Explanation Given: The volume is 25.0 mL . The density is 0.807 g/mL . Radius of cylinder is 10.0 cm and length of the cylinder is 20.0 cm .The pressure is 760.0 mmHg . Explanation: Calculate the mass of liquid nitrogen as follows: Mass = Volume × Density = 25 mL × 0.807 g/mL = 20.175 g Calculate the number of moles of liquid nitrogen as follows: Number of moles = Mass of liquid nitrogen Molar masss of liquid nitrogen = 20.175 g 28 g/mol = 0.720 mol Calculate the number of moles of air as follows: Number of moles = P V R T = 1 atm × 6.280 L 0.08206 L ⋅ atm mol − 1 K − 1 × 298 K = 0.2568 mol Calculate total number of moles as follows: Total number of moles = 0.2568 mol + 0.72 mol = 0.9768 mol Calculate the volume of container ( V Container ) as follows: V Container = π r 2 l e n g t h = ( 3.14 ) ( 10 cm ) 2 ( 20 cm ) = 6280 cm 3 ( 10 − 3 dm 3 1 cm 3 ) = 6.280 dm 3 The volume of the container is 6.280 L

Chemistry: A Molecular Approach, Books a la Carte Edition; Modified MasteringChemistry with Pearson eText -- ValuePack Access Card -- for Chemistry: A Molecular Approach, 4/e

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Chapter 5, Problem 112E

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To determine: The total force in the interior of the container.

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Chapter 5, Problem 111E

Chapter 5, Problem 113E

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Chemistry: A Molecular Approach, Books a la Carte Edition; Modified MasteringChemistry with Pearson eText -- ValuePack Access Card -- for Chemistry: A Molecular Approach, 4/e

Ch. 5 - Prob. 1SAQ Ch. 5 - Prob. 2SAQ Ch. 5 - Prob. 3SAQ Ch. 5 - Prob. 4SAQ Ch. 5 - Prob. 5SAQ Ch. 5 - Prob. 6SAQ Ch. 5 - Prob. 7SAQ Ch. 5 - Prob. 8SAQ Ch. 5 - Q9. A gas sample at STP contains 1.15 g oxygen and...Ch. 5 - Prob. 10SAQ

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