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 SIII4RE

(a)

Interpretation Introduction

Interpretation:

The solution with the greatest concentration out of 100 mL of given solutions needs to be determined.

Concept introduction:

The molecular mass of a 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 SIII4RE

Since the volume is the same for all the given solutions, therefore, the NaOH and KCl will have a higher concentration.

Explanation of Solution

The concentration mainly depends on the number of moles and volume. Since the volume is the same for all the given solutions, therefore, the NaOH and KCl will have a higher concentration than the PbCl2 solution which is only 0.5 M compared to 1.0 M NaOH and KCl.

(b)

Interpretation Introduction

Interpretation:

The solution with greatest mass out of 100 mL of given solutions needs to be determined.

Concept introduction:

The molecular mass of a 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 SIII4RE

PbCl2 solution has the greatest mass.

Explanation of Solution

Molar mass of NaOH = 39.9 g/mol

Molar mass of KCl = 74.5 g/mol

Molar mass of PbCl2 = 278 g/mol

Concentration = 1.0 M

Volume = 100 mL = 0.1 L

Calculate mass of each:

Concentration = massmolar mass×volumemass = Concentration×molar mass×volumemass of NaOH  = 1.0 M  ×39.9 g/mol × 0.1 L = 3.99 gmass of KCl  = 1.0 M  ×74.5 g/mol × 0.1 L = 7.45 gmass of PbCl2  = 0.5 M  × 278 g/mol × 0.1 L = 13.9 g

(c)

Interpretation Introduction

Interpretation:

The solution with maximum number of moles needs to be determined if volume is 100 mL for each; 0.1 M NaOH, 1.0 M KCl and 0.5 M pbCl2 solution.

Concept introduction:

The molecular mass of a 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 SIII4RE

NaOH and KCl will have most of the moles.

Explanation of Solution

Molar mass of NaOH = 39.9 g/mol

Molar mass of KCl = 74.5 g/mol

Molar mass of PbCl2 = 278 g/mol

Concentration = 1.0 M

Volume = 100 mL = 0.1 L

Calculate moles of each:

Concentration = molesvolumeMoles = Concentration× volumeMoles of NaOH  = 1.0 M  ×  0.1 L = 0.1 molesMoles of KCl  = 1.0 M  × 0.1 L =0.1 molesMoles of PbCl2  = 0.5 M  ×  0.1 L = 0.05 moles

(d)

Interpretation Introduction

Interpretation:

The solution with maximum number of moles of ions needs to be determined if volume is 100 mL for each; 0.1 M NaOH, 1.0 M KCl and 0.5 M pbCl2 solution.

Concept introduction:

The molecular mass of a 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

(d)

Expert Solution
Check Mark

Answer to Problem SIII4RE

PbCl2will have most of the moles of ions.

Explanation of Solution

Molar mass of NaOH = 39.9 g/mol

Molar mass of KCl = 74.5 g/mol

Molar mass of PbCl2 = 278 g/mol

Concentration = 1.0 M

Volume = 100 mL = 0.1 L

Calculate mass of each:

Concentration = molesvolumeMoles = Concentration× volumeMoles of NaOH  = 1.0 M  ×  0.1 L = 0.1 moles =  0.1 ×2 = 0.2 moles of ionsMoles of KCl  = 1.0 M  × 0.1 L =0.1 moles=  0.1 ×2 = 0.2 moles of ionsMoles of PbCl2  = 0.5 M  ×  0.1 L = 0.05 moles=  0.05 ×3 = 0.15 moles of ions

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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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY