Hmmmm that’s a question you just have to take the time to think about the more you think about it the more u will get it u can take as much time as u need

8 0

2 years ago

gThe atomic radii of a divalent cation and a monovalent anion are 0.085 nm and 0.125 nm, respectively. (a) Calculate the force o

frozen [14]

Answer:

1.05 × 10⁻⁸ N

Explanation:

Using Coulomb's law

F force of attraction between them = k $\frac{q^{2} }{r^{2} }$ = k $\frac{z1z2q^{2} }{r^{2} }$

K coulomb constant = 8.99 × 10⁹ N m²C⁻²

Z₁ = + 2 ( valency electron)

Z₂ = -1 ( valency electron)

r is the distance between then since they just touch one another = 0.085 nm + 0.125 nm = 0.21 nm = 0.21 × 10⁻⁹ m

q charge = 1.602 × 10⁻¹⁹ C of an electron

F = ( 2 × 1 × 8.99 × 10⁹ N m²C⁻² × (1.602 × 10⁻¹⁹ C)² / ( 0.21 × 10⁻⁹ m )² = 1046.34 × 10 ⁻¹¹ = 1.05 × 10⁻⁸ N

6 0

3 years ago