Given:
1 answer:
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
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Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.
For the compound, the electronic configuration of the atoms, carbon and hydrogen are:
Carbon (atomic number=6): In ground state=
In excited state:
Hydrogen (atomic number=1):
All the bonds in the compound is single bond(-bond) that is they are formed by head on collision of the orbitals.
The structure of the compound is shown in the image.
The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.
In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in of hydrogen will overlap to the 2p^{3}-orbital of carbon.
Thus, the hybridization of Hydrogen is -hybridization and the hybridization of Carbon is
-hybridization.
The hybridization of each atom is shown in the image.
