Lithosphere, asthenosfhere , lower mantle
1. The molar mass of the unknown gas obtained is 0.096 g/mol
2. The pressure of the oxygen gas in the tank is 1.524 atm
<h3>Graham's law of diffusion </h3>
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
<h3>1. How to determine the molar mass of the gas </h3>
- Rate of unknown gas (R₁) = 11.1 mins
- Rate of H₂ (R₂) = 2.42 mins
- Molar mass of H₂ (M₂) = 2.02 g/mol
- Molar mass of unknown gas (M₁) =?
R₁/R₂ = √(M₂/M₁)
11.1 / 2.42 = √(2.02 / M₁)
Square both side
(11.1 / 2.42)² = 2.02 / M₁
Cross multiply
(11.1 / 2.42)² × M₁ = 2.02
Divide both side by (11.1 / 2.42)²
M₁ = 2.02 / (11.1 / 2.42)²
M₁ = 0.096 g/mol
<h3>2. How to determine the pressure of O₂</h3>
From the question given above, the following data were obtained:
- Volume (V) = 438 L
- Mass of O₂ = 0.885 kg = 885 g
- Molar mass of O₂ = 32 g/mol
- Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
- Temperature (T) = 21 °C = 21 + 273 = 294 K
- Gas constant (R) = 0.0821 atm.L/Kmol
The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
Divide both side by V
P = nRT / V
P = (27.65625 × 0.0821 × 294) / 438
P = 1.524 atm
Learn more about Graham's law of diffusion:
brainly.com/question/14004529
Learn more about ideal gas equation:
brainly.com/question/4147359
Answer:
V of Sulfur tetrafluoride is 17.2 L
Explanation:
Given data;
T = -6°C = 267K [1° C = 273 K]
n = 786 mmol of SF4 which is 0.786 mol
P = 1 atm
from ideal gas law we have
PV = nRT
where n is mole, R is gas constant, V is volume
V of Sulfur tetrafluoride is 17.2 L
