Part A) a sample of 0.550 mole of O2 is added to the 5.80 g of O2 in the contain

Question

Part A) a sample of 0.550 mole of O2 is added to the 5.80 g of O2 in the container. (Volume in liters to 3 significant figures) Part B) a sample of 2.05 g of O2 is removed from the 5.80 g of O2 in the container. (Volume in liters to 3 significant figures) Part C) a sample of 4.10 g of O2 is added to the 5.80 g of O2 gas in the container. (Volume in liters to 3 significant figures) 8 Problem 8.42 with feedback hapter Problem 8.42 with feedback A sample containing 5.80 g of O2 gas has an initial volume of 12.5 L. What is the final volume, in liters, when each of the following changes occurs in the quantity of the gas at constant pressure and temperature? You may want to reference ( pages 304-307) Section 8.6 while completing this problem

Explanation / Answer

initial mol of O2= mass/MW = 5.80/32 = 0.18125 mol of O2

V1 = 12.5 L

apply

V1/n1 = V2/n2

then

V2= n2/n1*V1

V2 = n2/(0.18125)*12.5

V2 = 68.965*n2

a)

V2 = 68.965*n2

n2 = 0.18125 + 0.55

V2 = 68.965*(0.18125 + 0.55) = 46.98 L

b)

relate directly to:

V1/n1 = V2/n2

12.5/5.8 = V2/(5.8-2.05)

V2 = 12.5/5.8 *(5.8-2.05)

V2 = 8.081 Liters

c)

n2 = (4.10/32) + 0.18125 = 0.309375

0.309375

V2 = 68.965*n2 = 68.965*0.309375 = 21.336 L

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