Air containing 20.0 mol% water vapor at an initial pressure of 1 atm absolute is cooled in a 1- liter sealed vessel from 200°C t
o 15°C. (a) What is the pressure in the vessel at the end of the process? (Hint: The partial pressure of air in the system can be determined from the expression pair = nairRT/V and P = pair + pH2O. You may neglect the volume of the liquid water condensed, but you must show that condensation occurs.) (b) What is the mole fraction of water in the gas phase at the end of the process? (c) How much water (grams) condenses?
1 answer:
Answer:
This solution is quite lengthy
Total system = nRT
n was solved to be 0.02575
nH20 = 0.2x0.02575
= 0.00515
Nair = 0.0206
PH20 = 0.19999
Pair = 1-0.19999
= 0.80001
At 15⁰c
Pair = 0.4786atm
I used antoine's equation to get pressure
The pressure = 0.50
2. Moles of water vapor = 0.0007084
Moles of condensed water = 0.0044416
Grams of condensed water = 0.07994
Please refer to attachment. All solution is in there.
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Given :
Molarity of sulfuric acid solution is 3.0 M.
Amount of sulfuric acid present in solution is 9.809 g.
To Find :
The volume of solution.
Solution :
We know, molarity is given by :
Therefore, volume required is 33.33 ml .
Answer:
<u></u>
- <u>0.456M</u>
Explanation:
<u>1. Balanced molecular equation</u>
<u>2. Mole ratio</u>
<u>3. Moles of HNO₃</u>
- Number of moles = Molarity × Volume in liters
- n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃
<u>4. Moles Ba(OH)₂</u>
- n = 0.700M × 0.0310 liter = 0.0217 mol
<u>5. Limiting reactant</u>
Actual ratio:
Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.
Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.
<u>6. Final molarity of Ba(OH)₂</u>
- Molarity = number of moles / volume in liters
- Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M