Answer:
1.503 mM is the concentration of fructose-6-phosphate.
Explanation:
Glucose-6-phosphate ⇄ Fructose-6-phosphate
The equilibrium constant will be given by =
Relation between standard Gibbs free energy and equilibrium constant follows:
where,
= standard Gibbs free energy = 1.67 kJ/mol = 1670 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant =
T = temperature =
On solving above equation :
A
P 6 atoms
O 15 atoms
P = 2×3= 6
O = 5×3= 15
b.
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