A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres
1 answer:
Answer:
The partial pressure of argon in the flask = 71.326 K pa
Explanation:
Volume off the flask = 0.001
Mass of the gas = 1.15 gm = 0.00115 kg
Temperature = 25 ° c = 298 K
Gas constant for Argon R = 208.13
From ideal gas equation P V = m RT
⇒ P =
Put all the values in above formula we get
⇒ P = × 208.13 × 298
⇒ P = 71.326 K pa
Therefore, the partial pressure of argon in the flask = 71.326 K pa
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Answer:
32.6%
Explanation:
Equation of reaction
2KClO₃ (s) → 2KCl (s) + 3O₂ (g)
Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)
Molar volume of Oxygen at s.t.p = 22.4L / mol
since the gas was collected over water,
total pressure = pressure of water vapor + pressure of oxygen gas
0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C
pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1
P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p
Using ideal gas equation
=
V2 =
V2 = 664.1052 ml
245.2 yielded 67.2 molar volume of oxygen
0.66411 will yield = = 2.4232 g
percentage of potassium chlorate in the original mixture = = 32.6%