The total pressure when the new equilibrium is stabilized is half of the initial pressure of the system.
The given chemical reaction at a stable equilibrium is,
2H₂O(g)+O₂(g) = 2H₂O₂(g)
According to the ideal gas equation,
PV = nRT
P is pressure,
V is volume,
n is moles
R is gas constant,
T is temperature.
Assuming the temperature is constant.
If the volume of the system is twice the initial volume then the total pressure at the new equilibrium can be found out as,
P₁V₁ = P₂V₂
Where, P₁ and V₁ are initial volume and pressure while P₂ and V₂ are final pressure and volume.
If V₂ = 2V₁,
P₂ = P₁/2
So, the final total pressure will be half of the initial pressure.
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Answer:
b. transfer of electron(s).
Explanation:
An oxidation-reduction also called a redox reaction is a
chemical reaction in which electrons are transferred of between two species of reactants. It is a chemical reaction where the oxidation number of an atom, ion, or molecule, increases or decreases by losing or gaining electrons
Answer : The amount of oxygen gas collected are, 0.217 mol
Explanation :
Using ideal gas equation :
where,
P = pressure of gas = 
(1 atm = 760 torr)
V = volume of gas = 5 L
T = temperature of gas = 
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
Thus, the amount of oxygen gas collected are, 0.217 mol
Answer:
V = 81.14 L
Explanation:
Given data:
Volume of gas = ?
Number of moles = 3.30 mol
Temperature of gas = 25°C
Pressure of gas = 0.995 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
25+273 = 298 K
now we will put the values in formula:
V = 3.30 mol 0.0821 atm.L/ mol.K 298 K / 0.995 atm
V = 80.74 L. atm / 0.995 atm
V = 81.14 L
