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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The N2H4 bond angle will be probably 109 degrees. Since, well,<span> it has a bent </span>trigonal pyramidal<span> geometry.</span>
Answer:

Explanation:
Given:
A solution contains one or more of the following ions such as Ag,
and 
Here the Lithium bromide is added to the solution and no precipitate forms
Solution:
Since with LiBr no precipitation takes place therefore Ag+ is absent
Here on adding
to it precipitation takes place.
Precipitate is as follows,

Thus,
is present
When
is added again precipitation takes place.
Therefore the reaction is as follows,

Therefore,
are present in the solution
Answer: This is known as substitution cipher.
Explanation: