21.I°C, it In a mixture of helium and chlorine, occupying a volume of 23.3 L at

Question

21.I°C, it In a mixture of helium and chlorine, occupying a volume of 23.3 L at 744.6 mmlg and is found tha A) 1.97 g B) 34.2 g C) 70.9 g D) 32.2 g E) 1.82 g t the partial pressure of chlorine is 357 mmlHg. What is the total mass of the sample? Using the van der Wals equation, determine the pressure of 352.0 g of so(g) in a 5.30-L vessel at 751 K. For SO.(g), a-6.865 L2 atm/mol2 and b 0.05679 L/mol. (R-0.0821 L atm/(K. mol) 7. A) 63.9 atm B) 10.4 atm C) 75.3 atm D) 60.5 atm E) 10.5 atm 8. Using the following data, calculate the standard enthalpy of reaction for the coal gasification process 2C(s) + 2H20(g) CH4(g) + CO2(g). A) +15.3 kJ B) -378.6 kJ C) 378.6 k D) -157.2 kJ E) 116.0kJ 9, An unknown gaseous hydrocarbon consists of 85.63% carbon by mass. A 0.504-g sample of the gas occupies a volume of 0.268 L at STP. What is the identity of the gas? A) C4Hs B) C3H C) C2H D) CH 0 E) CH

Explanation / Answer

Q6

He +Cl mix

V = 23.3 L

P = 744.6 mm Hg

T = 21.1°C = 294.25 K

find total mass

PV = nRT

n = PV/(RT) = 744.6 * 23.3 / (62.4*294.25 )

n = 0.9448 mol

of which

x-Cl2 = 357/744.6 = 0.47945

mol of Cl2 = 0.47945*0.9448 = 0.4529 mol of Cl2

mol of He = 0.9448 - 0.4529 = 0.4919 mol of He

mass of Cl2 = mol*MW = 0.4529 *70.9 = 32.11 g

mass of He = mol*MW = 0.4919 *4= 1.9676 g

total masss = 1.9676+32.11

Total mass = 34.0776 g

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