A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to
2 answers:
Answer:The change in temperature of the gas mixture is -68.44 K.
Explanation:
1. Temperature of the bottle when only helium gas was present :T
Pressure inside the container =P= 1.83 atm
Volume of the container = V = 2.50 L
Moles of helium gas = n = 0.458 moles
Using an Ideal gas equation:
2. Temperature of the bottle when argon gas is added into the container:T'
Pressure inside the container = P' =2.05 atm
Volume of the container = V = 2.50 L
Moles of helium gas = n' = 0.458 mol + 0.713 mol = 1.171 mol
The change in temperature will be given as:
Final temperature = Initial temperature:T' - T
53.37 k - 121.81 K = -68.44 K
The change in temperature of the gas mixture is -68.44 K.
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Answer: The theoretical yield and percent yield for this experiment are 40 g and 82% respectively.
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number of particles.
To calculate the moles:
According to stochiometry:
2 moles of produce = 2 moles of
0.54 moles of should produce =
of
Thus theoretical yield is
But Experimental yield is 33 g.
The theoretical yield and percent yield for this experiment are 40 g and 82% respectively.