First, recognize that this is an ionic compound.

Identify the cation and the anion. The cation is lead, and the anion is nitrate.

Since lead can form more than one type of ion, figure out the charge associated with the given cation.

\(\displaystyle \text{NO}_3\) has a charge of \(\displaystyle -1\). Since there are \(\displaystyle 2\) nitrate ions, the charge for lead must be \(\displaystyle +2\). Remember to put the charge of the cation in parentheses when naming the compound.

The name of this compound is then \(\displaystyle \text{lead (II) nitrate}\).

\(\displaystyle HClO_{4}\) consists of a \(\displaystyle H^{+}\) ion bonded to a \(\displaystyle ClO_{4}^{-}\) ion, which is a perchlorate ion. Thus, the name of the ion's corresponding acid is perchloric acid. \(\displaystyle HCl\) is the molecular formula for hydrochloric acid, \(\displaystyle HClO_{3}\) is the formula for chloric acid, while \(\displaystyle HClO_{2}\) is the formula for chlorous acid. 

Because there are four oxygen atoms like in perchloric acid, the correct name is perbromic acid. Hypobromic acid is not a valid name for an acid since the "hypo-" prefix refers to one or two oxygen atoms but the "-ic" suffix refers to either three or four oxygen atoms. Hypobromous and bromous acid refer to one and two oxygen atoms, respectively. Hydrobromic acid has no oxygen atoms. 

The correct nomenclature for \(\displaystyle H_{2}SO_{3}\) is sulfurous acid. Oxyanions ending in -ate form oxyacids ending in -ic. For example, sulfate \(\displaystyle SO_{4}^{2-}\) becomes sulfuric acid \(\displaystyle H_{2}SO_{4}\). Oxyanions ending in -ite form oxyacids ending in -ous. In this problem, sulfite \(\displaystyle SO_{3}^{2-}\) becomes sulfurous acid \(\displaystyle H_{2}SO_{3}\).  

The correct molecular formula for cobalt (II) sulfite is \(\displaystyle CoSO_{3}\). The charge on cobalt (II) is \(\displaystyle Co^{2+}\). When combined with sulfite \(\displaystyle SO_{3}^{2-}\), the positive and negative charges cancel out to create a neutral compound \(\displaystyle CoSO_{3}\).  

When assigning oxidation states to elements of a given compound, non-transition metal elements are assigned specific oxidation states corresponding to their group number and valence relative to a complete octet.

Group 1 elements have 1 valence electron and an oxidation state of +1.

Group 2 elements have 2 valence electrons and an oxidation state of +2. 

Group 8 elements (the noble gases) have a complete octet, thus their assigned oxidation state is 0.

Group 7 elements (halogens) have 7 valence electrons and an oxidation state of -1.

Group 6 elements such as oxygen have 6 valence electrons and an oxidation state of -2.

Permanganate has 4 oxygen atoms and an overall charge of -1. The oxidation state of the \(\displaystyle Mn\) atom may be found by first calculating the combined oxidation state of the oxygen atoms:

\(\displaystyle 4 * (-2) = -8\) 

and finding the difference between their combined oxidation state and the overall charge of the ion (-1):

\(\displaystyle (-1) - (-8) = +7\)

Both the electron-pair and molecular geometries of ammonium are tetrahedral, as opposed to ammonia (\(\displaystyle NH_3\)), in which the electron-pair geometry is tetrahedral, but the molecular geometry is trigonal pyramidal. As evidenced below:

Ammonia:

Note that the difference in geometry arises in the presence of the lone pair on nitrogen in ammonia. The lone pair only contributes to molecular geometry, but does not contribute to electron-pair geometry. Bonding electrons, however, contribute to both electron and molecular geometries.

Ammonium:

In ammonium, the lone pair seen in ammonia is shared in a bond with an additional hydrogen atom. 

On average, the molar masses of carbon, hydrogen, and oxygen are approximately \(\displaystyle 12\frac{g}{mol}\),  \(\displaystyle 1\frac{g}{mol}\), and \(\displaystyle 16\frac{g}{mol}\), respectively.

The molar mass of a molecule of glucose is:

\(\displaystyle \left (6 * 12 \frac{g}{mol} \right ) + \left (12 * 1 \frac{g}{mol} \right ) + \left (6 * 16 \frac{g}{mol} \right ) = 180 \frac{g}{mol}\) 

Each term added in the previous step is the mass of each element present in a mole of glucose, so the mass percent of its components are found by dividing their respective contributions to the molar mass by the molar mass:

\(\displaystyle C: \frac{72\frac{g}{mol}}{180\frac{g}{mol}} * 100 \% = 40\%\)

\(\displaystyle H: \frac{12\frac{g}{mol}}{180\frac{g}{mol}} * 100 \% = 6.7\%\)

\(\displaystyle O: \frac{96\frac{g}{mol}}{180\frac{g}{mol}} * 100\% = 53.3\%\)

\(\displaystyle C_2H_3O_2^-\) (also written as \(\displaystyle CH_3COO^-\)) is the formula for the acetate ion. Most polyatomic ions tend to follow certain naming trends, but acetate is one of the ions that does not follow those naming trends, so memorize this one since it shows up frequently in chemistry.

Start by identifying the cation and the anion of this ionic compound.

\(\displaystyle \text{Cation: Lead}\)

\(\displaystyle \text{Anion: Iodine}\)

Next, deduce the charge found on the lead. Since we know that ion for iodine normally has a \(\displaystyle -1\) charge, we can figure out that the lead must have a \(\displaystyle +4\) charge in order to create the neutral compound.

Next, remember that the anion name of iodine is iodide.

Finally, put the name together by using the following formula:

\(\displaystyle \text{Name: Cation (charge of cation in Roman numerals) anion name}\)

Thus, the name of the compound is \(\displaystyle \text{Lead (IV) iodide}\).