Explanation Given: Mass of argon gas = 12.5 g Pressure of argon gas = 1.05 atm Temperature of argon gas = 322 K Formula used: According to ideal gas law , PV = nRT Here, P = pressure of a gas V = volume of the gas n = moles of the gas R = gas constant T = temperature of the gas in Kelvin scale Calculation: Here, P = pressure of argon gas = 1.05 atm V = volume of argon gas =? n = moles of argon gas = M a s s M o l a r m a s s = 12 . 5 g 39 . 948 g / m o l = 0 . 313 m o l [Since molar mass of argon = 39.948g/mol] R = 0.082 L.atm/K.mol T = temperature of the gas in Kelvin scale = 322 K PV = nRT V = n R T P = ( 0 . 313 m o l ) × ( 0 . 082 L . a t m / K . m o l ) × ( 322 K ) ( 1 ...

Chemistry: Structure and Properties (2nd Edition)

2nd Edition

ISBN: 9780134293936

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 10, Problem 38E

Interpretation Introduction

To determine: Volume of 12.5 g of argon gas at a pressure of 1.05 atm and a temperature of 322 K and the volume of 12.5 g of helium gas under the same conditions.

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Chapter 10, Problem 37E

Chapter 10, Problem 39E

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Chemistry: Structure and Properties (2nd Edition)

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Volume of 12.5 g of argon gas at a pressure of 1.05 atm and a temperature of 322 K and the volume of 12.5 g of helium gas under the same conditions.

Solution Summary: The author calculates the volume of 12.5 g of argon gas at a pressure of 1.05 atm, and the temperature of 322 K.

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To determine: Volume of 12.5 g of argon gas at a pressure of 1.05 atm and a temperature of 322 K and the volume of 12.5 g of helium gas under the same conditions.

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Chapter 10 Solutions