Chemistry: The force of attraction between a divalent cation and a divalent anion is 1.12 x 10-8 n. if the ionic radius of the cation is 0.071 nm, what is the anion radius?

The force of attraction between a divalent cation

Ответ:

smarty5187

06.01.2022

Chemistry

$r_{cation}$ = 0.2162 nm

Explanation:

The force of attraction can be written as:

$F=\frac {K\times |Z_{cation}|e\times |Z_{anion}|e}{r^2}$

Where,

K is the Coulomb's constant having value 9×10⁹ N. m²/C²

$Z_{cation}$ is the charge on the cation

$Z_{anion}$ is the charge on the anion

e is electronic charge = $1.602\times 10^{-19}\ C$

r is the distance between the cation and anion and in bond, it is the sum of the ionic radius of cation and anion

So,

Given, F = $1.12\times 10^{-8}\ N$

$Z_{cation}=Z_{anion}=2$

Applying in the equation,

$1.12\times 10^{-8}=\frac{9\times 10^9\times 2\times 1.602\times 10^{-19}\times 2\times 1.602\times 10^{-19}}{r^2}$

$r^2=\frac{9\times \:10^9\times \:2\times \:1.602\times \:10^{-19}\times \:2\times \:1.602\times \:10^{-19}}{1.12\times \:10^{-8}}$

$r^2=\frac{10^{-29}\cdot \:92.390544}{10^{-8}\cdot \:1.12}$

$r=0.2872\times 10^{-9}\ m$

Also, 1 m = $10^{-9}\ nm$

So, r = 0.2872 nm

Also,

$r=r_{cation}+r_{anion}$

$r_{cation}$ = 0.071 nm

Thus,

$0.2872\ nm=0.071\ nm+r_{anion}$

$r_{cation}$ = 0.2162 nm

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Ответ:

cesargarcia17671

28.02.2023

Chemistry

Heart

Not voluntary.

Explanation:

The heart is made up of special muscle called the cardiac muscle that contains interconnected muscle fibres which are supplied with blood through the coronary artery. Also so that the waves of contraction can travel throughout the mass of the muscle.

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The force of attraction between a divalent cation and a divalent anion is 1.12 x 10-8 n. if the ionic radius of the cation is 0.071 nm, what is the anion radius?

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