This uses something called the combined gas law. The combined gas law is as follows: (P1*V1/T1) = (P2*V2/T2)
According to question 2, you are given the following values initially:
P1 = 680 mm Hg * (1 atm/760 mm Hg) = 0.895 atm
V1 = 20.0 L
T1 = 293 K
STP or standard temperature and pressure implies that the other values we know are:
P2 = 1 atm
T2 = 273 K
Our unknown is V2
If we plug in our known values into the combined gas law:
(P1*V1/T1) = (P2*V2/T2)
(0.895 atm * 20.0 L)/293K = (1 atm * X liters)/273 K
0.0611 L*atm/K = (1 atm * X liters)/273 K
16.7 L = X liters
Therefore, the volume occupied at STP is 16.7 liters
This makes sense because the gas would occupy a smaller volume at a lower temperature, since the gas would have a lower average kinetic energy.
Answer:
67.5% ≅ 67.6%
Explanation:
Given data:
Mass of water = 17.0 g
Mass of oxygen produced (actual yield)= 10.2 g
Percent yield of oxygen = ?
Solution:
Chemical equation:
2H₂O → 2H₂ + O₂
Number of moles of water:
Number of moles = mass/ molar mass
Number of moles = 17.0 g/ 18.016 g/mol
Number of moles = 0.944 mol
Now we will compare the moles of oxygen with water to know the theoretical yield of oxygen.
H₂O : O₂
2 : 1
0.944 : 1/2×0.944 = 0.472 mol
Mass of oxygen:
Mass = number of moles× molar mass
Mass = 0.472 mol × 32 g/mol
Mass = 15.104 g
Percent yield:
Percent yield = [Actual yield / theoretical yield] × 100
Percent yield = [ 10.2 g/ 15.104 g] × 100
Percent yield = 0.675 × 100
Percent yield = 67.5%
There are 3 different elements
