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.
A. good conductors - copper, aluminium
b. fair conductors - carbon,human body
c. insulator - paper, wood
The correct answer is B. Low chemical reactivity
Δmc
2
For one reaction:
Mass Defect =Δm
=2(m
H
)−m
He
−m
n
=2(2.015)−3.017−1.009
=0.004 amu
1 amu=931.5 MeV/c
2
Hence,
E=0.004×931.5 MeV=3.724 MeV
E=3.726×1.6×10
−13
J=5.96×10
−13
J
For 1 kg of Deuterium available,
moles=
2g
1000g
=500
N=500N
A
=3.01×10
26
Energy released =
2
N
×5.95×10
−13
J
=8.95×10
13