<span>The ideal gas law.
PV=nRT
pressure x volume = moles x Faraday's constant x Temp Kelvin (C+273)
Original data
Pressure 1 atmosphere
Volume 1 liter
Temp 25C = 298K
New data
Volume 0.5 liter
pressure X
Temp 260C = 533K
P1v1T1 = P2v2T2
plug and chug.
(1)(1)(293) = (x)(0.5)(533)
Solve for X, which is the new pressure. </span>
Answer:
b
Explanation:
The reaction that is not a displacement reaction from all the options is 
In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.
<em>Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.</em>
Correct option = b.
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.
<u>B</u>
Hope this helped you, have a good day bro cya)