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.
B. It is not accelerating since equal and opposite forces cancel each other. Hope it helps
There are 1.923×10^23 formula units in 30.4g mgcl2
molar mass of mgcl2= 95.211g/mol
magnesium= 24.305g/mol
chlorine= 35.45g/mol times 2
24.305+(35.45*2)= 95.211g/mol
sample divided by molar mass times Avogadro's number gives u the formula units
30.4÷ 95.211= .31929
.3193*6.022=1.9288
