Number 2 is the correct answer
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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A cold front is the leading edge of a cooler mass of air, replacing at ground level a warmer mass of air, which lies within a fairly sharp surface trough of low pressure.
Answer:
Explanation:
The wavelength is the distance between two adjacent wavefronts. ... If the wave crosses to the new medium at an angle (not 90 degrees), the change ... When light enters a more optically dense medium, it is refracted closer to the normal. the same as the critical angle, light will travel along the boundary of the 2 mediums.
Ph= - log [H+] = -log 1.00× 10-7 = -(log 1 + log 10-7) = -( 0 + (-7log 10) = -( -7) = 7