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:
The pH is 11
Explanation:
Because pH + pOH = 14
and pOH = - log [OH⁻] = - log (1 x 10⁻³) = 3
we can now calculate pH by manipulating equation one above
pH = 14 - pOH = 14 - 3 = 11
Thus the pH of the solution is 11
Answer:contact forces or transfer energy(not sure)
Explanation:
Answer:
The dosage of the solution is 2.5 parts per million.
Explanation:
The solvent proportion in parts per million is equal to the ratio of solution in miligrams to ratio of water in miligrams multiplied by a million. (A kilogram is equivalent to a million miligrams) That is:
The dosage of the solution is 2.5 parts per million.
Because C-12 is fixed value of any organisms but the value of C-14 is constantly decreasing by time thats why every organism have their specific ratio of C14/C12
