Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
Question
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Chapter U1, Problem SIV6RQ
Interpretation Introduction

(a)

Interpretation:

The chemical formulas for sodium silicate, sodium chlorite and sodium bicarbonate are to be written.

Concept introduction:

In some ionic compound, the ions are composed of more than one atom. These are called polyatomic ions and the compound is known as polyatomic compound. The entire cluster of atoms shares the charge on a polyatomic ion.

To write the chemical formulas of such polyatomic compounds, one must apply the rule of zero charge which states that charges on metal cations and polyatomic anions must add up to zero.

Expert Solution
Check Mark

Answer to Problem SIV6RQ

  • Sodium silicate - Na2SiO3
  • Sodium chlorite - NaClO2
  • Sodium bicarbonate - NaHCO3

Explanation of Solution

When there is more than one of the same polyatomic ion in a formula, the ion is enclosed in parenthesis and a subscript number is written indicating the number of ions in the compound.

  • Sodium silicate - Na2SiO3 - The polyatomic ion here is silicate (SiO32-). Sodium has + 1 charge so 2 sodium atoms will combine with 1 silicate ion. Hence, sum of charges becomes zero. The formula for this compound is Na2SiO3.
  • Sodium chlorite - NaClO2 - The polyatomic ion here is chlorite (ClO2-) and sodium has + 1 charge so 1 sodium atom will combine with 1 chlorite ion. Hence, sum of charges becomes zero. The formula for this compound is NaClO2.
  • Sodium bicarbonate - NaHCO3 - The polyatomic ion here is bicarbonate (HCO3-). Sodium has 2 + charge so 1 sodium atom will combine with 1bicarbonate ion. Hence, sum of charges becomes zero. The formula for this compound is NaHCO3.
Interpretation Introduction

(b)

Interpretation:

The chemical formulas for calcium silicate, calcium chlorite and calcium bicarbonate are to be written.

Concept introduction:

In some ionic compound, the ions are composed of more than one atom. These are called polyatomic ions and the compound is known as polyatomic compound. The entire cluster of atoms shares the charge on a polyatomic ion.

To write the chemical formulas of such polyatomic compounds, one must apply the rule of zero charge which states that charges on metal cations and polyatomic anions must add up to zero.

Expert Solution
Check Mark

Answer to Problem SIV6RQ

  • Calcium silicate - CaSiO3
  • Calcium chlorite - Ca(ClO2)2
  • Calcium bicarbonate - Ca(HCO3)2

Explanation of Solution

When there is more than one of the same polyatomic ion in a formula, the ion is enclosed in parenthesis and a subscript number is written indicating the number of ions in the compound.

  • Calcium silicate - Ca2SiO3 - The polyatomic ion here is silicate (SiO32-). Calcium has + 2 charges so 1 calcium atom will combine with 1 silicate ion. Hence, sum of charges becomes zero. The formula for this compound is Ca2SiO3.
  • Calcium chlorite - Ca(ClO2)2 - The polyatomic ion here is chlorite (ClO2-)Calcium has + 2 charge so 1 calcium atom will combine with 2chlorite ions. Hence, sum of charges becomes zero. The formula for this compound is Ca(ClO2)2.
  • Calcium bicarbonate - Ca(HCO3)2 - The polyatomic ion here is bicarbonate (HCO3-). Calcium has + 2 charges so 1 calcium atom will combine with 2bicarbonate ion. Hence, sum of charges becomes zero. The formula for this compound is Ca(HCO3)2.

Chapter U1 Solutions

Living By Chemistry: First Edition Textbook

Ch. U1.2 - Prob. 1ECh. U1.2 - Prob. 2ECh. U1.2 - Prob. 5ECh. U1.3 - Prob. 1TAICh. U1.3 - Prob. 1ECh. U1.3 - Prob. 2ECh. U1.3 - Prob. 3ECh. U1.3 - Prob. 4ECh. U1.3 - Prob. 5ECh. U1.3 - Prob. 6ECh. U1.4 - Prob. 1TAICh. U1.4 - Prob. 1ECh. U1.4 - Prob. 2ECh. U1.4 - Prob. 3ECh. U1.4 - Prob. 4ECh. U1.4 - Prob. 5ECh. U1.4 - Prob. 6ECh. U1.4 - Prob. 7ECh. U1.4 - Prob. 8ECh. U1.4 - Prob. 9ECh. U1.4 - Prob. 10ECh. U1.4 - Prob. 11ECh. U1.5 - Prob. 1TAICh. U1.5 - Prob. 1ECh. U1.5 - Prob. 2ECh. U1.5 - Prob. 3ECh. U1.5 - Prob. 4ECh. U1.5 - Prob. 5ECh. U1.5 - Prob. 6ECh. U1.5 - Prob. 7ECh. U1.6 - Prob. 1TAICh. U1.6 - Prob. 1ECh. U1.6 - Prob. 2ECh. U1.6 - Prob. 3ECh. U1.6 - Prob. 4ECh. U1.6 - Prob. 5ECh. U1.6 - Prob. 6ECh. U1.7 - Prob. 1TAICh. U1.7 - Prob. 1ECh. U1.7 - Prob. 2ECh. U1.7 - Prob. 3ECh. U1.7 - Prob. 4ECh. U1.7 - Prob. 5ECh. U1.8 - Prob. 1TAICh. U1.8 - Prob. 1ECh. U1.8 - Prob. 2ECh. U1.8 - Prob. 4ECh. U1.8 - Prob. 5ECh. U1.8 - Prob. 6ECh. U1.8 - Prob. 7ECh. U1.9 - Prob. 1TAICh. U1.9 - Prob. 1ECh. U1.9 - Prob. 2ECh. U1.9 - Prob. 5ECh. U1.9 - Prob. 7ECh. U1.10 - Prob. 1TAICh. U1.10 - Prob. 1ECh. U1.10 - Prob. 2ECh. U1.10 - Prob. 3ECh. U1.10 - Prob. 4ECh. U1.10 - Prob. 5ECh. U1.10 - Prob. 6ECh. U1.10 - Prob. 7ECh. U1.10 - Prob. 8ECh. U1.11 - Prob. 1TAICh. U1.11 - Prob. 1ECh. U1.11 - Prob. 2ECh. U1.11 - Prob. 3ECh. U1.11 - Prob. 4ECh. U1.11 - Prob. 5ECh. U1.11 - Prob. 6ECh. U1.11 - Prob. 7ECh. U1.11 - Prob. 9ECh. U1.11 - Prob. 11ECh. U1.11 - Prob. 12ECh. U1.12 - Prob. 1TAICh. U1.12 - Prob. 1ECh. U1.12 - Prob. 2ECh. U1.12 - Prob. 3ECh. U1.12 - Prob. 4ECh. U1.12 - Prob. 5ECh. U1.12 - Prob. 6ECh. U1.12 - Prob. 7ECh. U1.12 - Prob. 8ECh. U1.13 - Prob. 1TAICh. U1.13 - Prob. 1ECh. U1.13 - Prob. 2ECh. U1.13 - Prob. 3ECh. U1.13 - Prob. 4ECh. U1.13 - Prob. 5ECh. U1.13 - Prob. 6ECh. U1.13 - Prob. 7ECh. U1.13 - Prob. 8ECh. U1.13 - Prob. 9ECh. U1.14 - Prob. 1TAICh. U1.14 - Prob. 1ECh. U1.14 - Prob. 2ECh. U1.14 - Prob. 3ECh. U1.14 - Prob. 4ECh. U1.14 - Prob. 5ECh. U1.14 - Prob. 6ECh. U1.14 - Prob. 7ECh. U1.14 - Prob. 8ECh. U1.14 - Prob. 9ECh. U1.14 - Prob. 10ECh. U1.14 - Prob. 11ECh. U1.14 - Prob. 12ECh. U1.14 - Prob. 13ECh. U1.14 - Prob. 14ECh. U1.15 - Prob. 1TAICh. U1.15 - Prob. 1ECh. U1.15 - Prob. 2ECh. U1.15 - Prob. 3ECh. U1.15 - Prob. 4ECh. U1.15 - Prob. 5ECh. U1.15 - Prob. 6ECh. U1.15 - Prob. 7ECh. U1.15 - Prob. 8ECh. U1.15 - Prob. 9ECh. U1.15 - Prob. 10ECh. U1.15 - Prob. 11ECh. U1.15 - Prob. 12ECh. U1.16 - Prob. 1TAICh. U1.16 - Prob. 1ECh. U1.16 - Prob. 2ECh. U1.16 - Prob. 3ECh. U1.16 - Prob. 4ECh. U1.16 - Prob. 5ECh. U1.16 - Prob. 6ECh. U1.17 - Prob. 1TAICh. U1.17 - Prob. 1ECh. U1.17 - Prob. 2ECh. U1.17 - Prob. 3ECh. U1.17 - Prob. 4ECh. U1.17 - Prob. 5ECh. U1.17 - Prob. 6ECh. U1.17 - Prob. 7ECh. U1.17 - Prob. 8ECh. U1.17 - Prob. 9ECh. U1.17 - Prob. 10ECh. U1.17 - Prob. 11ECh. U1.18 - Prob. 1TAICh. U1.18 - Prob. 1ECh. U1.18 - Prob. 2ECh. U1.18 - Prob. 3ECh. U1.18 - Prob. 4ECh. U1.18 - Prob. 5ECh. U1.18 - Prob. 6ECh. U1.18 - Prob. 7ECh. U1.18 - Prob. 8ECh. U1.18 - Prob. 9ECh. U1.18 - Prob. 10ECh. U1.19 - Prob. 1TAICh. U1.19 - Prob. 1ECh. U1.19 - Prob. 2ECh. U1.19 - Prob. 3ECh. U1.19 - Prob. 4ECh. U1.19 - Prob. 5ECh. U1.19 - Prob. 6ECh. U1.19 - Prob. 7ECh. U1.19 - Prob. 8ECh. U1.19 - Prob. 9ECh. U1.19 - Prob. 10ECh. U1.19 - Prob. 11ECh. U1.19 - Prob. 12ECh. U1.19 - Prob. 13ECh. U1.19 - Prob. 14ECh. U1.19 - Prob. 15ECh. U1.19 - Prob. 16ECh. U1.20 - Prob. 1TAICh. U1.20 - Prob. 1ECh. U1.20 - Prob. 2ECh. U1.20 - Prob. 3ECh. U1.20 - Prob. 4ECh. U1.20 - Prob. 5ECh. U1.20 - Prob. 6ECh. U1.20 - Prob. 7ECh. U1.21 - Prob. 1TAICh. U1.21 - Prob. 1ECh. U1.21 - Prob. 2ECh. U1.21 - Prob. 3ECh. U1.21 - Prob. 4ECh. U1.21 - Prob. 5ECh. U1.21 - Prob. 6ECh. U1.21 - Prob. 7ECh. U1.21 - Prob. 8ECh. U1.22 - Prob. 1TAICh. U1.22 - Prob. 1ECh. U1.22 - Prob. 2ECh. U1.22 - Prob. 3ECh. U1.22 - Prob. 4ECh. U1.22 - Prob. 5ECh. U1.22 - Prob. 6ECh. U1.22 - Prob. 7ECh. U1.23 - Prob. 1TAICh. U1.23 - Prob. 1ECh. U1.23 - Prob. 2ECh. U1.23 - Prob. 3ECh. U1.23 - Prob. 4ECh. U1.23 - Prob. 5ECh. U1.24 - Prob. 1TAICh. U1.24 - Prob. 1ECh. U1.24 - Prob. 2ECh. U1.24 - Prob. 3ECh. U1.24 - Prob. 4ECh. U1.24 - Prob. 5ECh. U1.24 - Prob. 6ECh. U1.24 - Prob. 7ECh. U1.24 - Prob. 8ECh. U1.24 - Prob. 9ECh. U1.24 - Prob. 10ECh. U1.24 - Prob. 11ECh. U1.24 - Prob. 12ECh. U1.24 - Prob. 13ECh. U1.25 - Prob. 1TAICh. U1.25 - Prob. 1ECh. U1.25 - Prob. 2ECh. U1.25 - Prob. 3ECh. U1.25 - Prob. 4ECh. U1.25 - Prob. 5ECh. U1.25 - Prob. 6ECh. U1.26 - Prob. 1TAICh. U1.26 - Prob. 1ECh. U1.26 - Prob. 2ECh. U1.26 - Prob. 3ECh. U1.26 - Prob. 4ECh. U1.26 - Prob. 5ECh. U1.26 - Prob. 6ECh. U1.26 - Prob. 7ECh. U1.26 - Prob. 8ECh. U1.26 - Prob. 9ECh. U1.26 - Prob. 10ECh. U1.27 - Prob. 1TAICh. U1.27 - Prob. 1ECh. U1.27 - Prob. 2ECh. U1.27 - Prob. 3ECh. U1.27 - Prob. 4ECh. U1.27 - Prob. 5ECh. U1.27 - Prob. 6ECh. U1.27 - Prob. 7ECh. U1 - Prob. SI1RECh. U1 - Prob. SI2RECh. U1 - Prob. SI3RECh. U1 - Prob. SI4RECh. U1 - Prob. SI5RECh. U1 - Prob. SI6RECh. U1 - Prob. SII1RQCh. U1 - Prob. SII2RQCh. U1 - Prob. SII3RQCh. U1 - Prob. SII4RQCh. U1 - Prob. SIII1RQCh. U1 - Prob. SIII2RQCh. U1 - Prob. SIII3RQCh. U1 - Prob. SIII4RQCh. U1 - Prob. SIV1RQCh. U1 - Prob. SIV2RQCh. U1 - Prob. SIV3RQCh. U1 - Prob. SIV4RQCh. U1 - Prob. SIV5RQCh. U1 - Prob. SIV6RQCh. U1 - Prob. SV1RQCh. U1 - Prob. SV2RQCh. U1 - Prob. SV3RQCh. U1 - Prob. SV4RQCh. U1 - Prob. 1RECh. U1 - Prob. 2RECh. U1 - Prob. 3RECh. U1 - Prob. 4RECh. U1 - Prob. 5RECh. U1 - Prob. 6RECh. U1 - Prob. 7RECh. U1 - Prob. 8RECh. U1 - Prob. 9RECh. U1 - Prob. 10RECh. U1 - Prob. 11RECh. U1 - Prob. 12RE
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