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 U4.10, Problem 5E

(a)

Interpretation Introduction

Interpretation:

Between 10.0 g of calcium ( Ca ) and 10.0 g of calcium chloride ( CaCl2 ), the molecule having more number of moles of metal atoms should be determined.

Concept Introduction:

Mole is SI unit which is used to measure the quantity of the substance. It is the quantity of a substance which contains same number of atoms as present in accurately 12.00 g of carbon-12 is known as mole

Number of moles is defined as the ratio of mass to the molar mass.

  Number of moles = Mass (g)Molar mass

The ratio of mass of the given compound to the number of moles of compound is known as molar mass.

The one mole of any substance is equal to the Avogadro number i.e. 6.022×1023 atoms.

(a)

Expert Solution
Check Mark

Answer to Problem 5E

Number of moles of calcium atom is more in 10 g of calcium in comparison to 10 g of calcium chloride.

Explanation of Solution

The given information is: 10.0 g of calcium ( Ca ) and 10.0 g of calcium chloride ( CaCl2 )

First step is to calculate the molar mass of each molecule.

Molar mass of calcium = 40.078 g/mole

Molar mass of calcium chloride = 1×40.078 g/mole + 2×35.45 g/mole

= 110.98 g/mole

Now,

Number of moles of calcium metal is calculated as:

  Number of moles = Mass (g)Molar mass

Put the values,

Number of moles of calcium metal = 10.0 g40.078 g/mole

Number of moles of calcium metal = 0.25 mole

Number of moles of calcium chloride is calculated as:

  Number of moles = Mass (g)Molar mass

Number of moles of calcium chloride = 10.0 g110.98 g/mole

Number of moles of calcium chloride = 0.090 mole

In one mole of calcium chloride, CaCl2 there is one mole of calcium ion and two moles chloride ions as shown:

  CaCl2  Ca2+ + 2Cl

So, the number of moles of calcium metal ion in 0.090 mole of calcium chloride is 0.090 mole .

Hence, number of moles of calcium atom is more in calcium in comparison to calcium chloride.

(b)

Interpretation Introduction

Interpretation:

Between 5.0 g of sodium chloride ( NaCl ) and 5.0 g of sodium fluoride ( NaF ), the molecule having more number of moles of metal atoms should be determined.

Concept Introduction:

Mole is SI unit which is used to measure the quantity of the substance. It is the quantity of a substance which contains same number of atoms as present in accurately 12.00 g of carbon-12 is known as mole

Number of moles is defined as the ratio of mass to the molar mass.

  Number of moles = Mass (g)Molar mass

The ratio of mass of the given compound to the number of moles of compound is known as molar mass.

The one mole of any substance is equal to the Avogadro number i.e. 6.022×1023 atoms.

(b)

Expert Solution
Check Mark

Answer to Problem 5E

Number of moles of sodium atom is more in 5.0 g of sodium fluoride in comparison to 5.0 g of sodium chloride.

Explanation of Solution

The given information is: 5.0 g of sodium chloride ( NaCl ) and 5.0 g of sodium fluoride ( NaF )

First step is to calculate the molar mass of each molecule.

Molar mass of sodium chloride = 1×23.0 g/mole + 1×35.45 g/mole

= 58.45 g/mole

Molar mass of sodium fluoride = 1×23.0 g/mole + 1×19.0 g/mole

= 42.0 g/mole

Now,

Number of moles of sodium chloride is calculated as:

  Number of moles = Mass (g)Molar mass

Put the values,

Number of moles of sodium chloride = 5.0 g58.45 g/mole

Number of moles of sodium chloride = 0.0855 mole

Number of moles of sodium fluoride is calculated as:

  Number of moles = Mass (g)Molar mass

Number of moles of sodium fluoride = 5.0 g42.0 g/mole

Number of moles of sodium fluoride = 0.119 mole

The number of metal atoms, Na+ in both the compounds is 1 as shown:

  NaClNa+ + Cl

  NaFNa+ + F

Hence, number of moles of sodium atom is more in sodium fluoride in comparison to sodium chloride.

(c)

Interpretation Introduction

Interpretation:

Between 2.0 g of iron oxide ( FeO ) and 2.0 g of iron sulfide ( FeS ), the molecule having more number of moles of metal atoms should be determined.

Concept Introduction:

Mole is SI unit which is used to measure the quantity of the substance. It is the quantity of a substance which contains same number of atoms as present in accurately 12.00 g of carbon-12 is known as mole

Number of moles is defined as the ratio of mass to the molar mass.

  Number of moles = Mass (g)Molar mass

The ratio of mass of the given compound to the number of moles of compound is known as molar mass.

The one mole of any substance is equal to the Avogadro number i.e. 6.022×1023 atoms.

(c)

Expert Solution
Check Mark

Answer to Problem 5E

Number of moles of iron atom is more in 2.0 g of iron oxide in comparison to 2.0 g of iron sulfide.

Explanation of Solution

The given information is: 2.0 g of iron oxide ( FeO ) and 2.0 g of iron sulfide ( FeS )

First step is to calculate the molar mass of each molecule.

Molar mass of iron oxide = 1×55.845 g/mole + 1×16 g/mole

= 71.845 g/mole

Molar mass of iron sulfide = 1×55.845 g/mole + 1×32.065 g/mole

= 87.91 g/mole

Now,

Number of moles of iron oxide is calculated as:

  Number of moles = Mass (g)Molar mass

Put the values,

Number of moles of iron oxide = 2.0 g71.845 g/mole

Number of moles of iron oxide = 0.028 mole

Number of moles of iron sulfide is calculated as:

  Number of moles = Mass (g)Molar mass

Number of moles of iron sulfide = 2.0 g87.91 g/mole

Number of moles of iron sulfide = 0.023 mole

The number of metal atoms, Fe2+ in both the compounds is 1 as shown:

  FeOFe2+ + O2

  FeSFe2+ + S2

Hence, number of moles of iron atom is more in iron oxide in comparison to iron sulfide.

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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