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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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Using the knowledge in computational language in JAVA it is possible to write a code that reads a list of words.

<h3>Writting the code in JAVA:</h3>

<em>import java.util.*;</em>

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<em>        //for loop for iteration </em>

<em>        for(int i=0; i<ListSize; i++){</em>

<em>            //if current word matches with words present in list</em>

<em>            //then increase the frequency by 1</em>

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<em>            }</em>

<em>        }</em>

<em>        //return frequency at last </em>

<em>        return frequency;</em>

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<em>    public static void main(String[] args) {</em>

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<em>        Scanner sc = new Scanner(System.in);</em>

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<em>        System.out.println("Enter the size of list :");</em>

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<em>        //creating array of same size as user input size</em>

<em>        String[] wordList = new String[size];</em>

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<em>        //asking user to input array elements </em>

<em>        System.out.println("Enter list elements one by one :");</em>

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<em>        }</em>

<em>    }</em>

<em>}</em>

See more about JAVA at brainly.com/question/12978370

#SPJ1

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