Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 8, Problem 109AP
Interpretation Introduction

(a)

Interpretation:

The number of atoms present in the sample of gold should be calculated.

Concept Introduction:

Moles of an element are calculated by dividing the given mass by its molar mass.

Moles of an element =MassMolarMass

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.

Expert Solution
Check Mark

Answer to Problem 109AP

9.03×1021 atoms are present in 2.89g of gold.

Explanation of Solution

Mass of gold = 2.89g

Molar mass of gold = 196.96 g/mol

Moles of gold =MassMolarMass=2.89 g196.96 g/mol=0.015 mol

Number of atoms present in 0.015 mole of gold (Au)=(0.015×6.023×1023)=9.03×1021 atoms.

Interpretation Introduction

(b)

Interpretation:

The number of atoms present in the sample of platinum should be calculated.

Concept Introduction:

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.

Expert Solution
Check Mark

Answer to Problem 109AP

1.56×1020 atoms are present in 0.000259 mole of platinum.

Explanation of Solution

Moles of platinum present in the sample = 0.000259

Number of atoms present in 0.000259 mole of platinum (Pt)=(0.000259×6.023×1023)=1.56×1020.

Interpretation Introduction

(c)

Interpretation:

The number of atoms present in the sample of platinum should be calculated.

Concept Introduction:

Moles of an element are calculated by dividing the given mass by its molar mass.

Moles of an element =MassMolarMass

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.

Expert Solution
Check Mark

Answer to Problem 109AP

8.01×1017 atoms are present in 0.000259 g of platinum.

Explanation of Solution

Mass of platinum = 0.000259g

Molar mass of platinum = 195.08g/mol

Moles of platinum =MassMolarMass=0.000259 g195.08 g/mol=1.33×106 mol

Number of atoms present in 1.33×106 mole of platinum (Pt)=(1.33×106×6.023×1023)=8.01×1017.

Interpretation Introduction

(d)

Interpretation:

The number of atoms present in the sample of magnesium should be calculated.

Concept Introduction:

Moles of an element are calculated by dividing the given mass by its molar mass.

Moles of an element =MassMolarMass

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.

Expert Solution
Check Mark

Answer to Problem 109AP

2.25×1025 atoms are present in 2.0lb of platinum.

Explanation of Solution

Mass of magnesium = 2.0lb = 907.184g [Since, 1lb = 453.592g]

Molar mass of magnesium = 24.3g/mol

Moles of magnesium =MassMolarMass=907.184 g24.3 g/mol=37.33 mol

Number of atoms present in 37.33 moles of magnesium (Mg)=(37.33×6.023×1023)=2.25×1025.

Interpretation Introduction

(e)

Interpretation:

The number of atoms present in the sample of mercury should be calculated.

Concept Introduction:

Mass can be calculated by multiplying the density by the volume.

Mass=Density×Volume

Moles of an element are calculated by dividing the given mass by its molar mass.

Moles of an element =MassMolarMass

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.

Expert Solution
Check Mark

Answer to Problem 109AP

7.7×1022 atoms are present in 1.90 mL of mercury.

Explanation of Solution

Volume of mercury = 1.90mL

Density of mercury = 13.6g/mL

Mass of mercury =(Density×Volume)=(13.6g/mL×1.90mL)=25.84 g

Molar mass of mercury = 200.59g/mol

Moles of mercury =MassMolar Mass=25.84 g200.59 g/mol=0.128 mol

Number of atoms present in 37.33 moles of

Mercury (Hg)=(0.128×6.023×1023)=7.7×1022.

Interpretation Introduction

(f)

Interpretation:

The number of atoms present in the sample of tungsten should be calculated.

Concept Introduction:

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.

Expert Solution
Check Mark

Answer to Problem 109AP

2.58×1023 atoms are present in 4.30 moles of tungsten.

Explanation of Solution

Moles of tungsten = 4.30

Number of atoms present in 4.30 moles of

Tungsten (W)=(4.30×6.023×1023)=2.58×1023.

Interpretation Introduction

(f)

Interpretation:

The number of atoms present in the sample of tungsten should be calculated.

Concept Introduction:

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.

Expert Solution
Check Mark

Answer to Problem 109AP

2.58×1023 atoms are present in 4.30 moles of tungsten.

Explanation of Solution

Moles of tungsten = 4.30

Number of atoms present in 4.30 moles of

Tungsten (W)=(4.30×6.023×1023)=2.58×1023.

Interpretation Introduction

(g)

Interpretation:

The number of atoms present in the sample of tungsten should be calculated.

Concept Introduction:

Moles of an element are calculated by dividing the given mass by its molar mass.

Moles of an element =MassMolarMass

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.

Expert Solution
Check Mark

Answer to Problem 109AP

1.38×1022 atoms are present in 4.30g of tungsten.

Explanation of Solution

Mass of tungsten = 4.30g

Molar mass of tungsten = 183.84g/mol

Moles of tungsten =MassMolar Mass=4.30 g183.84 g/mol=0.023 mol

Number of atoms present in 0.023 moles of

Tungsten (W)=(0.023×6.023×1023)=1.38×1022.

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Chapter 8 Solutions

Introductory Chemistry: A Foundation

Ch. 8.8 - trong>Exercise 8.9 Sevin, the commercial name for...Ch. 8.8 - e part of the problem-solving strategy for...Ch. 8.8 - ercise 8.10 The most common form of nylon...Ch. 8.9 - Exercise 8.11 A compound used as an additive for...Ch. 8 - n chemistry, what is meant by the term mole? What...Ch. 8 - hat is the difference between the empirical and...Ch. 8 - substance A2B is 60% A by mass. Calculate the...Ch. 8 - ive the formula for calcium phosphate and then...Ch. 8 - ow would you find the number of “chalk molecules”...Ch. 8 - 0.821 -mol sample of a substance composed of...Ch. 8 - ow many molecules of water are there in a 10.0-g...Ch. 8 - hat is the mass (in grams) of one molecule of...Ch. 8 - onsider separate 100.0-g samples of each of the...Ch. 8 - A molecule has a mass of 4.651023 g. Provide two...Ch. 8 - Differentiate between the terms atomic mass and...Ch. 8 - Consider Figure 4.19 in the text. 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