Answer:
F = -5.91 E-10 N
Explanation:
The force of attraction between the ions Ca²⁺ and O²⁻ is given by the Coulomb's Law:
[tex] F = K \frac{q_{1}q_{2}}{r^{2}} [/tex]
where K: is the Coulomb's constant, q₁ and q₂: are the charges of the electrons (e = 1.602E-19 C) of the ions, and r: is the distance between the charges.
With q₁ = charge of Ca = (2+)(1.602E-19 C), q₂ = charge of O = (2-)(1.602E-19 C), K= 9E09 N.m².C⁻² and r = 1.25nm = 1.25 E-09 m, the force of attraction between the ions is:
[tex] F = K \frac{q_{1}q_{2}}{r^{2}} = 9E09 N.m^{2}.C^{-2} \frac{(2)(-2)(1.602E-19 C)^{2}}{(1.25E-09 m)^{2}} = -5.91 E-10 N [/tex]
Therefore, the force of attraction between a Ca²⁺ ion and O²⁻ ion is -5.91 E-10 N.
I hope it helps you!
