A 0.109 g sample of an unknown gas occupies 112 mL at 100 degree C and 750 torr.
ID: 538509 • Letter: A
Question
A 0.109 g sample of an unknown gas occupies 112 mL at 100 degree C and 750 torr. What is the molecular weight of the compound? 22.7 amu 30.2 amu 35.4 amu 47.5 amu 53.6 amu Oxygen masks use canisters containing potassium superoxide. The superoxide consumes the CO_2 exhaled by a person and replaces it with O_2, which is summarized in the balanced reaction below. Calculate the mass of KO_2 required to consume 8.90 L of CO_2 at 22 degree C and 720 mm Hg. 4 KO_2 (s) + 2CO_2(G) rightarrow 2K_2CO_3 (s) + 3O_2 (g) 17.1 g 87.5 g 33.3 g 49.4 61.9 g A 0.109 g sample of an unknown gas occupies 112 mL at 100 degree C and 750 torr. What is the molecular weight of the compound? 22.7 g/mol 35.4 g/mol 30.2 g/mol 47.5 g/mol 53.6 g/mol A helium weather balloon is filled to a volume of 219,000 L at a weather station where the atmospheric pressure is 754 mm Hg and the temperature is 25 degree C. As the balloon rises, the pressure and temperature will decrease, so it is important to know how much the gas will expand to ensure the balloon can withstand the expansion. Calculate the new volume that the balloon will occupy at an altitude of 10.000 meters, where the atmospheric pressure has dropped to 210 mm Hg and the temperature is now-43 degree C. 331,000 L 419,000 L 608,000 L 735,000 L 813,000 LExplanation / Answer
As per the latest convention of IUPAC STP is 273 K and 1 bar pressure
750 torr = 1 bar
also, at STP one mole of a gas occupies 22.414 Volume
By Charles's law
V1/T1= V2/T2
the volume of this gas at STP
V1/273 = 0.112/373
V1= 0.08197 L
one mol occupies 22.4L
number of moles = 0.08197/22.4 = 0.0036 moles
molar mass = weight/number of moles = 0.109/0.0036 = 30.2 amu
option B is the correct naswer
23.
PV = nRT
n = PV/RT =
p = 0.947 atm
T = 295 K
R = 0.08206
V= 8.9 L
n = 0.947*8.9/(0.08206 *295) = 0.348
As per the balnced equation one mole needs 2 KO2
0.6963 moles KO2 is required
mass = 0.693*71.1 = 49.4 g (Note molar mass of KO2 is 71.1)
option d is the correct answer
As per the policy of Chegg only one question is answered.
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