Answer:
Explanation:
From the given information:
The density of O₂ gas =
here:
P = pressure of the O₂ gas = 310 bar
=
= 305.97 atm
The temperature T = 415 K
The rate R = 0.0821 L.atm/mol.K
molar mass of O₂ gas = 32 g/mol
∴
= 287.37 g/L
To find the density using the Van der Waal equation
Recall that:
the Van der Waal constant for O₂ is:
a = 1.382 bar. L²/mol² &
b = 0.0319 L/mol
The initial step is to determine the volume = Vm
The Van der Waal equation can be represented as:
where;
R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol
Replacing our values into the above equation, we have:
After solving;
V = 0.1152 L
∴
= 277.77 g/L
We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.
The answer is (3) methyl butane. The isomer means that they have the same chemical formula but different structure. Pentane has five carbon atoms and is alkane. Only methyl butane has five carbon atoms.
Speed (m/s) = distance (metres) ÷ time (seconds)
For Track 1: 0.2 ÷ 2 = 0.1m/s
For Track 2: 0.2 ÷ 2 = 0.1m/s
For Track 3: 0.6 ÷ 6 = 0.1m/s
For Track 4: 0.4 ÷ 6 = 0.07m/s
The train on Track 4 had the slowest speed because it's got the shortest speed and it's covering less distance per second therefore it is slower.