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:
A)The characteristic frequency to look out for is 1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.
B)Characteristic frequency present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.
C)If the product is wet with water there will be no change in the infrared spectrum
Explanation:
The characteristic frequency to look out for is 1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.
Characteristic frequency present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.
If the product is wet with water there will be no change in the infrared spectrum
Explanation:
<h2>The number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital.</h2>
1. A. All the elements in the column have similar chemical properties.
2. Substances on the periodic table cannot be broken down into other substances and are therefore elements.
The percentage error is the error of the measured value to the true value. To find he percent error, the equation is as follows:
Percent error = |Measured Value - True Value|/True Value * 100
The || is needed to get the absolute value of the difference. Substituting the values,
Percent error = |(10.085 g/10 mL) - 0.9975 g/mL|/<span>0.9975 g/mL * 100
</span><em>Percent error = 1.1% </em>
