doubles the pressure because the moles of gas are doubled 28 g N 2 is 1 mol of

Doubles the pressure because the moles of gas are

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doubles the pressure because the moles of gas are doubled (28 g N 2 is 1 mol of N 2 ). Process b won’t double the pressure since the absolute temperature is not doubled (303 K to 333 K). 7) An ideal gas at 7 o C is in a spherical flexible container having a radius of 1.00 cm. The gas is heated at constant pressure to 88 o C. Determine the radius of the spherical container after the gas is heated. Because the container is flexible, P is assumed constant. The moles of gas present are also constant.
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CHEM111 Week 5 Discussion Questions Fall 2014 8) Nitroglycerine, the explosive ingredient in dynamite, decomposes violently when shocked to form three gasses (N 2 , CO 2 , O 2 ) as well as water: C 3 H 5 (NO 3 ) 3 (l) → N 2 (g) + CO 2 (g) + O 2 (g) + H 2 O(l) a) Balance this equation b) Calculate how many moles of each gas are created in the explosion of 1kg of nitroglycerine. c)What volume would these gasses occupy at 1.0 atm? d)What are the partial pressures of each gas under these conditions? a) 4C 3 H 5 (NO 3 ) 3 (l) → 6N 2 (g) + 12CO 2 (g) + O 2 (g) + 10H 2 O(l) b) Molar Masses: C 3 H 5 (NO 3 ) 3 :227.09 g/mol N 2 , CO 2 , O 2 , H 2 O: Not needed Moles of C 3 H 5 (NO 3 ) 3 (l): (1000. g)(1 mol C 3 H 5 (NO 3 ) 3 /227.09 g) = 4.40 mol CO 2 : (4.40 mol C 3 H 5 (NO 3 ) 3 )(12 mol CO 2 /4 molC 3 H 5 (NO 3 ) 3 ) = 13.2 mol CO 2 N 2 :( 4.40 mol C 3 H 5 (NO 3 ) 3 )(6 mol N 2 /4 molC 3 H 5 (NO 3 ) 3 ) = 6.6 mol N 2 O 2 :( 4.40 mol C 3 H 5 (NO 3 ) 3 )(1 mol O 2 /4 molC 3 H 5 (NO 3 ) 3 ) = 1.1 mol O 2 Total: 13.2 mol + 6.6 mol + 1.1 mol = 20.9 mol gasses b) V = nRT/P V = (20.9 mol)(.08201 Latm/molK)(298K)/(1atm) V = 510 L d) P i = X i P = (n i /n)P P CO2 = (13.2 mol/20.9 mol)(1atm) = 0.63 atm P N2 = (6.6 mol/20.9 mol)(1atm) = 0.32 atm P O2 = (1.1 mol/20.9 mol)(1atm) = .053 atm 9)
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  • Spring '08
  • Kenney
  • Chemistry, Colum, Methode Champenoise

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