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.
Answer:
B) 7.7
Explanation:
For the reaction Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)
Kc = (CO₃²⁻) / (CrO₄²⁻)
and the Ksp given are
Ag₂CO₃ ⇒ 2 Ag⁺(aq) + CO₃²⁻(aq) Ksp₁ = (Ag⁺)²(CO₃²⁻)
Ag₂CrO₄ ⇒ 2 Ag⁺(aq)+ CrO₄²⁻(aq) Ksp₂ = (Ag⁺)²(CrO₄²⁻)
Where (...) indicate concentrations M
Notice if we divide the expressions for Ksp we get:
Ksp₁/Ksp₂ = (CO₃²⁻) / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7
which is the desired answer.
the answer is option D "people emphasized obtaining knowledge through scientific experiments" (on plato)