Type:
Solution Manual
Resource:
Chemistry The Central Science
Edition:
14th Edition
Author(s):
Theodore E. Brown Bruce E. Bursten Patrick Woodward H. Eugene LeMay Catherine Murphy Matthew E. Stoltzfus
1
Introduction: Matter and Measurement
Visualizing Concepts 1.1
1.2
Pure elements contain only one kind of atom. Atoms can be present singly or as tightly bound groups called molecules. Compounds contain two or more kinds of atoms bound tightly into molecules. Mixtures contain more than one kind of atom and/or molecule, not bound into discrete particles. (a)
pure element: i
(b)
mixture of elements: v, vi
(c)
pure compound: iv
(d)
mixture of an element and a compound: ii, iii
After a physical change, the identities of the substances involved are the same as their identities before the change. That is, molecules retain their original composition. During a chemical change, at least one new substance is produced; rearrangement of atoms into new molecules occurs. The diagram represents a chemical change because the molecules after the change are different than the molecules before the change.
1.3
To brew a cup of coffee, begin with ground coffee beans, a heterogeneous mixture, and water, a pure substance. Hot water contacts the coffee grounds and dissolves components of the coffee bean that are water-soluble. This creates a new heterogeneous mixture of undissolved coffee bean solids and liquid coffee solution; this mixture is separated by filtration. Undissolved grounds are left on the filter paper and liquid coffee, itself a homogeneous mixture, drips into the container below. Overall, two separations occur. Chemical differences among the components of the coffee bean allow certain compounds to dissolve in water, while other components remain insoluble. This kind of separation based on solubility differences is called extraction. The insoluble grounds are then separated from the coffee solution by filtration.
1.4
1.5
(a)
time
(b) mass
(c) temperature
(d) area
(f)
area
(g) temperature
(h) density
(i) volume
(e) length
Density is the ratio of mass to volume. For a sphere, size is like volume; both are determined by the radius of the sphere. (a)
For spheres of the same size or volume, the denominator of the density relationship is the same. The denser the sphere, the heavier it is. A list from
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1 Matter and Measurement
Solutions to Exercises
lightest to heaviest is in order of increasing density and mass. The aluminum 3 sphere (density = 2.70 g/cm ) is lightest, then nickel (density = 8.90 g/cm3), then silver (density = 10.49 g/cm3). (b)
1.6
For spheres of equal mass, the numerator of the density relationship is the same. The denser the sphere, the smaller its volume or size. A list from smallest to largest is in order of decreasing density. The platinum sphere (density = 21.45 g/cm3) is smallest, then gold (density = 19.30 g/cm3), then lead (density = 11.35 g/cm3).
Results (bullet holes) that are close to each other are precise. Results that are close to the “true value” (the bull’s-eye) are accurate. (a)
The results on target A are precise but not accurate. The individual shots landed close to each other, but not close to the bull’s-eye. The results on target B are both precise and accurate. The individual shots landed close together and in the center of the target. The results on target C are neither precise nor accurate. The individual shots are scattered around the target.
(b)
The precise grouping on A is high and to the right of center. To improve accuracy, the sighting mechanism on the gun should be adjusted down and slightly left from its current position. To improve the results on C, someone who can produce a precise grouping must shoot the student’s target rifle. Then, the position of the sighting mechanism can be adjusted to produce an accurate shot. After adjusting the sight, the student needs more practice to improve precision.
1.7
1.8
1.9
(a)
7.5 cm. There are two significant figures in this measurement; the number of cm can be read precisely, but there is some estimating (uncertainty) required to read tenths of a centimeter. Listing two significant figures is consistent with the convention that measured quantities are reported so that there is uncertainty in only the last digit.
(b)
The speed is 72 mi/hr (inner scale, two significant figures) or 115 km/hr (outer scale, three significant figures). Both scales are read with certainty in the “hundreds” and “tens” place, and some uncertainty in the “ones” place. The km/hr speed has one more significant figure because its magnitude is in the hundreds.
(a)
Volume = length × width × height. Because the operation is multiplication, the dimension with fewest significant figures (sig figs) determines the number of sig figs in the result. The dimension “2.5 cm” has 2 sig figs, so the volume is reported with 2 sig figs.
(b)
Density = mass/volume. Because the operation is division, again the datum with fewer significant figures determines the number of sig figs in the result. While mass, 104.72 g, has 5 sig figs, volume [from (a)] has 2 sig figs, so density is also reported to 2 sig figs.
When converting units, arrange the conversion factor so that the given unit cancels and the desired unit is in the correct position. For example, suppose a quantity is expressed in terms of centimeters, but the desired result is expressed in inches. If the given unit
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1 Matter and Measurement
Solutions to Exercises
has ‘cm’ in the numerator, then the conversion factor must have ‘cm’ in its denominator. However, if the original unit has ‘cm’ in the denominator, the conversion factor must have ‘cm’ in the numerator. Ideally, this will lead to the desired units in the appropriate location, numerator or denominator. However, the inverse of the answer can be taken when necessary. 1.10
Given: m/s Find: mi/hr. Both the given and desired units have distance in the numerator and time in the denominator. Use appropriate conversion factors to change ‘m’ to ‘mi’ in the numerator and ‘s’ to ‘hr’ in the denominator.
1.11
Given: masses of six jelly beans, mass of full jar, mass of empty jar. Find: number of jelly beans in the jar. The total mass of jelly beans is the mass of the jar full minus the mass of the jar empty. The mass of an “average” jelly bean is the average of the six masses. Then, the number of jelly beans is the total mass of beans divided by the average mass of a single bean. Total mass of beans = 2082 g – 653 g = 1429 g Average mass of a bean = (3.15 + 3.12 + 2.98 + 3.14 + 3.02 + 3.09) / 6 = 3.08 g Number of beans = 1429 g total / 3.08 g per bean = 463.96 = 464 beans By applying the significant figure rules for addition and subtraction, the total mass of beans has 0 decimal places and thus 4 significant figures. The mass of an average bean has 2 decimal places and 3 sig figs. The number of beans then has 3 sig figs, by the rules for multiplication and division. This makes sense, because we expect an integer number of beans in the jar. (Note that using an unrounded average bean mass of 3.0833 g predicts the number of beans to be 463.46, which rounds to 463 beans. The difference in these two values shows uncertainty in the last significant figure of the number of beans, as we expect in an experimental result.)
1.12
Compounds are pure substances, they have constant composition and properties throughout. The agate stone cannot be a compound, because materials with different properties appear as irregular rings in the stone. Ellen is correct.
Classification and Properties of Matter (Sections 1.2 and 1.3) 1.13
1.14
(a)
Pure substance
(b)
Heterogeneous mixture
(c)
Homogeneous mixture
(d)
Heterogeneous mixture
(a)
homogeneous mixture
(b)
heterogeneous mixture (particles in liquid)
(c)
pure substance
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1 Matter and Measurement 1.15 1.16 1.17
Solutions to Exercises
(d)
heterogeneous mixture
(a)
O
(b) Ag
(c) Ni
(g)
Iron
(h) Calcium
(i) Nitrogen
(j) Neon
(a)
C
(b) N
(c) Ti
(d) Zn
(e) Fe
(g)
calcium (h) helium
(i) lead
(j) silver
A(s)
(d) F
(e) Sn
(f) Plutonium (f) phosphorus
B(s) + C(g)
When solid carbon is burned in excess oxygen gas, the two elements combine to form a gaseous compound, carbon dioxide. Clearly substance C is this compound. Because C is produced when A is heated in the absence of oxygen (from air), both the carbon and the oxygen in C must have been present in A originally. A is, therefore, a compound composed of two or more elements chemically combined. Without more information on the chemical or physical properties of B, we cannot determine absolutely whether it is an element or a compound. However, few if any elements exist as white solids, so B is probably also a compound. 1.18
Gold, Au, is an element and “fool’s gold,” FeS 2 , is a compound; both are solids and pure substances. Take advantage of differences in physical and or chemical properties between the two substances. Density and melting point measurements are often used to identify solids. For these two substances, melting points are very high, but densities are easy to measure. Gold is much denser than “fool’s gold.” Gold is much less chemically reactive than FeS 2 , so relative reactivity with acids and bases can be observed. Of these experiments, density measurement is the most definitive and does not destroy the sample. (Note that neither substance is attracted to a magnet, so this test will not identify the gold.)
1.19
Physical properties: silvery white (color); lustrous; melting point = 649°C; boiling point = 1105°C; density at 20°C = 1.738 g/cm 3 ; pounded into sheets (malleable); drawn into wires (ductile); good conductor. Chemical properties: burns in air to give intense white light; reacts with Cl 2 to produce brittle white solid.
1.20
(a)
Physical properties: melting point = 420°C; hardness = 2.5 Mohs; density = 7.13 g/cm 3 at 25 °C. Chemical properties: granules react with dilute hydrochloric acid to produce hydrogen gas; at elevated temperatures, reacts slowly with oxygen gas to produce ZnO.
(b)
From the photo, we see that zinc is a shiny dark gray solid at atmospheric conditions, (physical state, color). These are physical properties.
1.21
(a) Physical (b) chemical (c) physical (d) chemical (e) chemical
1.22
(a)
chemical
(b)
physical
(c)
physical (The production of H 2 O is a chemical change, but its condensation is a physical change.)
(d)
physical (The production of soot is a chemical change, but its deposition is a physical change.)
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