To determine the mass of oxygen per gram of sulfur for sulfur dioxide, we simply obtain the ratio of the mass of oxygen and the mass of sulfur produced from the decomposition of sulfur dioxide. All other values given in the problem statement above are just to confuse us that the question is a difficult one. We do as follows:
mass of oxygen per gram sulfur = 3.45 g / 3.46 g
mass of oxygen per gram sulfur = 0.9971 g O2 / g S
First the theoretical yield of Nabr
by use of mole ratio between FeBr3 and NaBr which is 2:6 the theoretical yield
=2.36 x6/2= 7.08 moles
the % yield = actual yield/ theoretical yield x 100
that is 6.14/7.08 x100= 86.72%
Answer:
282.7KPa
Explanation:
Step 1:
Data obtained from the question.
Number of mole of (n) = 1.5 mole
Volume (V) = 13L
Temperature (T) = 22°C = 22 + 273°C = 295K
Pressure (P) =..?
Gas constant (R) = 0.082atm.L/Kmol
Step 2:
Determination of the pressure exerted by the gas.
This can be obtained by using the ideal gas equation as follow:
PV = nRT
P = nRT /V
P = 1.5 x 0.082 x 295 / 13
P = 2.79atm.
Step 3:
Conversion of 2.79atm to KPa.
This is illustrated below:
1 atm = 101.325KPa
Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa
Therefore, the pressure exerted by the gas in KPa is 282.7KPa
Increase.
I don't even know the point of the first sentence. Average Kinetic Energy *is* temperature. The higher AVE the higher the temp. The higher the temp the faster the AVE. They're directly correlated.
