I think the answer is C. Hope this helped.
Answer:
The equilibrium partial pressure of O2 is 0.545 atm
Explanation:
Step 1: Data given
Partial pressure of SO2 = 0.409 atm
Partial pressure of O2 = 0.601 atm
At equilibrium, the partial pressure of SO2 was 0.297 atm.
Step 2: The balanced equation
2SO2 + O2 ⇆ 2SO3
Step 3: The initial pressure
pSO2 = 0.409 atm
pO2 = 0.601 atm
pSO3 = 0 atm
Step 4: Calculate the pressure at the equilibrium
pSO2 = 0.409 - 2X atm
pO2 = 0.601 - X atm
pSO3 = 2X
pSO2 = 0.409 - 2X atm = 0.297
X = 0.056 atm
pO2 = 0.601 - 0.056 = 0.545 atm
pSO3 = 2*0.056 = 0.112 atm
Step 5: Calculate Kp
Kp = (pSO3)²/((pO2)*(pSO2)²)
Kp = (0.112²) / (0.545 * 0.297²)
Kp = 0.261
The equilibrium partial pressure of O2 is 0.545 atm
55.8 grams Would be the answer
Answer:
V = 5.17L
Explanation:
Mass of gas = 8.7g
T = 23°C = (23 + 273.15)K = 296.15K
P = 1.15 atm
V = ?
R = 0.082atm.L / mol.K
From ideal gas equation
PV = nRT
P = pressure of the gas
V = volume of the gas
n = no. Of moles
R = ideal gas constant
T = temperature of the gas
no of moles = mass / molar mass
Molar mass of Chlorine = 35.5g / mol
No. Of moles = 8.7 / 35.5
No. Of moles = 0.245 moles
PV = nRT
V = nRT / P
V = (0.245 * 0.082 * 296.15) / 1.15
V = 5.9496 / 1.15
V = 5.17L
The volume of the gas is 5.17L
This is a reduction half-reaction. Initially, gold exists as
, and after the reaction, it has been converted to elemental gold. To make this possible, three electrons must be added to <span>
. This fits within the definition of reduction reactions, where the reactant gains electrons in the reaction.</span>