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densk [106]
4 years ago
6

A _______ allosteric regulator _______ the rate of an enzyme-catalyzed reaction by inducing the active site of the enzyme to bec

ome _______.
Chemistry
1 answer:
inn [45]4 years ago
6 0
<span>Below are the choices that can be found from other sources:

a) negative; increases; hidden
b) positive; increases; exposed
c) positive; decreases; hidden
d) negative; decreases; exposed

The answer is </span><span>b) positive; increases; exposed.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
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Which is a proper description of chemical equilibrium?
serg [7]
D) because both reactions are occurring at the same rate. They are not equal but their concentrations are constant.
3 0
3 years ago
Calculate 6890/1230000 and conclude using the correct scientific figure.
Vesnalui [34]

Answer:

5.60162×10^−3

5 0
4 years ago
The osmotic pressure of an aqueoussolution of urea at 300 K is
den301095 [7]

<u>Answer:</u> The freezing point of solution is -0.09°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=imRT

where,

\pi = osmotic pressure of the solution = 120 kPa

i = Van't hoff factor = 1 (for non-electrolytes)

m = concentration of solute in terms of molality = ?

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

T = temperature of the solution = 300 K

Putting values in above equation, we get:

120kPa=1\times m\times 8.31\text{ L kPa }mol^{-1}K^{-1}\times 300K\\\\m=0.05m

  • To calculate the depression in freezing point, we use the equation:

\Delta T=i\times K_f\times m

where,

i = Vant hoff factor = 1 (for non-electrolytes)

K_f = molal freezing point depression constant = 1.86°C/m

m = molality of solution = 0.05 m

Putting values in above equation, we get:

\Delta T=1\times 1.86^oC/m\times 0.05m\\\\\Delta T=0.09^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.  

  • The equation used to calculate freezing point of solution is:

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

where,

\Delta T = Depression in freezing point = 0.09 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

0.09^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-0.09^oC

Hence, the freezing point of solution is -0.09°C

4 0
3 years ago
student in chemistry 150-02 weighed out 55.5g of octane c8h18 and allowed it to react with oxygen, o2 the product formed were ca
Tasya [4]

Answer:

C8H8 + 10O2 → 8CO2 + 4H2O

Explanation:

unbalanced reaction:

C8H8 + O2 → CO2 + H2O

balanced for semireactions:

(1) 16H2O + C8H8 → 8CO2 + 40H+

(2) 10(4H+ + O2 → 2H2O)

⇒ 40H+ + 10O2 → 20H2O

(1) + (2):

balanced reaction:

⇒ C8H8 + 10O2 → 8CO2 + 4H2O

                      8 - C - 8

                     20 - O2 - 20

                      8 - H - 8

3 0
4 years ago
If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% saline solution, how much of each sol
GenaCL600 [577]
The correct answer is A. <span>167 milliliters of 7% solution and 333 milliliters of 4% solution and here is how: 
</span><span>If x is the number of milliliters of the 7% saline solution and y is the number of milliliters of the 4% saline solution then add up to 500 milliliters total, so x + y = 500. 
</span>and if we do
x + y = 500 
<span>x = 500 - y </span>
<span>0.07x + 0.04y = 25 (substitute 500 - y for x) </span>
<span>0.07(500 - y) + 0.04y = 25 </span>
<span>35 - 0.07y + 0.04y = 25 </span>
<span>-0.03y + 35 = 25 </span>
<span>-0.03y = -10 </span>
<span>y = 333.333... </span>
<span>y = about 333 </span>
<span>x = 500 - y = 500 - 333 = 167 
</span>Then you know why the answer is A. 
4 0
3 years ago
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