(General Chemistry The Essential Concepts, 7e Raymond Chang, Kenneth Goldsby)
(Solution Manual, For Complete File, Download link at the end of this File)
CHAPTER 1
INTRODUCTION
1.1 (a) Matter is anything that occupies space and has mass. (b) Mass is a measure of the quantity of matter in an
object. (c) Weight is the force that gravity exerts on an object. (d) A substance is matter that has a definite or
constant composition and distinct properties. (e) A mixture is a combination of two or more substances in
which the substances retain their distinct identities.
1.2 The scientifically correct statement is, “The mass of the student is 56 kg.”
1.3 Table salt dissolved in water is an example of a homogeneous mixture. Oil mixed with water is an example of
a heterogeneous mixture.
1.4 A physical property is any property of a substance that can be observed without transforming the substance
into some other substance. A chemical property is any property of a substance that cannot be studied without
converting the substance into some other substance.
1.5 Density is an example of an intensive property. Mass is an example of an extensive property.
1.6 (a) An element is a substance that cannot be separated into simple substances by chemical means.
(b) A compound is a substance composed of atoms of two or more elements chemically united in fixed
proportions.
1.7 (a) Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity are
changed.
(b) Chemical property. The fertilizer is consumed by the growing plants; it is turned into vegetable matter
(different composition).
(c) Physical property. The measurement of the boiling point of water does not change its identity or
composition.
(d) Physical property. The measurement of the densities of lead and aluminum does not change their
composition.
(e) Chemical property. When uranium undergoes nuclear decay, the products are chemically different
substances.
1.8 (a) Physical change. Helium is not changed in any way by leaking out of the balloon.
(b) Chemical change in the battery.
(c) Physical change. The orange juice concentrate can be regenerated by evaporation of the water.
(d) Chemical change. Photosynthesis changes water, carbon dioxide, and so on, into complex organic
matter.
(e) Physical change. The salt can be recovered unchanged by evaporation.
1.9 (a) extensive (b) extensive (c) intensive (d) extensive
1.10 (a) extensive (b) intensive (c) intensive
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,2 CHAPTER 1: INTRODUCTION
1.11 (a) element (b) compound (c) element (d) compound
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
, CHAPTER 1: INTRODUCTION 3
1.12 (a) compound (b) element (c) compound (d) element
1.13 (a) m (b) m2 (c) m3 (d) kg (e) s (f) N (g) J (h) K
1.14 (a) 106 (b) 103 (c) 10−1 (d) 10−2 (e) 10−3 (f) 10−6 (g) 10−9
(h) 10−12
1.15 Density is the mass of an object divided by its volume. Chemists commonly use g/cm3 or equivalently g/mL
for density. Density is an intensive property.
5C
1.16 ? C = (F − 32F)
9F
9F
? F = C + 32F
5C
1.17 The density of the sphere is given by:
m 1.20 104 g
d = = = 11.4 g/cm3
V 1.05 103 cm3
1.18 Strategy: We are given the density and volume of a liquid and asked to calculate the mass of the liquid.
Rearrange the density equation, Equation (1.1) of the text, to solve for mass.
mass
density =
volume
Solution:
mass= density volume
13.6 g
mass of Hg = 95.8 mL = 1.30 103 g
1 mL
5C
1.19 (a) ? C = (105 − 32)F = 41C
9F
9F
(b) ? F = −11.5 C + 32F = 11.3 F
5C
9F
(c) ? F = 6.3 103 C + 32F = 1.1 104 F
5C
5C
(d) ? C = (451 − 32)F = 233C
9F
1.20 Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between
Celsius and Fahrenheit given in Section 1.5 of the text. Substitute the temperature values given in the
problem into the appropriate equation.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
, 4 CHAPTER 1: INTRODUCTION
Solution:
1K
(a) K = (C + 273C)
1C
(i) K= 113C + 273C =386 K
(ii) K= 37C + 273C =3.10 102 K
(iii) K= 357C +273C=6.30 102 K
1K
(b) K = (C + 273C)
1C
(i) C= K − 273 = 77 K − 273 =−196C
(ii) C= 4.2 K − 273 =−269C
(iii) C= 601 K − 273 =328C
1.21 (a) 2.7 10−8 (b) 3.56 102 (c) 9.6 10−2
1.22 Strategy: Writing scientific notation as N 10n, we determine n by counting the number of places that the
decimal point must be moved to give N, a number between 1 and 10.
If the decimal point is moved to the left, n is a positive integer, the number you are working with is larger
than 10. If the decimal point is moved to the right, n is a negative integer. The number you are working with
is smaller than 1.
(a) Express 0.749 in scientific notation.
Solution: The decimal point must be moved one place to give N, a number between 1 and 10. In this case,
N = 7.49
Since 0.749 is a number less than one, n is a negative integer. In this case, n = −1.
Combining the above two steps:
0.749 = 7.49 10−1
(b) Express 802.6 in scientific notation.
Solution: The decimal point must be moved two places to give N, a number between 1 and 10. In this case,
N = 8.026
Since 802.6 is a number greater than one, n is a positive integer. In this case, n = 2.
Combining the above two steps:
802.6 = 8.026 102
(c) Express 0.000000621 in scientific notation.
Solution: The decimal point must be moved seven places to give N, a number between 1 and 10. In this
case,
N = 6.21
Since 0.000000621 is a number less than one, n is a negative integer. In this case, n = −7.
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© 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
