=5.7 g
Thus Fuel value of substance = 4769.76 J/5.7 g = 836.8 J/g =836.8 KJ/Kg.
4) a) One-eighth of sample will remain after 3 times of half-life time.Thus 54 days= 3 *half-life of sample
Therefore, half-life of sample = 54/3 = 18 days.
b) rate of reaction = K[I]2
Say initial rate of reaction = 7*109[I]2 M/s
On doubling the [I] , rate of reaction increases by 4- fold. rate of reaction= 7*109[2I]2 M/s=4*7*109[I]2 M/s =28*109[I]2 M/s
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