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enot [183]
3 years ago
11

) Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2is 7

.78 × 10-1M/s, what is the rate of the loss of O3
Chemistry
1 answer:
Travka [436]3 years ago
6 0

Answer:

Rate of the reaction is 0.2593 M/s

-0.5186 M/s is the rate of the loss of ozone.

Explanation:

The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

2O_3\rightleftharpoons 3O_2

Rate of formation of oxygen : 7.78\times 10^{-1} M/s

Rate of the reaction(R) =\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}

R=\frac{1}{3}\frac{d[O_2]}{dt}

Rate of formation of oxygen=3 × (R)

7.78\times 10^{-1} M/s=3\times R

Rate of the reaction(R): 0.2593 M/s

Rate of the reaction is 0.2593 M/s

Rate of disappearance of the ozone:

R=-\frac{1}{2}\frac{d[O_3]}{dt}

\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s

-0.5186 M/s is the rate of the loss of ozone.

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ruslelena [56]

Mole percent of O2 = 10% = 0.1 moles
Mole percent of N2 = 10% = 0.1 moles
Mole percent of He = 80% = 0.8 moles
Molar Mass of O2 = (2 x 16) x 0.1 = 3.2
Molar Mass of N2 = (2 x 14) x 0.1 = 2.8
Molar Mass of He = 4 x 0.8 = 3.2
1. Molar Mass of the mixture = 3.2 + 2.8 + 3.2 = 9.2 grams
2. Since at constant volume density is proportional to mass, so the ratio of
mass will be the ratio of density.
Ratio = Molar Mass of the mixture / Molar Mass of O2 = 9.2 / 32 = 0.2875




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Westkost [7]

Explanation:

To solve this problem, follow these steps;

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The first choice is the correct answer:

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