Question

a) At what temperature will a 2.9 g sample of neon gas exert a pressure of 490. torr in a 4.6 L container?

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b) Suppose a 23.1 mL sample of helium gas at 25.°C and 1.21 atm is heated to 50°C and compressed to a volume of 14.0 mL. What will be the pressure of the sample?

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c) A tank contains a mixture of 53.4 g of oxygen gas and 62.0 g of carbon dioxide gas at 23°C. The total pressure in the tank is 8.96 atm. Calculate the partial pressure (in atm) of each gas in the mixture.

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Pcarbon dioxide WebAssign will check your answer for the correct number of significant figures. atm

Answer 1

Dear Student, Kindly refer to the solution of all the questions. Feel free to ask if you face difficulty in understanding any part of the solution attached so that we can assist you in a better way.

Dear student, Given: P = 490 Tove V=4.6L T= ? No. of moles of Ne= mass of Ne Atomie moss - 2-98 20. 18g mot = 0.144 mol Applying ideal gas equation, PV = nRT T= PV R = 490 Torr x4.64 0.144 molt 62.364 L Tovek mot T & 250.99k a) Given: V=23-1 me Ta= 50°C = 323.15k T, = 25°C -298-15% Y = 14.ome P = 1.21 atm Pzz! Applying combined gas law, Ti P.T.O.

substituting values, 1. glatm x 9301 me = Px 14.0me. 298.15 K. 323-15k Pa = 1.2latm x 23.1 me x 323.15 298.15KX 14.ome Partial Pa = 2.16atro ③ We will apply Dalton's lawo of pressure according to which 1 Pi = x; x Protal] Pi = Partial plessure of each component Protel Total pressure X; = mole fraction of each component. no. of moles of af atomic mass = 53.49 = 1.67 mol 3ag/mol Ja en gas no. of moles of a mass mowot = 62g 44olmal 1.41 mol Total moles= l. 67+ 1.41 - 3.08 moles Р.Т.o

mole fraction of a no Total = 1.67 3.08 = 0.54 mole feaction of co n ce nTotal = 1.41 3.08 = 0.46 of O2 = Xox ratal Partial pressure 8.96 atin = 0.54 X & 4.84 atm Partial pressure of co= Xeox Protel = 0.46% 8.96 atm = 4ola atm Best wishes !!