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
Mass defect for oxygen-16 = 0. 13261 amu, in the kilograms the mass defect equals to 2.20 × 10⁻²⁸ kg.
<h3>What is mass defect?</h3>
Mass defect is the difference between the mass of of an whole atom and the combined mass of its individual particles present in that atom.
We know that, 1 amu = 1.6 × 10⁻²⁷ kg
Given that, mass defect for oxygen-16 = 0.13261 amu
To calculate this defect in terms of kilograms, we have to convert into kg unit as:
0.13261 amu = 0.13261 amu × 1.6 × 10⁻²⁷ kg/amu
0.13261 amu = 2.20 × 10⁻²⁸ kg
Hence option (2) is correct.
To know more about Mass defect, visit the below link:
brainly.com/question/4334375
Answer:
Scientific theories are testable and make falsifiable predictions. They describe the causes of a particular natural phenomenon and are used to explain and predict aspects of the physical universe or specific areas of inquiry (for example, electricity, chemistry, and astronomy).
A good theory in the theoretical sense is (1) consistent with empirical observations; is (2) precise, (3) parsimonious, (4) explanatorily broad, and (5) falsifiable; and (6) promotes scientific progress (among others; Table 1.1).
Answer:
The answer is "2%"
Explanation:
Equation:
Formula:
Let
at equilibrium
therefore,
Calculating the % ionization: