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Chapter 2: Problem 39

Write the formulas of the following ionic compounds. (a) iron(III) carbonate (b) sodium azide \(\left(\mathrm{N}_{3}-\right)\) (c) calcium sulfate (d) copper(I) sulfide (e) lead(IV) oxide

Short Answer

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Question: Write the formulas for the following ionic compounds: a) iron(III) carbonate b) sodium azide c) calcium sulfate d) copper(I) sulfide e) lead(IV) oxide Answer: a) Fe2(CO3)3 b) NaN3 c) CaSO4 d) Cu2S e) PbO2

Step by step solution

01

(a) iron(III) carbonate

The iron(III) cation has a charge of +3, written as Fe3+. The carbonate anion has a charge of -2, which is written as CO3^2-. To balance the charges, we need two iron cations (2 x 3+) and three carbonate anions (3 x 2-). The formula for iron(III) carbonate is \(\mathrm{Fe}_{2}(\mathrm{CO}_{3})_{3}\).
02

(b) sodium azide

Sodium cation (Na+) has a charge of +1 and azide anion (N3-) has a charge of -1. These charges are already balanced, so the formula for sodium azide is \(\mathrm{Na}(\mathrm{N}_{3})\).
03

(c) calcium sulfate

Calcium cation (Ca2+) has a charge of +2 and the sulfate anion (SO4^2-) has a charge of -2. These charges are already balanced, so the formula for calcium sulfate is \(\mathrm{Ca}(\mathrm{SO}_{4})\).
04

(d) copper(I) sulfide

The copper(I) cation has a charge of +1, written as Cu+. The sulfide anion has a charge of -2, which is written as S2-. To balance the charges, we need two copper cations (2 x 1+) and one sulfide anion (1 x 2-). The formula for copper(I) sulfide is \(\mathrm{Cu}_{2}\mathrm{S}\).
05

(e) lead(IV) oxide

The lead(IV) cation has a charge of +4, written as Pb4+. The oxide anion has a charge of -2, which is written as O2-. To balance the charges, we need one lead cation (1 x 4+) and two oxide anions (2 x 2-). The formula for lead(IV) oxide is \(\mathrm{Pb}(\mathrm{O}_{2})\).

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