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.
<span>This question is simple, answers are D and C respectively</span>
Answer:
b. ICl experiences dipole-dipole interactions
Explanation:
The molecule with the <em>stronger intermolecular forces</em> will have the higher boiling point.
In Br₂, the Br-Br bond has a <em>dipole moment of zero</em>, because the two atoms are identical.
In ICl, the I-Cl bond, has two different atoms. One must be more electronegative than the other, so there will be a <em>non-zero bond dipole</em>.
ICl will have the higher boiling point.
a is <em>wrong</em>. Br₂ is nonpolar, so it has no dipole-dipole interactions.
c is <em>wrong</em>. Br₂ cannot form hydrogen bonds, because there is no hydrogen.
d is <em>wrong</em>. ICl has dipole-dipole interactions.
Answer:
only japanese and english I understand sorry your languages
Explanation:
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Answer:
Here's the answer in order
A
B
B
C
C
A
A
D
A
C
D
Explanation:
lol nvm
