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Answer:
36.92 mg of oxygen required for bio-degradation.
Explanation:
Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )
Moles benzene =
According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.
Then 0.0003846 mol of benzene will react with:
of oxygen gas
Mass of 0.0011538 moles of oxygen gas:
0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg
36.92 mg of oxygen required for bio-degradation.
Answer:
0.558mole of SO₃
Explanation:
Given parameters:
Molar mass of SO₃ = 80.0632g/mol
Mass of S = 17.9g
Molar mass of S = 32.065g/mol
Number of moles of O₂ = 0.157mole
Molar mass of O₂ = 31.9988g/mol
Unknown:
Maximum amount of SO₃
Solution
We need to write the proper reaction equation.
2S + 3O₂ → 2SO₃
We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.
So we simply compare the molar relationship between sulfur and product formed to solve the problem:
First, find the number of moles of Sulfur, S:
Number of moles of S =
Number of moles of S = = 0.558mole
Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:
2 mole of Sulfur produced 2 mole of SO₃
Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃
Answer:
5200 ppm
Explanation:
As per the definition, parts per million of a contaminant is a measure of the amount of mass of contaminant present per million amount of the solution. It is denoted by ppm.
Given in the question,
Water = 250 ml = 250 g
Lead = 1.30 g
So,
ppm of Lead = =
= 5200 ppm
So, as calculated above, there is 5200 ppm of lead present in 250 ml of water.