Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 3, Problem 3.124P

(a)

Interpretation Introduction

Interpretation:

The number of moles present in 0.588 g of ammonium bromide is to be calculated.

Concept introduction:

The number of moles is calculated by the formula:

  Number of moles=given mass  of sample(g)molecular mass (g/mol)        (1)

The formula to calculate mass is as follows:

  Mass of a given sample(g)=(molecular mass(g/mol))(number of moles)        (2)

(a)

Expert Solution
Check Mark

Answer to Problem 3.124P

The number of moles present in 0.588 g of ammonium bromide is 0.00600 mol.

Explanation of Solution

The chemical formula for ammonium bromide is NH4Br. The molar mass of ammonium bromide is 97.94 g/mol.

Substitute 0.588 g for mass and 97.94 g/mol for molar mass in equation (1).

  Moles of NH4Br(mol)=0.588 g of C7H897.94 g/mol C7H8= 0.006003675 mol0.00600 mol.

Conclusion

In 0.588 g of ammonium bromide, the number of moles present is found to be 0.00600 mol.

(b)

Interpretation Introduction

Interpretation:

The number of potassium ions in 88.5 g of potassium nitrate is to be calculated.

Concept introduction:

The number of moles is calculated by the formula :

  Number of moles=given mass  of sample(g)molecular mass (g/mol)        (1)

The formula to calculate mass from moles is as follows:

  Mass of a given sample(g)=(molecular mass(g/mol))(number of moles)        (2)

(b)

Expert Solution
Check Mark

Answer to Problem 3.124P

The number of potassium ions in 88.5 g of potassium nitrate is 5.27×1023.

Explanation of Solution

The chemical formula for potassium nitrate is KNO3. The molar mass of potassium nitrate is 101.11 g/mol.

Substitute 88.5 g for mass and 97.94 g/mol for molar mass in equation (1).

  Moles of KNO3(mol)=88.5 g 101.11 g/mol = 0.875284 mol

The formula to evaluate the number of ions is as follows:

  number of ions=(number of moles)(6.022×1023)        (3)

Substitute  0.875284 mol for the number of moles in equation (3).

  Number of  potassium ions=( 0.875284 mol)(6.022×1023ions/mol)=5.27096×10235.27×1023.

Conclusion

In 88.5 g of potassium nitrate, the number of potassium ions is 5.27×1023.

(c)

Interpretation Introduction

Interpretation:

The mass of 5.85 mol glycerol is to be calculated.

Concept introduction:

The formula to calculate the moles is as follows:

  Number of moles=given mass  of sample(g)molecular mass (g/mol)        (1)

The formula to calculate mass from moles is as follows:

  Mass of a given sample(g)=(molecular mass(g/mol))(number of moles)        (2)

(c)

Expert Solution
Check Mark

Answer to Problem 3.124P

The mass of 5.85 mol glycerol is 539 g.

Explanation of Solution

The molar mass of glycerol (C3H8O3) is calculated as:

  Molar mass=(3)(12.01 g/mol)+(8)(1.008 g/mol)+(3)(16.02 g/mol)=92.09 g/mol

Substitute 5.85 mol for the number of moles of C3H8O3 and 92.09 g/mol for molar mass in equation (2).

  Mass of C3H8O3 (g)=(92.09 g/mol)(5.85 mol)=538.7265 g539 g.

Conclusion

The mass of 5.85 mol glycerol is found to be 539 g.

(d)

Interpretation Introduction

Interpretation:

The volume of 2.85 mol chloroform is to be calculated.

Concept introduction:

The number of moles is calculated by the formula:

  Number of moles=given mass  of sample(g)molecular mass (g/mol)        (1)

The relation between density, mass and volume is as follows:

  Density(g/mL) =Mass(g)Volume(mL)        (2)

(d)

Expert Solution
Check Mark

Answer to Problem 3.124P

The volume of 2.85 mol chloroform is 230 mL.

Explanation of Solution

The chemical formula for chloroform is CHCl3. The molar mass of chloroform is 119.37 g/mol.

Substitute 2.85 mol for the number of moles of chloroform and 119.37 g/mol for molar mass of chloroform in equation (2).

  Mass of C3H8O3 (g)=( 2.85 g/mol)(119.37 mol)=340.2045 g

The density of chloroform is 1.48 g/mL.

Rearrange equation (2) for volume as follows:

  Volume(mL)=Mass(g)Density(g/mL)

Substitute 1.48 g/mL for density and 340.2045 g for mass in the above formula.

  Volume(mL)=340.2045(g)1.48(g/mL)=229.868 mL230 mL.

Conclusion

Hence the volume of 2.85 mol chloroform is found to be 230 mL.

(e)

Interpretation Introduction

Interpretation:

The number of sodium ions present in 2.11 mol sodium carbonate is to be calculated.

Concept introduction:

The number of moles is calculated by the formula:

  Number of moles=given mass  of sample(g)molecular mass (g/mol)        (1)

(e)

Expert Solution
Check Mark

Answer to Problem 3.124P

The number of sodium ions present in 2.11 mol sodium carbonate is 2.54×1024.

Explanation of Solution

One mole of Na2CO3 gives 2 mol of Na+1 ions hence 2.11 molNa2CO3 would yield:

  Moles of Na+1=(2.11 mol)(2 mol1 mol)=4.22 mol

The formula to evaluate the number of ions is as follows:

  number of ions=(number of moles)(6.022×1023)        (3)

Substitute 4.22 mol number of moles in equation (3).

  Number of Na+1 ions=( 4.22 mol)(6.022×1023ions/mol)=25.412×1023 ions2.54×1024 ions.

Conclusion

The number of sodium ions present in 2.11 mol sodium carbonate is found to be 2.54×1024.

(f)

Interpretation Introduction

Interpretation:

The number of atoms in 25.0 μg cadmium is to be calculated.

Concept introduction:

The symbol and molar mass of cadmium is Cd 112.4 g/mol respectively.

The number of moles is calculated by the formula:

  Number of moles=given mass  of sample(g)molecular mass (g/mol)        (1)

The formula to evaluate the number of atoms is as follows:

  Number of atoms=(number of moles)(6.022×1023)        (3)

The conversion factor for converting μg to g is as follows:

1 μg=106g

(f)

Expert Solution
Check Mark

Answer to Problem 3.124P

The number of atoms in 25.0 μg of cadmium is 1.34×1017.

Explanation of Solution

Since 1 μg=106 g so 25.0 μg is equal to 25.0×106 g.

Substitute 25.0×106g for mass and 112.4 g/mol for molar mass in equation (1).

  Moles of Cd(mol)=25.0×106 g112.4 g/mol = 2.54128×1024 mol2.54×1024 mol

In order to calculate the number of atoms from the number of moles substitute 2.54×1024mol for moles of cadmium in equation (3).

  Number of atoms=(2.224199×107 mol)(6.022×1023 atoms/mol)=1.3394126×1017 atoms1.34×1017 atoms.

Conclusion

The number of atoms in 25.0 μg of cadmium is found to be 1.34×1017.

(g)

Interpretation Introduction

Interpretation:

The number of atoms in 0.0015 mol of fluorine gas is to be calculated.

Concept introduction:

The number of moles is calculated by the formula:

  Number of moles=given mass  of sample(g)molecular mass (g/mol)        (1)

(g)

Expert Solution
Check Mark

Answer to Problem 3.124P

The number of atoms in 0.0015 mol of fluorine gas is 1.8×1021.

Explanation of Solution

The chemical formula of fluorine gas is F2. 1 mol fluorine gas has 2 mol F atoms.

Therefore, the number of fluorine atoms in 0.0015 mol of fluorine gas is calculated as,

  Moles of F atoms=(0.0015 mol)(2 mol1 mol)=3×103 mol

In order to calculate the number of atoms from the number of moles substitute 2.54×1024mol for moles in equation (3).

  Number of atoms=(3×103 mol)(6.022×1023 atoms/mol)=1.8066×1021 atoms1.8×1021 atoms.

Conclusion

The number of fluorine atoms present in 0.0015 mol of fluorine gas is found to be 1.8×1021.

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Chapter 3 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 3.1 - Prob. 3.6AFPCh. 3.1 - For many years, compounds known as...Ch. 3.1 - Use the information in Follow-up Problem 3.6A to...Ch. 3.1 - Prob. 3.7BFPCh. 3.2 - Prob. 3.8AFPCh. 3.2 - A sample of an unknown compound contains 6.80 mol...Ch. 3.2 - A sample of an unknown compound is found to...Ch. 3.2 - Prob. 3.9BFPCh. 3.2 - Prob. 3.10AFPCh. 3.2 - Prob. 3.10BFPCh. 3.2 - A dry-cleaning solvent (ℳ = 146.99 g/mol) that...Ch. 3.2 - Prob. 3.11BFPCh. 3.3 - Prob. 3.12AFPCh. 3.3 - Prob. 3.12BFPCh. 3.3 - Prob. 3.13AFPCh. 3.3 - Prob. 3.13BFPCh. 3.4 - Prob. 3.14AFPCh. 3.4 - The tarnish that forms on objects made of silver...Ch. 3.4 - Prob. 3.15AFPCh. 3.4 - In the reaction that removes silver tarnish (see...Ch. 3.4 - Prob. 3.16AFPCh. 3.4 - Prob. 3.16BFPCh. 3.4 - Prob. 3.17AFPCh. 3.4 - Prob. 3.17BFPCh. 3.4 - Prob. 3.18AFPCh. 3.4 - Prob. 3.18BFPCh. 3.4 - In the reaction in Follow-up Problem 3.18A, how...Ch. 3.4 - Prob. 3.19BFPCh. 3.4 - Prob. 3.20AFPCh. 3.4 - Prob. 3.20BFPCh. 3.4 - Marble (calcium carbonate) reacts with...Ch. 3.4 - Prob. 3.21BFPCh. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Why might the expression “1 mol of chlorine” be...Ch. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Calculate the molar mass of each of the...Ch. 3 - Prob. 3.10PCh. 3 - Prob. 3.11PCh. 3 - Calculate each of the following quantities: Mass...Ch. 3 - Calculate each of the following quantities: Amount...Ch. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Calculate each of the following: Mass % of H in...Ch. 3 - Calculate each of the following: Mass % of I in...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Oxygen is required for the metabolic combustion of...Ch. 3 - Cisplatin (right), or Platinol, is used in the...Ch. 3 - Allyl sulfide (below) gives garlic its...Ch. 3 - Iron reacts slowly with oxygen and water to form a...Ch. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - The mineral galena is composed of lead(II) sulfide...Ch. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - List three ways compositional data may be given in...Ch. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - Prob. 3.38PCh. 3 - Prob. 3.39PCh. 3 - What is the molecular formula of each...Ch. 3 - Prob. 3.41PCh. 3 - Prob. 3.42PCh. 3 - Find the empirical formula of each of the...Ch. 3 - An oxide of nitrogen contains 30.45 mass % N. (a)...Ch. 3 - Prob. 3.45PCh. 3 - A sample of 0.600 mol of a metal M reacts...Ch. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Prob. 3.66PCh. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Prob. 3.70PCh. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Prob. 3.74PCh. 3 - Elemental phosphorus occurs as tetratomic...Ch. 3 - Prob. 3.76PCh. 3 - Solid iodine trichloride is prepared in two steps:...Ch. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Prob. 3.86PCh. 3 - Prob. 3.87PCh. 3 - Prob. 3.88PCh. 3 - Prob. 3.89PCh. 3 - Prob. 3.90PCh. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - When 20.5 g of methane and 45.0 g of chlorine gas...Ch. 3 - Prob. 3.96PCh. 3 - Prob. 3.97PCh. 3 - Prob. 3.98PCh. 3 - Prob. 3.99PCh. 3 - Prob. 3.100PCh. 3 - Sodium borohydride (NaBH4) is used industrially in...Ch. 3 - Prob. 3.102PCh. 3 - The first sulfur-nitrogen compound was prepared in...Ch. 3 - Prob. 3.104PCh. 3 - Prob. 3.105PCh. 3 - Prob. 3.106PCh. 3 - Serotonin () transmits nerve impulses between...Ch. 3 - In 1961, scientists agreed that the atomic mass...Ch. 3 - Prob. 3.109PCh. 3 - Isobutylene is a hydrocarbon used in the...Ch. 3 - The multistep smelting of ferric oxide to form...Ch. 3 - Prob. 3.112PCh. 3 - Prob. 3.113PCh. 3 - Prob. 3.114PCh. 3 - Prob. 3.115PCh. 3 - Prob. 3.116PCh. 3 - Prob. 3.117PCh. 3 - Prob. 3.118PCh. 3 - Prob. 3.119PCh. 3 - Prob. 3.120PCh. 3 - For the reaction between solid tetraphosphorus...Ch. 3 - Prob. 3.122PCh. 3 - Prob. 3.123PCh. 3 - Prob. 3.124PCh. 3 - Prob. 3.125PCh. 3 - Prob. 3.126PCh. 3 - Prob. 3.127PCh. 3 - Prob. 3.128PCh. 3 - Ferrocene, synthesized in 1951, was the first...Ch. 3 - Prob. 3.130PCh. 3 - Prob. 3.131PCh. 3 - Citric acid (below) is concentrated in citrus...Ch. 3 - Prob. 3.133PCh. 3 - Nitrogen monoxide reacts with elemental oxygen to...Ch. 3 - Prob. 3.135PCh. 3 - Prob. 3.136PCh. 3 - Manganese is a key component of extremely hard...Ch. 3 - The human body excretes nitrogen in the form of...Ch. 3 - Aspirin (acetylsalicylic acid, C9H8O4) is made by...Ch. 3 - Prob. 3.140PCh. 3 - Prob. 3.141PCh. 3 - Prob. 3.142PCh. 3 - When powdered zinc is heated with sulfur, a...Ch. 3 - Cocaine (C17H21O4N) is a natural substance found...Ch. 3 - Prob. 3.145P
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