# Chapter 10 Nuclear Chemistry - 10–1 Chapter 10 Nuclear Chemistry Solutions to In-Chapter Problems…

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Chapter 101

Chapter 10 Nuclear Chemistry Solutions to In-Chapter Problems 10.1 Refer to Example 10.1 to answer the question.

The atomic number (Z) = the number of protons. The mass number (A) = the number of protons + the number of neutrons. Isotopes are written with the mass number to the upper left of the element symbol and the

atomic number to the lower left of the element symbol.

Atomic Number Mass Number Number of

Protons Number of Neutrons

Isotope Symbol

Cobalt-59 27 59 27 32 27Co59

Cobalt-60 27 60 27 33 27Co

60

10.2 Fill in the table as in Example 10.1, using the rules in Answer 10.1.

Atomic Number Mass Number Number of Protons Number of Neutrons

Sr3885a.

38 85 38 47

Ga3167b.

31 67 31 36 Selenium-75c. 34 75 34 41

10.3 An particle has no electrons around the nucleus and a +2 charge, whereas a helium atom has

two electrons around the nucleus and is neutral. 10.4 Use Table 10.1 to identify Q in each of the following symbols.

Q42Q01 Qa. b. c.0

+1

! particle " particle positron 10.5 Write a balanced nuclear equation as in Example 10.2.

[1] Write an incomplete equation with the original nucleus on the left and the particle emitted on the right. Radon-222 has an atomic number of 86.

222 He2

4 + ?86Rn [2] Calculate the mass number and the atomic number of the newly formed nucleus on the

right. Radon-222 emits an particle. Mass number: Subtract the mass of an particle (4) to obtain the mass of the new

nucleus; 222 4 = 218. Atomic number: Subtract the two protons of an particle to obtain the atomic number of

the new nucleus; 86 2 = 84.

[3] Use the atomic number to identify the new nucleus and complete the equation.

Nuclear Chemistry 102

From the periodic table, the element with an atomic number of 84 is polonium, Po. Write the mass number and the atomic number with the element symbol to complete the

equation.

2He + 84Po222 4 21886Rn

10.6 Work backwards to write the equation that produces radon-222.

?

22688Ra

2He + 86Rn2224

2He + 86Rn2224Mass number = 222 + 4 = 226

Atomic number = 86 + 2 = 88 10.7 Write the balanced nuclear equation for each isotope as in Example 10.2 or Answer 10.5.

84Po218 4

2He + 82Pb214a.

b. 90Th230

2He + 88Ra4 226

c. 99Es252

2He + 97Bk4 248

10.8 Write a balanced nuclear equation for the emission of iodine-131 as in Example 10.2.

[1] Write an incomplete equation with the original nucleus on the left and the particle emitted on the right. Use the identity of the element to determine the atomic number; iodine has an atomic

number of 53.

I53131 +e1

0 ?

[2] Calculate the mass number and the atomic number of the newly formed nucleus on the

right. Mass number: Since a particle has no mass, the masses of the new particle and the

original particle are the same, 131. Atomic number: Since emission converts a neutron into a proton, the new nucleus has

one more proton than the original nucleus; 53 = 1 + ?. Thus, the new nucleus has an atomic number of 54.

[3] Use the atomic number to identify the new nucleus and complete the equation.

From the periodic table, the element with an atomic number of 54 is xenon, Xe. Write the mass number and the atomic number with the element symbol to complete the

equation.

I53131 +e1

0 Xe54131

Chapter 103

10.9 Write a balanced nuclear equation for the emission of each isotope as in Example 10.2 and Answer 10.8.

F9

20 +e10 Ne

1020a.

Sr3892 +e1

0 Y3992b.

c. Cr2455 +e1

0 Mn2555

10.10 Write a balanced nuclear equation for positron emission as in Example 10.3. a. [1] Write an incomplete equation with the original nucleus on the left and the particle

emitted on the right. Use the identity of the element to determine the atomic number; arsenic has an atomic

number of 33.

As3374 + ?e+1

0

[2] Calculate the mass number and the atomic number of the newly formed nucleus on the

right. Mass number: Since a + particle has no mass, the masses of the new particle and the

original particle are the same, 74. Atomic number: Since + emission converts a proton into a neutron, the new nucleus has

one fewer proton than the original nucleus; 33 1 = 32. Thus, the new nucleus has an atomic number of 32.

[3] Use the atomic number to identify the new nucleus and complete the equation.

From the periodic table, the element with an atomic number of 32 is germanium, Ge. Write the mass number and the atomic number with the element symbol to complete the

equation.

As3374 +e+1

0 Ge3274

b. Use the same steps as in part (a).

O815 +e+1

0 N715

10.11 When and emission occur together, the atomic number increases by one (due to the

particle), and a ray is released.

Ir77192 + e1

0Pt78192 +

Both ! particles and " rays are emitted.

"

10.12 When emission occurs alone, there is no change in the atomic number or the mass number.

When and emission occur together, the atomic number increases by one (due to the particle), and a ray is released.

Nuclear Chemistry 104

Both ! particles and " rays are emitted.

40 + "K1940 + e1

0B511 + " 20Cab.B5

11

" emission aloneno change in atomic number or mass number

Atomic number increases, but no change in the mass number.

a.

10.13 To calculate the amount of radioisotope present after the given number of half-lives, multiply the

initial mass by for each half-life.

xinitial mass

12

x 12

1.00 g = 0.250 g of phosphorus-32 remains.

The mass is halved two times.

a.

xinitial mass

12

x 12

x 12

x 12

1.00 g = 0.0625 g of phosphorus-32 remains.

The mass is halved four times.

b.

xinitial mass

12

x 12

x 12

x 12

1.00 g

= 0.00391 g of phosphorus-32 remains.The mass is halved eight times.

c. x12

x 12

x 12

x 12

xinitial mass

12

x 12

x 12

x 12

1.00 g

= 9.54 x 107 g of phosphorus-32 remains.The mass is halved twenty times.

d. x12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

10.14 To calculate the amount of radioisotope present, first determine the number of half-lives that

occur in the given amount of time. Then multiply the initial mass by for each half-life to determine the amount present.

6.0 hours 1 half-life

6.0 hoursx = 1.0 half-life x

initial mass

12

160. mg = 80. mg of Tc-99ma.

18.0 hours 1 half-life6.0 hours

x = 3.0 half-lives xinitial mass

12

160. mg = 20.0 mg of Tc-99mb. x 12x 1

2

24.0 hours 1 half-life

6.0 hoursx = 4.0 half-lives x

initial mass

12

160. mg = 10.0 mg of Tc-99mc. x 12x 1

2x 1

2

2 days 1 half-life

6.0 hoursx = 8 half-lives

xinitial mass

12

160. mg = 0.6 mg of Tc-99m

d.

x 12

x 12

x 12

24 hours1 day

x

x 12

x 12

x 12

x 12

Chapter 105

10.15 If an artifact has 1/8 of the amount of C-14 compared to living organisms, it has decayed by three half-lives ( ).

1 half-life

5,730 yearsx =3 half-lives 17,200 years

10.16 Use the amount of radioactivity (mCi/mL) as a conversion factor to convert the dose of radioactivity from millicuries to a volume in milliliters.

x =110 mCi dose 1 mL25 mCi

Millicuries cancel.

4.4 mLAnswer

mCimL conversion factor

10.17 Use the conversion factor given to convert mrem to rem.

x =200 mrem 1 rem1000 mrem

0.2 rema. x =0.014 rem1 rem

1000 mrem 14 mremb.

larger dose 10.18 Calculate the number of half-lives that pass in nine days.

9 days 1 half-life3 days

x = 3 half-lives 12

=x 12

x 12

18

10.19

b. Sm62153 + e10Eu63

153

a. Sm62153

62 protons, 62 electrons153 62 = 91 neutrons

One ! particleis emitted.

Atomic number increases.

c. xinitial activity

12

150 mCi = 9.4 mCix 12x 1

2x 1

2

10.20 The emission of a positron decreases the atomic number by one, but the mass number stays the

same. Use Example 10.3.

N713 +e+1

0 C613

Nuclear Chemistry 106

10.21 Nuclear fission splits an atom into two lighter nuclei. Write the equation with the information given, and then balance the equation.

92U235 + 3 0n51Sb +

133 1+ 0n1

235 + 3 0n51Sb +133

41Nb100 1+ 0n

1

?

The atomic number must be 41 = niobium:92 = 51 + X

The mass number of niobium must be 100:235 + 1 = 133 + X + 3, X = 100

92U 10.22 Balance each reaction.

+ +12 H1

1 H e+10a. 1

1 H b. 1

1 H + 12 H 2

3 He c. 1

1 H + 23 He 2

4 He + e+10

Solutions to End-of-Chapter Problems 10.23 Refer to Example 10.1 to answer the question.

The atomic number (Z) = the number of protons. The mass number (A) = the number of protons + the number of neutrons. Isotopes are written with the mass number to the upper left of the element symbol and the

atomic number to the lower left of the element symbol.

a. Atomic Number b. Number of Protons

c. Number of Neutrons d. Mass Number Isotope Symbol

Fluorine-18 9 9 9 18 F918

Fluorine-19 9 9 10 19 F9

19

10.24 Refer to Example 10.1 to answer the question.

The atomic number (Z) = the number of protons. The mass number (A) = the number of protons (Z) + the number of neutrons. Isotopes are written with the mass number to the upper left of the element symbol and the

atomic number to the lower left of the element symbol.

a. Atomic number b. Number of protons

c. Number of neutrons d. Mass number Isotope symbol

Nitrogen-13 7 7 6 13 N713

Nitrogen-14 7 7 7 14 N7

14

Chapter 107

10.25 Fill in the table using Example 10.1 and the definitions in Answer 10.23.

Atomic Number Mass Number Number of

Protons Number of Neutrons

Isotope Symbol

a. Chromium-51 24 51 24 27 Cr2451

b. Palladium-103 46 103 46 57 Pd46103 c. Potassium-42 19 42 19 23 K1942 d. Xenon-133 54 133 54 79 Xe54133 10.26 Fill in the table using Example 10.1 and the definitions in Answer 10.23.

Atomic number Mass number Number of

protons Number of neutrons Isotope symbol

a. Sodium-24 11 24 11 13 Na1124

b. Strontrium-89 38 89 38 51 Sr3889 c. Iron-59 26 59 26 33 Fe2659 d. Samarium-153 62 153 62 91 Sm62153 10.27 Use the definitions in Section 10.1B and Table 10.1 to fill in the table.

Change in Mass Change in Charge a. particle 4 2 b. particle 0 +1 c. ray 0 0 d. positron 0 1

10.28 a. rays have the highest speed and particles have the slowest speed. b. rays have the highest penetrating power and particles have the lowest penetrating power.

c. Lab coats and gloves should be worn when working with substances that give off particles. Heavy lab coats and gloves must be worn when working with substances that give off particles. A lead shield is required to protect those working with substances that give off rays.

10.29 Use the definitions in Section 10.1B and Table 10.1 to fill in the table.

Mass Charge a. 4 +2 b. n 1 0 c. 0 0 d. 0 1

Nuclear Chemistry 108

10.30 Use the definitions in Section 10.1B and Table 10.1 to fill in the table.

Mass Charge e1

0a. 0 1

e+1 0b.

0 +1

He 2 4c.

4 +2

!d. + 0 +1

10.31 Complete the equation. The white circles represent neutrons and the black circles represent

protons.

+positron

7N13

6C13 + e+1

0

seven protons =atomic number

of nitrogen

six protons =atomic number

of carbon

10.32 Complete the equation. The white circles represent neutrons and the black circles represent

protons.

+

5B11

3Li 7 + He 2

4

five protons =atomic number

of boron

three protons =atomic number

of lithium

10.33 Complete the nuclear equation as in Example 10.2.

He24

e10Fe26

59 +

Pt78190 +

e+10Hg80

178 +a.

b.

c.Co2759

Os76186

Au79178

no change no change

+ 1 proton26 + 1 = 27

78 2 = 76

190 4 = 186

1 proton80 1 = 79

Chapter 109

10.34 Complete the nuclear equation as in Example 10.2.

He24

e+10Rb37

77 +

No102251 +

e10Cu29

66 +a.

b.

c.Kr3677

Fm100247

Zn30 66

no change no change

1 proton37 1 = 36

102 2 = 100

251 4 = 247

+ 1 proton29 + 1 = 30

10.35 Complete each nuclear equation.

He24Y39

90 +

+ e+10

+a.

b.

c.Zr4090

Pr59135

Bi83210e1

0

Nd60135

Tl81206

+ 1 proton ! particle

Work backwards: 59 + 1 = 60

positron

83 2 = 81

210 4 = 206

" particle

10.36 Complete each nuclear equation.

He24Sr38

90 +

+ e+10

+a.

b.

c.Y3990

Si14 29

Po84214e1

0

P15 29

Pb82210

+ 1 proton

Work backwards: 39 1 = 38

positron

84 2 = 82

214 4 = 210

! particle" particle

10.37 Write the two nuclear equations for the decay of bismuth-214.

! particle " particle

+ e10 He2

4+Bi83214 Tl81

210Bi83214 Po84

214

Nuclear Chemistry 1010

10.38 Write the two chemical equations for the formation of lead-210.

! particle" particle

+ e10 He2

4+Po84214 Pb82

210Tl81210 Pb82

210

10.39 Write the chemical equation for each nuclear reaction.

! particle

" particle

90Th232 2He + 88Ra4 228a.

+ e10Na11

25 Mg1225b.

c. e+10I53118Xe54118 +

d. 96Cm243 2He + 94Pu4 239

" particle

positron

10.40 Write the chemical equation for each nuclear reaction.

! particle

" particle

16S 35

2He4

a.

+

e10

Th90225 Ra88

221b.

c. e+10Ru44 93Rh45

93 +

d. 47Ag114 1e + 48Cd 0 114

! particle

positron

+ Cl1735

10.41

initial sample A16 black spheres 4 black spheres

16 is halved twice.

x 12

x 12

16 = 4

2 half-lives 10.42

initial sample A8 black spheres 1 black spheres

8 is halved 3 times.

x 12

x 12

8 = 1

3 half-lives

x 12

3 x 10 minutes = 30 minutes

Chapter 1011

10.43 Determine the number of half-lives that have passed in 12 days to determine the length of each half-life.

x

initial mass

12

x 12

x 12

2.4 g = 0.30 g

The mass is halved three times.

12 days3 half-lives

= 4.0 days in one half-life

10.44 Determine the number of half-lives that have passed in 22 minutes to determine the length of each

half-life.

xinitial mass

12

x 12

0.36 g = 0.090 g

The mass is halved two times.

22 minutes2 half-lives

= 11 minutes in one half-life

10.45 To calculate the amount of iodine present after the given number of half-lives, multiply the initial

mass by for each half-life. The total mass is always 64 mg.

xinitial mass

12

64 mg = 32 mg of iodine-131 64 32 = 32 mg of xenon-131

a. 8.0 days = 1 half-life

xinitial mass

12

64 mg = 16 mg of iodine-131 64 16 = 48 mg of xenon-131

b. 16 days = 2 half-lives

x 12

xinitial mass

12

64 mg = 8.0 mg of iodine-131 64 8.0 = 56 mg of xenon-131

c. 24 days = 3 half-lives

x 12

x 12

xinitial mass

12

64 mg = 4.0 mg of iodine-131 64 4.0 = 60. mg of xenon-131

d. 32 days = 4 half-lives

x 12

x 12

x 12

10.46 To calculate the amount of phosphorus present after the given number of half-lives, multiply the

initial mass by for each half-life. The total mass is always 124 mg.

x

initial mass

12

124 mg = 62.0 mg of phosphorus-32 124 62.0 = 62 mg of sulfur-32

a. 14 days = 1 half-life

xinitial mass

12

124 mg = 31.0 mg of phosphorus-32 124 31.0 = 93 mg of sulfur-32

b. 28 days = 2 half-lives

x 12

Nuclear Chemistry 1012

xinitial mass

12

124 mg = 15.5 mg of phosphorus-32 124 15.5 = 109 mg of sulfur-32

c. 42 days = 3 half-lives

x 12

x 12

xinitial mass

12

124 mg = 7.75 mg of phosphorus-32 124 7.75 = 116 mg of sulfur-32

d. 56 days = 4 half-lives

x 12

x 12

x 12

10.47 If the half-life of an isotope is 24 hours, then two half-lives have passed in 48 hours. Therefore,

25% of the initial amount of isotope still remains.

xiniti...

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