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
B. 1 mole of beryllium, 2 moles of oxygen, 2 moles of hydrogen
Surface runoff
Explanation:
The water that flows back to the streams and oceans are called surface runoff.
Surface runoff is a component of the water cycle usually composed of water in the liquid form that flows back into oceans that are nearby.
- The hydrologic cycle shows the cyclic process by which water passes in nature.
- Water passes through different forms, solid, liquid and gases.
- Surface runoff is water usually after rainfall that flows rapidly.
- They move to the final basin of deposition usually joining up with other water sources.
- This can be nearby streams, lakes or oceans.
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A) Ca(OH)2 + CO2 —> CaCO3 + H2O
B) when Ca(OH)2 is reacted with CO2, the CaCO3 produced is a precipitate which turns the solution milky