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

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.