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:
sulfur
Explanation:
In oxygen family sulfur has yellow color and also having stinky smell. Thus given statements are about sulfur.
It is present in oxygen family.
It has six valance electrons.
Its atomic number is 16.
Its atomic weight is 32 amu.
The electronic configuration of sulfur is given below,
S₁₆ = 1s² 2s² 2p⁶ 3s² 3p⁴
We can see the valance shell is third shell and it have six electrons thus sulfur have six valance electrons. (3s² 3p⁴ )
Sulfur is used in vulcanisation process.
It is used in bleach and also as a preservative for many food.
it is used to making gun powder.
If the.pressure exerted by a gas at [math]25^{\circ} \mathrm{C}[/math] in a volume of 0.044 L is 3.81 atm, how many moles of gas are present
Answer:
44.8 L
Explanation:
Using the ideal gas law equation:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
At Standard temperature and pressure (STP);
P = 1 atm
T = 273K
Hence, when n = 2moles, the volume of the gas is:
Using PV = nRT
1 × V = 2 × 0.0821 × 273
V = 44.83
V = 44.8 L
The answer is A: Areas where the geologic process occurred now have major petroleum reserves