What volume of carbon dioxide is produced when 189 L of butane (C_4H_10) react w
ID: 958061 • Letter: W
Question
What volume of carbon dioxide is produced when 189 L of butane (C_4H_10) react with excess oxygen, according 10 the following reaction? (all gases are al the same temperature and pressure) 2 C_4H_10(g) + 13 O_2(g) rightarrow 8 CO_2(g) + 10 H_2O(g) 151 L 378 L 526 L 756 L 993 L A 9.70-g piece of solid CO_2 (Molar Mass - 44.01 g/mol) is allowed to sublime in a balloon. The final volume of the balloon is 1.00 L at 298 K. What is the pressure of the gas made the balloon 5.39 atm 2.37 times 10^2 atm 2.52 atm 0.186 atm none of these A 2 00-L glass soda bottle filled only with air is tightly capped at 24 degree C and 744 mm Hg. If the bottle is placed in water at 75 degree C, what is the pressure in the bottle? 238 mm Hg 872 mm Hg 2330 mm Hg 635 mm Hg 383 mm Hg A Boltzmann distribution gives the: orientations of molecules in a liquid sample pressures of individual gas molecules with a particular volume rates of diffusion of a range of gas pressures relative number of molecules in a gas with a particular speed voting preferences for a sample of RepublicansExplanation / Answer
28. 2 moles of butane gives 8 moles of carbon dioxide
189 L of butane gives 189 x 4 L of carbon dioxide
answer is d.
29. data given :
molar mass of CO2 = 44.01 g/mol
given mass of CO2 = 9.70 g
volume, V = 1 L
Temperature, T = 298 K
gas constant, R = 0.082 L atm/mol/K
for an ideal gas, PV = nRT
where n is number of moles of CO2,
(n = 9.70/44..01 = 0.22 moles)
Pressure, P = nRT/V
P = 0.22 x 0.082 x 298/1
P = 5.39 atm
answer is a
30. PV = nRT
P = 744 mmHg
V = 2 L
R = 62.363 L mmHg/K/mol
T = 24oC = 297K
n = PV/RT
n = (744 x 2) / (62.363 x 297.15)
n = 0.080 mole
pressure in 75oC,
P = nRT/V
T = 75oC = 348 K
P = (0.080 x 62.363 x 348)/2
P = 872 mmHg
answer is b.
31. Answer is d.
A boltzmann distibution gives the relative number of molecules in a gas with a particular speed.
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