The balanced chemical equation for the combustion of butane is: 2C2H2 + 5O2 CO2 + 2H2O 2CH4 + 5O2 2CO2 + 4H2O 2C4H10 + 13O2 8CO2 + 10H2O C4H10 + 12O2 4CO2 + 5H2O
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The balanced equation is 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O.
The balanced chemical equation for the combustion of butane is:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O. Therefore, option C is correct.
This equation represents the combustion of butane in the presence of oxygen to produce carbon dioxide and water.
The equation shows that 2 molecules of butane react with 13 molecules of oxygen gas. On the left side of the equation, we have a total of 8 carbon atoms and 20 hydrogen atoms.
On the right side of the equation, 8 molecules of carbon dioxide (each with 1 carbon atom) and 10 molecules of water (each with 2 hydrogen atoms).
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The carbon atoms bind together to create the compound's framework. Therefore, the name of the hydrocarbon CH(CH
)
CH
is n-octane.
What is hydrocarbon?
Any member of a group of organic compounds comprised only of the atoms carbon (C) as well as hydrogen (H) (H). The carbon atoms bind together to create the compound's framework, as well as the hydrogen atoms connect to them in a variety of ways.
Petroleum as well as natural gas are mostly composed of hydrocarbons. They are used as fuels and lubricants, as well as raw materials to make polymers, fibers, rubbers, solvents, explosives, various industrial chemicals. The name of the hydrocarbon CH(CH
)
CH
is n-octane.
Therefore, the name of the hydrocarbon CH(CH
)
CH
is n-octane.
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Answer is: n-octane.
Octane is the straight-chain alkane (an acyclic saturated hydrocarbon) with eight carbon atoms.
Carbons in octane have sp3 hybridization (carbon’s 2s and three 2p orbitals combine into four identical sp3 orbitals).
Orbitals in sp3 hybridization have a tetrahedral arrangement and form single (sigma) bonds.
Burning octane fuel is very fast chemical reaction, it happens in immediately.
Main products of the combustion of fossil fuels are carbon(IV) oxide and water.
Chemical reaction: C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O; ΔH = -5500 kJ/mol.
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a. 1
b. 2
c. 4
d. 8
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There are 4 orbitals in the second shell, and the number increases in further shells.
Answer:
4
Explanation:
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primary consumer
tertiary consumer
secondary consumer