Two springs have different spring constants. How could you identify the spring with the greater spring constant value?
2 answers:
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Answer:
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
Explanation:given values
Half life of lipase t_1/2 = 8 min x 60s/min = 480 s
Rate constant for first order reaction
k_d = 0.6932/480 = 1.44 x 10^-3 s-1
Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L
rate of hydrolysis Vm0 = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3
K_m = 5mmol/L
time take is given by
all values are given and putting these value we get
t=1642.83 secs
which is equal to
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]