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alexandr1967 [171]
3 years ago
9

Consider the chemical system co + cl2 cocl2; k = 4.6 ? 109 l/mol. 2. how do the equilibrium concentrations of the reactants comp

are to the equilibrium concentration of the product?
a.they are much smaller.

b.they are much bigger.

c.they are about the same.

d.they have to be exactly equal.

e.you can't tell from the information given.
Chemistry
1 answer:
ikadub [295]3 years ago
3 0
<span>they are much bigger because </span><span>equilibrium constant shows ratio of  </span><span>concentrations of the reactants to the equilibrium concentration of the product.</span>
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The total amount of solute decreases.
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Photosynthesis can be represented by6CO2(g)+6H2O(l)⇌C6H12O6(s)+6O2(g)Which of the following will be false when the photosynthesi
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Answer:

The concentration of O2 will begin decreasing and The concentrations of CO2 and O2 will be equal.

Explanation:

Equilibrium occurs when the velocity of the formation of the products it's equal to the velocity of the formation of the reactants, thus the concentrations of the compounds remain constant.

Analyzing the information and the reaction given, we can notice that in equilibrium the rate (velocity) of formation of O2 (product) is equal to the rate of formation of CO2 (reactant).

As the CO2 and H2O are placed in the reaction, the Le Chateliêr's principle states that the equilibrium must shift to reestablish the equilibrium, thus, they must be consumed, and the concentration of O2 must increase.

As state above, in equilibrium, the concentrations didn't change, thus, the concentrations of CO2 and O2 will not change.

The concentrations of CO2 and O2 depends on the rate of the reaction and the initial quantities presented, so it's not possible to affirm they'll be equal.

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Calculate the pH of the solutions: [H^+]= 1.6 x 10^-3 M
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Answer:

A) pH = 2.8

B) pH = 5.5

C) pH = 8.9

D) pH = 13.72

Explanation:

a) [H⁺]  = 1.6 × 10⁻³ M

pH = -log [H⁺]

pH = -log [1.6 × 10⁻³ ]

pH = 2.8

b) [H⁺]  = 3 × 10⁻⁶

pH = -log [H⁺]

pH = -log [3 × 10⁻⁶ ]

pH = 5.5

c) [OH⁻] = 8.2 × 10⁻⁶

pOH = -log[OH]

pOH = -log[8.2 × 10⁻⁶]

pOH = 5.1

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 5.1

pH = 8.9

d) [OH⁻] = 0.53 M

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pOH = -log[0.53]

pOH = 0.28

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 0.28

pH = 13.72

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