Midterm Review Problem Set 2 Key

Transcription

Midterm Review Problem Set 2 Key
Dr. Casagrande’s Chemistry 1 Midterm Exam Review Sheet
Problem Set 2: Chapters 4, 25, 5
Ch. 4: pg. 113 #60, 62, 63, 64, 65, 66, 67
Ch. 25: pg. 836-837 #38, 44, 45, 69, 71, 74, 76, 78, 79
Ch. 5: pg. 147 #65, 66, 70, 73, 79(a,c,e,g,i), 80(b,d,f)
Chapter 4
60. For each of the following chemical symbols, determine the element name and the number of protons and
electrons an atom contains.
a. V Vanadium; 23 protons & 23 electrons
c. Ir Iridium; 77 protons & 77 electrons
b. Mn Manganese; 25 protons & 25 electrons
d. S Sulfur; 16 protons & 16 electrons
62. An isotope of mercury has 80 protons and 120 neutrons. What is the mass number of this isotope?
80 + 120 = 200
63. An isotope of xenon has an atomic number of 54 and contains 77 neutrons. What is the xenon isotope’s mass
number?
54 + 77 = 131
64. How many electrons, protons, and neutrons are contained in each of the following atoms?
Atom
Electrons
Protons
Neutrons
132
55
55
77
a.
55 Cs
b.
59
27
27
27
32
c.
163
69
Tm
69
69
94
d.
70
30
Zn
30
30
40
Co
65. How many electrons, protons, and neutrons are contained in each of the following atoms?
Atom
Electrons
Protons
Neutrons
a. gallium-64
31
31
33
b. fluorine-23
9
9
14
c. titanium-48
22
22
26
d. helium-8
2
2
6
66. Chlorine, which has an atomic mass of 35.453 amu, has two naturally occurring isotopes, Cl-35 and Cl-37.
Which isotope occurs in greater abundance? Explain.
Cl-35; the average atomic mass is closer to 35 than to 37, indicating that there must be more Cl-35 atoms.
67. Silver has two isotopes, 107
47 Ag has a mass of 106.905 amu (52.00%) and
(48.00%). What is the atomic mass of silver?
(0.5200)(106.905 amu) = 55.59 amu
(0.4800)(108.905 amu) = 52.27 amu
55.59 amu + 52.27 amu = 107.86 amu
109
47
Ag has a mass of 108.905 amu
Chapter 25
38. Match each numbered choice on the right with the correct radiation type on the left.
a. alpha
1. high speed electrons
b
b. beta
2. 2+ charge
a
c. gamma
3. no charge
c
4. helium nucleus
a
5. blocked very easily
a
6. electromagnetic radiation
c
44. Explain the relationship between an atom’s neutron-to-proton ratio and its stability.
For smaller atoms, the n/p ratio is about 1:1. As atoms become larger, the n/p ratio increases to about 1.5:1.
Outside these ranges, an isotope is likely to be unstable.
45. What is the significance of the band of stability?
It shows the range of stable nuclei; nuclei outside its range are likely to be radioactive.
69. Complete the following equations:
4
210
214=4+ A; A=210
a. 214
83 Bi → 2 He + 81Tl 83=2+Z ; Z=81
b.
239
93
Np →
239
94
Pu +
0
–1
e
239=239+ A; A=0
93=94+Z ; Z=−1
71. Write a balanced nuclear equation for the beta decay of bromine-84.
84
0
84
35 Br → –1 e + 36 Kr
74. Write a balanced nuclear equation for the alpha particle bombardment of
is a neutron.
253
4
256
1
99 Es + 2 He → 101 Md + 0 n
253
99
Es . One of the reaction products
76. Write the balanced nuclear equation for the alpha particle bombardment of plutonium-239. The reaction
products include a hydrogen atom and two neutrons.
239
4
240
1
1
94 Pu + 2 He →
95 Am + 1 H + 2 0 n
78. Technetium-104 has a half-life of 18.0 minutes. How much of a 165.0 g sample remains after 90.0 minutes?
n= 90.0 min/18.0 min = 5 half lives; A0 = 165.0 g; A = 165.0 g ( 12 ) = 5.156 g or
165.0 g ⎯1⎯⎯
→ 82.50 g ⎯2⎯⎯⎯
→ 41.25g ⎯3⎯⎯⎯
→ 20.625 g ⎯4⎯⎯⎯
→ 10.3125 g ⎯5⎯⎯⎯
→ 5.165 g
half-life
half-lives
half-lives
half-lives
half-lives
79. Manganese-56 decays by beta emission and has a half-life of 2.6 hours. How many half-lives are there in 24
hours? How many mg of a 20.0 mg sample will remain after five half-lives?
n = 24 hrs/2.6 hrs = 9.23 half-lives; A0 = 20.0 mg
5
A = 20.0 mg ( 12 ) = 0.625 mg; or
20.0 mg ⎯1⎯⎯
→ 10.0 mg ⎯2⎯⎯⎯
→ 5.00 mg ⎯3⎯⎯⎯
→ 2.50 mg ⎯4⎯⎯⎯
→ 1.25 mg ⎯5⎯⎯⎯
→ 0.625 mg
half-life
half-lives
half-lives
half-lives
half-lives
5
Chem 1 Midterm Review Problem Set 2 p. 2
Chapter 5
65. What is the wavelength of electromagnetic radiation having a frequency of 5.00×1012 Hz? What kind of
electromagnetic radiation is this?
3.00 × 108 m/s
8
12
–5
c = λν; 3.00 × 10 m/s = λ(5.00 × 10 Hz); λ =
= 6.00 × 10 m ; infrared radiation
12
5.00 × 10 Hz
66. What is the frequency of electromagnetic radiation having a wavelength of 3.33×10–8 m? What type of
electromagnetic radiation is this?
8
3.00 × 10 m/s
8
−8
c=λν; 3.00 × 10 m/s = (3.33 × 10 m)ν; ν =
= 9.01 × 1015 Hz ; ultraviolet radiation
–8
3.33 × 10 m
70. What is the energy of a photon of red light having frequency of 4.48×1014 Hz?
E = hν; E = (6.626×10–34 J⋅s)(4.48×1014 Hz) = 2.97×10–19 J
73. A photon has an energy of 2.93×10–25 J. What is its frequency? What type of electromagnetic radiation is the
photon?
2.93 × 10 –25 J
E = h ν; 2.93×10–25 J = (6.626×10–34 J·s)ν; ν =
= 4.42 × 108 Hz ; TV or FM wave
–34
6.626 × 10 J ⋅s
79. Use noble-gas notation to describe the electron configurations of the elements represented by the following
symbols.
a. Mn [Ar]4s23d5
g. Pb [Xe]6s24f145d106p2
c. P [Ne]3s23p3
i. Sm [Xe]6s25d14f5
2
2
e. Zr [Kr]5s 4d
80. What elements are represented by each of the following electron configurations?
b. [Ar]4s2 Ca
f. 1s22s22p63s23p64s23d104p5 Br
2
10
4
d. [Kr]5s 4d 5p Te
Chem 1 Midterm Review Problem Set 2 p. 3