Answer:
3.94 L
Explanation:
From the question given above, the following data were obtained:
Mass of O₂ = 5.62 g
Volume of O₂ =?
Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:
Mass of O₂ = 5.62 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mole of O₂ =?
Mole = mass / molar mass
Mole of O₂ = 5.62 / 32
Mole of O₂ = 0.176 mole
Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.
Thus 5.62 g (i.e 0.176 mole) of O₂ occupied 3.94 L at STP
Answer:
0.02moles
Explanation:
To answer this question, the general gas law equation is used. The General gas law is:
Pv = nRT
Where; P = standard atmospheric pressure (1 atm)
V = volume (L)
n = number of moles
R = Gas law constant
T = Temperature
For this question; volume = 1.00L, atmospheric pressure (P) = 1 atm, R = 0.0821 L-atm / mol K, T = 600K, n = ?
Therefore; Pv = nRT
n = PV/RT
n = 1 × 1/ 0.0821 × 600
n = 1/49.26
n = 0.0203moles
Hence, there are 0.02 moles of gas.
