The value of Δ G ° ' for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is + 1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 2.95 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0 ° C .

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 :

1.503 mM is the concentration of fructose-6-phosphate.

can you attach a picture of the lettuce? : )

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6