Putting it together, Chapters 1-4
1.
1.00 cm is equal to how many meters?
(a) 2.54 (b) 100. (c) 10.0
2.
1.00 cm is equal to how many inches?
(a) 0.394
(b) 0.10 (c) 12
(d) 0.0100
(d) 2.54
1.00 cm x (1 in / 2.54 cm) = 0.394 in.
3.
4.50 ft is how many centimeters? (a) 11.4 (b) 21.3 (c) 454 (d) 137
4.50 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 137.16 cm
4.
The number 0.0048 contains how many significant figures?
(a) 1 (b) 2 (c) 3 (d) 4 (Leading zeros are not significant)
5.
Express 0.00382 in scientific notation. (a) 3.82 x 10
3
3.82 x 10
-2
(d) 3.82 x 10 -3
(b) 3.8 x 10
-3
(c)
(move decimal point 3 times to the right)
6.
42.0 o
C is equivalent to: (a) 273 K (b) 5.55 o
F (c) 108
( o
F = 1.8 o
C +32 = 1.8 x 42.0 o
C + 32 = 107.6 o
F) o F (d) 53.3 o
F
7.
267 oF is equivalent to: (a) 404 K (b) 116 o
C (c) 540 K (d) 389 K
( o
C = ( o
F – 32) / 1.8 = (267 o
F – 32) / 1.8 = 130.555 o
C);
(K = o
C + 273.15 = 130.555 + 273.15 = 403.7 K)
8.
An object has a mass of 62 g and a volume of 4.6 mL. Its density is:
(a) 0.074 mL/g (b) 285 g/mL (c) 7.4 g/mL (d) 13 g/mL
(Density is in g/mL, thus (a) is obviously wrong! Plug the numbers: d = 62g / 4.6 mL = 13.478 g/mL)
9.
The mass of a block is 9.43 g and its density 2.35 g/mL. The volume of the block is: (a) 4.01 mL (b) 285 g/mL (c) 22.2 mL (d) 2.49 mL
(Since d = m / V,

V = m / d; V = 9.43 g / 2.35 g/mL = 4.013 mL)
10.
The density of copper is 8.92 g/mL. The mass of a piece of copper that has a volume of 9.5 mL is: (a) 2.6 g (b) 85 g (c) 0.94 g (d) 1.1 g.
(density = mass/volume

m = d x V = 8.92 g/mL x 9.5 mL = 84.74 g)
11.
An empty graduated cylinder has a mass of 54.772 g. When filled with 50.0 mL of an unknown liquid, it has a mass of 101.074 g. The density of the liquid is:
(a) 0.926 g/mL (b) 1.08 g/mL (c) 2.02 g/mL (d) 1.85 g/mL
(Calculate the mass of the liquid, and then plug into density: 101.074 g – 54.772 g
= 46.302 g. d = m / V = 46.302 g / 50.0 mL = 0.92604 g/mL)
12.
The conversion factor to change grams into milligrams is: (a) 100 mg / 1 g
(b) 1 g / 100 mg (c) 1 g / 1000 mg (d) 1000 mg / 1 g .
(The idea is to introduce mg and cancel the grams. Hence, mg / g)
13.
What Fahrenheit temperature is twice the Celsius temperature? (a) 64 o
F
(b) 320 o F (c) 200 o
F (d) 546 o
F

(must do by trial an error: o
C = ( o
F – 32) / 1.8; 64 o
F = 17.8 o
C; 320 o
F = 160 o
C)
14.
A gold alloy has a density of 12.41 g/mL and contains 75.0 % gold by mass. The volume of this alloy that can be made from 255 g of pure gold is:
(a) 4.22 x 10
3
mL (b) 2.37 x 10
3
mL (c) 27.4 mL (d) 20.5 mL
(Assume 100 g alloy, its volume is V = m/d = 100 g / 12.41 g/mL = 8.058 mL.
This volume contains 75% by mass gold. Thus, 8.058 mL of the alloy contains 75 g of gold. From proportion: 8.058 mL : 75 g = X: 255 g

X = 27.4 mL)
15.
A lead cylinder (V =
 r
2 h) has radius 12.0 cm and length 44.0 cm, and a density of 11.4 g/mL. The mass of the cylinder is: (a) 2.27 x 10 5 g (b) 1.89 x 10
5
g
(c) 1.78 x 10
3
g (d) 3.50 x 10
5
g
V =
 r 2 h = 3.14 x (12.0 cm) 2 x 44.0 cm = 19905 cm 3 = 19.9 x 10 3 cm 3 = 19.9 x 10 3 mL ( 1 cm 3 = 1 mL !!) . m = V x d = 19.9 x 10
3
mL x 11.4 g/mL = 2.27 x 10
5
g)
16.
The following units can be used for density EXCEPT: (a) g / cm 3 (b) kg/m 3
(c) g/L (d) kg/m 2 .
(density MUST have units of mass/volume; m
2
is the unit for area , not volume!)
17.
37.4 cm x
2.2 cm equals: (a) 82.28 cm 2 (b) 82.3 cm 2 (c) 82 cm 2 (d) 82.2 cm 2 .
(The number 2.2 cm has two significant figures, so the result must have two)
18.
The following elements are among the five most abundant by mass in Earth crust, seawater and atmosphere EXCEPT: (a) oxygen (b) hydrogen (c) silicon
(d) aluminum (Hydrogen is the most abundant in space, not on Earth)
19.
Which of the following is a compound? (a) lead (b) wood (c) potassium
(d) water (lead and potassium are elements, wood is a mixture).
20.
Which of the following is a mixture? (a) water (b) chromium (c) wood
(d) sulfur (water is a compound, chromium and sulfur are elements)
21.
How many atoms are represented in the formula Na
2
CrO
4
? (a) 3 (b) 5
(c) 7 (d) 8. (There are 2 sodium, one chromium and four oxygen atoms, total 7)
22.
Which of the following is a characteristic of a metal? (a) ductile (b) easily shattered (c) extremely strong (d) dull. (ductile = drawn into wires)
23.
Which of the following is a characteristic of nonmetals? (a) always a gas
(b) poor conductor of electricity (c) shiny (d) combines only with metals.
24.
When a pure substance is analyzed, it was found to contain carbon and chlorine.
This substance must be classified as: (a) an element (b) a mixture
(c) a compound (d) both a mixture and a compound.
(Since it is a pure substance and can be decomposed into smaller substances, it must be a compound)
25.
Chromium, fluorine and magnesium have the symbols: (a) Ch, F, Ma
(b) Cr, Fl, Mg (d) Cr, F, Mg (e) Cr, F, Ma

26.
Sodium, carbon and sulfur have the symbols: (a) Na, C, S
(b) So, C, Su (c) Na, Ca, Su (d) So, Ca, Su
27.
Coffee is an example of (a) an element (b) a compound
(c) a homogeneous mixture (d) a heterogeneous mixture
28.
The number of oxygen atoms in Al(C
2
H
3
O
2
)
3
is: (a) 2 (b) 3 (c) 5 (d) 6 .
29.
Which of the following is a mixture? (a) water
(c) sugar solution (d) iodine
(b) iron (II) oxide
30.
Which is the most compact state of matter? (a) solid (b) liquid (c) gas
(d) amorphous
(Solid state of most matter has the highest density. Water is an exception).
31.
Which is not characteristic of a solution? (a) a homogeneous mixture (b) a heterogeneous mixture (c) one that has two or more substances (d) one that has a variable composition
(Solution is another name for homogeneous mixture)
32.
A chemical formula is a combination of (a) symbols (b) atoms (c) elements (d) compounds (see p. 59!)
33.
The number of nonmetal atoms in Al
2
(SO
3
)
3
is (a) 5 (b) 7 (c) 12
34.
Which of the following is not a physical property? (a) boiling point physical state (c) bleaching action (d) color
(d) 14
(b)
35.
Which of the following is a physical change? (a) a piece of sulfur is burned
(b) a firecracker explodes (c) a rubber band is stretched (d) a nail rusts
36.
Which of the following is a chemical change? (a) water evaporates melts (c) rocks are ground to sand (d) a penny tarnishes .
(b) ice
37.
When 9.44 g of calcium are heated in air, 13.22 g of calcium oxide is formed. The percent my mass of oxygen in the compound is (a) 28.6% (b) 40.0 %
(c) 71.4 % (d) 13.2 %
(%Ca in the oxide = (9.44 / 13.22) x 100% = 71.4 %. The rest, 28.6 % is oxygen.)
38.
Barium Iodide, BaI
2
, contains 35.1 % Ba by mass. An 8.50 g sample of BaI
2 contains what mass of iodine? (a) 5.52 g (b) 2.98 g (c) 3.51 g (d) 6.49 g
(In 100 g BaI
2
there is 35.1 g Ba and 64.9 g iodine; Use proportion:
100 g BaI
2
: 64.9 g Iodine = 8.50 g BaI
2
: X Iodine

X Iodine = 64.9 g Ba x 8.50 g BaI
2
/ 100 g BaI
2
= 5.52 g Iodine.)
39.
Mercury (II) sulfide, HgS, contains 86.2 % mercury by mass. The mass of HgS that can be made from 30.0 g of mercury is: (a) 2586 g (b) 2.87 g (c)
25.9 g (d) 34.8 g . (Assume 100 g sample of HgS, there is 86.2 g mercury. By proportion: 100 g HgS : 86.2 g Hg = X g HgS : 30.0 g Hg

X g HgS = 34.8 g)

40.
The changing of liquid water to ice is known as a (a) chemical change (b) heterogeneous change (c) homogeneous change (d) physical change .
41.
Which of the following does not represent a chemical change? (a) heating of copper in air (b) combustion of gasoline (c) cooling of red-hot iron (d) digestion of food (heating of Cu in air produces copper oxide!)
42.
Heating 30. g of water from 20. o
C to 50. o
C requires:
(b) 50. cal (c) 3.8 x 10
3
J (d) 6.3 x 10
3
J
(a) 30. cal
43.
The specific heat of aluminum is 0.900 J/g o
C. How many Joules of energy are required to raise the temperature of 20.0 g of aluminum from 10.0 to 15.0 o
C?
(a) 78 J
(
 t = 5.0 o
C
 
(b) 90. J (c) 100. J (d) 112 J

H = mC
 t = 20.0 g x 0.900 J/g o
C x 5.0 o
C = 90. J)
44.
A 100. g iron ball (specific heat = 0.473 J / g o
C) is heated to 125 o
C and is placed into a calorimeter holding 200. g of water at 25.0 o
C. What will be the highest temperature reached by water? (a) 43.7 o
C (b) 30.4 o C (c) 65.3 o
C (d)
35.4 o C
(Assuming the calorimeter is an isolated system (no heat loss to the surroundings), the heat lost by the iron ball is given to the water, so the highest temperature of water will be when it is in thermal equilibrium with the iron ball i.e. when iron ball and water have the same temperature, T f
. The change of heat of iron ball,

H of water,

H w
Fe
is negative because iron loses heat, whereas the change of heat is positive because water gains heat.
Write down the equation for heat change for iron (subscript Fe) and water (w):

H
Fe
= m
Fe
C
Fe
(T f
– ti
Fe
) ti
Fe
= 125 o
C; C
Fe
= 0.473 J /g o
C; m
Fe
= 100. g

H w
= m w
C w
(T f
– ti w
)
Realizing that -

H
Fe
=

H w
: ti w
= 25 o
C Cw = 4.184 J / g o
C; m
Fe
= 200. g

H
Fe
= m
Fe
C
Fe
(T f
– ti
Fe
) = m
Fe
C
Fe
(ti
Fe
- T f
) = m w
C w
(T f
– ti w
)
Final temperature T f
is the only unknown. Multiply the terms in parenthesis: m
Fe
C
Fe
ti
Fe
m
Fe
C
Fe
T f
= m w
C w
T f
m w
C w
t iw
T f
( m w
C w
+ m
Fe
C
Fe
) = m
Fe
C
Fe
ti
Fe
+ m w
C w
t iw arranging the terms for T f
:
T f
= (m
Fe
C
Fe
ti
Fe
+ m w
C w
t iw
) / (m w
C w
+ m
Fe
C
Fe
)
T f
= (5912.5 J + 20920 J) / (836.8 J/ o
C + 47.3 J/ o
C) = 26832.5 J / 884.1 J/ o
C =
30.35 o C, with 3 significant figures 30.4 o C.
45.
What has the highest specific heat? (a) ice (b) lead (c) water (d) aluminum. (See table 4.3, p.73!)
46.
When 20.0 g of mercury are heated from 10.0 o
C to 20.0 o
C, 27.6 J of energy are absorbed. What is the specific heat of mercury? (a) 0.726 J/g o
C
(b) 0.138 J/g o C (c) 2.76 J/g o C (d) no correct answer is given.
 t = 20.0 o
C – 10.0 o
C = 10.0 o
C

H = m C
 t

C =

H / m
 t
C = 27.6 J / (20.0 g x 10.0 o
C) = 0.138 J/g o
C.)
47.
Changing hydrogen and oxygen into water is a: (a) physical change
(b) chemical change (d) conservation reaction (d) no correct answer is given

Free response questions (1-6 only)
1.
Convert 1.5 meters into feet, 4.0 meters into feet, and 27 o
C into Fahrenheit.
1.5
m x (1.0936 yards / 1 m) x (3 ft / 1 yard) = 4.9 ft
4.0 m x (1.0936 yards / 1 m) x (3 ft / 1 yard) = 13 ft o
F = 1.8 o
C + 32 = 1.8 x 27 o
C + 32 = 80.6 o
F = 81 o F
2.
What is needed to determine how much heat is absorbed if the amount of heat released per minute by the heater is known? ( Time in minutes )
3.
75 g of CaCO
3
is heated, producing 42 g of CaO and a quantity of CO
2
. How many liters of CO2 is collected if 44 g CO
2
occupies 24 dm
3
? mass CO
2
= 75 g – 42 g = 33 g. 24 dm
3
= 24 L
Proportion: 44 g : 24 L = 33 g : X  X = 18 L .
4.
Use the pictures at p. 83 to answer the following:
(a) Which picture best describes a homogeneous mixture? ( 2 )
(b) How would you classify the contents of other containers? (1) and (3) are heterogeneous mixtures, (4) is a compound.
(c) Which picture contains a compound? ( 4 ) It is composed of a large and a smaller atom.
5.
Use the picture to answer the following;
(a) Which picture best represents fluorine gas? (c) because it is diatomic molecule
(b) Which other elements could that picture also represent ( H
2
, O
2
, N
2
, etc.)
(c) Which of the pictures could represent SO
3
gas? ( 2 ) because it has one large atom and three smaller atoms.
6.
Two one-quart bowls, Bowl A filled with water and covered, Bowl B empty, are left outside. The snow fell and filled the empty one, and then both bowls were frozen at 12 o F.
(a) Which bowl would require less energy to bring its contents to room temperauture (25 o
C) and why? Bowl B , because snow has low packing density and would produce less ice when frozen than water in Bowl A.
(b) What temperature change (
 t, in o
C) is required to warm the bowls to 25 o
C?
Initial temp: o
C = ( o
F – 32) / 1.8 = (12 – 32) / 1.8 = -11 o
C.
 t= 36 o
C.
(c) How much heat, (in kJ) is required to raise the temperature of the contents of
Bowl A to 0 o
C, without converting ice into water? Before the water in Bowl
A was frozen, its volume was: 1 qt = 0.946 L; since the density of water is 1 g/mL, the mass of water in Bowl A was 946 g; after freezing, the mass remained the same. The specific heat of ice is 2.059 J/g o
C (see Table 4.3) and the temperature change
 t = 11 o C (see part question 6b).

H = m C
 t = 946 g x 2.059 J/g o
C x 11 o
C = 21425 J = 21 kJ .
(d) Did the water in Bowl A undergo chemical or physical changes ? physical .