Ksp of AgCl= 1.6×10⁻¹⁰
AgCl=Ag⁺ +Cl⁻
Ksp=[Ag⁺][Cl⁻]
Assume [Ag⁺]=[Cl⁻]=x
Ksp=x²
1.6×10⁻¹⁰=x²
x=0.000012
In FeCl₃:
FeCl₃------>Fe⁺³+ 3Cl⁻
as there is 0.010 M FeCl₃
So there will be ,
[Cl⁻]= 0.030
So
[Ag⁺]=Ksp/[Cl⁻]
=1.6×10⁻¹⁰/0.030
=5.3×10⁻⁹
so solubility of AgCl in FeCl₃ will be 5.3×10⁻⁹.
Answer:
Performance Tasks 188; Instructional Videos 75
Explanation:
Answer:
The maximum length of the specimen before deformation is 240.64 mm
Explanation:
Strain = stress ÷ elastic modulus
stress = load ÷ area
load = 2130 N
diameter = 3.4 mm = 3.4×10^-3 m
area = πd^2/4 = 3.142 × (3.4×10^-3)^2/4 = 9.08038×10^-6 m^2
stress = 2130 N ÷ 9.08038×10^-6 m^2 = 2.35×10^8 N/m^2
elastic modulus = 126 GPa = 126×10^9 Pa
Strain = 2.35×10^8 ÷ 126×10^9 = 0.00187
Length = extension ÷ strain = 0.45 mm ÷ 0.00187 = 240.64 mm
B I hope it’s right I don’t really help a lot but yeah lol
