1. Calculate the pressure of argon gas at 22 C ito 1.39 g of this gas is placed
ID: 883507 • Letter: 1
Question
1. Calculate the pressure of argon gas at 22 C ito 1.39 g of this gas is placed in a 825 ml. container. What is the molecular weight of a gas if 0.223 g of it occupies 149 ml. under 756 torr at 90°C? 2. 3. A sample of ideal gas under 1.70 atm at 22'C has a volume of 6.20 L. Calculate the volume of gas at SATP The volume of a sample of gas at 23°C and 1 .20 atm is V. Calculate its volume in terms of V at 90°C and 1.10 atm. 4. The pressure of a sample of gas when its volume is 1.20 L is 1.95 atm. What will be its pressure if this sample of gas placed in a 1.30 L container at the same temperature? 5. 6. How many liters of oxygen will be produced at STP when 1.03 g of KC10, is decomposed? 2KCIO49 2KCI(s) + 302(g). 7, A sample of hydrogen was collected over water at 25 when the atmospheric pressure is 758.0 torr. If the volume of gas is 250 ml, calculate (a) the partial pressure of hydrogen, and (b) mol of hydrogen. 8. The mass fraction of oxygen in the atmosphere is 0233 and that of nitrogen is 0.767. What is the partial pressure of each gas when the total pressure is 754 torr?Explanation / Answer
1) T = 22 + 273 = 295K
weight = 1.39 g
Ar molar mass = 40 g / mol
n = moles = weight / molar mass = 1.39 / 40 = 0.0348
volume (V) = 825 ml = 0.825 L
R = universal gas constant = 0.0821 L-atm / K mol
ideal gas equation PV = nRT
P = nRT/V
= 0.0348 x 0.0821 x 295 / 0.825
= 1.02 atm
pressure = 1.02 atm
(2)
V = 149ml = 0.149 L
T = 90 + 273 = 363 K
P = 756 torr
1 atm = 760 torr
? =756 torr
P = 756 / 760 = 0.9947 atm
PV = nRT
n = PV / RT
= 0.9947 x 0.149 / 0.0821 x 363
= 0.00497
moles = weight / molecular weight
0.00497 = 0.223 /molecular weight
molecular weight = 44.84 g / mol
(3)
P1 = 1.70 atm
T1 = 22+ 273 = 295K
V1 = 6.2 L
at STP
P2 = 1 atm
T2 = 298K
V2 = ?
P1V1/T1 = P2V2/T2
1.7 x 6.2 / 295 = 1 x V2 / 298
V2 = 10.65 L
volume at STP = 10.65 L
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