Explanation Volume of the room is calculated as follows, V = length×breadth×height = ( 7 .00 m ) ( 3 .00 m ) ( 3 .00 m ) = 63 m 3 Volume of air will be 63 m 3 . Concentration of C O is 1000 ppm . Hence, number of moles of C O is calculated as follows, n C O = 63 m 3 air × ( 100 cm 1 m ) 3 × 1 mL 1 cm 3 × 1 L 1000 mL × 1000 L CO 10 6 L air × 22 .414 mol CO 1 L = 1 .41 × 10 3 mol CO The given reaction is shown below, C 8 H 18 ( l ) + 8

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ISBN: 9781285460369

Author: STANITSKI

Publisher: Cengage Learning

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Kinetic-Theory of Gases

Kinetic theory of gases is the model or method which was developed in the 19th century by Maxwell and Boltzmann. This is possible as the interatomic force which is of short-range forces that are important for solids and liquids can be neglected for gases…

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Chapter 8, Problem 90QRT

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Interpretation:

Mass of gasoline that must be burned has to be calculated.

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Chapter 8, Problem 89QRT

Chapter 8, Problem 91QRT

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Chapter 8 Solutions

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:

Ch. 8.1 - Prob. 8.1PSP Ch. 8.1 - Prob. 8.1E Ch. 8.1 - Prob. 8.2E Ch. 8.2 - Prob. 8.3CE Ch. 8.2 - Prob. 8.4CE Ch. 8.3 - Prob. 8.6CE Ch. 8.3 - Prob. 8.2PSP Ch. 8.3 - Prob. 8.3PSP Ch. 8.3 - Prob. 8.4PSP Ch. 8.3 - Prob. 8.7CE

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Mass of gasoline that must be burned has to be calculated.

Solution Summary: The author calculates the mass of gasoline that must be burned and the concentration of CO.

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Interpretation: Mass of gasoline that must be burned has to be calculated.

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Chapter 8 Solutions

Interpretation:

Mass of gasoline that must be burned has to be calculated.