Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
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Chapter U4, Problem SIII3RE

(a)

Interpretation Introduction

Interpretation:The number of lead ions in the 0.002 M of 10 mL solution of lead nitrate needs to be calculated.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

  Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023

(a)

Expert Solution
Check Mark

Answer to Problem SIII3RE

  1.20 ×1019 lead ions

Explanation of Solution

Molar mass of lead nitrate = 331.2 g/mol

Concentration = 0.002 M

Volume = 10 mL = 0.01 L

1 mole = 6.022 x 1023atoms

Calculate number of lead ions:

  0.01L×0.002  mole1 L × 6.022×1023molecules1 mole× 1 mole of lead ions 1 mole of lead nitrate = 1.20 ×1019 lead ions

(b)

Interpretation Introduction

Interpretation:The mass of lead nitrate in the 0.002 M of 10 mL solution of lead nitrate needs to be determined.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

  Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023

(b)

Expert Solution
Check Mark

Answer to Problem SIII3RE

  6.62 ×103 g

Explanation of Solution

Molar mass of lead nitrate = 331.2 g/mol

Concentration = 0.002 M

Volume = 10 mL = 0.01 L

1 mole = 6.022 x 1023atoms

Calculate mass of lead nitrate:

  0.01L×0.002  mole1 L × 331.2 g1 mole =  6.62 ×103 g

(c)

Interpretation Introduction

Interpretation:The mass of lead ion in the 0.002 M of 10 mL solution of lead nitrate needs to be determined.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

  Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023

(c)

Expert Solution
Check Mark

Answer to Problem SIII3RE

   4.14 ×103 g lead ions.

Explanation of Solution

Molar mass of lead = 207 g/mol

Concentration = 0.002 M

Volume = 10 mL = 0.01 L

1 mole = 6.022 x 1023atoms

Calculate mass of lead nitrate:

  0.01L×0.002  mole1 L = 2 ×105 moles of lead nitrates

1 mole of lead ion= 1 mole of lead nitrate

Calculate mass of lead ion:

  2 ×105 moles of lead ions×207 g1 mole = 4.14 ×103 g

Chapter U4 Solutions

Living By Chemistry: First Edition Textbook

Ch. U4.2 - Prob. 3ECh. U4.2 - Prob. 4ECh. U4.2 - Prob. 5ECh. U4.3 - Prob. 1TAICh. U4.3 - Prob. 1ECh. U4.3 - Prob. 2ECh. U4.3 - Prob. 3ECh. U4.3 - Prob. 4ECh. U4.3 - Prob. 6ECh. U4.4 - Prob. 1TAICh. U4.4 - Prob. 1ECh. U4.4 - Prob. 2ECh. U4.4 - Prob. 3ECh. U4.4 - Prob. 4ECh. U4.4 - Prob. 5ECh. U4.4 - Prob. 6ECh. U4.4 - Prob. 7ECh. U4.5 - Prob. 1TAICh. U4.5 - Prob. 1ECh. U4.5 - Prob. 2ECh. U4.5 - Prob. 3ECh. U4.5 - Prob. 4ECh. U4.6 - Prob. 1TAICh. U4.6 - Prob. 1ECh. U4.6 - Prob. 2ECh. U4.6 - Prob. 3ECh. U4.6 - Prob. 5ECh. U4.6 - Prob. 6ECh. U4.6 - Prob. 7ECh. U4.7 - Prob. 1TAICh. U4.7 - Prob. 1ECh. U4.7 - Prob. 2ECh. U4.7 - Prob. 4ECh. U4.7 - Prob. 5ECh. U4.8 - Prob. 1TAICh. U4.8 - Prob. 1ECh. U4.8 - Prob. 2ECh. U4.8 - Prob. 4ECh. U4.8 - Prob. 5ECh. U4.8 - Prob. 6ECh. U4.8 - Prob. 7ECh. U4.8 - Prob. 8ECh. U4.8 - Prob. 9ECh. U4.8 - Prob. 10ECh. U4.8 - Prob. 11ECh. U4.9 - Prob. 1TAICh. U4.9 - Prob. 1ECh. U4.9 - Prob. 2ECh. U4.9 - Prob. 3ECh. U4.9 - Prob. 4ECh. U4.9 - Prob. 5ECh. U4.9 - Prob. 6ECh. U4.9 - Prob. 7ECh. U4.9 - Prob. 8ECh. U4.10 - Prob. 1TAICh. U4.10 - Prob. 1ECh. U4.10 - Prob. 2ECh. U4.10 - Prob. 3ECh. U4.10 - Prob. 4ECh. U4.10 - Prob. 5ECh. U4.10 - Prob. 6ECh. U4.10 - Prob. 7ECh. U4.10 - Prob. 8ECh. U4.11 - Prob. 1TAICh. U4.11 - Prob. 1ECh. U4.11 - Prob. 2ECh. U4.11 - Prob. 3ECh. U4.11 - Prob. 4ECh. U4.11 - Prob. 5ECh. U4.11 - Prob. 6ECh. U4.11 - Prob. 7ECh. U4.11 - Prob. 8ECh. U4.12 - Prob. 1TAICh. U4.12 - Prob. 1ECh. U4.12 - Prob. 2ECh. U4.12 - Prob. 3ECh. U4.12 - Prob. 4ECh. U4.13 - Prob. 1TAICh. U4.13 - Prob. 1ECh. U4.13 - Prob. 2ECh. U4.13 - Prob. 3ECh. U4.13 - Prob. 4ECh. U4.13 - Prob. 5ECh. U4.13 - Prob. 6ECh. U4.13 - Prob. 7ECh. U4.13 - Prob. 8ECh. U4.13 - Prob. 9ECh. U4.13 - Prob. 10ECh. U4.14 - Prob. 1TAICh. U4.14 - Prob. 1ECh. U4.14 - Prob. 2ECh. U4.14 - Prob. 3ECh. U4.14 - Prob. 4ECh. U4.14 - Prob. 5ECh. U4.14 - Prob. 6ECh. U4.14 - Prob. 7ECh. U4.14 - Prob. 8ECh. U4.14 - Prob. 9ECh. U4.15 - Prob. 1TAICh. U4.15 - Prob. 1ECh. U4.15 - Prob. 2ECh. U4.15 - Prob. 3ECh. U4.15 - Prob. 4ECh. U4.15 - Prob. 5ECh. U4.15 - Prob. 6ECh. U4.15 - Prob. 7ECh. U4.15 - Prob. 8ECh. U4.16 - Prob. 1TAICh. U4.16 - Prob. 1ECh. U4.16 - Prob. 2ECh. U4.16 - Prob. 3ECh. U4.16 - Prob. 4ECh. U4.16 - Prob. 5ECh. U4.16 - Prob. 6ECh. U4.16 - Prob. 7ECh. U4.17 - Prob. 1TAICh. U4.17 - Prob. 1ECh. U4.17 - Prob. 2ECh. U4.17 - Prob. 3ECh. U4.17 - Prob. 5ECh. U4.17 - Prob. 6ECh. U4.17 - Prob. 7ECh. U4.18 - Prob. 1TAICh. U4.18 - Prob. 1ECh. U4.18 - Prob. 2ECh. U4.18 - Prob. 3ECh. U4.18 - Prob. 4ECh. U4.18 - Prob. 5ECh. U4.18 - Prob. 6ECh. U4.18 - Prob. 7ECh. U4.18 - Prob. 8ECh. U4.19 - Prob. 1TAICh. U4.19 - Prob. 1ECh. U4.19 - Prob. 2ECh. U4.19 - Prob. 3ECh. U4.19 - Prob. 4ECh. U4.19 - Prob. 6ECh. U4.19 - Prob. 7ECh. U4.19 - Prob. 8ECh. U4.20 - Prob. 1TAICh. U4.20 - Prob. 1ECh. U4.20 - Prob. 2ECh. U4.20 - Prob. 3ECh. U4.20 - Prob. 4ECh. U4.20 - Prob. 5ECh. U4.20 - Prob. 6ECh. U4.20 - Prob. 7ECh. U4.20 - Prob. 8ECh. U4.21 - Prob. 1TAICh. U4.21 - Prob. 1ECh. U4.21 - Prob. 2ECh. U4.21 - Prob. 4ECh. U4.21 - Prob. 5ECh. U4.21 - Prob. 6ECh. U4.21 - Prob. 7ECh. U4.21 - Prob. 8ECh. U4.22 - Prob. 1TAICh. U4.22 - Prob. 1ECh. U4.22 - Prob. 2ECh. U4.22 - Prob. 3ECh. U4.22 - Prob. 4ECh. U4.22 - Prob. 5ECh. U4.22 - Prob. 6ECh. U4.23 - Prob. 1ECh. U4.23 - Prob. 2ECh. U4.23 - Prob. 3ECh. U4.23 - Prob. 4ECh. U4.23 - Prob. 5ECh. U4.23 - Prob. 6ECh. U4.23 - Prob. 7ECh. U4.24 - Prob. 1ECh. U4.24 - Prob. 2ECh. U4.24 - Prob. 3ECh. U4.24 - Prob. 5ECh. U4.24 - Prob. 6ECh. U4.25 - Prob. 1TAICh. U4.25 - Prob. 1ECh. U4.25 - Prob. 2ECh. U4.25 - Prob. 3ECh. U4.25 - Prob. 4ECh. U4.26 - Prob. 1TAICh. U4.26 - Prob. 1ECh. U4.26 - Prob. 2ECh. U4.26 - Prob. 4ECh. U4.26 - Prob. 5ECh. U4.26 - Prob. 6ECh. U4 - Prob. SI3RECh. U4 - Prob. SI4RECh. U4 - Prob. SII1RECh. U4 - Prob. SII2RECh. U4 - Prob. SII3RECh. U4 - Prob. SII5RECh. U4 - Prob. SII6RECh. U4 - Prob. SIII1RECh. U4 - Prob. SIII2RECh. U4 - Prob. SIII3RECh. U4 - Prob. SIII4RECh. U4 - Prob. SIII5RECh. U4 - Prob. SIII6RECh. U4 - Prob. SIII7RECh. U4 - Prob. SIII8RECh. U4 - Prob. SIV1RECh. U4 - Prob. SIV2RECh. U4 - Prob. SIV3RECh. U4 - Prob. SIV4RECh. U4 - Prob. SV1RECh. U4 - Prob. SV2RECh. U4 - Prob. SV3RECh. U4 - Prob. 1RECh. U4 - Prob. 4RECh. U4 - Prob. 5RECh. U4 - Prob. 6RECh. U4 - Prob. 7RECh. U4 - Prob. 8RECh. U4 - Prob. 9RECh. U4 - Prob. 10RECh. U4 - Prob. 11RECh. U4 - Prob. 12RE

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