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3 years ago

Consider the following reaction: 2 SO2(g) + O2(g) → 2SO3(g)

Chemistry

1 answer:

matrenka [14]3 years ago

7 0

Answer:

B) -1.00 M/min

Explanation:

Based on the reaction:

2 SO2(g) + O2(g) → 2SO3(g)

<em>2 moles of SO2 are consumed per mole of O2 producing 2 moles SO3</em>

<em />

In the problem, the sulfur dioxide has a rate of change of -2.00M/min. As the moles of oxygen that reacted are half the moles of SO2, its rate of change is halved, that is:

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In a chemical reaction, substrate molecule A is broken down to form one molecule of product B and one molecule of product C. The

MAVERICK [17]

Answer:

The answer to your question is d. 0.5 M

Explanation:

Data

[A] = 1M

K = 0.5

Concentration of B and C at equilibrium = x

Concentration of A at equilibrium = 1 - x

Equation of equilibrium

k = $\frac{[B][C]}{A}$

Substitution

$0.5 = \frac{[x][x]}{1 - x}$

Simplification

0.5 = $\frac{x^{2}}{1 - x}$

Solve for x

0.5(1 - x) = x²

0.5 - 0.5x = x²

x² + 0.5x - 0.5 = 0

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