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:
Answer:
3. 75.0%
Explanation:
2 ClO2(g) + F2(g) → 2 FClO2(g)
First order with respect to ClO2 and F2.
This means the rate equation is given as;
Rate = k [ClO2][F2]
When the initial concentrations of ClO2 and F2 are equal?
Let's assume an initial value of 1 for both reactants, so rate equation is given as;
Rate = k * 1 * 1 = k
The rate after 25% of the F2 has reacted is what percent of the initial rate?
The concentration left of F2 is 75% ( 100% - 25%) = 0.75
Concentration of ClO2 remains 1.
So rate equation is given as;
Rate = k * 1 * 0.75 = 0.75 k
Comparing 0.75k and k.
This means our answer is;
3. 75.0%
True due to the fact that eight electrons are what applies directly to the octet rule
Carbon is very special because it can form so many compounds. ... Carbon atoms can form strong links with four other atoms. This dramatically increases the number of patterns that carbon atoms can make. Carbon can also link together in long chains or rings, carbon to carbon to carbon to carbon and so on.