Question

1. crcl₃ = chromium(iii) chloride
2. f₂o₃ = iron(iii) oxide
3. co = carbon monoxide
4. sio₂ = silicon dioxide
5. nahco₃ = sodium bicarbonate
6. cu(no₃)₂ = copper(ii) nitrate
7. al₂(so₄)₃ =
8. sncl₄ =

Explanation:

Step1: Analyze the compound \(Al_2(SO_4)_3\)

The cation is \(Al^{3 +}\) (aluminum ion) and the anion is \(SO_4^{2 -}\) (sulfate ion). When naming ionic compounds, we name the cation first followed by the anion.

Step2: Analyze the compound \(SnCl_4\)

The cation is \(Sn^{4+}\) (tin(IV) ion as tin can have multiple oxidation states) and the anion is \(Cl^{-}\) (chloride ion). For elements with variable oxidation states, we use Roman numerals in the name.

Answer:

1. \(Al_2(SO_4)_3=\text{Aluminum sulfate}\)
2. \(SnCl_4=\text{Tin(IV) chloride}\)