Answer:
1.06 × 10^5 J
Explanation:
We have to use the formula;
log(k2/k1) = Ea/2.303 ×R (1/T1 - 1/T2)
Given that;
k1 = 1.10 × 10-4 s-1
k2= 1.02 × 10-3 s-1
Ea=?
R=8.314 Jmol-1K-1
T1= 470. °C
T2= 510. °C
Hence;
log(1.02 × 10-3 s-1/1.10 × 10-4 =Ea/2.303 ×8.314(1/470 -1/510)
0.9672 = Ea/19.122(2.13 - 1.96) × 10^-3
0.9672= Ea/19.122(0.17) × 10^-3
0.9672= 1.7 ×10^-4 Ea/19.122
19.122 × 0.9672 = 1.74 × 10^-4Ea
Ea= 19.122 × 0.9672/1.74 × 10^-4
Ea= 18.495/1.74 × 10^-4
Ea= 1.06 × 10^5 J
well , it's true because they are of sp3d type occur on sets of four
Mass of KNO₃ : = 40.643 g
<h3>Further explanation</h3>
Given
28.5 g of K₃PO₄
Required
Mass of KNO₃
Solution
Reaction(Balanced equation) :
2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃
mol K₃PO₄(MW=212,27 g/mol) :
= mass : MW
= 28.5 : 212,27 g/mol
= 0.134
Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :
= 6/2 x mol K₃PO₄
= 6/2 x 0.134
= 0.402
Mass of KNO₃ :
= mol x MW KNO₃
= 0.402 x 101,1032 g/mol
= 40.643 g
No mutations that neither benefit nor harm the organism has no effect on the organism's survival.
Explanation:
The neutral mutations do not cause any effect on the organism they occur. They are called silent point mutations because they do not code for the amino acid change in the proteins.
They have no effect on the organism's survival neither beneficial nor harmful effect.
The neutral mutations do not affect the process of Natural selection.