The correct answer is 12.2% BaO.
The solution is found by dividing the mass of the BaO, which is 25.8 grams, by the total mass of the solution, which is 212 grams, then multiplying it by 100 to get the percentage:
Answer:
The molar mass of the gas is 36.25 g/mol.
Explanation:
- To solve this problem, we can use the mathematical relation:
ν = 
Where, ν is the speed of light in a gas <em>(ν = 449 m/s)</em>,
R is the universal gas constant <em>(R = 8.314 J/mol.K)</em>,
T is the temperature of the gas in Kelvin <em>(T = 20 °C + 273 = 293 K)</em>,
M is the molar mass of the gas in <em>(Kg/mol)</em>.
ν = 
(449 m/s) = √ (3(8.314 J/mol.K) (293 K) / M,
<em>by squaring the two sides:</em>
(449 m/s)² = (3 (8.314 J/mol.K) (293 K)) / M,
∴ M = (3 (8.314 J/mol.K) (293 K) / (449 m/s)² = 7308.006 / 201601 = 0.03625 Kg/mol.
<em>∴ The molar mass of the gas is 36.25 g/mol.</em>
<span>b.mixtures are made up of more than one component</span>