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aalyn [17]
3 years ago
14

HELP with these 2 questions asap

Chemistry
2 answers:
Allisa [31]3 years ago
8 0

Answer: For question one...B. Because cells are too small to see without a microscope. It would have been impossible because cells are hard to deal with without a microscope because they are so tiny. So, without it, he wouldn't have been able to get the answer out of his discovery.

For question two...D. The nucleus is the control center of the cell. The cell theory states that, for 1., All living things are made of cells and their product., 2. New cells are created by only old cells dividing into two, and 3. Cells are he basic units of life. Cells having a nucleus is not mentioned and will never be because, cells don't need a nucleus to keep moving or keep the cell in control.

I hope this helps you! I really do hope it does.

katovenus [111]3 years ago
5 0

Answer:

number 1 is b and and number two d

Explanation:

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Answer:

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The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat
Dmitriy789 [7]

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

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}


Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

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

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 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)]

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
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

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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