Answer:
Wetlands prevent flooding by temporarily storing and slowly releasing stormwater. Wetlands also reduce water flow, thus allowing sediments and associated pollutants to settle out. Beneficial microor- ganisms (called biofilm) live on wetland plants and process some forms of pollution.
Explanation:
Answer:
<em>249 kJ</em>
Explanation:
To obtain the energy change of the reaction:
H₂O → H₂ + ¹/₂ O₂
It is necessary to obtain the difference between bond energy of the products and bond energy of the reactant, thus:
Energy of products:
1 mol of H-H bond × 436 kJ/mol = 436 kJ
¹/₂ mol of O=O bond × 498 kJ/mol = 249 kJ
Energy of reactant:
2 mol of H-O bond × 467 kJ/mol = 934 kJ
Energy change of the reaction is:
934 kJ - (436 kJ + 249 kJ) = <em>249 kJ</em>
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I hope it helps!
The molecular weight is between 70 and 85. Let's accidentally pick some quantity in amid, like 78.
7.74% of 78 is how many 'u' there are of H:
78 u * 7.74% = 78 u * .0774 = 6.0372 u
The only hydrocarbon below that has 6 hydrogens is C6H6. We can even check this answer:
Total u = 6C + 6H
= 6(12.01) + 6(1.0079)
= 78.11 u
6(1.0079) / 78.11 * 100 = 7.74%
Answer:
Percentage yield = 25%
Explanation:
Given data:
Moles of C₈H₁₈ = 4 mol
Moles of O₂ = 4 mol
Actual yield of CO₂ = 28.16 g
Percentage yield = ?
Solution:
Chemical equation:
2C₈H₁₈ + 25 O₂ → 16CO₂ + 18H₂O
Now we will compare the moles of CO₂ with O₂ and C₈H₁₈.
C₈H₁₈ : CO₂
2 : 16
4 : 16/2×4=32 mol
O₂ : CO₂
25 : 16
4 : 16/25×4=2.56 mol
Mass of CO₂:
Mass = number of moles× molar mass
Mass = 2.56 mol ×44 g/mol
Mass = 112.64 g
Percentage yield:
Percentage yield = Actual yield /theoretical yield × 100
Percentage yield = 28.16 g / 112.64 g× 100
Percentage yield = 0.25× 100
Percentage yield = 25%
