How many significant figures are in 31400

Mandarinka [93]

C. The answer is c.

4 0

3 years ago

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.

6 0

3 years ago

The average person in the United States is exposed to the following amount of radiations annually. Rank the following source of

Papessa [141]

<span>#1 is air radon, #2 is x-ray, #3 is ground, #4 is cosmic radiation, #5 is TV tube, #6 is weapons test fallout . That's all I got hope I helped!</span>

8 0

3 years ago

Read 2 more answers

A hydrogen halide diffuses 1.49 times faster than HBr. This hydrogen halide is

marysya [2.9K]

To solve this problem, we must assume ideal gas behaviour so that we can use Graham’s law:

vA / vB = sqrt (MW_B / MW_A)

where,

<span>vA = speed of diffusion of A (HBR)</span>

vB = speed of diffusion of B (unknown)

MW_B = molecular weight of B (unkown)

MW_A = molar weight of HBr = 80.91 amu

We know from the given that:

vA / vB = 1 / 1.49

So,

1/1.49 = sqrt (MW_B / 80.91)

MW_B = 36.44 g/mol

Since this unknown is also hydrogen halide, therefore this must be in the form of HX.

HX = 36.44 g/mol , therefore:

x = 35.44 g/mol

From the Periodic Table, Chlorine (Cl) has a molar mass of 35.44 g/mol. Therefore the hydrogen halide is:

HCl

6 0

3 years ago

Answer both 8 and 9 will give u brain list with explanation as well so don’t just answer

goldfiish [28.3K]

Answer:

8. the answer is B.

9. the answer is A.

Explanation:

Hello!

8. In this case, by bearing to mind that the limiting reactant is always completely consumed and the excess one remain as a leftover at the end of the reaction, we can also infer that as all the limiting reactant is consumed, it must determine the maximum amount of product as the excess reactant will hypothetically produce a greater mass than expected; thus, the answer to this question is B.

9. In this case, since the mole ratio of oxygen to water is 1:2, the following proportional factor is used to calculate the produced mass of water:

$3molO_2*\frac{2molH_2O}{1molO_2}=6molH_2O$

Thus, the answer is this case is A.

Best regards!

6 0

3 years ago