- The molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol.
- If this sample was placed under extreme pressure, the volume of the sample will decrease.
<h3>How to calculate molar mass?</h3>
The molar mass of a substance can be calculated by first calculating the number of moles using ideal gas law equation:
PV = nRT
Where;
- P = pressure
- V = volume
- T = temperature
- R = gas law constant
- n = no of moles
0.98 × 1.2 = n × 0.0821 × 287
1.18 = 23.56n
n = 1.18/23.56
n = 0.05moles
mole = mass/molar mass
0.05 = 0.458/mm
molar mass = 0.458/0.05
molar mass = 9.15g/mol
- Therefore, the molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol
- If this sample was placed under extreme pressure, the volume of the sample will decrease.
Learn more about gas law at: brainly.com/question/12667831
The answer would be 0.55 moles! good luck!
1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).
ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).
ΔrH = 180.6 kJ.
2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
ΔS = 24.8 J/K.
3) ΔG = ΔH - TΔS.
55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.
2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.
3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.
Answer:
The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s
Explanation:
Step 1: Data given
Molar mass helium = 4.0 g/mol
Molar mass O2 = 32 g/mol
Step 2: Graham's law
Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of the molecular mass : 1/(Mr)^0.5
Rate of escape for He = 1/(4.0)^0.5 = 0.5
Rate of escape for O2 = 1/(32)^0.5 = 0.177
The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s
