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:
C
Explanation:
Silt is very benifical for plant growth. So the one with the most silt would be the best for gardening soil.
Answer:
[CO₂] = 375 ppm
Explanation:
We can determine the concentration in value of ppm
ppm = mass of solute (mg) / mass of solution (kg)
Solution: atmosphere
Solute: CO₂
We convert the mass of CO₂ from g to mg → 0.1875 g . 1000 mg / 1g = 187.5 mg
We convert the mass of the atmosphere from g to kg: 500 g . 1 kg/1000g = 0.500 kg
ppm = 187.5 mg / 0.500kg = 375
Answer:
B. Similar fossils were found in different continents.
Bacteria, because the reproduce at fast rates the characteristics of them change and the emergence of antibiotic resistance is what causes the natural selection process to emerge at a high rate.
