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
Answer:
The basic essential activities performed by an organism to withstand its life are called as life processes. These include nutrition, respiration, circulation, excretion and reproduction. Organisms obtain energy from food to perform these life processes which are essential for survival.
Complete balanced equation: 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O
Ionized equation (with spectator ions):
2H⁺ + 2NO₃⁻ + Ca²⁺ + 2OH⁻ → Ca²⁺ + 2NO₃⁻ + 2H₂O
By eliminating the ions that are the same of both sides of the equation (spectator ions):
2H⁺ + 2OH⁻ → 2H₂O [Net Ionic Equation]
Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.
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