Question

write the formulas:

1. calcium bromide
2. copper (i) phosphate
3. barium sulfate
4. phosphorus trisulfide
5. francium phosphide
6. antimony (v) carbide
7. potassium hydrogen carbonate
8. cesium sulfide
9. mercury (ii) nitrate
10. aluminum fluoride

Explanation:

Step1: Determine the ions

For ionic compounds, identify the cation and anion. For example, calcium (\(Ca^{2+}\)) and bromide (\(Br^-\)) in calcium bromide.

Step2: Balance the charges

Use the criss - cross method. For \(Ca^{2+}\) and \(Br^-\), the formula is \(CaBr_2\) (2 bromide ions balance the + 2 charge of calcium).
For covalent compounds like phosphorus trisulfide, the prefixes indicate the number of atoms. "Tri -" means 3, so \(P_2S_3\) (2 phosphorus and 3 sulfur atoms based on common valences).
For compounds with polyatomic ions (e.g., phosphate \(PO_4^{3 -}\), carbonate \(CO_3^{2 -}\), nitrate \(NO_3^-\)):

• In copper (I) phosphate (\(Cu^+\) and \(PO_4^{3 -}\)), 3 \(Cu^+\) ions balance the - 3 charge of phosphate (\(Cu_3PO_4\)).
• In potassium hydrogen carbonate (\(K^+\) and \(HCO_3^-\)), the formula is \(KHCO_3\) (1:1 ratio as charges balance).
• In mercury (II) nitrate (\(Hg^{2+}\) and \(NO_3^-\)), 2 nitrate ions balance the + 2 charge of mercury (\(Hg(NO_3)_2\)).

For metals with specific oxidation states (e.g., antimony (V) \(Sb^{5+}\) and carbide \(C^{4 -}\) in antimony (V) carbide), use the criss - cross method: \(Sb_4C_5\) simplifies to \(SbC_3\) (dividing by 1, since 4 and 5 have no common factors other than 1, but considering common chemical formulas and valences).
For simple ionic compounds like cesium sulfide (\(Cs^+\) and \(S^{2 -}\)), 2 \(Cs^+\) balance the - 2 charge (\(Cs_2S\)).
For aluminum fluoride (\(Al^{3+}\) and \(F^-\)), 3 \(F^-\) balance the + 3 charge (\(AlF_3\)).

Answer:

1. \(CaBr_2\)
2. \(Cu_3PO_4\)
3. \(BaSO_4\)
4. \(P_2S_3\)
5. \(Fr_3P\)
6. \(SbC_3\)
7. \(KHCO_3\)
8. \(Cs_2S\)
9. \(Hg(NO_3)_2\)
10. \(AlF_3\)