The word is 'Observation'. It fits the given definition, begins with the letter 'O' and is 11 letters long.
Answer:
67.5% ≅ 67.6%
Explanation:
Given data:
Mass of water = 17.0 g
Mass of oxygen produced (actual yield)= 10.2 g
Percent yield of oxygen = ?
Solution:
Chemical equation:
2H₂O → 2H₂ + O₂
Number of moles of water:
Number of moles = mass/ molar mass
Number of moles = 17.0 g/ 18.016 g/mol
Number of moles = 0.944 mol
Now we will compare the moles of oxygen with water to know the theoretical yield of oxygen.
H₂O : O₂
2 : 1
0.944 : 1/2×0.944 = 0.472 mol
Mass of oxygen:
Mass = number of moles× molar mass
Mass = 0.472 mol × 32 g/mol
Mass = 15.104 g
Percent yield:
Percent yield = [Actual yield / theoretical yield] × 100
Percent yield = [ 10.2 g/ 15.104 g] × 100
Percent yield = 0.675 × 100
Percent yield = 67.5%
Answer:
3300.85 g
Explanation:
Given data:
Mass of ZnCl₂ produced = ?
Mass of H₂ produced = 49.8 g
Solution:
Chemical equation:
Zn + 2HCl → ZnCl₂ + H₂
Number of moles of H₂:
Number of moles = mass/molar mass
Number of moles = 49.8 g/ 2.056 g/mol
Number of moles = 24.22 mol
Now we will compare the moles of H₂ with ZnCl₂ form balance chemical equation.
H₂ : ZnCl₂
1 : 1
24.22 : 24.22
Mass of ZnCl₂:
Mass = number of moles × molar mass
Mass = 24.22 × 136.286 g/mol
Mass = 3300.85 g
Answer: Opaque hope this is helpful
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Moles of Hydrogen present: 100 / 2 = 50 moles
Moles of Nitrogen present: 200 / 28 = 7.14 moles
Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles
Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.
Molar ratio of Nitrogen : Ammonia = 1 : 2
Moles of ammonia = 7.14 x 2 = 14.28 moles
