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
<span>Now consider a low pressure area on a disk as shown below.A parcel of air at point A would move toward the center of the low pressure area. That movement would take it farther away from the center of the disk and therefore it would move to the west. A parcel of air at B would move toward the center of the low pressure area which would also take it closer to the center of the spinning disk where its speed is greater than the surrounding points. It would appear to move to the east. With A moving to the west and B moving to the east the line from A to B is rotating counterclockwise.</span>
A compound is a substance that can be separated into simpler substances only by chemical means.
The correct answer is <span>ball-and-stick model I just take it</span>
