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Norma-Jean [14]
3 years ago
11

When a suspension of algae is incubated in a flask in the presence of light and CO2 and then transferred to the dark, the reduct

ion of 3-phosphoglycerate to glyceraldehyde 3-phosphate is blocked. This reaction stops when the algae are placed in the dark because?
Chemistry
1 answer:
Aloiza [94]3 years ago
8 0

Answer:

The photosynthesis process is interrupted.

Explanation:

Algae produce energy using the photosynthesis process. The reduction of 3-phosphoglycerate to glyceraldehyde 3-phosphate is part of this process. Despite this reduction reaction being light-independent (Calvin Cycle), the precursors of this reaction are synthesized in light-dependent steps.

This is the reason why the reduction is blocked when the algae is placed in the dark.

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Determine the equilibrium constant, Keq, at 25°C for the reaction
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Explanation:

The given chemical reaction is:

2Br^- (aq) + I_2(s)  Br_2(l) + 2I^- (aq)

E^ocell=oxidation potential of anode + reduction potential of cathode\\

The relation between Eo cell and Keq is shown below:

deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

E^ocell= (-1.07+0.53)V\\=-0.54V

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3

Answer:

Keq=6.13x10^33

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3 years ago
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Valentin [98]

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