Download - Chapter 10 Nuclear Chemistry - websites.rcc.eduwebsites.rcc.edu/grey/files/2012/02/Chapter-10-Solutions-Smith.pdfChapter 10–1 Chapter 10 Nuclear Chemistry Solutions to In-Chapter

Chapter 10–1

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 α 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.

Q42Q0

–1 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 α particle. • Mass number: Subtract the mass of an α particle (4) to obtain the mass of the new

nucleus; 222 – 4 = 218. • Atomic number: Subtract the two protons of an α 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.

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Nuclear Chemistry 10–2

• 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 42He + 82Pb214a.

b. 90Th2302He + 88Ra4 226

c. 99Es2522He + 97Bk4 248

10.8 Write a balanced nuclear equation for the β 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 +e–1

0 ?

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

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

original particle are the same, 131. • Atomic number: Since β 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.


• 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 +e–1

0 Xe54131


Chapter 10–3

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

F9

20 +e–10 Ne

1020a.

Sr3892 +e–1

0 Y3992b.

c. Cr2455 +e–1

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 β+ particle has no mass, the masses of the new particle and the

original particle are the same, 74. • Atomic number: Since β+ 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.


• 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 β and γ emission occur together, the atomic number increases by one (due to the β

particle), and a γ ray is released.

Ir77192 + e–1

0Pt78192 +

Both ! particles and " rays are emitted.

"

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

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



Both ! particles and " rays are emitted.

40 + "K1940 + e–1

0B511 + "

20Cab.B511

" 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 ½ 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. x 12

x 12

x 12

x 12

xinitial mass

12

x 12

x 12

x 12

1.00 g

= 9.54 x 10–7 g of phosphorus-32 remains.The mass is halved twenty times.

d. 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

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 ½ 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 12

x 12

24.0 hours 1 half-life6.0 hours

x = 4.0 half-lives xinitial mass

12

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

x 12

x 12

2 days 1 half-life6.0 hours

x = 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 10–5

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 (½ × ½ × ½).

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

mCi–mL 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 + e–1

0Eu63153

a. Sm62153

62 protons, 62 electrons153 – 62 = 91 neutrons

One ! particleis emitted.

Atomic number increases.

c. xinitial activity

12

150 mCi = 9.4 mCix 12

x 12

x 12

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



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+ 0n

1

235 + 3 0n51Sb +13341Nb100 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


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


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 10–7

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 Pd46

103

c. Potassium-42 19 42 19 23 K1942

d. Xenon-133 54 133 54 79 Xe54

133

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 Sr38

89

c. Iron-59 26 59 26 33 Fe2659

d. Samarium-153 62 153 62 91 Sm62

153

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. α particle –4 –2 b. β particle 0 +1 c. γ ray 0 0 d. positron 0 –1

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

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


Mass Charge a. α 4 +2 b. n 1 0 c. γ 0 0 d. β 0 –1




Mass Charge e–1

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

7N136C13 + 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.

+

5B113Li 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

e–10Fe26

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 10–9

10.34 Complete the nuclear equation as in Example 10.2.

He24

e+10Rb37

77 +

No102251 +

e–10Cu29

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

Bi83210e–1

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

Po84214e–1

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

+ e–10 He2

4+Bi83214 Tl81

210Bi83214 Po84

214



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

! particle " particle

+ e–10 He2

4+Po84214 Pb82

210Tl81210 Pb82

210

10.39 Write the chemical equation for each nuclear reaction.

! particle

" particle

90Th2322He + 88Ra4 228a.

+ e–10Na11

25 Mg1225b.

c. e+10I

53118Xe54

118 +

d. 96Cm2432He + 94Pu4 239

" particle

positron

10.40 Write the chemical equation for each nuclear reaction.

! particle

" particle

16S 35

2He4

a.

+

e–10

Th90225 Ra88

221b.

c. e+10Ru44

93Rh4593 +

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 10–11

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 ½ 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 ½ 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


b. 28 days = 2 half-lives

x 12



xinitial mass

12


c. 42 days = 3 half-lives

x 12

x 12

xinitial mass

12


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.

xinitial mass

12

100% = 25%x 12

10.48 The half-life of carbon-14 can be used to date objects because the ratio of radioactive carbon-14

to stable carbon-12 is a constant value in living organisms. Once the organism dies, the radioactive carbon-14 decays and thus its concentration decreases. Scientists compare the ratio of C-14 to C-12 in an artifact to that in a living organism in order to determine the age of the artifact (i.e., the number of C-14 half-lives since death).

10.49 In artifacts over 50,000 years old, the percentage of carbon-14 is too small to accurately measure. 10.50 Radiocarbon dating can’t be used to estimate the age of rocks because the technique can only be

used for substances that are made of carbon. 10.51 Calculate the amount of radioactivity present after each amount of time has elapsed.

xinitial activity

12

20 mCi = 10 mCia.

6 h = one half-life

xinitial activity

12

20 mCi = 5 mCib.

12 h = two half-lives

x 12

xinitial activity

12

20 mCi = 1 mCic.

24 h = four half-lives

x 12

x 12

x 12

10.52 Calculate the amount of radioactivity present after each amount of time has elapsed.

xinitial activity

12

200. mCi = 100. mCia.

8.0 days = one half-life

xinitial activity

12

200. mCi = 25.0 mCib.

24 days = three half-lives

x 12

x 12


Chapter 10–13

xinitial activity

12

200. mCi = 3.13 mCic.

48 days = six half-lives

x 12

x 12

x 12

x 12

x 12

10.53 Use the conversion factors in Table 10.3 to convert mCi to disintegrations/second.

= 1.9 x 108 disintegrations/second5.0 mCi x3.7 x 1010 disintegrations/s

x 1000 mCi1 Ci

1 Ci 10.54 The average amount of background radiation is generally higher at higher elevations because

there is less atmosphere to act as a shield against cosmic rays. 10.55 Use the concentration as a conversion factor as in Example 10.4.

2.3 mL solution28 mCi x1 mL

12 mCi=

10.56 Use the concentration as a conversion factor as in Example 10.4.

0.89 mL solution12 mCi x

8.0 mL

108 mCi=

10.57 Use conversion factors to solve the problem; 1 mCi = 1,000 µCi.

2.6 mL solution68 kg x190 µCi

1 kg=x

1 mCi

1000 µCix

1 mL

5 mCi 10.58 Use conversion factors to solve the problem; 1 mCi = 1,000 µCi.

22 mL solution15 mCi x mL

670 µCi=x

1000 µCi1 mCi

10.59 Curie refers to the number of disintegrations per second and the number of rads indicates the

radiation absorbed by one gram of a substance. 10.60 The millicurie is a measure of the amount of radioactivity in a sample, whereas the rem is a

measure of the amount of radioactivity absorbed from a radioactive sample. 10.61 This represents a fatal dose because 600 rem is uniformly fatal.

2,000 rem20 Sv x100 rem

1 Sv=

10.62 This does not represent a fatal dose, but individuals would most likely have had a temporary

decrease in white blood cell count.

25 rem0.25 Sv x

100 rem

1 Sv=



10.63 Nuclear fission is the splitting of nuclei and nuclear fusion is the joining of small nuclei to form larger ones.

10.64 The difference between the nuclear fission process that takes place in a nuclear reactor and the

nuclear fission that occurs in an atomic bomb is the amount of uranium-235 present. When the amount present is equal to the critical mass (the amount required for a chain reaction to take place) an atomic explosion occurs. When the amount present is less than the critical mass, as in a nuclear reactor, the energy is produced in a manner that can be controlled.

10.65 a. Nuclear fusion is a reaction that occurs in the sun.

b. Nuclear fission occurs when a neutron is used to bombard a nucleus. c. In both nuclear fission and nuclear fusion a large amount of energy is released. d. Nuclear fusion reactions require very high temperatures.

10.66 a. Nuclear fission splits a nucleus into lighter nuclei. b. Nuclear fusion joins two lighter nuclei into a heavier nucleus. c. Nuclear fission is used to generate energy in a nuclear power plant. d. Nuclear fission generates radioactive waste with a long half-life. 10.67 Nuclear fission splits an atom into two lighter nuclei. Balance each equation.

U92235 n0

1+ n01+ + 2Mo42

97a. ?

U92235 n0

1+ n01+ + 2Mo42

97Sn50137

92 – 42 = 50 = atomic number for Sn

[235 + 1] – [97 + 2(1)] = 137 = mass number

U92235 n0

1+ +b. Ba56140 n0

1+ 3

Kr3693U92

235 n01+ + Ba56

140 n01+ 3

?

92 – 56 = 36 = atomic number for Kr

[235 + 1] – [140 + 3(1)] = 93 = mass number

10.68 Nuclear fission splits an atom into two lighter nuclei. Balance each equation.


Chapter 10–15

U92235 n0

1+ n01+ + 2La57

139a. ?

U92235 n0

1+ n01+ + 2La57

139Br35 95

92 – 57 = 35 = atomic number for Br

[235 + 1] – [139 + 2(1)] = 95 = mass number

U92235 n0

1+ +b. Ce58140 n0

1+ 2

Zr4094U92

235 n01+ + Ce58

140 +

?

92 – 58 + 6= 40 = atomic number for Zr

[235 + 1] – [140 + 2(1)] = 94 = mass number

e–10+ 6

n012 e–1

0+ 6

10.69

+ + +

10.70

+ +

10.71 Write out the equation from the information given. Then balance the equation.

H12 + H1

2 H13 + H1

1

tritium 10.72 Write out the equation from the information given. Then balance the equation.

He23 + He2

3 He24 + H1

12

10.73 Two problems with generating electricity from nuclear power plants include the containment of

radiation leaks and the disposal of radioactive waste. 10.74 There are no nuclear power plants that use nuclear fusion to produce electricity because the

process requires very high temperatures (> 100,000,000 °C) and pressures (> 100,000 atm).



10.75 Balance the nuclear equation.

83 + 26 = 109 = atomic number

209 + 58 – 1 = 266 = mass number

Fe2658 n0

1Bi83209 + ? +

Fe2658 n0

1Bi83209 + +109Mt266

meitnerium-266 10.76 Balance the nuclear equation.

92 + 7 = 99 = atomic number

235 + 14 – 5 = 244 = mass number

N 714 n0

1U92235 + ? +

N 714 n0

1U92235 + +99Es244

einsteinium-244

5

5

10.77

As3374 +e+1

0 Ge3274a.

90 days 1 half-life18 days

x = 5 half-lives xinitial mass

12

120 mg = 4 mg of As-74b. x 12

x 12

x 12

x 12

7.5 mCi 2.0 mL10.0 mCi

x = 1.5 mLc.

10.78

Na1124 +e–1

0 Mg1224a.


Chapter 10–17

2.5 days 1 half-life15 h

x = 4.0 half-lives

xinitial mass

12

84 mg = 5.3 mg of Na-24

b.

x 12

x 12

x 12

24 h1day

x

6.5 mCi 5.0 mL

10.0 mCix = 3.3 mLc.

10.79

b. Sr3889 + e–1

0Y3989

c. xinitial acitivity

12

10.0 µg = 0.625 µgx 12

x 12

x 12a. Sr38


10.80

b. Re75186 + e

–10Os76

186

c. xinitial acitivity

12

200. µg = 12.5 µgx 12

x 12

x 12a. Re75


10.81 a. Iodine-131 is used for the treatment and diagnosis of thyroid diseases. b. Iridium-192 is used for the treatment of breast cancer. c. Thallium-201 is used for the diagnosis of heart disease.

10.82 a. Iodine-125 is used for the treatment of prostate cancers.

b. Technetium-99 is used to evaluate the functioning of the gall bladder and bile ducts, to determine sites of internal bleeding, or to find the location of metastatic cancers. c. Cobalt-60 is used in radiation treatment for cancer.

10.83 a. Iodine-125, with its longer half-life, is used for the treatment of prostate cancers with

implanted radioactive seeds. b. Iodine-131, with its shorter half-life, is used for the diagnostic and therapeutic treatment of

thyroid diseases and tumors. A patient is administered radioactive iodine-131, which is then incorporated into the thyroid hormone, thyroxine. Since its half-life is short, the radioactive iodine isotope decays so that little remains after a month or so.

10.84 Food is irradiated with γ rays in order to kill harmful bacteria, thus extending its shelf life. 10.85 Radiology technicians must be shielded to avoid exposure to excessive and dangerous doses of

radiation. 10.86 A lead apron is placed over a patient’s body when dental X-rays are taken in order to protect the

patient from exposure to the radiation. 10.87 High doses of stable iodine will prevent the absorption and uptake of the radioactive iodine-131.



10.88 Exposure to strontium-90 is especially hazardous to children, because they have fast-growing, dividing cells and their bodies readily take up calcium. Any substance with properties similar to calcium would also be readily absorbed.

10.89 Use conversion factors to solve the problem.

1.0 g coal 1 kg

1000 gx = 370 kcal2.2 lb

1 kgx 3.4 x 108 kcal

2000 lbx

10.90

Po84218 +

e–10

Pb82214 He 2

4

Pb82214 +Bi83

214

X

Y

Bi83214 +Tl81

210

Z

He 24


Download - Chapter 10 Nuclear Chemistry - websites.rcc.eduwebsites.rcc.edu/grey/files/2012/02/Chapter-10-Solutions-Smith.pdfChapter 10–1 Chapter 10 Nuclear Chemistry Solutions to In-Chapter

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