The force of attraction between a divalent cation and a divalent anion is 1.91 x 10-8 n. if the ionic radius of the cation is 0.074 nm, what is the anion radius?

The force of attraction (F) is given by the formula:

F = (1/4π∈r²)(Zc*e)(Za*e)(1)

where:

∈ = permittivity of free space = 8.85*10^-15 F/m

Zc = charge on the cation = +2

Zc = charge on the anion = -2

e = charge on an electron = 1.602*10^-19 C

r = interionic distance = rc + ra

where rc and ra are the radius of cation and anion respectively

F = 1.91 * 10^-8 N

therefore based on equation (1) we have:

1.91 *10^-8 = [1/4π(8.85*10^-15)r²](2*1.602*10^-19)²

r² = 0.04832 * 10^-15

r = 6.951 nm

Now: r = rc + ra

where rc = 0.074 nm

thus, ra = r - rc = 6.951 - 0.074 = 6.877 nm

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

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~ trueboss72