Calculate the force of attraction between a cation with a valence of +3 and an anion with a valence of -2, the centers of which are separated by a distance of 3.4 nm.\

Answer is: attraction force is 1,185·10⁻¹⁰N.
Coulomb's law: F = k·q₁·q₂ ÷ r².
F - attraction force.
k - Coulomb's law constant, 8,9·10⁹ N·m² / C².
q₁ -  quantity of charge of cation.
q₂ - quantity of charge of anion.
r - distance between anions.
q₁ = 3·1,6·10⁻¹⁹ = 4,8·10⁻¹⁹ C.
q₂ = -3,2·10⁻¹⁹ C.
F = 8,9·10⁹ N·m²/C²  · 4,8·10⁻¹⁹ C · (-3,2·10⁻¹⁹ C) ÷ (3,4·10⁻⁹ m)².
F =  -1,37·10⁻²⁷ N·m² ÷ 1,156·10⁻¹⁷ m².
F = -1,185·10⁻¹⁰N.

The correct answer is c