(Solution Manual) Chemistry The Molecular Nature of Matter

Jespersen, Hyslop, 8

Chemistry

Solution Manual

8th Edition

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Author(s): Jespersen, Hyslop

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Chapter 0 0-1 Chapter Zero A Very Brief History of Chemistry Practice Exercises 0.1 This answer is student dependent. For example Chapter 1: Measurements – Careful observations on the laboratory scale Chapter 2: Elements – Atomic theory Chapter 3: The Mole – Careful obser vations on the laboratory scale Chapter 4: Reactions – Atomic theory Chapter 5: Reduction Oxidation Reactions – Atomic theory Chapter 6: Thermochemistry – Energy changes Chapter 7: Quantum Chemistry – Atomic theory Chapter 8: Bonding I – Atomic theory Chap ter 9: Bonding II – Geometric shapes of molecules and Atomic theory Chapter 10: Gases – Careful observations on the laboratory scale Chapter 11: Solids and liquids - Careful observations on the laboratory scale Chapter 12: Solutions – Careful observations on the laboratory scale and Energy changes Chapter 13: Kinetics – Careful observations on the laboratory scale and Energy changes Chapter 14: Equilibrium – Careful observations on the laboratory scale and Energy changes Chapter 15: Acids and Bases I – Care ful observations on the laboratory scale and Geometric shapes of molecules Chapter 16: Acids and Bases II – Careful observations on the laboratory scale and Geometric shapes of molecules Chapter 17: Solubility – Careful observations on the laboratory scale and Energy changes Chapter 18: Thermodynamics – Energy changes Chapter 19: Electrochemistry – Energy changes Chapter 20: Nuclear chemistry – Atomic theory and Energy changes Chapter 21: Metal complexes – Atomic theory and Geometric shapes of molecules Chapter 22: Organic Chemistry – Geometric shapes of molecules and Energy changes 0.2 Nucleosynthesis would occur at temperatures around 1 billion degrees and a high density of nucleons . 0.3 Only light elements were synthesized during the big bang because the temperat ure was too high for the heavier elements to form. 0.4 Core and enriched layers or nuclei are needed for nucleosynthesis in stars because the enriched layers are the lighter layers , which react to form the heavier nuclei. The heavier nuclei are more dense an d move to the center of the star. Experimental evidence for the concept of nucleosynthesis lies in the layers of the stars and that stars have different compositions and that the most abundant elements are the lightest ones. Iron is the heaviest element in stars since the nucleosynthesis of iron absorbs heat and causes the collapse of the red giant and a supernova, thus preventing larger nuclei from forming . Supernovas provide for the synthesis of heavier elements because the extremely high temperature of the supernova makes the atomic collisions Chapter 0 0-2 strong enough for the nuclei to combine. 94240Pu , 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. 0.5 Most of the hydrogen that exists today was formed within the first three minutes of the formation of the universe. 0.6 Such elements are formed i n supe rnovas and collisions between two neutron stars. 0.7 Supernovas allow atomic nuclei to capture free neutrons. 0.8 Pu has an atomic number of 94; i.e., 94 protons. A Pu isotope that has 146 neutrons thus has atomic mass of 94 + 146 = 250. Assuming tha t the atom is neutral, it has the name number of electrons as protons; i.e., 94. 0.9 35 17Cl contains 17 protons, 17 electrons, and 18 neutrons. We can discard the subscript 17 on the symbol since the 17 tells the number of protons which is infor mation 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 i n question and therefore the number of neutrons in the atom. 0.10 2.24845 × 12 u = 26.9814 u Copper is 63.546 u 12 u = 5.2955 times as heavy as carbon Any other atom could have been used as the standard for atomic mass, for example oxygen used to be the standa rd, and on e atomic mass unit was 1/16 the mass of oxygen, but that was before they took into account the isotopes of oxygen. Scientists needed to use one isotope of an element. (0.199 × 10.0129 u) + (0.801 × 11.0093 u) = 10.8 u (0.90483 × 19.992 u) + (0.00271 × 20.994 u) + (0.09253 × 21.991 u) = 20.18 u 0.11 Average mass of copper: 63.546 u. Mass of 12C = 12 u. Thus, 63.546/12 = 5.3 times as heavy. 0.12 Based on a need to avoid confusion between disciplines, hydrogen and oxygen are no longer used as the standard for the definition of the size of the atomic mass unit. Wikipedia has a brief overview: https://en.wikipedia.org/wiki/Dalton_(unit)#History . If scientists were to choose 1/2 of a C -12 atom s as the baseline measure, that would match the data for naturally occurring water: 18 u. 0.13 (0.199 × 10.0129) u + (0.801 × 11.0093) u = 10.8110 u. 0.14 (0.90483 × 19.992) u + (0.00271 × 20.994) u + (0.09253 × 2 1.991 ) u = 20.181 u Review Questions 0.1 This answer will be student dependent. 0.2 #2 In any sample of a pure element, all the atoms are identical in mass . #5 Atoms are indestructible . 0.3 This answer will be student dependent. 0.4 This answer will be student dependent. The cubic shape of the c rystals hint s that the atoms are arranged in a cubic pattern. Chapter 0 0-3 0.5 This answer will be student dependent. Some answers might be the shape of salt crystals or the sharp edges of quartz. 0.6 If a reaction gives off heat, the substances must have less energy than whe n the reaction s tarted. 0.7 This answer will be student dependent. Some answers might be vinegar mixing with sodium bicarbonate. The atoms would have to give off heat to lose energy. 0.8 This answer will be student dependent. An answer might be an enzyme. 0.9 (a) Hy drogen, helium, an d lithium; (b) Carbon, argon, oxygen, silicon, and elements up to iron; (c) elements heavier than iron 0.10 The formation of iron causes the core of a red giant to cool and then the star collapses creating a very high density core which heats up and explodes cr eating a supernova. 0.11 Hydrogen 0.12 The earth's core is made of iron and nickel, they are denser than aluminum and more abundant than lead. 0.13 The different melting points, densities, and crystallization of the elements. The separation of elements and minerals is st ill occurring in the form of earthquakes and volcanic eruptions. 0.14 Elements that result from atomic fission of uranium include bari um, iodine, caesium, xenon, and strontium . Hydrogen molecules and helium atoms move so quickly in the atmosphere that they ’re lost to space more quickly than other elements. 0.15 A balance to measure mass and graduated cylinders 0.16 Conservation of mass derives fro m the postulate that atoms are not destroyed in normal chemical reactions. The Law of Definite Proportions derives from the not ion that compound substances are always composed of the same types and numbers of atoms of the various elements in the compound. 0.17 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. 0.18 Protons, 11p+ , +1 charge Electron, 00e- , –1 charge Neutron, 01n , no charge 0.19 Nearly all of the mass is located in the nucleus, because this is the port ion of the atom where the protons and the neutrons are located. 0.20 A nucleon is a subatomic particle found in the atomic nucleus. W e have studied neutrons and protons. 0.21 The charge -to-mass ratio of the electron was determined b y J.J. Thomson using a cathode r ay tube. Electrons were generated and passed through the tube to a detection screen. The ratio of e/m was determined by applying an el ectric and magnetic field, at 90° to each other, to the electrons and observing to what degree they were bend when the fie lds were applied. Chapter 0 0-4 0.22 Millikan generated a fine mist of oil in a container. Electrons were then generated in the container and attach ed themselves to the oil drops. He applied an electric field to the oil drops and observed their motion on the electric field. From his observations of the effect of an electric field on the oil drops he was able to determine the charge of a single electro n. Using the ratio, e/m, of an electron determined by Thompson, he was able to determine the mass of an electron. 0.23 The proton w as discovered by passing hydrogen gas through a mass spectrometer. The hydrogen was ionized leaving the nucleus. Three different species were observed corresponding to the mass of a proton, a proton with one neutron, and a proton with two neutrons. 0.24 Rutherf ord passed alpha particles through a thin sheet of gold metal foil and observed their deflection to determine the existence and n ature of the nucleus. 0.25 The mass of the electrons is small compared to the mass of protons and neutrons in the atom . 0.26 The atomic n umber is equal to the number of protons in the nucleus of the atom, and the mass number is the sum of the number of neutrons and the number of protons. The atomic number (symbol Z) is designated by a subscript preceding the chemical symbol and the mass num ber (symbol A) is a superscript preceding the chemical symbol. 0.27 (a) mass number (b) atomic number 0.28 (a) 131 53I (b) 90 38Sr (c) 137 55Ce (d) 18 9F 0.29 The statement does not indicate w hich isotop e corresponds to the given mass, and 35.453 u is the weighted average of all of the isotopes of chlorine. 0.30 Since the average atomic mass of the element depends on the weighted average of all of the isotopes, we do not know which isotopes make up silver. Review Problems 0.31 Using the ratio of the number of atoms of O and X and the atomic mass of O, we can compare that to th e ratio of the masses of O and X to calculate the atomic mass of X: 1.125 g X 1.000 g O=2 atoms X 3 atoms Oæ èçö ø÷uX 15.9994u Oæ èçö ø÷ u X = 27.00 u The element is alum inum. 0.32 Since we know that the formula is M3N, we know that one third of the total mass due to the metal atom constitutes the ma ss that may be compared to the nitrogen. Hence we have 1.486 g M  3 = 0.4953 g M and 1.00 g assigned to the amount of N in the c ompound. Then it is necessary to realize that the ratio 1.00 g N  14.0067 for nitrogen (we use the weighted average atomic mas s for N since no specific isotope was designated) is equal to the ratio 0.4953 g N  X, where X equals the relative atomic mass o f the metal. 1.000 g N 0.4953 6.938 u14.0067 u N==  The metal is lithium, Li 0.33 (a) 226Ra has 8 8 protons, 88 electrons, and 138 neutrons Chapter 0 0-5 (b) 206Pb has 82 protons, 82 electrons, and 124 neutrons (c) 14C has 6 protons, 6 electrons, and 8 neutrons (d) 23Na has 11 protons, 11 electrons, and 12 neutrons electrons protons neutrons (a) Cesium -137 55 55 82 (b) 238U 92 92 146 (c) Tellurium -125 52 52 73 (d) 285Cn 112 112 173 0.35 131I has 53 protons, 53 electrons, and 78 neutrons. 0.36 99Tc has 43 protons, 36 electrons in the +7 state, and 56 neutrons 0.37 Since we know that the formula is CH 4, we know that one fourth of the total mass due to the hydrogen atom constitutes the mass that may be compared to the carbon. Hence we have 0.33597 g H  4 = 0.083993 g H and 1.00 g assigned to the amount of C -12 in the compound. Then it is necessary to realize that the ratio 1.00 g C  12 for carbon is equal to the ratio 0.083993 g H  x, where x equals the relative atomic mass of hydrogen. 1.000 g C 12 u Cæ èçö ø÷ = 0.083993 g H xæ èçö ø÷ =1.008 u 0.38 Since we know that the formula is CCl 4, we know that one fourth of the total mass due to the chlorine atom constitut es the mass that may be compared to the carbon. Hence we have 11.818 g Cl  4 = 2.9545 g Cl and 1.00 g assigned to the amount of C-12 in the compound. Then it is necessary to realize that the ratio 1.00 g C  12 for carbon is equal to the ratio 2.9545 g Cl  x, where x equals the relative atomic mass of hydrogen. 1.000 g C 12 u Cæ èçö ø÷ = 2.9545 g Cl xæ èçö ø÷ =35.454 u 0.39 Regardless of the definition, the ratio of the mass of hydrogen to that of carbon would be the same. If C–12 were assigned a mass of 24 (twice its accepted value), then hydrogen would also have a mass twice its current value, or 2.01588 u. 0.40 Taking the mass ratio of 109Ag to 12C and multiplying it by 12 we see that the mass of Ag -109 is 108.90 u. 0.41 (0.6917 × 62.9 296 u) + (0.3083 × 64.9278 u) = 63.55 u 0.42 (0.7899 × 23.9850 u) + (0.1000 × 24.9858 u) + (0.1101 × 25.9826 u) = 24.31 u Additional Exercises 0.43 The ratio of N in N 2O 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 2O:NO 2 0.44 (a) Chromium (b) 52 24Cr (c) 28 neutrons (d) 24 electrons Chapter 0 0-6 (e) 51.99964.33312.000= times heavier than a C -12 atom 0.45 (0.0580 × 53.9396 u) + (0.91 72 × 55.9349 u) + (0.0220 × 56.9354 u) + (0.0028 × 57.9333 u) = 55.85 u 0.46 The protons are red. The neutrons are gray. The el ectrons are the gold cloud. There a three protons in the nucleus so the element is lithium, Li. There are three protons and four neut rons on the nucleus so the mass number is 7. 37Li Chapter 1 1-1 Chapter One Scientific Measurements Practice Exercises 1.1 The scientific method is an iterative process of gathering information through making observations and colle cting data and then formulating explanations that lead to a conclusion. 1.2 (a) element (b) mixt ure, homogeneous (c) compound (d) mixture, heterogeneous (e) element 1.3 (a) chem ical change (b) physical change (c) physical change (d) physical change 1.4 (a) intensive (b) extensive (c) intensive (d) extensive 1.5 V = 4 3pr3 , the SI unit for rad ius, r, is meters, the numbers 4 3 and  do not have units. Therefore, the SI unit for volume is meter3 or m3. 1.6 Force equals mass × acceleration ( F = ma ), and acceleration equals change in velocity divided by change in time ( a = change in v change in t ), and velocity equals distance divided by time ( v = d t ). Put the equations together: F = m change in v change in tæ èçö ø÷ F= m change in d t change in tæ èç ç ççö ø÷ ÷ ÷÷ = m change in d change in t2æ èççö ø÷÷ The unit for mass is kilogram ( kg); the unit for distance is meter (m) and the unit for time is second (s). Substitute the units into the equation above: Unit for force in SI base units = kg m s2æ èçö ø÷ or kg m s–2 1.7 tF=9 °F 5 °Cæ èçö ø÷tC + 32 °F = 9 °F 5 °Cæ èçö ø÷355 °C ( ) + 32 ° F =671 °F 1.8 tC=tF-32 °F ( )5 °C 9 °Fæ èçö ø÷=55 oF-32 °F ( )5 °C 9 °Fæ èçö ø÷=13 °C To convert from °F to K we first convert to °C. tC=tF-32 °F ( )5 °C 9 °Fæ èçö ø÷=68 °F-32 °F ( )5 °C 9 °Fæ èçö ø÷=20 °C TK = (273 °C + tC) 1 K 1 °Cæ èçö ø÷ = (273 °C + 20 °C) 1 K 1 °Cæ èçö ø÷ = 293 K 1.9 (a) 21.0233 g + 21.0 g = 42.0233 g: rounded correctly to 42.0 g (b) 10.0324 g / 11. 7 mL = 0.8574 g / mL: rounded correctly to 0.857 g/ mL Chapter 1 1-2 (c) 14.25 cm ´ 12.334 cm 2.223 cm - 1.04 cm ( ) = 148.57 cm: rounded correctly to 149 cm 1.10 (a) 54.183 g – 0.0278 g = 54.155 g (b) 10.0 g + 1.03 g + 0.243 g = 11.3 g (rounded after adding) (c) 43.4 in × 1 ft 12 in.æ èçö ø÷ = 3.62 ft (1 and 12 are exact numbers) (d) 1.03 m ´ 2.074 m ´ 2.9 m 12.46 m +4.778 m = 0.36 m2 (e) 32.6 mm × 117.5 mm = 383 mm2 (32.6 is certain to three digits) (f) 15.7 mL + 25 mL = 41 mL (25 is certain to the ones decimal place) 1.11 m2 = 124 ft2( )30.48 cm 1 ftæ èçö ø÷21 m 100 cmæ èçö ø÷2 = 11.5 m2 1.12 (a) in.3 = 3.00 yd3( )3 ft 1 ydæ èçö ø÷312 in. 1 ftæ èçö ø÷3 = 140,000 in.3= 1.40 ´ 105 in3 (b) cm = 1.25 km ( )1000 m 1 kmæ èçö ø÷100 cm 1 mæ èçö ø÷ = 1.25 ´ 105 cm (c) g = 3.27 oz ( )28.35 g 1 ozæ èçö ø÷ = 92.7 g (d) km L = 20.2 mile 1 galæ èçö ø÷1.609 km 1 mileæ èçö ø÷1 gal 3.785 Læ èçö ø÷ = 8.59 km L 1.13 Density = mass volume=0.244 g 15.0 mL = 0.0163 g/mL = 0.0163 g cm–3 1.14 Density = mass volume=mass Vf-Vi = 0.547 g (5.95 mL – 5.70 mL) =2.2 g mL–1 = 2.2 g cm–3 1.15 Density = mass volume Density of the object = 365 g 22.12 cm3 = 16.5 g/cm3 The object is not composed of pure gold since the density of gold is 19.3 g/cm3. 1.16 The density of the alloy is 12.6 g/cm3. To determine the mass of the 0.822 ft3 sample of the alloy, first convert the density from g/cm3 to lb/ft3, then find the weight. Density in lb/ft3 = 12.6 g cm31 lb 453.6 gæ èçö ø÷30.48 cm 1 ftæ èçö ø÷3 = 787 lb/ft3 Mass of sample alloy = (0.822 ft3) (787 lb/ft3) = 647 lb 1.17 specific gravity = density of substance density of water 1.090 = density of wine 62.4 lb ft3 Chapter 1 1-3 density of wine = 1.090  62.4 lb ft3 = 68.02 lb ft3 mass of wine = 79 gallons  density of wine  ft3 7.481 gallons = 79 gallons  68.02 lb 1 ft3´ft3 7.481 gallons = 720 lb 1.18 specific gravity = density of substance density of water density o f water = 1.00 g 1 mL1 oz 28.3495 gæ èçö ø÷29.574 mL 1 liquid ozæ èçö ø÷ = 1.043 oz/liquid oz specific gravity of urine = 1.008 oz/liquid oz 1.043 oz/liquid oz = 0.966 The specific gravity of urine is below the normal range. Review Questions 1.1 This answer will be student dependent. 1.2 Observation, testing an d 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 , and therefore, disproves the theory. 1.5 Matter has mass and occupies space. All items , except (b) an idea, in the question are examples of matter. 1.6 (a) An element is a pure substance that cannot be deco mposed 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 mixtur e 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 r egions of the mixture. (g) A solution is a homogeneous mixture. 1.7 Changing a compound into its element is a chemical change. 1.8 (a) F (b) Se (c) Ni (d) Ar (e) Li (f) P (g) I (h) Ga (i) Hg (j) Mn Chapter 1 1-4 1.9 (a) sodium (b) zinc (c) silicon (d) tin (e) magnes ium (f) tungsten (g) cobalt (h) aluminum (i) oxygen (j) nitrogen 1.10 (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 2O. 1.11 (a) Diagrams (a) and (d) contain pure elements (b) Diagram (c) contains a compound (c) Diagram (a) and (b) contain diatomic molecules 1.12 Two everyday examples of a physical change are ice melting and water freezing. Two everyday examples of a chemical change are milk going sour and digestion of food. A physical change involves a change of state (no formation of a new substance) whereas a chemical change involves a change in chemical bonds (formation of a new substance). 1.13 The reaction of calcium metal with water is a chemical change resulting in the fo rmation 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. 1.14 These are all physical changes . 1.15 A chemical property describes a property that changes th e 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 composit ion of water. 1.16 Extensive properties, such as volume, and si ze, 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. 1.17 (a) Extensive Mass is a mass dependent property. Example: Mass of a car. (b) Intensive The boiling point of a substance is the same for a mL as it is for a L of the compoun d so it is mass independent. Example: Boiling point of wa ter. (c) Intensive The color of a substance does not change when you change the amount of substance. Example: Color of a leaf. (d) Extensive Surface area depends on the amount of substance. It also depends on the nature of the substance. A bar of metal ha s a smaller surface area than that of the same bar ground into fine particles. Example: Surface area of a rug. (e) Intensive The physical state, gas, liquid, or solid, depends on temperature and pr essure but not on the mass of the substance. Example: Phys ical state of liquid nitrogen. (f) 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 . Example: Densi ty of milk. (g) Extensive The volume occu pied by a substance is dependent on the mass of substance. Example: Volume of a swimming pool. Chapter 1 1-5 1.18 (a) Gas Temperature, density, volume, viscosity (b) Liquid Temperature, density, volume, viscosity (c) Solid Temperature, density, volume 1.19 Density and temperature do not depend on the quantity; for example, the density of liquid nitrogen is the same regardless of the mass. Length and mass depend on the quantity; for example, a 5 -m length is different from a 9-m length. 1.20 Pressure and chemical potential are intensive properties (do not depend on the amount). Volume and area are extensive properties (depend on the amount). 1.21 Measureme nts involve a comparison. The unit gives the number meaning. 1.22 Kilogram 1.23 Kilogram, meter, second, kelvin 1.24 Derived units are a result of multiplying or dividing a unit by 1, by a multiplier, or by another unit. Examples include m2 for area, m/s for velocity , kg m/s2 for energy. 1.25 (a) 0.01 10–2 c (b) 0.001 10–3 m (c) 1000 103 k (d) 0.00000 1 10–6  (the Greek letter mu) (e) 0.000000001 10–9 n (f) 0.000000000001 10–12 p (g) 1,000,000 106 M 1.26 The melting points and boiling points of water at 1 atmosphere pres sure. On the Celsius scale these points correspond to 0 °C and 100 °C respectively . 1.27 (a) 1 Fahrenheit degree < 1 Celsius degree (b) 1 Celsius degree = 1 Kelvin (c) 1 Fahrenheit degree < 1 Kelvin 1.28 The digits that are significant figures in a quantity are tho se that are known (measured) with certainty plus the last digit, which contains som e uncertainty. 1.29 Rounding numbers: if the number after the significant figure is less than 5, keep the number. If the number after the significant figure is 5 or more, and if it is 5 and the 5 is followed by nonzero digits, raise the number by 1. If the nu mber after the significant figure is 5, and followed by a zero, drop the 5 if the preceding digit is even, add 1 if it is odd. 1.30 The accuracy of a measured value is the closen ess of that value to the true value of the quantity. The precision of a number of repeated measurements of the same quantity is the closeness of the measurements to one another. Accuracy can be described using error because it involves a comparison with an exact value . Precision can be described using uncertainty because it involves a comparison with a range of values. 1.31 The minimum uncertainty that is implied in this measurement is ± 0.01 cm. 1.32 In addition, the significant figures are the least precise number . In multiplication, the number of significant figures in the answer depends on sm allest number of significant figures. Chapter 1 1-6 1.33 A conversion factor changes the units and usually the numerical value, but not the scientific meaning of the quantity. For example, the conversion factor 12 in2 / 1 ft2 converts 5.00 ft2 to 720 in2. Although 5.00 is a different numerical value than 720, 5.00 ft2 expresses the same quantity at 720 in2. 1.34 You use the conversion factor 103 mL / L because you wish to convert the unit of mL in the denominator to units of L. 1.35 The problem with using the fraction 3 yd/1 ft as a co nversion factor is that there are 3 feet in one yard. The conversion factor should be 1 yd/3 ft. For the second part of the question, it is not possible to construct a valid conversion factor relating centimeters to meters from the equation 1 cm = 1000 m , since 100 cm = 1 m. 1.36 To convert 250 seconds to hours multiply 250 by: 1 h 3600 s To convert 3.84 hours to seconds multiply 3.84 hours by: 3600 s 1 h 1.37 Four significant figures would be correct because the conversion factor co ntains exact values. The measured value determines the number of significant figur es. 1.38 d = m V : d = density; m = mass; V = volume 1.39 Density is the ratio of the mass of a substance divided by its volume and is an intensive property. Specific gravity is the ratio of the density of a substance divided by the density of water using the same units. Specific gravity does not have any units. Specific gravity is useful since it avoids units, and only the density of water with the desired uni ts needs to be tabulated. 1.40 The answer will be student dependent, but some answers mi ght be g/mL, lb/ gallon, kg/L, ft3/lb. These would have to be divided by the density of water with the same units: 1.00 g/mL, 8.34 lb/gallon, 1.00 kg/L, 62.4 lb/ft3 at 25 °C. 1.41 10.5 g silver = 1 cm3 silver 10.5 g Ag 1 cm3 and 1 cm3 10.5 g Ag Review Problems 1.42 (a) Physical change. Copper does not change chemically when electricity flows through it: It remains copper. (b) Physical change. Gallium is change s 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) Physical change. Dissolving solid baking soda , a covalent compound, in water to obtain a uniform mixture changes the state of the baking soda from solid to liquid but does not break any chemical bonds. Chapter 1 1-7 1.43 (a) Chemical change. Iron reacts with oxygen to form the rust. (b) 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 co rn. (c) Physical change. Generally alloys are mixtures of s ubstances and no chemical change occurs. On occasion, a chemical change can occur during the production of an alloy. (d) Physical change. During the production of butter fat molecules aggregate, d ue to the agitation of whipping, and separate from the wate r. (e) Physical and chemical change. Mixing baking soda with vinegar results in a chemical reaction (chemical change) that results in gas production (physical change). 1.44 (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 . 1.45 (a) Potassi um chloride is a solid at room temperature. (b) Carbon dioxide is a gas at room temperature. (c) Methane is a gas at room temperature. (d) Sucrose is a soli d at room temperature. 1.46 (a) 1 × 1021 (b) 1000 (c) 1012 (d) 0.1 (e) 0.001 (f) 0.01 1.47 (a) 10–9 (b) 10–6 (c) 103 (d) 106 (e) 1 × 1024 (f) 0.1 1.48 (a) tF = 9 °F 5 °Cæ èçö ø÷ (tC) + 32 F = 9 °F 5 °Cæ èçö ø÷ (57 °C) + 32 °F = 135 F when round ed to the proper number of significant figures. (b) tC = 5 °C 9 °Fæ èçö ø÷ (tF – 32 F) = 5 °C 9 °Fæ èçö ø÷ (–25.5 °F – 32 °F) = –31.9 C (c) tC = (TK – 273 K) 1 °C 1 Kæ èçö ø÷ = (378 K – 273 K) 1 °C 1 Kæ èçö ø÷ = 105 C (d) TK = (tC + 273 °C) 1 K 1 °Cæ èçö ø÷ = (–31 + 273 ) 1 K 1 °Cæ èçö ø÷ = 242 K 1.49 (a) tC = 5 °C 9 °Fæ èçö ø÷ (tF – 32 °F) = 5 °C 9 °Fæ èçö ø÷ (98 °F – 32 °F) = 3 7 C (b) tF = 9 °F 5 °Cæ èçö ø÷ (tC) + 32 °F = 9 °F 5 °Cæ èçö ø÷ (–55 °C) +32 °F = –67 F (c) tC = (TK – 273 K) 1 °C 1 Kæ èçö ø÷ = (299 K – 273 K) 1 °C 1 Kæ èçö ø÷ = 26 C (d) TK = (tC + 273 °C) 1 K 1 °Cæ èçö ø÷ = (40 °C + 273 °C) 1 K 1 °Cæ èçö ø÷ = 313 K 1.50 Temperatu re in °C: Chapter 1 1-8 tC = (TK – 273 K) 1 °C 1 Kæ èçö ø÷ = (15.7 × 106 K – 273 K) 1 °C 1 Kæ èçö ø÷ = 15.7 × 106 C Temperature in °F: tF = 9 °F 5 °Cæ èçö ø÷ (C) + 32 °F = 9 °F 5 °Cæ èçö ø÷ (15.7 × 106 C) + 32 °F = 2.83 × 107 F 1.51 Temp erature in °C: tC = (TK – 273 K) 1 °C 1 Kæ èçö ø÷ = (109 K – 273 K) 1 °C 1 Kæ èçö ø÷ = –164 C Temperature in °F: tF = 9 °F 5 °Cæ èçö ø÷ (C) + 32 °F = 9 °F 5 °Cæ èçö ø÷ (–164 C) + 32 °F = –263.2 F 1.52 tC=tF-32 °F ( )5 °C 9 °Fæ èçö ø÷=103.5 oF-32 °F ( )5 °C 9 °Fæ èçö ø÷=39.7 °C This dog has a fever; the temperature above is out of normal canine range. 1.53 tC=tF-32 °F ( )5 °C 9 °Fæ èçö ø÷=120.0 oF-32 °F ( )5 °C 9 °Fæ èçö ø÷=48.89 °C The temperature in Death Valley, 56.7 °C, was warmer. 1.54 9.2 cm, 2 significant figures; 9.15 cm, 3 significant figures 1.55 24.25 °C, 4 significant figures; 18.9 °C, 3 significant figures 1.56 (a) 4 significant figures (b) 5 significant figures (c) 4 significant figures (d) 2 significant figures (e) 2 significant figures 1.57 (a) 3 significant figures (b) 6 significant figures (c) 1 significant figures (d) 5 signific ant figures (e) 1 significa nt figure 1.58 (a) 0.72 m2 (b) 84.24 kg (c) 4.19 g/cm3 (dividing a num ber with 4 sig. figs by one with 3 sig. figs) (d) 19.42 g/mL (e) 858.0 cm2 1.59 (a) 2.06 g/mL (b) 4.02 mL (c) 12.4 g/mL (d) 0.276 g/mL (e) 0.0006 m/s2 1.60 (a) finite number of significant figures (b) exact number (c) finite number of significant figures (d) finite number of significant figures 1.61 (a) finite number of significant figures (b) finite number of significant figures (c) finite number of significant figures (d) exact number Chapter 1 1-9 1.62 (a) km/hr = 32.0 dm/s ( )1 m 10 dmæ èçö ø÷1 km 1000 mæ èçö ø÷3600 s 1 hæ èçö ø÷ = 11.5 km/h (b) mg/L = 8.2 mg/mL ( )1 g 1000 mgæ èçö ø÷1 ´ 106 mg 1 gæ èççö ø÷÷1000 mL 1 Læ èçö ø÷ = 8.2 ´ 106 mg/L (c) kg = 75.3 mg ( )1 g 1000 mgæ èçö ø÷1 kg 1000 gæ èçö ø÷ = 7.53 ´ 10-5 kg (d) L = 137.5 mL ( )1 L 1000 mLæ èçö ø÷ = 0.1375 L (e) mL = 0.025 L ( )1000 mL 1 Læ èçö ø÷ = 25 mL (f) dm = 342 pm2( )1´10-12 m 1 pmæ èççö ø÷÷2 10 dm 1 mæ èçö ø÷2 = 3.42 ´ 10-20 dm 1.63 (a) m3 = (92 dL) 1 L 10 dLæ èçö ø÷1000 cm3 1 Læ èççö ø÷÷1 m 100 cmæ èçö ø÷3106 mm 1 mæ èççö ø÷÷3 = 9.2 × 1015 m3 (b) g = (22 ng) 1 g 109 ngæ èççö ø÷÷106 mg 1 gæ èççö ø÷÷ = 0.022 g (c) nL = (83 pL) 1 L 1012 pLæ èççö ø÷÷109 nL 1 Læ èççö ø÷÷ = 0.083 nL (d) m3 = (230 km3) 1000 m 1 kmæ èçö ø÷3 = 2.3 × 1011 m3 (e) km hr-2 = (87.3 cm s–2) 1 m 100 cmæ èçö ø÷1 km 1000 mæ èçö ø÷3600 s 1 hræ èçö ø÷2 = 1.13 × 104 km hr-2 (f) nm2 = (238 mm2) 1 m 1000 mmæ èçö ø÷21 nm 10-9 mæ èçö ø÷2 = 2.38 × 1014 nm2 1.64 (a) cm = 36 in. ( )2.54 cm 1 in.æ èçö ø÷ = 91 cm (b) kg = 5.0 lb ( )1 kg 2.205 lbæ èçö ø÷ = 2.3 kg (c) mL = 3.0 qt ( )946.4 mL 1 qtæ èçö ø÷ = 2800 mL (d) mL = 8 oz( )29.6 mL 1 ozæ èçö ø÷ = 200 mL (e) km/hr = 55 mi/hr ( )1.609 km 1 miæ èçö ø÷ = 88 km/hr (f) km = 50.0 mi ( )1.609 km 1 miæ èçö ø÷ = 80.4 km Chapter 1 1-10 1.65 (a) qt = 250 mL ( )1 qt 946.4 mLæ èçö ø÷ = 0.26 qt (b) m = 3.0 ft ( )12 in. 1 ftæ èçö ø÷2.54 cm 1 in.æ èçö ø÷1 m 100 cmæ èçö ø÷ = 0.91 m (c) lb = 1.62 kg ( )2.205 lb 1 kgæ èçö ø÷ = 3.57 lb (d) oz = 1.75 L ( )1000 mL 1 Læ èçö ø÷1 oz 29.6 mLæ èçö ø÷ = 59.1 oz (e) mi/hr = 35 km/hr ( )1 mi 1.609 kmæ èçö ø÷ = 22 mi/hr (f) mi = 80.0 km ( )1 mi 1.609 kmæ èçö ø÷ = 49.7 mi 1.66 (a) cm2 = 8.4 ft2( )30.48 cm 1 ftæ èçö ø÷2 = 7,800 cm2 (b) km2 = 223 mi2( )1.609 km 1 miæ èçö ø÷2 = 577 km2 (c) cm3 = 231 ft3( )30.48 cm 1 ftæ èçö ø÷3 = 6.54´106 cm3 1.67 (a) m2 = (2.4 yd2) 0.9144 m 1 ydæ èçö ø÷2 = 2.0 m2 (b) mm2 = (8.3 in2) 2.54 cm 1 inæ èçö ø÷2 10 mm 1 cmæ èçö ø÷2 = 5400 mm2 (c) L = (9.1 ft3) 1 yd 3ftæ èçö ø÷30.9144 m 1 ydæ èçö ø÷3100 cm 1 mæ èçö ø÷31 mL 1 cm3æ èçö ø÷1 L 1000 mLæ èçö ø÷ = 260 L 1.68 mL = 4.2 qt ( )946.35 mL 1 qtæ èçö ø÷ = 4.0 ´ 103 mL (stomach volume) 4.0 × 103 mL  0.9 mL = 4,000 pistachios (don’t try this at home) 1.69 To determine if 5 0 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 eggæ èçö ø÷1 L 1000 mLæ èçö ø÷1.057 qt 1 Læ èçö ø÷ = 2.8 qt 2.8 qt < 4.2 qt Luke can eat 50 eggs. 1.70 2m 190 mi 5280 ft 30.48 cm 1 10 m 1 hr 1 min m 84.9 s 1 hr 1 mi 1 ft 1 cm 60 min 60 s s−       ==                 Chapter 1 1-11 1.71 km/h = 738 ft 1 yd 0.9144 m 1 km 3600 s s 3 ft 1 yd 1000 m 1 h                   = 810 km/h 1.72 mi 2700 ft 1 mi 60 s 60 min mi 1841 hr 1 s 5280 ft 1 min 1 hr hr    ==         1.73 tons/day = 2.05´105 ft3 s62.4 lb 1 ft3æ èçö ø÷1 ton 2000 lbæ èçö ø÷3600 s 1 hæ èçö ø÷24 h 1 dæ èçö ø÷ = 5.53 × 108 tons/day 1.74 1 light year = 1 y365.25 d 1 yæ èçö ø÷24 h 1 dæ èçö ø÷3600 s 1 hæ èçö ø÷3.00´108 m 1 sæ èççö ø÷÷ = 9.47 × 1015 m miles = 8.7 light years 9.47 ´ 1015m 1 light yearæ èççö ø÷÷1 km 1000 mæ èçö ø÷1 mi 1.609 kmæ èçö ø÷ = 5.1 × 1013 mi 1.75 There are 360 degrees of latitude around the circumference of the earth. statute miles = 360 degree latitude60 nautical miles 1 degree latitudeæ èçö ø÷1.151 statute miles 1 nautical mileæ èçö ø÷ = 2.49 × 104 statute miles 1.76 density = mass/ volum e = 36.4 g/45.6 mL = 0.798 g/mL 1.77 density = mass volume d = 14.3 g 8.46 cm3 = 1.69 g/cm3 1.78 mL = 25.0 g1 mL 0.791 gæ èçö ø÷ = 31.6 mL 1.79 mL = 26.223 g ( )1 mL 0.99704 gæ èçö ø÷ = 26.301 mL 1.80 g = 185 mL1.492 g 1 mLæ èçö ø÷ = 276 g 1.81 kg = 34 L( )1000 mL 1 Læ èçö ø÷0.65 g 1 mLæ èçö ø÷1 kg 1000 gæ èçö ø÷ = 22.1 kg lbs=22.1 kg( )2.2 lbs 1 kgæ èçö ø÷=49 lbs 1.82 mass of silver = 62.00 g – 27.35 g = 34.65 g volume of silver = 18.3 mL –15 mL = 3.3 mL or 3.3 cm3 density of silver = (mass of silver)/(volume of silver) = (34.65 g)/(3.3 cm3) = 11 g/cm3 1.83 volume of titanium = (1.84 cm)(2.24 cm)(2.44 cm) = 10.1 cm3 density of titanium = 45.7 g/10.1 cm3 = 4.54 g/cm3 1.84 density = 227,641 lb 385,265 galæ èçö ø÷=0.591 lb gal–1 Chapter 1 1-12 density = 0.591 lb gal–1 453.6 g 1 lbæ èçö ø÷3785 mL–1 1 gal–1æ èçö ø÷=0.0708 g mL–1 1.85 10.1 ft ´ 32.3 ft ´ 4.00 in ´1 ft 12 in ´0.3048 m fté ëêù ûú3 = 3.08 m3 3.08 m3 ´ 0.686 kg L3 ´ 1000 L m3= 2112 kg Additional Exercises 1.86 CN$ 1 US$ 3.785 L1.039 = 2.98 US$/galL 1.321 CN$ gal 1.87 28.5 km L´0.6214 mi 1.00 kmæ èçö ø÷´3.785 L 1.00 galæ èçö ø÷ = 67.0 mi gal–1 The first car has a mileage of 67.0 mi gal–1 and is more efficien t than 31 mi gal–1. 1.88 The density of rod A is (37.8 g / 4.2 cm3) = 9.0 g/cm3 (close to the density of nickel). The density of rod B is (26.2 g / 3.7 cm3) = 7.1 g/cm3 (clos e to the density of zinc). 1.89 Divide the experimental mass by the experimental volume, and then compare with the known density of each metal. 1.90 If the density is in metric tons… g = 1 teaspoon5.00 mL 1 tspæ èçö ø÷1 cm3 1 mLæ èçö ø÷1´108 tons 1 cm3æ èçö ø÷1000 kg 1 tonæ èçö ø÷1´103 g 1 kgæ èçö ø÷ = 5.00 × 1014 g If the density is in English tons… g = 1 teaspoon5.00 mL 1 tspæ èçö ø÷1 cm3 1 mLæ èçö ø÷1´108 tons 1 cm3æ èçö ø÷2000 lbs 1 tonæ èçö ø÷453.59 g 1 lbæ èçö ø÷ = 4.54 × 1014 g 1.91 days = 3.50 × 1014 km 1000 m 1 kmæ èçö ø÷1 s 3.00 ´ 108 mæ èçö ø÷1 h 3600 sæ èçö ø÷1 d 24 hæ èçö ø÷ = 1.35 × 104 d 1 light year = 3.00 × 108 m/s 3600 s 1 hæ èçö ø÷24 h 1 dæ èçö ø÷365 d 1 yæ èçö ø÷ = 9.46 × 1015 m/yr light years = 3.50 × 1014 km 1000 m 1 kmæ èçö ø÷1 light year 9.46 ´ 1015 mæ èçö ø÷ = 37.0 light years 1.92 (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 = 36.842 g – 27.314 g = 9.528 g volume = (9.528 g) 1 mL 0.99704 gæ èçö ø÷ = 9.556 mL Chapter 1 1-13 (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 pycnomet er – mass of empty pycnometer = 41.428 g – 27.314 g = 14.114 g Density of chloroform = 14.114 g 9.556 mLæ èçö ø÷ = 1.477 g/mL 1.93 For the message to get to Mars : time = 225,000,000 miles ( )1.609 km 1 mileæ èçö ø÷1000 m 1 kmæ èçö ø÷1 s 3.00 ´ 108 mæ èçö ø÷ = 1210 s The reply would take the same amount of time, so the minimum time would be: 1210 s × 2 = 2420 s 1.94 (a) $7.35 = 30 min $7.35 30 min=$0.245 min 30 min $7.35=1 min $0.245 (b) $ = 1 hr ´ 60 min hræ èçö ø÷ + 45 minæ èççö ø÷÷$0.245 minæ èçö ø÷ = $25.73 (c) min = $333.50 ( )1 min $0.245æ èçö ø÷ = 1360 min 1.95 dsea water = 1.025 g cm3æ èçö ø÷1 lb 453.59 gæ èçö ø÷30.48 c m 1 ftæ èçö ø÷3 = 64.0 lb ft–3 ft3 = 4255 tons ( )2000 lbs 1 tonæ èçö ø÷1 ft3 64.0 lbæ èççö ø÷÷ = 1.330 ´ 105 ft3 1.96 g = 2510 c m3 1 in 2.54 c mæ èçö ø÷30.00011 lbs 1 in3æ èçö ø÷453.6 g 1 lbæ èçö ø÷ = 7.6 g 1.97 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 . 1.98 The density of osmium (Os) at room temperature is 22.6 g/cm3 and the density of tin (Sn) at room temperature is 5.8 g/cm3. Proceed with dimensional analysis as follows , based on a 1:1 mass ratio: ()333 31 cm Os 1 g Os 5.8 g Sn 9.25 cm Sn = 2.4 cm Os22.6 g Os 1 g Sn 1 cm Sn    1.99 g/mL = 69.22 lb/ft3 453.59 g 1 lbæ èçö ø÷1 ft 30.48 cmæ èçö ø÷31 cm3 1 mLæ èççö ø÷÷ = 1.109 g/mL Since t he density closely matches the known value, we conclude that this is an authentic sample of ethylene glycol . 1.100 (0.200 g / 1.00 carat) × 1.75 carat = 0.35 g 1.101 We solve by combining two equations: Chapter 1 1-14 tF = 9 °F 5 °Cæ èçö ø÷ (tC) + 32 F tF = tC If tF = tC, we can use the same variable for both temperatures: tC = 9 °F 5 °Cæ èçö ø÷ (tC) + 32 F 5 5tc = 9 °F 5 °Cæ èçö ø÷ (tC) + 32 F -4 5tc = 32 tc = 32 5 4− = –40, therefore the answe r is –40 °C. 1.102 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 °Fæ èçö ø÷ (tF – 32 °F) and tC = (TK – 273 K) 1 °C 1 Kæ èçö ø÷ therefore (TK – 273 K) 1 °C 1 Kæ èçö ø÷ = 5 °C 9 °Fæ èçö ø÷ (tF – 32 °F) at TK = 0 K = 0 R (0 K – 273 K) 1 °C 1 Kæ èçö ø÷ = 5 °C 9 °Fæ èçö ø÷ (tF – 32 °F) –273 °C = 5 °C 9 °Fæ èçö ø÷ (tF – 32 °F) –491 °F = tF – 32 °F tF = –459 °F at ab solute zero Also, TR at absolute zero is 0 R and TR = (tF + 459 °F) 1 R 1 °Fæ èçö ø÷ So, the boiling point of water is 212 °F and in TR: TR = (212 °F + 459 °F) 1 R 1 °Fæ èçö ø÷ = 671 R 1.103 Sand dsand = 2.84 g/mL Gold dgold = 19. 3 g/mL Mixture dmixture = 3.10 g/mL 1.00 kg mixture 1000 g 1 kgæ èçö ø÷ = 1.00 × 103 g of mixture 1.00 × 103 g of mixture = msand + mgold msand = (dsand)(Vsand) mgold = (dgold)(Vgold) 1.00 × 103 g of mixture = ( dsand)(Vsand) + (dgold)(Vgold) 1.00 × 1 03 g of mi xture = (2.84 g/mL)( Vsand) + (19.3 g/mL)( Vgold) Vmixture = Vsand + Vgold Chapter 1 1-15 d = m Væ èçö ø÷ 1.00´ 103 g 3.10 g mL–1æ èççö ø÷÷ = 323 mL Vsand + Vgold = 323 mL Vsand = 323 mL – Vgold 1.00 × 103 g of mixture = (2.84 g/mL)(323 mL – Vgold) + (19 .3 g/mL)( Vgold) 1.00 × 103 g of mixture = 917 g sand – (2.84 g/mL)( Vgold) + (19.3 g/mL)( Vgold) 1.00 × 103 g of mixture – 917 g sand = (16.5 g/mL)( Vgold) 5.0 mL = Vgold 1.00 × 103 g of mixture – 917 g sand = 83 g gold % mass of gold = 83 g gold 1000 g total´100% =8.3% gold 1.104 Area of gold in cm2 = 14.6 ft2 30.48 cm 1 ftæ èçö ø÷2 = 1.36 × 104 cm2 Volume of gold in cm3 = 1.36 × 104 cm2 × 2.50 m × 1 cm 1´104 mmæ èççö ø÷÷ = 3.39 cm3 Cost of gold = 3.39 cm3 × 31 mL 19.3 g 1 troy ounce $1910.50 1 mL 31.1035 g troy ounce 1 cm                   = $4019 1.105 Volume o f diamond = 3. 45 carat 200 mg 1 caratæ èçö ø÷1 g 1000 mgæ èçö ø÷1 cm3 3.51 gæ èçö ø÷ = 0.197 cm3 V = 4 3pr3 0.197 cm3 = 4 3pr3 r = 0.361 cm 2r = 0.722 cm Chapter 2 2-1 Chapter Two Elements, Compounds, and the Periodic Table Practice Exercises 2.1 (a) SF6 contains 1 S and 6 F atoms per molecule (b) (C2H5)2N2H2 contains 4 C, 12 H, and 2 N per molecule (c) Ca3(PO 4)2 contains 3 Ca, 2 P, and 8 O atoms per formula unit (d) Co(NO 3)2·6H 2O contains 1 Co, 2 N, 12 O, and 12 H per formula unit 2.2 (a) NH 4NO 3 contains 2 nitrogen, 4 hydrogen, 3 oxygen atoms per formula unit (b) FeNH 4(SO 4)2 contains 1 iron, 1 nitrogen, 4 hydrogen, 2 sulfur, 8 oxygen atoms per formula unit (c) Mo(NO 3)2·5H 2O con tains 1 molybdenum, 2 nitrogen, 11 oxygen, and 10 hydrogen atoms per formula unit (d) C6H4ClNO 2 contains 6 carbon, 4 hydrogen , 1 chlorine, 1 nitrogen, and 2 oxygen atoms per molecule 2.3 Reactants: 4 N, 12 H, and 6 O; Products: 4 N, 12 H, and 6 O. 2.4 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 . Thus the equation obeys the law of conservation of mass. 2.5 The number of protons is equal to the atomic number of an element. Th e 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 a tom 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 . 2.6 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 charg e, the number of electrons is determined by subtracting the charge from the atomic number. If the a tom 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 2.7 (a) NaF (b) Na2O (c) MgF 2 (d) Al4C3 2.8 (a) Ca3N2 (b) AlBr 3 (c) K2S (d) CsCl 2.9 (a) CrCl 3 and CrCl 2, Cr 2O3 and CrO (b) CuCl, CuCl 2, Cu 2O and CuO 2.10 (a) Au 2S and Au 2S3, Au 3N and AuN (b) TiS, Ti 2S3 and TiS2; Ti3N2, TiN and Ti3N4 2.11 (a) Na2CO 3 (b) (NH 4)2SO 4 Chapter 2 2-2 2.12 (a) KC 2H3O2 (b) Sr(NO 3)2 (c) Fe(C 2H3O2)3 2.13 (a) K2S (b) BaBr 2 (c) NaCN (d) Al(OH) 3 (e) Ca3P2 2.14 (a) aluminum chlorid e (b) barium hydroxide (c) sodium bromide (d) calcium fluoride (e) potassium phosphide 2.15 lithium sulfide, magnesium phosphide , nickel(II) chloride , titanium(II) chloride , iron(III) oxide 2.16 (a) Al2S3 (b) SrF 2 (c) TiO 4 (d) CoO (e) Au 2O3 2.17 (a) lithium carbonate (b) potassium permanganate (c) iron(III) hydroxide 2.18 (a) KClO 3 (b) NaOCl (c) Ni3(PO 4)2 2.19 The term "octa' means eight, therefore there are 8 carbon atoms in octane. The formula for an alkane i s C nH2n+2, so octane has 8 carbons and ((2 × 8) + 2) = 18 H. The molecular formula is C8H18 and the condensed structural formula is CH 3CH 2CH 2CH 2CH 2CH 2CH 2CH 3. 2.20 (a) Propanol: C3H8O, CH 3CH 2CH 2OH (b) Butanol: C 4H10O, CH 3CH 2CH 2CH 2OH 2.21 (a) phosphorus t richloride (b) sulfur dioxide (c) dichlorine heptoxide (d) hydrogen sulfide 2.22 (a) AsCl 5 (b) SCl 6 (c) S2Cl2 (d) H2Te 2.23 diiodine pentoxide 2.24 chromium(III) acetate Review Questions 2.1 For all group 1A elements (the alkali metals), the formula is MX, that is one Cl per atom of metal. For all group 2A elements (the alkaline earth metals), the formula is MX 2, that is two Cl atoms per atom of metal. The correspondence in formula and the similarities in chemical behavior allowed Mendeleev to locate theses two series into their sepa rate groups on the periodic table. 2.2 Mendeleev constructed his periodic table by arranging the elements in order of increasing atomic weight, and grouping the elements by their recurring properties. The modern periodic table is arranged in order of increasin g number of protons. 2.3 Strontium and calcium are in the same Group of the periodic table, so they are expected to have similar chemical properties. Strontium should therefore form compounds that are similar to those of calcium, including the sorts of compou nds found in bone. 2.4 Silver and gold are in the same periodic table group as copper, so they might well be expected to occur together in nature, because of their similar properties and tendencies to form similar compounds. Chapter 2 2-3 2.5 Cadmium is in the same periodic tab le group as zinc, but silver is not. Therefore cadmium would be expected to have properties similar to those of zinc, whereas silver would not. 2.6 The superscript before the symbol indicates the mass number; the superscript after the symbol indicates the char ge on the atom; the subscript before the symbol indicates the atomic number; and the subscript after the symbol indicates the number of atoms in the compound. For example: charge mass number 14 atomic number 7 2 number of atomsN N Nitrogen has 7 protons and 7 electrons for a neutral atom . The molecule has two atoms in it, and the isotope with 7 neutrons gives it a mass number of 14 2.7 See Figure of the periodic table in the margin in Section 2.1 2.8 (a) K (b) Br (c) Os (d) Kr (e) Pr (f) Am (g) Sr 2.9 Luster, electrical conductivity, the rmal conductivit y, ductility, and malleability are the characteristic properties of metals. 2.10 Mercury was used in thermometers because it is a liquid, and tungsten is used in light bulbs because is has such a high melting point. Mercury is no longer used because of its to xicity. 2.11 The noble gases: He, Ne, Ar, Kr, Xe, and Rn 2.12 Mercury and bromine 2.13 They are semiconductors. 2.14 See Fi gure 2.3. 2.15 The heavy line separates the metals from the nonmetals, and the metalloids border the line. 2.16 Metals which are used to make j ewelry are those tha t do not corrode, silver, gold, and platinum. Iron would be useless for jewelry because it is susceptible to rusting. Potassium reacts violently with water to form hydrogen and potassium hydroxide. 2.17 Luster, malleability, color, and brit tleness are some tre nds that are mentioned in terms of moving from the metals to the nonmetals across the periodic table, or moving down a group from nonmetals to metals. 2.18 (a) In general, melting points decrease from left to right across the periodic table and increases from t op to bottom. (b) In general, boiling points decrease from left to right across the periodic table and increases from top to bottom. (c) In general, density has a maximum in the middle of the periodic table and falls off to the right an d left. Also, the d ensity increases moving down a group. 2.19 This may stand for the name of an element or for the name of one atom of an element. 2.20 The smallest particle that is representative of a particular element is the atom of that element. A molecule is a representative uni t that is made up of two or more atoms linked together. Chapter 2 2-4 2.21 H2, hydrogen N2, nitrogen, O2, oxygen F2, fluorine Cl2, chlorine Br2. bromine I2, iodine Se8, selenium P4, phosphorus S8, sulfur 2.22 (a) nitrogen, N (b) bromine, Br (c) chlorine, Cl (d) oxygen, O 2.23 (a) carbon, C (b) hydrogen, H (c) sulfur, S (d) iodine, I 2.24 DNA contains hydrogen, carbon, nitrogen, oxygen, and phosphorus atoms. 2.25 A chemical reaction is balanced when there is the same number of each kind of atom on both the reactant and produc t side of the equati on; and the total charges on both the reactant and product sides of the equation are the same. These conditions must be met due to the law of conservation of matter. 2.26 Reactants are the substances to the left of the arrow in a reaction t hat are present befo re the reaction begins. Products are the substances to the right of the arrow in a reaction and they are formed during the reaction and are present when the reaction is over. 2.27 (a) Magnesium reacts with oxygen to give (yield) magnesium o xide. (b) The react ants are Mg and O 2. (c) The product is MgO. 2.28 An ionic compound is formed by the transfer of electrons, and it is accompanied by the formation of ions of opposite charge. 2.29 Metals react with nonmetals. 2.30 An ion is a charged particle. It can be monatomic or poly atomic, and it can have either a positive or a negative charge. It is derived from an atom or a molecule by gain or loss of electrons. Atoms and molecules are neutral. 2.31 In ionic substances, no molecules exist. Rather we have a continuous array of cations an d anions, which are present in a constant ratio. The ratio is given by the formula unit. 2.32 (a) Na, Na+ (b) These particles have the same number of nuclei. (c) These particles have the same number of protons (d) These particles could have different numbers o f neutrons, if they are different isotopes. (e) These particles do not have the same number of electrons; Na+ has one less electron. 2.33 A cation is a positively charged ion with one or fewer electrons than its neutral atom. An anion is a negatively charged i on with one or more electrons than its neutral atom. A polyatomic ion is made up of more than one atom; the whole unit is the ion. 2.34 Titanium lost four electrons to form Ti4+; it has 22 protons and 18 electrons. 2.35 Negative ; anion 2.36 Nitrogen gained 3 electrons t o form N3–; it has 7 protons and 10 electrons. 2.37 (a) Fe2+, Fe3+ (b) Co2+, Co3+ (c) Hg2+, Hg 22+ (d) Cr2+, Cr3+ (e) Sn2+, Sn4+ (f) Mn2+, Mn4+