6.73 L
This one requires the combined gas law.
P1V1/T1 = P2V2/T2
P1 = 150 kPa = 1.48 atm
V1 = 5 L
T1 = 300 K
P2 = 1 atm
V2 = This is what we’re solving for.
T2 = 273 K
P2 and T2 are based on the conditions of STP. I also changed P1 to atm to keep the units consistent.
Solve the equation for V2:
V2 = P1V1T2 / T1P2
V2 = (1.48 atm)(5 L)(273 K) / (300 K)(1 atm) = 6.73 L
Answer:
38.36 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of H₂ in atm (P = 1.0 atm, STP conditions).
V is the volume of H₂ in L (V = ??? L).
n is the no. of moles of H₂ in mol (n = mass/molar mass = (10.0 g)/(2.0 g/mol) = 5.0 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of H₂ in K (T = 0.0°C + 273 = 273.0 K, STP conditions).
<em>∴ V = nRT/P</em> = (5.0 mol)(0.0821 L.atm/mol.K)(273.0 K)/(1.0 atm) = <em>38.36 L.</em>
Answer:
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Answer:
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Explanation: