Write the formula for each of the following compounds: a. chromium(III) hydroxide b. magnesium cyanide c. lead(IV) carbonate d. ammonium acetate

For each compound, we first need to identify the symbols and charges of the elements or polyatomic ions making up the compound. We can use a periodic table to find the symbols of individual elements and a table of polyatomic ions for finding the symbols and charges of polyatomic ions. a. chromium(III) hydroxide - Chromium(III) has the symbol Cr with a charge of +3, as indicated by the Roman numeral (III) in its name. - Hydroxide is a polyatomic ion with the symbol OH and a charge of -1. b. magnesium cyanide - Magnesium has the symbol Mg and a charge of +2. - Cyanide is a polyatomic ion with the symbol CN and a charge of -1. c. lead(IV) carbonate - Lead(IV) has the symbol Pb with a charge of +4, as indicated by the Roman numeral (IV) in its name. - Carbonate is a polyatomic ion with the symbol CO₃ and a charge of -2. d. ammonium acetate - Ammonium is a polyatomic ion with the symbol NH₄ and a charge of +1. - Acetate is a polyatomic ion with the symbol CH₃COO (or C₂H₃O₂) and a charge of -1.

Next, we will combine the symbols of the elements or polyatomic ions and their charges to get the final chemical formula. We do this by balancing the charges and adding subscripts to indicate the number of each ion or element present in the compound. a. chromium(III) hydroxide - The charges of Cr(+3) and OH(-1) need to be balanced. To do this, we need to have three OH ions for every Cr ion. - The formula is Cr(OH)₃. b. magnesium cyanide - The charges of Mg(+2) and CN(-1) need to be balanced. To do this, we need to have two CN ions for every Mg ion. - The formula is Mg(CN)₂. c. lead(IV) carbonate - The charges of Pb(+4) and CO₃(-2) need to be balanced. To do this, we need to have two CO₃ ions for every Pb ion. - The formula is Pb(CO₃)₂. d. ammonium acetate - The charges of NH₄(+1) and CH₃COO(-1) are already balanced. - The formula is NH₄CH₃COO (or NH₄C₂H₃O₂).