How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 103 mm Hg and
ID: 906049 • Letter: H
Question
How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 103 mm Hg and 23°C?
) How many moles of Ar are contained in a 5.00 L tank at 155°C and 2.80 atm? A) 0.399 moles B) 0.455 moles C) 1.10 moles D) 2.51 moles E) 0.289 moles
4.00-L flask contains nitrogen gas at 25 °C and 1.00 atm pressure. What is the final pressure in the flask if an additional 2.00 g of N2 gas is added to the flask and the flask cooled to -55°C?
The density of krypton gas at 1.21 atm and 50.0°C is ________ g/L. A) 0.295 B) 7.65 C) 3.82 D) 0.262 E) 0.0456
Thanks
Explanation / Answer
(1) How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 103 mm Hg and 23°C?
We know that PV = nRT
Where
T = Temperature = 23 oC = 23+273 = 296 K
P = pressure = 4000 mmHg = 4000 / 760 atm = 5.26 atm
n = No . of moles = ?
R = gas constant = 0.0821 L atm / mol - K
V= Volume of the gas=5.00 L
n = PV / RT
= ( 5.26 x 5.00) / (0.0821 x 296)
= 1.08 mol
So mass of F2 , m = Number of moles x molar mass
= 1.08 mol x 38 (g/mol)
= 41.5 g
(2) How many moles of Ar are contained in a 5.00 L tank at 155°C and 2.80 atm?
We know that PV = nRT
Where
T = Temperature = 155 oC = 155+273 = 428 K
P = pressure = 2.08 atm
n = No . of moles = ?
R = gas constant = 0.0821 L atm / mol - K
V= Volume of the gas=5.00 L
n = PV / RT
= ( 2.80 x 5.00) / (0.0821 x 428)
= 0.399 mol
(3)The density of krypton gas at 1.21 atm and 50.0°C is ________ g/L
We have PV = nRT
= (w/M) RT
PM = (w/V) RT
PM = dRT
Where d = density = mass(m) / Volume (V)
So d = PM / RT
P = pressure = 1.21 atm
M = molar mass = 83.8 (g/mol)
R = gas constant = 0.0821 Latm/mol-K
T = temperature = 50 oC = 50+273 = 323 K
Plug the values we get
d = PM / RT
= (1.21 x 83.8) / (0.0821 x323)
= 3.82 g/L
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