At what temperature, in °C, would 1.50 moles of Ar gas at a pressure of 2.3 atm, occupy a volume of 25.8 L? Round your answer to
1 answer:
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Answer:
Explanation:
Hello,
In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
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Answer:
Explanation:
The expression for the work done is:
Where,
W is the amount of work done by the gas
R is Gas constant having value = 8.314 J / K mol
T is the temperature
P₁ is the initial pressure
P₂ is the final pressure
Given that:
T = 300 K
P₁ = 10 bar
P₂ = 1 bar
Applying in the equation as: