Download solution manual brady chemistry 6th edition and more Exercises Chemistry in PDF only on Docsity!
Practice Exercises
1.1 (a) SF 6 contains 1 S and 6 F atoms per molecule (b) (C 2 H 5 ) 2 N 2 H 2 contains 4 C, 12 H, and 2 N per molecule (c) Ca 3 (PO 4 ) 2 contains 3 Ca, 2 P, and 8 O atoms per formula unit (d) Co(NO 3 ) 2 ·6H 2 O contains 1 Co, 2 N, 12 O, and 12 H per formula unit
1.2 (a) NH 4 NO 3 contains 2 N nitrogen, 4 H hydrogen, 3 O oxygen atoms per formula unit (b) FeNH 4 (SO 4 ) 2 contains 1 Fe iron, 1 N nitrogen, 4 H hydrogen, 2 S sulfur, 8 O oxygen atoms per formula unit (c) Mo(NO 3 ) 2 ·5H 2 O contains 1 Mo molybdenum, 2 N nitrogen, 11 O oxygen, and 10 H hydrogen atoms per formula unit (d) C 6 H 4 ClNO 2 contains 6 C carbon, 4 H hydrogen, 1 Cl chlorine, 1 N nitrogen, and 2 O oxygen atoms per molecule
1.3 C 2 H 7 N, CH 3 NHCH 3.
1.4 Reactants: 4 N, 12 H, and 6 O; Products: 4 N, 12 H, and 6 O.
1.5 Reactants: 6 N, 42 H, 2 P, 20 O, 3 Ba, 12 C; Products: 3 Ba, 2 P, 20 O, 6 N, 42 H, 12 C; The reaction is balanced.
Review Questions
1.1 This answer will be student dependent.
1.2 Observation, testing and explanation.
1.3 (a) A law is a description of behavior based on the results of many experiments which are true while a theory is a tested explanation of the results of many experiments. (b) An observation is a statement that accurately describes something we see, hear, taste, feel or smell while a conclusion is a statement that is based on a series of observations. (c) Data are the observations made while performing experiments.
1.4 A theory is valid as long as there is no experimental evidence to disprove it. Any experimental evidence that contradicts the theory therefore, disproves the theory.
1.5 Matter has mass and occupies space. All items in the question are examples of matter.
1.6 A physical change does not change the chemical composition of matter. Melting, boiling, change of shape, or mass, and the formation of a mixture are examples of physical changes to matter.
A chemical change changes the chemical composition of matter. Formation of new compounds from the reaction of other substances is an example.
A chemical changes involves the change in composition while a physical change does not change in the composition of matter.
1.7 The reaction of calcium metal with water is a chemical change resulting in the formation of new compounds, hydrogen gas and calcium hydroxide. It is not stated in the problem, but the water also increases in temperature, which is a physical change.
1.8 These are physical changes.
1.9 (a) An element is a pure substance that cannot be decomposed into something simpler. (b) A compound is a pure substance that is composed of two or more elements in some fixed or characteristic proportion. (c) Mixtures result from combinations of pure substances in varying proportions. (d) A homogeneous mixture has one phase. It has the same properties throughout the sample. (e) A heterogeneous mixture has more than one phase. The different phases have different properties. (f) A phase is a region of a mixture that has properties that are different from other regions of the mixture. (g) A solution is a homogeneous mixture.
1.10 Changing a compound into its element is a chemical change.
1.11 (a) Cl (b) S (c) Fe (d) Ag (e) Na (f) P (g) I (h) Cu (i) Hg (j) Ca
1.12 (a) potassium (b) zinc (c) silicon (d) tin (e) manganese (f) magnesium (g) nickel (h) aluminum (i) carbon (j) nitrogen
1.13 (a) This is a heterogeneous mixture. (b) This is a pure substance and is an element, such as H 2 , O 2 , N 2 or a halogen. (c) This is a homogeneous mixture. (d) This is a pure substance and is a molecule such as H 2 O.
1.14 (a) Diagrams (a) and (d) contain pure elements (b) Diagram (c) contains a compound (c) Diagram (a) and (b) contain diatomic molecules
1.15 The first law of chemical combination is the law of conservation of mass: no detectable gain or loss of mass occurs in chemical reactions. The other law is the law of definite proportions: in a given chemical compound, the elements are always combined in the same proportions by mass.
1.16 Conservation of mass derives from the postulate that atoms are not destroyed in normal chemical reactions. The Law of Definite Proportions derives from the notion that compound substances are always composed of the same types and numbers of atoms of the various elements in the compound.
1.17 This is the Law of Definite Proportions, which guarantees that a single pure substance is always composed of the same ratio of masses of the elements that compose it.
1.18 The law of multiple proportions states that when elements combine to make a molecule the ratio of atoms is always a small, whole number. When elements combine to form more than one compound the ratio of one element in the compound compared to a fixed amount of a second element is always a small, whole number.
The first compound in the diagram has two blue and three red atoms. The second compound has two blue and one red atom. Since each compound has two blue atoms we can compare the ratio of red atoms between the
1.34 CH 3 COOH or C 2 H 4 O 2
1.35 CH 3 SCH 3 , or C 2 H 6 S; using parentheses we can write the formula as (CH 3 ) 2 S
1.36 NH 3
1.37 HOCH 2 CH 2 OH, or C 2 H 6 O 2
1.38 b H
H
H
N
1.39 (a) is the proper representation for C 2 H 6 O 2
1.40 (a) 1 Ca calcium, 2 N nitrogen, 6 O oxygen (b) 3 H hydrogen, 1 P phosphorus, 4 O oxygen (c) 6 C carbon, 14 H hydrogen (d) 1 H hydrogen, 1 C carbon, 1 N nitrogen (e) 1 Cu copper, 1 S sulfur, 4 O oxygen
1.41 (a) 3 H, 1 P, 4 O, hydrogen, phosphorus, oxygen (b) 1 Ca, 4 H, 2 P, 8 O, calcium, hydrogen, phosphorus, oxygen (c) 4 C, 9 H, 1 Br, carbon, hydrogen, bromine (d) 3 Fe, 2 As, 8 O, iron, arsenic, oxygen (e) 3 C, 8 H, 3 O, carbon, hydrogen, oxygen
1.42 (a) 1 Sr strontium, 1 Cr chromium, 4 O oxygen (b) 1 K potassium, 1 Mn manganese, 4 O oxygen (c) 2 N nitrogen, 8 H hydrogen, 2 S sulfur, 3 O oxygen (d) 1 Mg magnesium, 1 S sulfur, 11 O oxygen, 14 H hydrogen (e) 2 Fe iron, 3 S sulfur, 12 O oxygen
1.43 (a) 6 C, 12 H, 2 O, carbon, hydrogen, oxygen (b) 1 Mg, 1 S, 14 H, 11 O, magnesium, sulfur, hydrogen, oxygen (c) 1 K, 1 Al, 2 S, 20 O, 24 H, potassium, aluminum, sulfur, oxygen, hydrogen (d) 1 Cu, 2 N, 6 O, copper, nitrogen, oxygen (e) 4 C, 10 H, 1 O, carbon, hydrogen, oxygen
1.44 (a) 6 C, 18 H (b) 4 N, 16 H, 2 S, 8 O (c) 4 Cu, 8 Cl, 16 H, 8 O
1.45 (a) 14 C, 28 H, 14 O (b) 4 N, 8 H, 2 C, 2 O (c) 15 C, 40 H, 15 O
1.46 C 2 H 6 O 2 , C 3 H 6 O These two compounds illustrate that different masses of carbon combine with the same mass of hydrogen and those masses are in small, whole number ratios. It is also true for the ratio of oxygen to a fixed mass of hydrogen.
1.47 NH 3 , N 2 H 4
The ratio of H to a fixed mass of N in the two compounds is 3 to 2. Also, the ratio of N to a fixed mass of H in the two compounds is 2 to 3.
1.48 (a) 2 N (b) 10 O (c) 2 Na (d) 1 S
1.49 (a) 12 C (b) 28 H (c) 38 O
1.50 2 S(CH 3 ) 2 , 9 O 2 , 2 SO 2 , 4 CO 2 , 6 H 2 O 2 S(CH 3 ) 2 + 9 O 2 ( g ) → 2 SO 2 ( g ) + 4 CO 2 ( g ) + 6 H 2 O( l )
1.51 1 CH 3 OH, 3 N 2 O, 1 CO 2 , 2 H 2 O, 3 N 2
CH 3 OH( l ) + 3 N 2 O( g ) → CO 2 ( g ) + 2 H 2 O( l ) + 3 N 2 ( g )
1.52 Reaction is not balanced as written.
The proper balanced reaction is:
C 3 H 8 ( l ) + 5O 2 ( g ) → 3CO 2 ( g ) + 4H 2 O( l )
1.53 The reaction is not balanced as written.
The proper balanced reaction is:
3 NO 2 ( g ) + H 2 O( l ) → 2 HNO 3 ( aq ) + NO( g )
Additional Exercises
1.54 The ratio of N in N 2 O to N in NO 2 compared to a fixed mass of O is 4 to 1.
Examine the ratio of nitrogen to a fixes mass of oxygen: 2 N 2 O:NO 2
1.55 Ethanol, C 2 H 6 O CH 3 CH 2 OH
Diethyl ether, C 2 H 2 O (CH 3 ) 2 O
1.56 A repeat experiment, using exactly the same conditions, though possibly different masses of reactants, would result in a sample having the same composition, namely 3 CO 2 to 2 N 2 molecules. Also, the molecules produced would always have the same ratio of atoms in the molecules, namely CO 2 and N 2. The law of definite composition states that elements combine in the same mass ratio, or proportion, for a given compound.
1.57 (a) A physical change will not separate the molecules into their respective constituent chemical elements.
(b) A physical change could be used to separate the molecules. By cooling the gas mixture, one gas would liquefy before the other and thus would allow one to separate the mixture.
(c) A chemical change would be required to separate the molecules into their component chemical elements.
(d) You would need to do a chemical reaction on CO 2 to reduce it to carbon and oxygen. N 2 is already an element.
(e) reducing CO 2 would result in one element in molecular form, namely oxygen (O 2 ).
2 2 m^2 = 124 ft 2 30.48 cm^ 1 m = 11.5 m^2 1 ft 100 cm
2.8 (a) ( )
3 ft 12 in. in. = 3.00 yd = 108 in. 1 yd 1 ft
^
(b) ( ) 5
1000 m 100 cm cm = 1.25 km = 1.25 10 cm 1 km 1 m
×
(c) ( )
1 m 100 cm 1 in. 1 ft ft = 3.27 mm = 0.0107 ft 1000 mm 1 m 2.54 cm 12 in.
(d) km 20.2 mile 1.609 km 1 gal (^) km = = 8. L 1 gal 1 mile 3.785 L L
^ ^
2.9 Density = mass volume
Density of the object = (^3) 365 g 22.12 cm
= 16.5 g/cm^3
The object is not composed of pure gold since the density of gold is 19.3 g/cm^3.
2.10 The density of the alloy is 12.6 g/cm^3. To determine the mass of the 0.822 ft^3 sample of the alloy, first convert the density from g/cm^3 to lb/ft^3 , then find the weight.
Density in lb/ft^3 =
3 3
12.6 g 1 lb 30.48 cm cm^ 453.6 g^ 1 ft
^ ^
= 787 lb/ft^3
Mass of sample alloy = (0.822 ft^3 ) (787 lb/ft^3 ) = 647 lb
2.11 density = mass/volume = (1.24 × 10^6 g)/(1.38 × 10^6 cm^3 ) = 0.899 g/cm^3 .
2.12 volume of one-carat diamond = 225 mg 1 g 1 cm^3 1000 mg 3.52 g
^ ^
^
^
= 0.0639 cm^3
Review Questions
2.1 Physical properties include boiling point, melting point, density, color, refractive index, mass and volume.
2.2 A chemical property describes a property that changes the chemical nature of a substance while physical properties describe properties that do not change the chemical nature of a substance. For example, boiling water does not change the chemical composition of water.
2.3 (a) Physical change. Copper does not change chemically when electricity flows through it: It remains copper.
(b) Physical change. Gallium is changes its state, not its chemical composition when it melts.
(c) Chemical change. This is an example of the Maillard reaction describing the chemical reaction of sugar molecules and amino acids.
(d) Chemical change. Wine contains ethanol which can be converted to acetic acid.
(e) Chemical change. Concrete is composed of many different substances that undergo a chemical process called hydration when water is added to it.
2.4 (a) Physical change. When corn is popped water is turned into steam by heating the corn. The pressure of the steam caused the kernel to pop open resulting in popped corn.
(b) Physical change. Generally alloys are mixtures of substances and no chemical change occurs. On occasion, a chemical change can occur during the production of an alloy. An example is when iron and carbon are mixed together to make steel. During this process compounds of iron and carbon such as cementite, Fe 3 C , are produced.
(c) Physical change. During the whipping process air is mixed with cream to increase its volume.
(d) Physical change. During the production of butter fat molecules aggregate, due to the agitation of whipping, and separate from the water.
(e) Physical change. The aluminum is not chemically altered during recycling.
2.5 Extensive properties, such as volume, and size, are properties that depend on the amount of substance or mass of substance while intensive properties, such as density, are not dependent on the amount of substance. The density of a milliliter of water is the same as the density of a liter of water at the same temperature.
2.6 (a) Extensive Obviously, mass is a mass dependent property.
(b) Intensive The boiling point of a substance is the same for a mL as it is for a L of the compound so it is mass independent.
(c) Intensive The color of a substance does not change when you change the amount of substance.
(d) Intensive The physical state, gas, liquid, or solid, depends on temperature and pressure but not on the mass of the substance.
2.7 (a) Intensive The melting point of 1.0 g of water is that same as 100.0 thus melting point is not mass Dependent.
(b) Intensive The density of 1.0 g of water is the same as 100.0 g if both samples are at the same Temperature. Thus, density is not dependent on the mass of substance.
(c) Extensive The volume occupied by a substance is dependent on the mass of substance.
(d) Extensive Surface area depends on the amount of substance. It also depends on the nature of the Substance. A bar of metal has a smaller surface area than that of the same bar ground Into fine particles.
2.8 (a) Gas Temperature, density, volume, viscosity
(b) Liquid Temperature, density, volume, viscosity
(c) Solid Temperature, density, volume
2.9 (a) Hydrogen is a gas at room temperature
(b) Aluminum is a solid at room temperature
(c) Nitrogen is a gas at room temperature
(d) Mercury is a liquid at room temperature
1 h 3600 s
To convert 3.84 hours to seconds multiply 3.84 hours by: 3600 s 1 h
2.23 Four significant figures would be correct because the conversion factor contains exact values. The measured value determines the number of significant figures.
2.24 d = m v
: d = density; m = mass; v = volume
2.25 10.5 g silver = 1 cm^3 silver
3
10.5 g Ag 1 cm
and 1 cm^3 10.5 g Ag
Review Problems
2.26 (a) 100 cm (b) 0.001 km (c) 1 × 10 −^12 m
(d) 10 dm (e) 1000 g (f) 100 cg
2.27 (a) 10–9^ (b) 10–6^ (c) 10^3 (d) 10^6 (e) 10–3^ (f) 0.
2.28 (a) TC = (TF − 32)( 59 )
TC = (135 − 32)( 59 )
TC = (103)( 59 ) = 57.2°C
(b) TC = (61 − 32)( 59 )
TC = (29)( 59 ) = 16°C
(c) TF = ( 95 )TC + 32
TF = ( 95 )( −3.6) + 32
TF = − 6.48 + 32 = 26°F
(d) TF = ( 95 )(15) + 32
TF = 27 + 32 = 59°F
(e) TK = TC + 273.
236 = TC + 273.
−37°C = TC
(f) TK = 39 + 273.
TK = 312 K
2.29 (a) tC =
5 C
9 F
tF – 32 °F) =
5 C
9 F
(96 °F – 32 °F) = 36 °C
(b) tC =
5 C
9 F
(tF – 32 °F) =
5 C
9 F
(–6 °F –32 °F) = –21 °C
(c) tF =
9 F
5 C
(tC) + 32 °F =
9 F
5 C
(–55 °C) +32 °F = –67 °F
(d) tC = (TK – 273 K)
1 C
1 K
= (273 K – 273 K)
1 C
1 K
= 0 °C
(e) tC = (TK – 273 K)
1 C
1 K
= (299 K – 273 K)
1 C
1 K
= 26 °C
(f) TK = (tC + 273 °C)
1 K
1 C
= (40 °C + 273 °C)
1 K
1 C
= 313 K
2.30 C ( F ) ( )
5 C 5 C
t t 32 F 104.5 F 32 F 40.3 C 9 F (^9) F
° ^ ^
° ^
This dog has a fever; the temperature is out of normal canine range.
2.31 Convert –96 °F to tc:
tC =
5 C
9 F
(tF – 32 °F) =
5 C
9 F
(–96 °F – 32 °F) = –71 °C
2.32 Range in Kelvins:
6 K= 10 MK 1 10 K = 1.0 107 K 1 MK
×
×
6 K= 25 MK 1 10 K = 2.5 107 K 1 MK
×
×
Range in degrees Celsius:
tC = (TK – 273 K)
1 C
1 K
= (1.0 × 10^7 K – 273 K)
1 C
1 K
≈ 1.0 × 10^7 °C
tC = (TK – 273 K)
1 C
1 K
= (2.5 × 10^7 K – 273 K)
1 C
1 K
≈ 2.5 × 10^7 °C
Range in degrees Fahrenheit:
tF =
9 F
5 C
(°C) + 32 °F =
9 F
5 C
(1.0 × 10^7 °C) + 32 °F ≈ 1.8 × 10^7 °F
tF =
9 F
5 C
(°C) + 32 °F =
9 F
5 C
(2.5 × 10^7 °C) + 32 °F ≈ 4.5 × 10^7 °F
2.33 Convert 111 K to tC:
tC = (TK – 273 K)
1 C
1 K
= (111 K – 273K)
1 C
1 K
= –162 °C
Convert –162 °C to tF
tF =
9 F
5 C
(tC) + 32 °F =
9 F
5 C
(–162°C) +32 °F = –260 °F
2.34 tC = (TK – 273 K)
1 C
1 K
= (4 K – 273 K)
1 C
1 K
= –269 °C
2.35 Convert 6000 K to tC:
(f) nm^2 = (238 mm^2 )
2 2 9
1 m 1 nm 1000 mm 10 − m
^
^
= 2.38 × 10^14 nm^2
2.42 (a) ( )
2.54 cm cm = 27 in. = 69 cm 1 in.
(b) ( )
1 kg kg = 7.6 lb = 3.4 kg 2.205 lb
(c) ( )
946.4 mL mL = 2.7 qt = 2600 mL 1 qt
(d) ( )
29.6 mL mL = 12 oz = 360 mL 1 oz
(e) ( ) 2
1.609 km km/hr = 65 mi/hr = 1.0 10 km/hr 1 mi
×
(f) ( )
1.609 km km = 68.0 mi = 109 km 1 mi
2.43 (a) ( )
1 qt qt = 250 mL = 0.26 qt 946.4 mL
(b) ( )
12 in. 2.54 cm 1 m m = 3.0 ft = 0.91 m 1 ft 1 in. 100 cm
(c) ( )
2.205 lb lb = 1.62 kg = 3.57 lb 1 kg
(d) ( )
1000 mL 1 oz oz = 1.75 L = 59.1 oz 1 L 29.6 mL
(e) ( )
1 mi mi/hr = 35 km/hr = 22 mi/hr 1.609 km
(f) ( )
1 mi mi = 80.0 km = 49.7 mi 1.609 km
2.44 (a) ( )
2 cm 2 9.8 ft 2 30.48 cm 9100 cm^2 1 ft
= ^ =
(b) ( )
2 km^2 546 mi^2 1.609 km 1410 km^2 1 mi
= ^ =
(c) ( )
3 cm^3 176 ft 3 30.48 cm 4.98 106 cm^3 1 ft
= = ×
2.45 (a) m^2 = (2.4 yd^2 )
2 0.9144 m 1 yd
= 2.0 m^2
(b) mm^2 = (8.3 in^2 )
2 2.54 cm 1 in
2 10 mm 1 cm
= 5400 mm^2
(c) L = (9.1 ft^3 )
3 3 3 3
1 yd 0.9144 m 100 cm 1 mL 1 L 3ft 1 yd 1 m (^) 1 cm 1000 mL
^ ^ ^
^ ^ ^
= 260 L
2.46 (^) ( ) 3 946.35 mL mL = 4.2 qt = 4.0 10 mL 1 qt
×
(stomach volume)
4.0 × 10^3 mL ÷ 0.9 mL = 4,000 pistachios (don’t try this at home)
2.47 To determine if 50 eggs will fit into 4.2 quarts, calculate the volume of fifty eggs, then compare the answer to the volume of the stomach:
Volume of 50 eggs = (50 eggs) 53 mL 1 L 1.057 qt 1 egg 1000 mL 1 L
^ ^ ^
= 2.8 qt
2.8 qt < 4.2 qt Luke can eat 50 eggs.
m 200 mi 5280 ft 30.48 cm 1 10 2 m 1 hr 1 min m 90 s 1 hr 1 mi 1 ft 1 cm 60 min 60 s s
^ ×^ −
^
2.49 km/h = 2435 ft 1 yd 0.9144 m 1 km 3600 s s 3 ft 1 yd 1000 m 1 h
^
= 2672 km/h
mi 2230 ft 1 mi 60 s 60 min mi 1520 h 1 s 5280 ft 1 min 1 hr hr
2.51 tons/day =
3 5 3
ft 62.4 lb 1 ton 3600 s 24 h 2.05 10 s (^) 1 ft 2000 lb 1 h 1 d
×
= 5.53 × 10^8 tons/day
2.52 1 light year = 365.25 d 24 h 3600 s 3.00 108 m 1 y 1 y 1 d 1 h 1 s
^ ×
^
^ ^ ^
= 9.47 × 10^15 m
miles = 8.7 light years 9.47 1015 m 1 km 1 mi 1 light year 1000 m 1.609 km
×
= 5.1 × 10^13 mi
2.53 There are 360 degrees of latitude around the circumference of the earth.
statute miles = 60 nautical miles 1.151 statute miles 360 degree latitude 1 degree latitude 1 nautical mile
^ ^
= 2.49 × 10^4 statute miles
2.54 meters = 6033.5 fathoms 6 ft 1 yd 0.9144 m 1 fathom 3 ft 1 yd
^
= 11,034 m
2.55 pounds/in^2 = 11,034 m 14.7 lb/in^2 10 m
= 16,200 lb/in^2
tons/in^2 = 162,000 lb/in^2
1 ton 2000 lb
= 8.10 ton/in
2
2.56 density = mass/ volume = 27.7 g/34.8 mL = 0.796 g/mL
2.67 Hausberg Tarn 4350 m 1 yd 0.9144 m
= 4760 yd
Mount Kenya 4600 m 1 yd 0.9144 m
= 5000 yd 4700 m 1 yd 0.9144 m
= 5100 yd
Temperature ∆tF =
9 F
5 C
(tC) =
9 F
5 C
(4.0 °C) = 7.2 °F
2.68 If the density is in metric tons… 3 8 3 3
4.93 mL 1 cm 1 10 tons 1000 kg 1 10 g g 1 teaspoon 1 tsp 1 mL (^) 1 cm 1 ton 1 kg
^ ^ × ^ ^ ×
^ ^ ^ ^ ^
= 4.93 × 10^14 g
If the density is in English tons…
3 8 3
4.93 mL 1 cm 1 10 tons 2000 lbs 453.59 g g 1 teaspoon 1 tsp 1 mL (^) 1 cm 1 ton 1 lb
^ ^ ×
^ ^ ^ ^ ^
= 4.47 × 10^14 g
2.69 1 light year = 3.00 × 10^8 m/s 3600 s 24 h 365 d 1 h 1 d 1 y
^ ^
^
= 9.46 × 10^15 m
Distance to Arcturus:
days = 3.50 × 10^14 km (^8)
1000 m 1 s 1 h 1 d 1 km (^) 3.00 10 m 3600 s 24 h
^
^
×
= 1.35 × 10^4 d
light years = 3.50 × 10^14 km 15
1000 m 1 light year 1 km (^) 9.46 10 m
^ ^
×
= 37.0 light years
2.70 (a) In order to determine the volume of the pycnometer, we need to determine the volume of the water that fills it. We will do this using the mass of the water and its density. mass of water = mass of filled pycnometer – mass of empty pycnometer = 32.954 g – 23.426 g = 9.528 g
volume = (9.528 g) 1 mL 9.556 mL 0.99704 g
(b) We know the volume of chloroform from part (a). The mass of chloroform is determined in the same way that we determined the mass of water. mass of chloroform = mass of filled pycnometer – mass of empty pycnometer = 37.540 g – 23.426 g = 14.114 g
Density of chloroform = 14.114 g 1.477 g/mL 9.556 mL
2.71 For the message to get to the moon:
s = (^) ( ) (^8) 1.609 km 1000 m 1 s 239, 000 miles 1.28 s 1 mile 1 km (^) 3.00 10 m
^
^ =
×
The reply would take the same amount of time, so the total time would be: 1.28 s × 2 = 2.56 s
2.72 (a) $5.75 = 30 min
30 min min
30 min 1 min $5.75 $0.
(b) $ = 60 min $0. 2 hr 15 min $25. hr min
×^ +^ =
^ ^ ^
(c) min = (^) ( ) 1 min $29.75 157 min $0.
3 sea water (^33)
1.025 g 1 lb 30.48 cm (^) lb d 64. cm 453.59 g^ 1 ft^ ft
^
( )
3 ft^3 3785 tons 2000 lbs^ 1 ft 1.183 105 ft^3 1 ton 64.0 lb
^ ^
= = ×
3 3 3
1 in 0.00011 lbs 453.59 g g 2510 cm 7.6 g 2.54 cm (^) 1 in 1 lb
2.75 The experimental density most closely matches the known density of methanol (0.7914 g/mL). The density of ethanol is 0.7893 g/mL. Melting point and boiling point could also distinguish these two alcohols, but not color.
2.76 g/mL = 69.22 lb/ft^3
(^3 ) 453.59 g 1 ft 1 cm 1 lb 30.48 cm 1 mL
^
^
^
= 1.1088g/mL
Since the density closely matches the known value, we conclude that this is an authentic sample of ethylene glycol.
2.77 tc = (TK – 273 K)
1 C
1 K
= (5800 K – 273 K)
1 C
1 K
= 5500 ºC
2.78 We solve by combining two equations:
tF =
9 C
5 F
(tC) + 32 °F
tF = tC If tF = tC, we can use the same variable for both temperatures:
tC =
9 C
5 F
(tC) + 32 °F
c
t 5
9 C
5 F
(tC) + 32 °F
c
t 5
tc = 32
= –40, therefore the answer is –40 °C.
2.79 Both the Rankine and the Kelvin scales have the same temperature at absolute zero: 0 R = 0 K. For converting from tF to TR:
tC =
5 C
9 F
(tF – 32 °F) and tC = (TK – 273 K)
1 C
1 K
2.82 Volume of cylindrical metal bar = π × r^2 × h = π × (^) ( )
2 1 0.828 cm 2
^
× 2.12 cm = 1.14 cm^3
Density of cylindrical metal bar = mass volume
9.276 g 1.14 cm
= 8.14 g/mL
Density in lb/ft^3 =
3 3
8.14 g 1 lb 1 mL 30.48 cm 1 mL 453.59 g (^) 1 cm 1 ft
^ ^
^ ^
= 508 lb/ft^3
2.83 Volume of diamond = 2.25 carat 200 mg 1 g 1 mL 1 carat 1000 mg 3.51 g
^ ^ ^
= 0.128 mL
2.84 Concentration of lead in blood in μg/dL = 3.2 10 4 g 106 g 1 L L 1 g 10 dL
(^) × − (^) μ (^) (^)
= 32 μg/dL
This person is in danger of exhibiting the effects of lead poisoning since the 32 μg/dL is above the threshold of 10 μg of lead/dL.
2.85 Radius of ball bearing = 2.000 mm × (1/2) = 1.000 mm Volume of ball bearing= 4/3 × π × r^3 = 4/3 × π × (1.000 mm)^3 = 4.189 mm^3 Radius of ball bearing + gold = 1.000 mm + 0.500 mm = 1.500 mm Volume of ball bearing + gold = 4/3 × π × r^3 = 4/3 × π × (1.500 mm)^3 = 14.14 mm^3 Volume of gold = (volume of ball bearing + gold) – (volume of ball bearing) = 14.14 mm^3 – 4.189 mm^3 = 9.95 mm^3
Mass of gold = 9.95 mm^3
1 cm 3 19.31 g 10 mm 1 cm
= 0.192 g
2.86 The question is asking to calculate the number of mile/gallon/person for a jet airliner and a car. The answer is:
Rate of fuel consumption =
1 mile 5.0 gallons of jet fuel 568 people
(^) = 3.5 × 10–4 (^) mile/gallon/person
Rate of fuel consumption for car =
21.5 miles 1 gallon 2 people
(^) = 11 mile/gallon/person
But a more insightful answer would be to calculate the number of gallons/person/mile which would give the number of gallons each person uses per mile.
Rate of fuel consumption =
5.0 gallons of jet fuel 568 people 1 mile
(^) = 0.0088 gallon/person/mile
Rate of fuel consumption for car =
1 gallon 2 people 21.5 miles
(^) = 0.75 gallon/person/mile
This would indicate that the jet airliner has better fuel consumption.
Pounds of jet fuel = 3470 miles 5.0 gallons 3.785 L 1000 mL 0.803 g 1 lb 1 mile 1 gallon 1 L 1 mL 453 g
^ ^ ^
^ ^ ^
= 1.16 × 10^5 lb
Practice Exercises
3.1 24094 Pu (^) , 94 electrons
The bottom number is the atomic number, found on the periodic table (number of protons). The top number is the mass number (sum of the number of protons and the number of neutrons). Since it is a neutral atom, it has 94 electrons.
3.2 1735 Cl contains 17 protons, 17 electrons, and 18 neutrons.
3.3 We can discard the 17 since the 17 tells the number of protons which is information that the symbol "Cl" also provides. In addition, the number of protons equals the number of electrons in a neutral atom, so the symbol "Cl" also indicates the number of electrons. The 35 is necessary to state which isotope of chlorine is in question and therefore the number of neutrons in the atom.
3.4 2.24845 × 12 u = 26.9814 u
3.5 Copper is 63.546 u ÷12 u = 5.2955 times as heavy as carbon
3.6 (0.199 × 10.0129 u) + (0.801 × 11.0093 u) = 10.8 u
3.7 (0.90483 x 19.992 u) + (0.00271 x 20.994 u) + (0.09253 x 21.991 u) = 20.2 u
3.8 The number of protons is equal to the atomic number of an element. The number of electrons is equal to the atomic number for a neutral atom. If the atom has a positive charge, the number of electrons is determined by subtracting the charge from the atomic number. If the atom has a negative charge the number of electrons is determined by adding the charge to the atomic number. (a) Fe has 26 protons, and 26 electrons (b) Fe3+^ has 26 protons, and 23 electrons (c) N3-^ has 7 protons, and 10 electrons (d) N has 7 protons, and 7 electrons.
3.9 The number of protons is equal to the atomic number of an element. The number of electrons is equal to the atomic number for a neutral atom. If the atom has a positive charge, the number of electrons is determined by subtracting the charge from the atomic number. If the atom has a negative charge the number of electrons is determined by adding the charge to the atomic number. (a) O has 8 protons, and 8 electrons (b) O2-^ has 8 protons, and 10 electrons (c) Al3+^ had 13 protons, and 10 electrons (d) Al has 13 protons, and 13 electrons
3.10 (a) NaF (b) Na 2 O (c) MgF 2 (d) Al 4 C 3
3.11 (a) Ca 3 N 2 (b) AlBr 3 (c) K 2 S (d) CsCl
3.12 (a) CrCl 3 and CrCl 2 , Cr 2 O 3 and CrO
(b) CuCl, CuCl 2 , Cu 2 O and CuO
3.13 (a) Au 2 S and Au 2 S 3 , Au 3 N and AuN
(b) TiS, Ti 2 S 3 and TiS 2 ; Ti 3 N 2 ,TiN and Ti 3 N 4
3.14 (a) KC 2 H 3 O 2 (b) Sr(NO 3 ) 2 (c) Fe(C 2 H 3 O 2 ) 3
