What is the energy required to evaporate two moles of liquid water given the following equations? 2H₂(g) + O₂(g) → 2H₂O(g) 2H2(g) + O₂(g) → 2H₂O(1) -88,0 kJ -1055.2 KJ O 1055.2 kJ 44.0 kJ O88.0 kJ ΔΗ = -483.6 kJ ΔΗ = -571,6 kJ

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
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Chapter5: Thermochemistry
Section: Chapter Questions
Problem 5.51QE: The enthalpy change when 1 mol methane (CH4) is burned is 890 kJ. It takes 44.0 kJ to vaporize 1 mol...
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What is the energy required to evaporate two moles of liquid water given the following equations?
2H₂(g) + O₂(g) → 2H₂O(g)
2H2(g) + O₂(g) → 2H₂O(1)
-88,0 kJ
-1055.2 KJ
1055.2 kJ
O 44.0 kJ
O88.0 kJ
ΔΗ = -483.6 kJ
ΔΗ = -571,6 kJ
Transcribed Image Text:What is the energy required to evaporate two moles of liquid water given the following equations? 2H₂(g) + O₂(g) → 2H₂O(g) 2H2(g) + O₂(g) → 2H₂O(1) -88,0 kJ -1055.2 KJ 1055.2 kJ O 44.0 kJ O88.0 kJ ΔΗ = -483.6 kJ ΔΗ = -571,6 kJ
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