1 is b 2 is a 3 is d 4 is a 5 is c
Answer:
This is a coal combustion process and we will assume
Inlet coal amount = 100kg
It means that there are
15kg of H2O, 2kg of Sulphur and 83kg of Carbon
Now to find the mole fraction of SO2(g) in the exhaust?
Molar mass of S = 32kg/kmol
Initial moles n of S = 2/32 = 0.0625kmols
Reaction: S + O₂ = SO₂
That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂
Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust
The mole fraction of SO2(g) in the exhaust=0.0625kmols
Explanation:
The number of moles of b2o3 that will be formed is determined as 4 moles.
<h3>
Limiting reagent</h3>
The limiting reagent is the reactant that will be completely used up.
4 b + 3O₂ → 2b₂O₃
from the equation above;
4 b ------------> 2 b₂O₃
2b ------------> b₂O₃
2 : 1
3O₂ -------------> 2b₂O₃
3 : 2
b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;
4 b ------------> 2 moles of b2o3
8 moles -------> ?
= (8 x 2)/4
= 4 moles
Thus, the number of moles of b2o3 that will be formed is determined as 4 moles.
Learn more about limiting reactants here: brainly.com/question/14222359
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The reaction is endothermic as energy is absorbed to break a bond.
There are 4 half lives of radium-226 in 6396 years. 15.0 divided by half 4 times (15.0/2= 7.5, 7.5/2 = 3.75, 3.75/2 = 1.875) equals 1.875 mg of radium-226.
