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Nimfa-mama [501]
3 years ago
10

Question:

Chemistry
1 answer:
Alecsey [184]3 years ago
5 0

Answer:

The laboratory demonstration consists of the following;

The compounds present in the combustion reaction = Methane, CH₄ and Oxygen, O₂

The chemical equation for the combustion reaction is given as follows;

CH₄ + 2O₂ → CO₂ + 2H₂O

Therefore;

A. The equation given as CH₄ + O → CO₂ is not correct because;

1) Oxygen gas exist as diatomic molecules, O₂, and given that the experiment involves the mixture of gases, the oxygen gas present which can exist as a separate compound, should be represented as O₂

2) The number of oxygen molecules in the reaction is two rather than one

3) The product also includes two molecules of water (vapor) H₂O

B. The correct equation for the reaction should be given as follows;

CH₄ + 2O₂ → CO₂ + 2H₂O

B i) The constituents of the equation is obtained by the knowledge of the fact that the combustion reaction of an organic substance such as methane in the presence of oxygen yields, carbon dioxide and water (vapor)

The equation showing the relative amounts the reacting compounds is by balancing the basic equation of the combustion of methane in the presence of oxygen

Explanation:

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a 4.18 g sample of a hydrocarbon is combusted in a bomb calorimeter that contains 974 g of water. the temperature of the water i
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The heat of the reaction, in kJ, when 4.18 g of the hydrocarbon are combusted 775.70 kJ.

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Thus, A 4.18 g sample of a hydrocarbon is combusted in a bomb calorimeter that contains 974 g of water. the temperature of the water increases by 6.9 °C when the hydrocarbon is combusted. the calorimeter constant for the calorimeter was determined to be 624 J/°C. what is the heat of the reaction is 775.70 kJ.

To learn more about calorimeter here

brainly.com/question/28943378

#SPJ4

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