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Pachacha [2.7K]
2 years ago
12

The combustion of ethene, C2H4 , occurs via the reaction C2H4(g)+3O2(g)→2CO2(g)+2H2O(g) with heat of formation values given by t

he following table: Substance ΔH∘f ( kJ/mol ) C2H4 (g) 52.47 CO2(g) − 393.5 H2O(g) − 241.8 Calculate the enthalpy for the combustion of 1 mole of ethene.
Chemistry
1 answer:
tekilochka [14]2 years ago
3 0
The combustion of ethene, C2H4 , occurs via the reaction C2H4(g)+3O2(g)→2CO2(g)+2H2O(g) with heat of formation values given by the following table: ...
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Answer:

The initial volume in mL is 5959.2 mL

Explanation:

As the number of moles of a gas increases, the volume also increases.  Hence, number of moles and volumes are directly proportional i.e

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Hence n₁/V₁ = n₂/V₂

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From the question,

n₁ = 0.693 moles

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n₂ = 0.928 moles

V₂ = 7.98 L

Putting the values into the equation

n₁/V₁ = n₂/V₂

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V₁ = 5.9592 L

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