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julia-pushkina [17]
3 years ago
9

Consider two different ions. The anion has a valence of -2. The cation has a valence of +2. The two ions are separated by a dist

ance of 1 nm. Please calculate the force of attraction between the anion and cation. The force of attraction is given by: (9 x 109 V/C) (Z)(2)(e2) FA valence of the ions, e = charge of an electron 1.602 x 10-19 C Where Z1 and Z2 r distance between ions 1N 1 (V C/m)
Chemistry
1 answer:
strojnjashka [21]3 years ago
6 0

Answer:

Force of attraction = 35.96 \times 10^{27}N

Explanation:

Given: charge on anion = -2

Charge on cation = +2

Distance = 1 nm = 10^{-9} m

To calculate: Force of attraction.

Solution: The force of attraction is calculated by using equation,

F = \dfrac{k \times q_1 q_2}{ \r^2} ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k = 8.99 \times 10^9 \ Nm^{2}C^{-2}

F = force of attraction

r = distance between ions.

Substituting all the values in the equation (1) the equation becomes

F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }

Force of attraction = 35.96 \times 10^{27}N

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