The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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Answer:
B. water displacement is used "graduated cylinder is filled with water (100 mL) and the object is then put inside. ... If the new water level is (120 mL) we now know that the object has a volume of 20 mL."
Explanation:
No He believed tiny particles were invisible and couldn't be changed....So No The person that believed in this was Dalton .
Answer:
Since HCl is a strong acid, it completely ionizes, and the pH of HCl in solution can be found from the concentration (molarity) of the H+ ions, by definition equal to 0.100 M. (The conjugate base of the acid, which is the chloride ion Cl–, would also have a concentration of 0.100 M.) The pH is thus –log(0.100) = 1.000.
Explanation: