The atomic radii of a divalent cation and a monovalent anion are 0.35 nm and 0.129 nm, respectively.(a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i. e., when the ions just touch one another).Enter your answer for part (a) in accordance to the question statement N(b) What is the force of repulsion at this same separation distance?

a) The force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another) is - 2.01 × 10⁻⁹ N

b) The force of repulsion at this same separation distance is 2.01 × 10⁻⁹ N

Explanation:

F = kq₁q₂/r²

r = 0.35 + 0.129 (since the ions are just touching each other)

r = 0.479 nm = 4.79 × 10⁻¹⁰ m

Since the first ion is a divalent cation, Z₁ = +2 and the monovalent anion, Z₂ = -1

q = Ze; e = 1.602 × 10⁻¹⁹ C

K = 8.99 × 10⁹ Nm²/C²

F = (8.99 × 10⁹)(1.602 × 10⁻¹⁹)²(2)(-1)/(4.79 × 10⁻¹⁰)² = - 2.01 × 10⁻⁹ N

b) At equilibrium,

Force of attraction + Force of repulsion = 0

Force of repulsion = -(Force of attraction) = 2.01 × 10⁻⁹ N

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.