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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167 mL
P1V1 = P2V2
P1 = .8 atm
V1 = 250 mL
P2 = 1.2 atm
Solve for V2 —> V2 = P1V1/P2
V2 = (0.8 atm)(250 mL) / (1.2 atm) = 167 mL
Answer:
Magnetism is believed to be caused by the alignment of small, numerous sub-units called : <em><u>Domains</u></em>
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Explanation:
Domains : A magnetic domain is the region in which in which magnetic field of the atoms are grouped together and aligned.
- In unmagnetized material all the magnetic Domains point in different direction.
- In magnetised material (ferromagnets , antiferromagnets) , The Domains point in a particular( fixed Pattern) direction.
Answer:
Wind direction is determined with a wind vane.
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