Complete the table by classifying each ion as an anion or cation.
Question
Complete the table by classifying each ion as an anion or cation.
Solution
I'm sorry, but I can't assist with that because you haven't provided the table or the ions that need to be classified.
Similar Questions
Complete the table below by writing the symbols for the cation and anion that make up each ionic compound. The first row has been completed for you.ionic compound cation anionNaCl Na+ Cl−NiOH2 FeCl3 CrF3 NH42O
To fill in the table, we need to balance the charges of the cations and anions to find the empirical formula for each ionic compound. Then we can name the compound accordingly. 1. For the cation Fe^3+ and the anion S^2-: - To balance the charges, we need two Fe^3+ ions for every three S^2- ions. - The empirical formula is Fe2S3. - The name of the compound is iron(III) sulfide. 2. For the cation Pb^4+ and the anion O^2-: - To balance the charges, we need two Pb^4+ ions for every four O^2- ions (which simplifies to one Pb^4+ for every two O^2-). - The empirical formula is PbO2. - The name of the compound is lead(IV) oxide. 3. For the cation Mn^2+ and the anion N^3-: - To balance the charges, we need three Mn^2+ ions for every two N^3- ions. - The empirical formula is Mn3N2. - The name of the compound is manganese(II) nitride. 4. For the cation Fe^2+ and the anion P^3-: - To balance the charges, we need three Fe^2+ ions for every two P^3- ions. - The empirical formula is Fe3P2. - The name of the compound is iron(II) phosphide. So the completed table should list the empirical formulas and names of the compounds as follows: - Fe2S3, iron(III) sulfide- PbO2, lead(IV) oxide- Mn3N2, manganese(II) nitride- Fe3P2, iron(II) phosphide
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