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4 years ago

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What are some examples of monomers and oligomers?

1 answer:

aleksandr82 [10.1K]4 years ago

7 0

<span>What are some examples of monomers and oligomers?
</span>Organic molecules, such as proteins, carbohydrates, lipids and nucleic acids, are made of simple subunits called monomers. <span>Plasticizers are </span>oligomeric esters widely used to soften thermoplastics such as PVC and <span>urethane acrylate </span>.
<span>
</span><span>If a chemical compound accelerates and regulates metabolic reactions, which type of role does it play - structural or physiological?
</span>I believe the function that it plays would be physiological since it focuses more on the regulation of the reactions inside the body.

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The length of the flower garden is 529 centimeters.what is its length in meter?

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_Cl2 + _NaBr → _NaCl + _Br2<br><br> Answer: 1,2,2,1

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3 years ago

An asian elephant has a weight of 52,900 newtons. this enormous weight is balanced on four feet. in a circus act, these elephant

Ira Lisetskai [31]

<h3><u>Answer;</u></h3>

≈ 18,116,438 N/m² or 18,116,438 Pascals

<h3><u>Explanation;</u></h3>

Pressure is the force exerted on a unit area.

Thus pressure is given by; Force/ Area. Therefore, pressure is measured in N/m² or pascals.

In this case, the force is 52,900 Newtons

The area = 2,920 cm²

But; 1 m² = 1000000 cm², thus;

Area = 2,920 cm²/ 1000000 cm²

= 0.00292 m²

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4 years ago

Read 2 more answers

He density of water is 1 g/mL.

balandron [24]

#1

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2 years ago

Consider the following equilibrium process at 686°C. CO2(g) + H2(g) equilibrium reaction arrow CO(g) + H2O(g) The equilibrium co

coldgirl [10]

Answer:

a) The value of the $K_c$ is 0.52.

b)Concentration of all species when equilibrium reestablishes:

$[CO_2] = 0.4748 M$

$[H_2] = 0.0198 M$

$[CO] = 0.0752 M$

$[H_2O] =0.0652 M$

Explanation:

a) $CO_2(g) + H_2(g)\rightleftharpoons CO(g) + H_2O(g)$

Equilibrium concentration of species :

$[CO] = 0.050 M, [H_2] = 0.045 M, [CO_2] = 0.086 M, and [H_2O] = 0.040 M$

The expression of equilibrium constant is given as :

$\K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

$=\frac{0.050 M\times 0.040 M}{0.086 M\times 0.045 M}=0.517\approx 0.52 M$

b)

$CO_2(g) + H_2(g)\rightleftharpoons CO(g) + H_2O(g)$

Initially:

0.50 M 0.045 M 0.050 M 0.040 M

At equilibrium ;

(0.50-x) M (0.045-x) M (0.050+x) M (0.040+x) M

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

$0.52=\frac{(0.050+x) M\times (0.040+x) M}{(0.50-x) M\times (0.045-x) M}$

Solving fro x;

x = 0.0252

Concentration of all species when equilibrium reestablishes:

$[CO_2] = (0.50-x) M=(0.50-0.0252)M = 0.4748 M$

$[H_2] = (0.045-x) M= (0.045-0.0252) M=0.0198 M$

$[CO] = (0.050+x) M=(0.050+0.0252)M = 0.0752 M$

$[H_2O] = (0.040+x) M=(0.040+0.0252) M=0.0652 M$

8 0

4 years ago