PaMastering Chemistry Chapter 11 Questions—Part 2 Question #14 Part A: What is t

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PaMastering Chemistry Chapter 11 Questions—Part 2

Question #14 Part A: What is the volume occupied by 26.9 g of argon gas at a pressure of 1.47 atm and a temperature of 370 K?

Question #15 Part A: A weather balloon is inflated to a volume of 29.4 L at a pressure of 742 mmHg and a temperature of 27.5 C. The balloon rises in the atmosphere to an altitude where the pressure is 380. mmHg and the temperature is -14.4 C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

Question #16 Part A: Use the molar volume of a gas at STP to determine the volume (in L) occupied by 34.4 g of neon at STP.

Question #17: A gas mixture contains each of the following gases at the indicated partial pressures: N2, 226 torr ; O2, 149 torr ; and He, 143 torr .

Part A: What is the total pressure of the mixture?

Part B: What mass of each gas is present in a 1.20 L sample of this mixture at 25.0 C? Enter your answers numerically separated by commas

Question #18: A heliox deep-sea diving mixture contains 2.0 g of oxygen to every 98.0 g of helium.

Part A: What is the partial pressure of oxygen when this mixture is delivered at a total pressure of 7.7 atm ? Express your answer using two significant figures.

Question #19: Ammonium nitrate decomposes explosively upon heating according to the following balanced equation:
2NH4NO3(s)2N2(g)+O2(g)+4H2O(g)

Part A: Calculate the total volume of gas (at 129 C and 768 mmHg ) produced by the complete decomposition of 1.43 kg of ammonium nitrate.

Explanation / Answer

1) Molar mass of Ar = 40 g/mole

Thus, moles of Ar in 29.6 g of it = mass/molar mass = 29.6/40 = 0.74

Applying Ideal Gas Equation , we get , Volume, V = (n*R*T)/P = (0.74*0.0821*370)/1.47 = 15.292 litres

2) Applying the combined gas law equation we get

P1*V1/T1 = P2*V2/T2

or, (742*29.4)/300.5 = (380*V2)/258.6

or, volume at higher altitude,V2 = 49.4 litres

3) 1 mole of each gas occupies 22.4 litres at STP

Now, molar mass of Ne = 20 g/mole

Thus, moles of Ne in 34.4 g of it = mass/molar mass = 1.72

volume of Ne = moles*22.4 = 38.528 litres

4) Total pressure of the gas mixture = sum of the partial pressure of the individual gases = 518 torr

5) Total moles of the gas mixture = (P*V)/(R*T) = (518*1.2/760)/(0.0821*298) = 0.033

Mole fraction of N2 = partial presure/total pressure = 226/518 = 0.436

Thus, moles of N2 = mole fraction*total moles = 0.0144

mass of N2 = moles*molar mass = 0.0144*28 = 0.403 g

Mole fraction of He = 143/518 = 0.276

moles of He = 0.276*0.033 = 0.0091

mass of He = 0.0091*4 = 0.0364 g

Mole fraction of O2 = 149/518 = 0.288

moles of O2 = 0.288*0.033 = 0.0095

mass of O2 = 0.0095*32 = 0.304 g

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