Give the systematic names for the formulas or the formulas for the names:

a. CoO;

b.Mercury(I) chloride

C.$\mathbf{Pb}{\mathbf{\left(}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{.}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{;}$

d. chromic oxide

The compound is an ionic compound. The name does not depend on the number of elements. Cobalt ions can have many forms so the charge should be known. O has a charge of -2 and the subscripts will just interchange as charges of these elements. The charge of cobalt is also +2 to balance the charges

$\mathbf{CoO}\mathbf{}\mathbf{\to }{\mathbf{Co}}^{\mathbf{2}\mathbf{+}}{\mathbf{O}}^{\mathbf{2}\mathbf{-}}$

The cation is cobalt (II) and the anion is oxide. So it is cobalt(II) oxide.

Mercury has a charge of 1+ and chloride has a charge of -1. The charges will just interchange as the subscripts of these elements

${\mathbf{Hg}}^{\mathbf{1}\mathbf{+}}{\mathbf{CI}}^{\mathbf{1}\mathbf{-}}\mathbf{\to }\mathbf{HgCI}$

There is an exception for this compound and HgCl only acts like an empirical formula for mercury(I) chloride. The molecular formula of mercury (I) oxide is ${\mathbf{Hg}}_{\mathbf{2}}{\mathbf{CI}}_{\mathbf{2}}$.

The compound is an ionic compound. The name does not depend on the number of the elements

Lead ion can have many forms so the charge should be known acetate charge is 2- the subscripts will just interchange as charges of these elements

$\mathbf{Pb}{\mathbf{\left(}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{.}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{;}$

The cation is lead (II) and the anion is acetate. So it is lead (II) acetate here it is three hydrates it makes a trihydrate so the name is lead acetate trihydrate.

Chromic has a charge of 3+ and oxide has a charge of 2-. The charges will just interchange as the subscripts of these elements:

${\mathbf{Cr}}^{\mathbf{3}\mathbf{+}}{\mathbf{O}}^{\mathbf{2}\mathbf{-}}\mathbf{}\mathbf{\to }{\mathbf{Cr}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$