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Chapter 2: Problem 67

Write the chemical formulas for the following compounds: (a) aluminum hydroxide, (b) potassium sulfate, (c) copper(I) oxide, (d) zinc nitrate, (e) mercury(II) bromide, (f) iron(III) carbonate, (g) sodium hypobromite.

Short Answer

Expert verified
The chemical formulas for the given compounds are: (a) Al(OH)_3, (b) K_2SO_4, (c) Cu_2O, (d) Zn(NO_3)_2, (e) HgBr_2, (f) Fe_2(CO_3)_3, and (g) NaBrO.

Step by step solution

01

(a) aluminum hydroxide

Aluminum has a charge of +3 (Al^3+) and hydroxide has a charge of -1 (OH^-). To form a neutral compound, we need three hydroxide ions (OH^-) for each aluminum ion (Al^3+). So, the formula for aluminum hydroxide is Al(OH)_3.
02

(b) potassium sulfate

Potassium has a charge of +1 (K^+) and sulfate has a charge of -2 (SO_4^2-). To form a neutral compound, we need two potassium ions (K^+) for each sulfate ion (SO_4^2-). Thus, the formula for potassium sulfate is K_2SO_4.
03

(c) copper(I) oxide

Copper(I) has a charge of +1 (Cu^+) and oxide has a charge of -2 (O^2-). To balance the charges, we need two copper(I) ions (Cu^+) for each oxide ion (O^2-). The formula for copper(I) oxide is Cu_2O.
04

(d) zinc nitrate

Zinc has a charge of +2 (Zn^2+) and nitrate has a charge of -1 (NO_3^-). To form a neutral compound, we need two nitrate ions (NO_3^-) for each zinc ion (Zn^2+). The formula for zinc nitrate is Zn(NO_3)_2.
05

(e) mercury(II) bromide

Mercury(II) has a charge of +2 (Hg^2+) and bromide has a charge of -1 (Br^-). To balance the charges, we need two bromide ions (Br^-) for each mercury(II) ion (Hg^2+). The formula for mercury(II) bromide is HgBr_2.
06

(f) iron(III) carbonate

Iron(III) has a charge of +3 (Fe^3+) and carbonate has a charge of -2 (CO_3^2-). To form a neutral compound, we need two iron(III) ions (Fe^3+) for every three carbonate ions (CO_3^2-). The formula for iron(III) carbonate is Fe_2(CO_3)_3.
07

(g) sodium hypobromite

Sodium has a charge of +1 (Na^+) and hypobromite has a charge of -1 (BrO^-). To balance the charges, we need one sodium ion (Na^+) for each hypobromite ion (BrO^-). The formula for sodium hypobromite is NaBrO.

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How does Dalton's atomic theory account for the fact that when \(1.000 \mathrm{~g}\) of water is decomposed into its elements, \(0.111 \mathrm{~g}\) of hydrogen and \(0.889 \mathrm{~g}\) of oxygen are obtained regardless of the source of the water?

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