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ladessa [460]
3 years ago
10

C6H12O6 + 6O2 ---> 6H2O + 6CO2

Chemistry
1 answer:
Ivahew [28]3 years ago
3 0

Answer:

24e⁻ are transferred by the reaction of respiration.

Explanation:

C₆H₁₂O₆   +  6O₂   →   6 H₂O   +  6CO₂

This is the reaction for the respiration process.

In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side -  REDUCTION

Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION

6C →  6C⁴⁺  +  24e⁻

In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻

6O⁻² +  6O₂  + 24e⁻  →  6O₂²⁻  +  6O⁻²

In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.

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From the question given above, we obtained the following information:

Diameter in micrometer = 2.25 μm

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The answer for the following problem is mentioned below.

  • <u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

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Final moles (n_{2})

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According to the ideal gas equation;

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So;

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\frac{V_{1} }{V_{2} } = \frac{n_{1} }{n_{2} }

where,

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(n_{1}) represents the initial  moles of the gas

(n_{2}) represents the final moles of the gas

Substituting the above values;

   \frac{230}{860} = \frac{3.8 * 10^-4}{n_{2} }

  n_{2} = 14.2 × 10^{-4} moles

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