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2 years ago

Please help me ASAP I’ll mark Brainly

Chemistry

1 answer:

mixer [17]2 years ago

5 0

Answer:

cell

chloroplast and cell wall

nucleus

life processes

cell membrane

shape and size

vacuole

Hope it helps

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If the stocks in a system always decrease and never increase, the system has what type of feedback loop? (Select the best answer

AlekseyPX

A psitive feedback
rawmaterial waste

6 0

3 years ago

M=<br> V=432.6 mL<br> D=8.4 g/ml

Nitella [24]

Answer:

Explanation:

M=D times V

Answer-3,633.84g

Rounded Answer (correct sig figs)- 3600g

6 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

2 years ago

In this reaction, _____.

ella [17]

Answer:

AB + CD ----> AC + BD

Explanation:

If you think this reaction:

AB + CD ----> AC + BD

(Reactants) (Products)

All the statements are true.

7 0

3 years ago

Gallium has two naturally occurring isotopes: 69ga with a mass of 68.9256 amu and a natural abundance of 60.11% and 71ga. use th

Nataly [62]

There are two naturally occurring isotopes of gallium: mass of Ga-69 isotope is 68.9256 amu and its percentage abundance is 60.11%, let the mass of other isotope that is Ga-71 be X, the percentage abundance can be calculated as:

%Ga-71=100-60.11=39.89%

Atomic mass of an element is calculated by taking sum of atomic masses of its isotopes multiplied by their percentage abundance.

Thus, in this case:

Atomic mass= m(Ga-69)×%(Ga-69)+X×%(Ga-71)

From the periodic table, atomic mass of Ga is 69.723 amu.

Putting the values,

$69.723 amu=(68.9256 amu)(\frac{60.11}{100})+X(\frac{39.89}{100})$