Homework:
Chemistry Chapter 2: 4, 5, 7, 8, 10, 11, 12, 14, 15
#4
A) centimeter
B) micrometer
C) kilogram
D) deciliter
#5
A) Millimeter
B) microsecond
C) centigram
D) picosecond
#7
No, the balance would subtract out.
#8
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(10.95 – 11.32 g/cm3 )÷ 11.342 =
•
3.5%
#10
A) 1
B) 2
C) 2
D) 2
E) 3
F) 1
G) 2
H) 3
I) 4
J) infinite
#11
8.76g / 3.07 cm3 = 2.85 g/cm3
#12
26.8g / 14.5 cm3 = 1.85 g/cm3
#14
24.9 cm3 X 2.72 g/cm3 =67.7 g
#15
7.91 g / 2.50 g/cm3 = 3.16 cm3
Chapter 2: 36, Chapter 8: 21-27, 29
36) Accuracy is the closeness of a measurement to the true value. Precision tells how
close a set of measurements (for a quantity) are to another set, regardless of whether the
measurements are correct. Precision also is used to explain the degree of exactness. For
instance, the triple beam balance measures to the .01 so we can estimate to the .001
which is more precise than the balance that measures to the .1.
21) 0753 m2
22) 1.1 X 106 mm2
23) 740 g/dm3
24) 763 g
25) 9.40 kg
26) 1.33 X 10-5 cm2
27) 3.59 X 1012 cm
29a) 1700 m
29b) 3.13 X 104 °C
29c) 16.19 kg/m3
29d) 0.46 cm4
29e) 32,413 kg/m2
Chapter 3: 1, 2, 3, 5, 6, 7, 8
#1
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A)air = solution
B) india ink = heterogeneous mix
C) paper = het mix
D) table salt = compound
E) wood alcohol = compound
F) apple = het mix
G) milk = het mix (colloid)
H) plutonium = element
I) water = compound
•
Examine under microscope
•
•
Copper is composed of only one type of atom so it is an element
Water is a compound because it is composed of two types of atoms: hydrogen and
oxygen
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A) 38g/100g H2O
C) 46g/100g H2O
#2
#3
#5
B) 230g/100g H2O
D) 95g/100g H2O
#6
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A) Extensive
B) Intensive
C) Intensive
D) Extensive
E) Intensive
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A) chemical
B) chemical
C) physical
D) chemical
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A) physical
B) chemical
C) chemical
D) physical
E) physical
F) chemical
#7
E) physical
F) chemical
G) physical
H) physical
#8
G) physical
H) physical
I) physical
J) physical
K) chemical
Chapter 15 13, 30, 35
15) Temperature is a measure of the average KE of the molecules in a sample. However,
all the molecules in the sample do not have the exact same KE. Some have more and
some have less than the average. This helps to explain why a cup of water can evaporate
at room temperature.
30) All molecular motion has ceased at absolute zero.
35) Intermolecular forces and the kinetic energy
CHAPTER 3: 10a-c, 11-16
10)
a)
1980J 1calorie

 473calories
1
4.184J
b)
1.1Cal 1000cal 4.184J


 4640J
1
1Cal
1cal
c)
800cal
1Cal

 0.8Cal
1
1000cal
11) mass=854g
Ti=23.5°C,Tf= 85.0°C, ΔT=61.5°C
q = Cp · m · ΔT
Cp =4.184J/g°C
= 4.184J/g°C x 854g x 61.5°C
= 220000J
12) mass=96.7g
Ti=31.7°C,Tf= 69.2°C, ΔT=37.5°C
q = Cp · m · ΔT
= 0.874J/g°C x 96.7g x 37.5°C
= 3170J
Cp =0.874J/g°C
13) mass=10.35g
Ti=32.1°C,Tf= 56.4°C, ΔT=24.3°C
q = Cp · m · ΔT
= 0.85651J/g°C x 10.35g x 24.3°C
= 215J
Cp =0.85651J/g°C
14) Aluminum: m=3.90g, Ti=99.3°C, Cp =0.903J/g°C
Water: m=10.0g, Ti=22.6°C, Cp =4.184J /g°C
qlost =qgained
Cp · m · ΔT = Cp · m · ΔT
(0.903J/g°C)(3.90g)(99.3°C- Tf) = (4.184J/g°C)(10 g)( Tf -22.6°C)
Tf = 28.6°C
15) Cadmium: m=65.6g, Ti=100.0°C, Cp =0.231J/g°C
Water: m=25.0g, Ti=23.0°C, Cp =4.184J /g°C
qlost =qgained
Cp · m · ΔT = Cp · m · ΔT
(0.231J/g°C)(65.6g)(100.0°C- Tf) = (4.184J/g°C)(25.0 g)( Tf -23.0°C)
Tf = 32.7°C
16)Unknown: m=23.8g, Ti=100.0°C, Tf=32.5°C
Water: m=50.0g, Ti=24.0°C, Tf=32.5°C, Cp =4.184J /g°C
qlost =qgained
Cp · m · ΔT = Cp · m · ΔT
(Cp)(23.8g)(100.0°C- 32.5°C) = (4.184J/g°C)(50.0 g)(32.5°C -23.0°C)
(Cp)(23.8g)(67.5°C) = (4.184J/g°C)(50.0 g)(8.5°C)
Cp = 1.1J/g°C