What transfers the electrons generated during the light cycle to co2 during the calvin cycle?

Chemistry

1 answer:

natta225 [31]4 years ago

6 0

Answer:

The correct answer is NADPH.

Explanation:

It takes 6 molecules of CO2 to generate one molecule of glucose. Each of the CO2 molecules is attached to a RuBP-acceptor molecule, which is then divided into two 3-phosphoglycerate molecules.

The ATP produced during the light reactions of photosynthesis yields phosphate groups to these molecules, resulting in 1,3-diphosphoglycerate. Simultaneously, NADPH yields electrons to these three-carbon molecules, creating glyceraldehyde-3-phosphate molecules. Glyceraldehyde-3-phosphate is used to make the sugar from 6 glucose carbons. Another part of the glyceraldehyde-3-phosphate is used together with an ATP molecule, to generate the CO2 acceptor ribulose-1,5-bisphosphate and start the cycle again.

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1. Complete the following

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Answer:

$\large \boxed{\text{0.603 mol}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of O₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 18.1

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{18.1 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1005 mol C$_{6}$H$_{12}$O}_{6}$

b) Moles of O₂

$\text{Moles of O}_{2} =\text{0.1005 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol O}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.603 mol O}_{2}\\\\\text{The reaction requires $\large \boxed{\textbf{0.603 mol}}$ of oxygen}$