How many electron shells/orbitals will be around the nucleus of an atom of aluminum?
2 answers:
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Answer:
The concentration of fructose-6-phosphate F6P ≅ 1.35 mM
Explanation:
Given that:
ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol
concentration of glucose-6-phosphate at equilibrium = 2.65 mM
Assuming temperature = 25.0°C
=( 25 + 273)K
= 298 K
We are to find the concentration of fructose-6-phosphate
Using the relation;
ΔG' = -RT In K_c
where;
R = 8.314 J/K/mol
1670 = - (8.314 × 298 ) In K_c
1670 = -2477.572 × In K_c
1670/ 2477.572 = In K_c
0.67 = In K_c
0.511
Now using the equilibrium constant
F6P = 0.511 × 2.65
F6P = 1.35415
F6P ≅ 1.35 mM