A sample of an unknown substance has a mass of 42g and a volume of
2 answers:
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Answer:
The number of moles of oxygen present in the crew cabin at any given time is 615.309 moles
Explanation:
The given parameters are;
Volume of the crew cabin = 74,000 L
Pressure of the crew cabin = 1.00 atm
Percentage of nitrogen in the mixture of gases in the cabin = 80%
Percentage of oxygen in the mixture of gases in the cabin = 20%
Temperature of the cabin = 20°C = 293.15 K
Therefore, volume of oxygen in the crew cabin = 20% of 74,000 L
Hence, volume of oxygen in the crew cabin =
From the universal gas equation, we have;
Where:
n = Number of moles of oxygen
P = Pressure = 1.00 atm
V = Volume of oxygen = 14,800 L
T = Temperature = 293.15 K
R = Universal Gas Constant = 0.08205 L·atm/(mol·K)
Plugging in the values, we have;
The number of moles of oxygen present in the crew cabin at any given time = 615.309 moles.
Answer:
- <u>104 g/mol</u>
Explanation:
Use the ideal gas equation:
Where:
- p is pressure: 0.950atm
- V is volume: unknown
- n is number of moles: unknown
- R is the universal constat of gases: 0.08206 atm.liter/ (K.mol)
- T is the absolute temperature: 345K
Use the <em>molar mass</em> of the gas to include the density in the formula:
- molar mass = mass in grams / number of moles
- ⇒ mass in grams = number of moles × molar mass
- density = mass in grams / volume
- ⇒ density = number of moles × molar mass / volume
- density = (n/V) × molar mass
- ⇒ n/V = density / molar mass
Clear n/V from the gas ideal equation and subsittute with density/molar mass:
- n/V = p/(RT)
- density / molar mass = n/V
- density/molar mass = p/(RT)
- molar mass = density × RT / p
Now you can subsitute the data:
molar mass = (3.50g/liter) × 0.08206 atm.liter/(K.mol) × 345K / 0.950 atm
- molar mass = 104.3 g/mol
- Round to the nearest whole number: 104g/mol ← answer