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:
The air in the atmosphere consists of nitrogen, oxygen, which is the life-sustaining substance for animals and humans, carbon dioxide, water vapour and small amounts of other elements (argon, neon, etc.). Higher in the atmosphere air also contains ozone, helium and hydrogen.
There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.
Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
Special Structures in Plant Cells. Most organelles are common to both animal and plant cells. However, plant cells also have features that animal cells do not have: a cell wall, a large central vacuole, and plastids such as chloroplasts.
The primary structure is the amino acids' unique sequence. The polypeptide's local folding to form structures such as the α-helix and β-pleated sheet constitutes the secondary structure. The overall three-dimensional structure is the tertiary structure
