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3 years ago

Pls help me with my work!! PLEASEEEE

Chemistry

1 answer:

Eddi Din [679]3 years ago

5 0

Answer:

a c b a d a

Explanation:

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Explanation:

According to the given data, we will calculate the following.

Half life of lipase $t_{1/2}$ = 8 min x 60 s/min

= 480 s

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$

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Rate of hydrolysis $V_m_{o}$ = 0.07 mmol/L/s

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Therefore, time taken will be calculated as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

Now, putting the given values into the above formula as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

= $-\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
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= $1642.83 s \times \frac{1 min}{60 sec}$

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Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

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