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Ludmilka [50]
3 years ago
9

In a laboratory demonstration, a balloon filled with methane and oxygen was exposed to a flame. The result was a brief, large fl

ame. The students were asked to formulate an equation for the reaction. One answer is below. (3 points) Is this equation correct? Explain your reasoning.
Chemistry
1 answer:
adell [148]3 years ago
6 0

Answer:

CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g)

Explanation:

Even though the question is incomplete because the equations are missing. However, one thing that is clear from the question is that the reaction is a combustion reaction.

A combustion reaction is a type of reaction that produces heat and light. The mixture of methane and oxygen yielded a brief, large flame which shows that combustion has taken place according to the equation shown above.

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The given statement is false.

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What is the molarity of 50.84 g of Na2CO3 dissolved in 0.400 L solution?
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Answer:

1.20 M

Explanation:

Convert grams of Na₂CO₃ to moles.  (50.84 g)/(105.99 g/mol) = 0.4797 mol

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What is nuclear fusion what role does it play in the sun
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2 years ago
Read 2 more answers
Ammonia and oxygen react to form nitrogen and water.
Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

4 0
3 years ago
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