C. 28 KJ
AMU of H2O2 = 2(1) + 2(16) = 34 g/mol
10 g / 34 g/mol = 0.294 mol H2O2
0.294 mol / H = 2 mol / 190 KJ
H = 28.9 KJ
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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Wavelength of the light is 2.9 × 10⁻⁷ m.
<u>Explanation:</u>
Planck - Einstein equation shows the relationship between the energy of a photon and its frequency, and they are directly proportional to each other and it is given by the equation as E = hν,
where E is the energy of the photon
h is the Planck's constant = 6.626 × 10⁻³⁴ J s
ν is the frequency
From the above equation, we can find the frequency by rearranging the equation as,
ν =
= 
Now the frequency and the wavelength are in inverse relationship with each other.
ν × λ = c
It can be rearranged to get λ as,
λ = c / ν
= 
So wavelength is 2.9 × 10⁻⁷ m.
Answer:
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Explanation:
i just did it and got it correct