Answer:
Keq = 0.053
7.3 kJ/mol
Explanation:
Let's consider the following isomerization reaction.
glucose 6‑phosphate ⇄ glucose 1 - phosphate
The concentrations at equilibrium are:
[G6P] = 0.19 M
[G1P] = 0.01 M
The concentration equilibrium constant (Keq) is:
Keq = [G1P] / [G6P]
Keq = 0.01 / 0.19
Keq = 0.053
We can find the standard free energy change, ΔG°, of the reaction mixture using the following expression.
ΔG° = -R × T × lnKeq
ΔG° = -8.314 J/mol.K × 298 K × ln0.053
ΔG° = 7.3 × 10³ J/mol = 7.3 kJ/mol
Answer: 0.5 g/cm^3
Density equals mass divided by volume so..
60/120 is 0.5 g/cm^3
Gain enough kinetic energy to get past each other. Ad you heat up a substance, the temperature increases as does the kinetic energy of the particles. At a point the temperature of the substance will stop increasing. The energy is now being used to increase the potential and move the particles further apart.
Answer:
Double replacement is the answer
