Ethanol, CzH5OH, is an organic chemical compound used in medical wipes and most commonly in antibacterial
2 answers:
Answer:
13.04 %
Explanation:
First, find the Molecular mass of the individual elements present in the compound.
Carbon = 12
Hydrogen = 1
Oxygen = 16
Then, find the total mass of C₂H₅OH
(12 × 2) + (1 × 5) + 16 + 1 = 46
In order to find the percentage mass of hydrogen, you must take the Molecular mass of hydrogen in the compound, divide it by the Molecular mass of the compound and times that by 100.
% by mass of hydrogen = × 100
[it is 1 × 6 because 1 is the mass of hydrogen and there are 6 hydrogen atoms in C₂H₅OH]
% by mass of hydrogen = 13.04 %
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Answer:
118.22 atm
Explanation:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
KP = 0.13 =
Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.
- With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
- With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.
The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:
- XSO₂ = 0.58/3.29 = 0.176
- XO₂ = 1.29/3.29 = 0.392
- XSO₃ = 1.42/3.29 = 0.432
The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.
- p(SO₂) = 0.176 * PT
- p(O₂) = 0.392 * PT
- p(SO₃) = 0.432 * PT
Rewriting KP and solving for PT: