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?

= 0.2162 nm

Explanation:

The force of attraction can be written as:

Where,  

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

 is the charge on the cation

 is the charge on the anion

e is electronic charge =

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 =

Applying in the equation,

Also, 1 m =

So, r = 0.2872 nm

Also,

= 0.071 nm

Thus,

= 0.2162 nm

it would be d. henry moseley

hope this !

and sorry if i'm late. have a good rest of your day! : d

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.