Answer:
The value of he change in Gibbs free energy ΔG = - 18.083 KJ
Explanation:
Given data
The concentration of glucose inside a cell is (P) = 0.12 m M
The concentration of glucose outside a cell is (R) = 12.9 m M
No. of moles = 1.5 moles
The change in Gibbs free energy
ΔG = RT ㏑
ΔG = 8.314 × 310 ㏑
ΔG = - 12.055 
Since No. of moles = 1.5 moles
Therefore
ΔG = - 12.055 × 1.5
ΔG = - 18.083 KJ
This the value of he change in Gibbs free energy.
The answer is letter A definitively .
None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this correct.
D, the rate increases as concentrations increase.
Typically, reaction rates decrease with time because reactant concentrations decrease as reactions are converted to products. Reaction rates generally increase when reactant concentrations are increased.
<span>The addition of a catalyst to a chemical reaction provides an alternate pathway that c</span>atalysts lowers the activation energy.
