A sample of gas occupies 0.04 L at 150 K. What volume does that sample occupy at 300 K?
1 answer:
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Answer:
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
Explanation:
Moles of silver nitrate = n
Volume of the solution = 425 mL = 0.425 L (1 mL = 0.001 L)
Molarity of the silver nitrate solution = 0.397 M
Moles of sodium phosphate = n'
Volume of the sodium phosphate solution = 427 mL = 0.427 L (1 mL = 0.001 L)
Molarity of the sodium phosphate solution = 0.459 M
According to reaction, 3 moles of silver nitrate reacts with 1 mole of sodium phosphate, then 0.1687 moles of silver nitrate will recat with :
of sodium phosphate
This means that only 0.05623 moles of sodium phosphate will react with all the 0.1687 moles of silver nitrate , making silver nitrate limiting reagent and sodium phosphate as an excessive reagent.
So, zero moles of silver nitrate will be left in the solution after mixing of the both solutions and hence zero moles of silver ions will left in the resulting solution.
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
Answer: The entropy change of the surroundings will be -17.7 J/K mol.
Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol
Amount of Acetone given = 10.8 g
Number of moles is calculated by using the formula:
Molar mass of acetone = 58 g/mol
Number of moles =
If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then
0.1862 moles will have =
To calculate the entropy change for the system, we use the formula:
Temperature = 56.2°C = (273 + 56.2)K = 329.2K
Putting values in above equation, we get
(Conversion Factor: 1 kJ = 1000J)
At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,
Answer:
Explanation: A yellow precipitate o lead iodide is formed. see equation of reaction below: →
Th PbI2 is the insoluble yellow precipitate