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k0ka [10]
3 years ago
15

Consider the equation below. Upper C a upper C upper O subscript 3 (s) double-headed arrow upper C a upper O (s) plus upper C up

per O subscript 2 (g). What is the equilibrium constant expression for the given reaction? K subscript e q equals StartFraction bracket upper C a upper O EndBracket over upper C a upper C upper O subscript 3 EndBracket EndFraction. K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction. K subscript e q equals StartBracket upper C upper O subscript 2 EndBracket. K subscript e q equals StartFraction 1 over StartBracket upper C upper O subscript 2 EndBracket EndFraction.
Chemistry
2 answers:
stiks02 [169]3 years ago
8 0

Answer:K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction

Explanation: the answer has it's root in Law of mass action which states that; the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients.

Len [333]3 years ago
8 0

Answer: is A

Explanation:pij

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Answer:

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Make sure that the equation for this reaction is balanced.

Coefficient of \rm C_2H_2 in this equation: 2.

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In other words, for every two moles of \rm C_2H_2 that this reaction consumes, two moles of \rm H_2O would be produced.

Equivalently, for every mole of \rm C_2H_2 that this reaction consumes, one mole of \rm H_2O would be produced.

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Apply this ratio to find the number of moles of \rm H_2O that this reaction would have produced:

\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}.

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