Question

can you help me figure out this table second row.

Answer 1

In all this table, we can solve on the same steps:

- find out the anion and its charge.

- find out the cation and its charge.

- Combine them so that the total charge is zero.

The acetate is CH₃CO₂⁻, so its charge is 1-. This means that we will use as many acetates as the charge of the cation and we will use only one cation everytime.

Zinc is Zn and its charge is 2+, so it is Zn²⁺, so we will combine it with two acetates:

$Zn(CH_3CO_2)_2$

The copper is indicated as copper(I), so its charge is 1+, so Cu⁺, so we combine with 1 acetate:

$CuCH_3CO_2$

Iron is indicated as iron(III), so its charge is 3+, so Fe³⁺, and we will combine with 3 acetates:

$Fe(CH_3CO_2)_3$

Potassium can only be 1+, so K⁺, and we will combine with 1 acetate:

$KCH_3CO_2_{}_{}$

Strontium can only be 2+, so Sr²⁺, and we will combine with 2 acetates:

$Sr(CH_3CO_2)_2$

Lead is indicated as lead(IV) so 4+ so Pb⁴⁺, and we will combine with 4 acetates:

$Pb(CH_3CO_2)_4$