ENERGY WOULD BE RELALISED, MEANING BONDS ARE BEING BROKEN, SO IT IS AN EXOTHERMIC REACTION
Answer:
The answer is below
Explanation:
Sulfuric acid is used for various purposes, some of which include the following:
1. It is used in the production of various manufactured goods.
2. It is used in the manufacturing of chemicals
3. It is also used in the making of fertilizer
4. It is used in the refining process of petroleum products
5. It is used in the processing of metals
Explanation:
According to the given data, we will calculate the following.
Half life of lipase
= 8 min x 60 s/min
= 480 s
Rate constant for first order reaction is as follows.
=
Initial fat concentration
= 45
= 45 mmol/L
Rate of hydrolysis
= 0.07 mmol/L/s
Conversion X = 0.80
Final concentration (S) =
= 45 (1 - 0.80)
= 9
or, = 9 mmol/L
It is given that
= 5mmol/L
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)]](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7BK_%7Bd%7D%7Dln%5B1%20-%20%5Cfrac%7BK_%7Bd%7D%7D%7BV%7D%7BK_%7BM%7D%20ln%20%28%5Cfrac%7BS_%7Bo%7D%7D%7BS%7D%29%20%2B%20%28S_%7Bo%7D%20-%20S%29%5D)
Now, putting the given values into the above formula as follows.
t =
= ![-\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s }{K_{M} ln (\frac{45 mmol/L }{9 mmol/L }) + (45 mmol/L - 9 mmol/L )]](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B1.44%20%5Ctimes%2010%5E%7B-3%7Ds%5E%7B-1%7D%7Dln%5B1%20-%20%5Cfrac%7B1.44%20%5Ctimes%2010%5E%7B-3%7Ds%5E%7B-1%7D%7D%7B0.07%20mmol%2FL%2Fs%0A%7D%7BK_%7BM%7D%20ln%20%28%5Cfrac%7B45%20mmol%2FL%0A%7D%7B9%20mmol%2FL%0A%7D%29%20%2B%20%2845%20mmol%2FL%20-%209%20mmol%2FL%0A%29%5D)
= 
= 27.38 min
Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.
Answer A: Connect a wire coil to an ammeter. Move a bar magnet into and out of the wire coil as you observe the ammeter.
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