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A 5.40 mol sample of solid A was placed in a sealed 1.00 L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.20 M, where it remained constant. A(s) = B(g) + C(g) Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain?

A 5.40 mol sample of solid A was placed in a sealed 1.00 L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.20 M, where it remained constant. A(s) = B(g) + C(g) Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain?

Trigonometry (MindTap Course List)
Trigonometry (MindTap Course List)
10th Edition
ISBN:9781337278461
Author:Ron Larson
Publisher:Ron Larson
Chapter6: Topics In Analytic Geometry
Section: Chapter Questions
Problem 33CT
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A 5.40 mol sample of solid A was placed in a sealed 1.00 L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.20 M, where it remained constant. A(s) = B(g) + C(g) Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain? moles of A: 4.20 mol Incorrect
Transcribed Image Text:A 5.40 mol sample of solid A was placed in a sealed 1.00 L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.20 M, where it remained constant. A(s) = B(g) + C(g) Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain? moles of A: 4.20 mol Incorrect
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A 5.40 mol sample of solid A was placed in a sealed 1.00 L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.20 M, where it remained constant. 
A(s)⇔B(g)+c(g)
Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain? 

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