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madam [21]
3 years ago
15

What is the excess reactant in the combustion of 211 g of methane in an open atmosphere?.

SAT
1 answer:
MAXImum [283]3 years ago
6 0

The excess reactant in the combustion of 211 g of methane in an open atmosphere is oxygen, O₂

<h3>What is a combustion reaction? </h3>

A combustion reaction is a reaction in which a substance burns in air (oxygen) to produce carbon dioxide and water.

<h3>Combustion of methane </h3>

Methane combust in air (oxygen) to produce carbon dioxide and water according to the following equation:

CH₄ + 2O₂ —> CO₂ + 2H₂O

From the question given above, we were told that 211 g of methane combust in an open atmosphere. This simply means that the amount of oxygen will be very high.

Therefore, we can conclude that oxygen will be the excess reactant since the reaction is carried out in an open atmosphere.

Learn more about stoichiometry:

brainly.com/question/14735801

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Suppose that the annual federal deficit is $350 billion. Gross domestic product 'gdp', a measure of the size of the economy is $
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Based on the information the ratio between the deficit and gdp as a percentage is 2.41%.

<h3>Ratio between the deficit and gdp:</h3>

Using this formula

Ratio between the deficit and gdp=Annual federal deficit/Gross domestic product×100

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Let plug in the formula

Ratio between the deficit and gdp=$350 billion/$14,500 billion×100

Ratio between the deficit and gdp=2.41%

Inconclusion the ratio between the deficit and gdp as a percentage is 2.41%.

Learn more about ratio between the deficit and gdp here:brainly.com/question/4731320

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