Answer:
a. the mole fraction of CO in the mixture of CO and O2.
mole fraction = moles of CO/ Total moles of the mixture
Mole fraction of CO = 10/(10+12.5)=0.444
b. Reaction - CO(g)+½O2(g)→CO2(g)
Stoichiometry: 1 mole of CO react with 0.5mole of O2 to give 1 mole of CO2
So given,
At a certain point in the heating, 3.0 mol CO2 is present. Determine the mole fraction of CO in the new mixture.
3mol of CO2 is produced from 3 mols of CO and 1.5mol of O2
This means that unused mols are : 7mols of CO and 11mols of O2
Total product mixture = 3 + 7 + 11 = 21mols
mole fraction of CO = 7/21 = 0.33
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Answer:
c a cold air mass and a warm air mass meet with neither moving
Answer:
The reactant/reagent that would be most atom economical is EtI (Ethy Iodide) and KOH (potassium oxide) as base
This is because the iodo group are weak base hence they have a good leaving character (i.e they are unstable on their own ) which would increase the rate of reaction and the strong base KOH give the most atom economical
Explanation:
