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Degger [83]
1 year ago
5

How many grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline?

Chemistry
1 answer:
egoroff_w [7]1 year ago
5 0

Grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline is 45.5g

Octane is a hydrocarbon which burns in gasoline in presence of oxygen according to the given balanced chemical equation,

2C₈H₁₈ + 25O₂------> 16CO₂ + 18H₂0

Molar mass of octane = 114.23g/mol

Molar mass of Oxygen = 32g/mol

According to the stiochiometry of the balanced equation the mole ratio of Octane and Oxygen is 2:25

2 mole of octane needs 25 mole of oxygen

1 mole of octane needs 12.5 moleof oxygen

114.23g of octane needs 400g of oxygen

13g   of octane  needs 45.5g of oxygen

Mass of oxygen needed =45.5g

Hence, the Mass of oxygen needed is 45.5g for the combustion of octane in gasoline.

Learn more about Octane here, brainly.com/question/21268869

#SPJ4

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0.1 mL of the stock solution of the enzyme is taken and made up to 5.0 mL with 0.001M HCl in order to prepare a 50-fold diluted enzyme.

<h3>What is dilution?</h3>

Dilution is a process of making a solution of lower concentration from a solution of higher concentration by the addition of solvent to a given volume of the solution of higher concentration.

Dilution of solutions is done using the dilution formula in order to determine the given volume of diluent or stock solution required. The dilution formula is given below:

  • C1V1 = C2V2

where:

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  • C2 = Final concentration of enzyme
  • V1 = Initial volume
  • V2 = Final volume

For the enzyme dilution;

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V= ?

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Therefore, 0.1 mL of the stock solution of the enzyme is taken and made up to 5.0 mL with 0.001M HCl in order to prepare a 50-fold diluted enzyme.

Learn more about dilution at: brainly.com/question/24881505

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