(Solution Manual) Chemistry Principles and Practice (International)

Reger, Goode, Ball, 3

Chemistry

Solution Manual

3rd Edition

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Solution Manual for Chemistry Principles and Practice (International), The downloadable file consists of 346 pages, and the total size of the file is 3.55 MB. Purchase now only for highly discounted price.

Author(s): Reger, Goode, Ball

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1Chapter 1: Introduction to Chemistry 1.2 See Section 1.1. Scientists: “Hypothesis” is a possible explanation for an event, even if it is an untested assumption. “Theory” is an explanation for a law of nature. General public: “Theory” is an educated hunch or guess. 1.4 See Section 1.1. Guess Model T h e o r y Data Hypothesis 1.6 See Section 1.2. Intensive properties of air: temperature, pressure, specific volume, and density Extensive properties of air: mass, volume, colorless, odorless 1.8 See Section 1.2. The astronauts in space are still composed of matter. They still have the same mass as they do on earth but weight less because they are further from the earth’s gravitational attraction. 1.10 See Section 1.2 Heterogeneous mixtures are typically easier to separate than homogeneous mixtures. These types of mixture can usually be separated using physical techni ques (i.e. filtration, distillation, etc.). However, because homogeneous mixtures have the same consistenc y throughout, they are more difficult to separate. 1.12 See Section 1.2. The light given by an electric bulb is an intensive property for a given wattage. 1.14 See Section 1.2 and Figure 1.4. A sample of matter that cannot be separated into co mponents by a physical process is called a substance. Compounds and elements are types of substances. Co mpounds are substances that can be decomposed into their constituent elements by chemical means whereas elements are substances that cannot be further decomposed by chemical means. 1.16 See Section 1.3 The term accuracy refers to how close a perform ed measurement is to the accepted value for that measurement. Precision refers to how close several measured values are to each other. A measurement that includes more significant figures indicates that the measurement was performed with an instrument calibrated in smaller increments allowing a more precise measurement to be made. 2 1.18 See Section 1.5. km hr ⎯→⎯⎯ m hr ⎯→⎯⎯ m min ⎯→⎯⎯ m sStep 1 Step 2 Step 3 Step 1. Use km to m conversion factor; mutiply by m/km. Step 2. Use hr to min conversion factor; multiply by hr/min, since hr is in the denominator. Step 3. Use min to s conversion factor; multiply by min/s, since min is in the denominator. 1.20 See Section 1.4. No, all measured quantities have some uncertainty which depends on the measuring device and/or the person estimating the distance between units. 1.22 See Section 1.5. The length of a King’s foot is easy to comprehend, but the size of a foot would vary from King to King. 1.24 See Section 1.1. meter (m): based on the distance that light travels in 1/299,792,458 of a second kilogram (kg): based on an internationa lly accepted weight with a mass of 1 kg second (s): based on the electronic transitions in Cs-133 isotope ampere (A): based on the cu rrent required to produce 2 × 10-7 N of force between two wires that are placed 1 m apart Kelvin (K): based on the triple point of water (1 K = 1/273.16 the thermodynamic temperature of water mole (mol): the amount of a substance which cont ains as many particles as there are atoms in 0.012 kilograms of the isotope C-12 candela (cd): based on the luminous intensity of an object that em its monochromatic radiation of a frequency of 540 × 1012 Hz and that has a radiant intensity of 1/683 watt 1.26 See Section 1.2. (a) extensive, physical (b) intensive, chemical (c) extensive, physical (d) intensive, physical (e) extensive, chemical 1.28 See Section 1.2. (a) physical (b) chemical (c) physical (d) chemical 1.30 See Section 1.2. (a) chemical (b) physical (c) chemical (d) physical 1.32 See Section 1.2. (a) chemical (b) chemical (c) physical (d) chemical 3 1.34 See Section 1.2. Rarely found as a free element ( Chemical ); Found combined with oxygen ( Chemical ); Can be obtained by reacting ore with carbon ( Chemical ); Silvery colored metal ( Physical ); Conducts heat and electricity (Physical ); Hardness and mechanical strength ( Physical ); Steel does not corrode ( Chemical ) 1.36 See Section 1.2. Liquid at room temperature ( Physical ); Acrid smelling ( Chemical ); Reacts readily with most metals (Chemical ); Evaporates easily ( Physical ); Obtained from sodium bromide ( Chemical ) 1.38 See Section 1.2. (a) compound (b) homogeneous mixture (c) homogeneous mixture (d) element 1.40 See Section 1.2. (a) element (b) homogeneous mixture (c) homoge neous mixture (d) heterogeneous mixture 1.42 See Section 1.2. Salt water should be classified as a solution; wood , champagne and cloudy tea are not because they are not uniform throughout. 1.44 See Section 1.3. (a) Both accurate and precise (b) Accurate (c) Neither (d) Precise 1.46 See Section 1.3. (a) 3 (b) 4 (c) 4 (d) 1 (e) 5 1.48 See Section 1.3. (a) 5 (b) 3 (c) 5 (d) 4 1.50 See Section 1.3. (a) 0.0821 kg (b) 1.01 m (c) 18.998 g (d) 19.00 g 1.52 See Section 1.3. (a) 39.50 ºC (b) 21.47 mL (c) 13.0 s 1.54 See Section 1.3. (a) 0.0821 kg (b) 1.01 m (c) 18.998 g (d) 19.00 g 4 1.56 See Section 1.3. (a) 21.9 (b) 2.0 x 103 (c) 0.58 (d) 8.0 1.58 See Section 1.3. (a) 80.0 (b) 0.7615 (c) 14.712 (d) 0.0286 1.60 See Section 1.4. × = 2.05 ×10−65 3.4×1051 + 1.9 × 10−3 = 1.9 × 10−3 ± 1 x 10-4 One decimal place in answ er, since there is just one decimal place in the quantity 1.9 × 10−3 1.62 See Table 1.5. (a) mass, kilogram (b) distance, meter (c ) temperature, kelvin (d) time, second 1.64 See Section 1.4. 1 g = 1 x 10-6 Mg 1 Mg = 1 x 106 g 1.66 See Section 1.4. 1 ns = 1 x 10-6 ms 1 ms = 1 x 106 ns 1.68 See Section 1.4. 1 g/L = ? lb/ft3 33 31g 1lb 1l 1mL 2.54cm 12inxx x x x1L 453.6g 1000mL 1in 1ft 1cm⎛⎞ ⎛ ⎞= ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠30.0624lb / ft 1.70 See Section 1.4. Volume = area x height. Convert the area to units of mi2 and the height to ft and calculate volume. Then convert from ft3 to liters. 2 62 1 3 2 5280 ftarea = 3.02 10 mi = 8.42 10 ft1 mi⎛⎞ ⎜⎟⎝⎠xx x 1 ftheight = 2 in = 0.167 ft12 inx 13 2 13 3volume = 8.42 10 ft 0.167 ft = 1.41 10 ft xx x 513 3 14 328.316 L1.41 10 ft 3.97 10 L1 ft= x xx 1.72 See Section 1.4. (a) ? km = 25.5 m x 1 km 1000 m = 2.55 x l0 −2 km (b) ? m = 36.3 km x 1000 m 1 km = 3.63 x 10 4 m (c) ? g = 487 kg x 1000 g 1 kg = 4.87 x l0 5 g (d) ? mL = 1.32 L x 1000 mL 1L = 1.32 x l0 3 mL (e) ? L = 55.9 dL x 1 L 10 dL = 5.59 L (f) ? cm 3 = 6251 L x 1000 cm 3 1 L = 6.251 x l0 6 cm3 (Note: 1 cm 3 = 1 mL) 1.74 See Section. 1.4. (a) 45 s = 4.5 x 104 ms = 0.75 minutes (b) 550 nm = 5.5 x 10-5 cm = 5.5 x 10-7 m (c) 4 ºC = 277.15 K = 39.2 ºF (d) 2.00 L = 2000 cm3 = 2.00 x 10-3 m3 = 2.11qt 1.76 See Sections 1.4, 1.5, and Table 1.9. ? area in cm2 = (8.5 in × 11 in) 2.54 cm 1 in⎛ ⎝⎜⎞ ⎠⎟2 = 6.0 × 102 cm2 1.78 See Sections 1.4, 1.5, and Table 1.9. ? speed limit in mi/hr = 100 km hr×1 mi 1.609 km = 62.2 mi/hr ? speed limit in m/s = 100 km hr×103m km×1 hr 60 min×1 min 60 s = 27.8 m/s 1.80 See Section.1.5 ? volume in fluid ounces = 1 qt 32 fl oz2.00 L × ×0.946 L 1 qt= 67.7 fl. oz. 1.82 See Section 1.5. 6 TK = TC + 273.15 and T C = 1.0DC 1.8oF⎡ ⎣⎢⎤ ⎦⎥ (TF − 32º) yield TK = 1.0DC 1.8oF⎡ ⎣⎢⎤ ⎦⎥ (T F − 32ºF) + 273.15 1.84 See Section 1.5. (a) T F = Tc 1.8DF 1.0oC⎡ ⎣⎢⎤ ⎦⎥ + 32ºF = 80ºC 1.8DF 1.0oC⎡ ⎣⎢⎤ ⎦⎥ + 32ºF = 176ºF T K = Tc + 273.15 = 80 + 273.15 = 353 K (b) T C = (T F - 32 ºF) 1.0 C 1.8 F⎡⎤ ⎢⎥ ⎣⎦D D = (350 ºF - 32 ºF) 1.0 C 1.8 F⎡ ⎤ ⎢ ⎥ ⎣ ⎦D D = 177 ºC 1.86 See Section 1.5. Let y = TF = Tc in T F = Tc 1.8DF 1.0oC⎡ ⎣⎢⎤ ⎦⎥ + 32ºF giving y = y 1.8 1.0⎛ ⎝⎜⎞ ⎠⎟ + 32. This yields y = 1.8y + 32, − 0.8y = 32, and y = 40. Hence, −40ºF = −40ºC. 1.88 See Section 1.5. Density = mass volume Rearranging this equation gives us: Volume = mass density Volume = 325.0 g ethyl acetate 0.9006 cm = 27.8 cm3 1.90 See Section 1.5. Volume of copper sphere = 3.73 × 103 g × 1 cm3 8.92 g = 420 cm3 Hence, r3 = 3 V 4∏ = 3×420 cm3 4×3.1416 = 100 cm3 and r = (100 cm3)1/3 = 4.64 cm 1.92 See Section. 1.3 Calculate the volume of the object from the change in the volume of the graduated cylinder, then calculate the density. Volume of object = 42.3 mL – 35.0 mL = 7.3 mL 35.0 mL42.3 mL 7 1.94 See Sections 1.4, 1.5. Since thickness = volume/area and volume = mass/density, thickness = mass/(area × density). ? thickness in nm = 12 oz 75 ft2×1 lb 16 oz×453.6 g 1 lb×1 ft 12 in⎛ ⎝⎜⎞ ⎠⎟2 ×1 in 2.54 cm⎛ ⎝⎜⎞ ⎠⎟2 ×1 cm3 2.70 g×1 m 102 cm×109 nm 1 m = 1.8 × 10−4 nm 1.96 See Sections 1.4, 1.5. Calculate the volume of the object from the change in the volume of the graduated cylinder, then calculate the density. Volume of object = 48.8 mL – 30.0 mL = 18.8 mL mass (g) 147.8 gdensity = volume (mL) 18.8 mL== 7.86 g/mL 1.98 See Section 1.5. (a) 8 1.609 km 1000 m 1 s 1 min 1 hr 1 day 1 yr93,000,000 mi mi km 60 s 60 min 24 hr 365.25 days 31 0 m×× × × × × × × = 1.58203628 × 10-5 light years (b) ? minutes = 93,000,000 mi × 1.609 km 1 mi × 103 m 1 km × 1 s 3.00 × 108 m × 1 min 60 s = 8.3 min 1.100 See Sections 1.4, 1.5. ? mL A = 9.9132 g A × 1 mL A 0.98 g A = 10 mL A ? mL B = 9.9132 g B × 1 mL B 1.03 g B = 9.62 mL B To obtain range of volumes based on uncertainties ± 1 in last digit of the densities, ? mL A max = 9.9132 g A × 1 mL A 0.97 g A = 10 mL A ? mL A min = 9.9132 g A × 1 mL A 0.99 g A = 10 mL A g/mL 1.6 = mL 7.3g 11.33 (mL) volume(g) mass density = = 8 ? mL B max = 9.9132 g B × 1 mL A 1.02 g B = 9.72 mL B ? mL B min = 9.9132 g B × 1 mL B 1.04 g B = 9.53 mL B The uncertainty in the density of A does not contribute to an uncertainty in the calculated volume of A having a mass of 9.9132 g because the result of the vol ume calculation is limited to just two significant figures. However, the uncertainty in the density of B causes the result of the calculated volume to be 9.62±0.10 mL, since the density of B was known to three significant figures. 1.102 See Section 1.5. Tc for 80.0ºF = 1.0D C 1.8o F⎛ ⎝⎜⎞ ⎠⎟ (80.0ºF − 32ºF) = 26.7ºC Tc for 215.0ºF = 1.0D C 1.8o F⎛ ⎝⎜⎞ ⎠⎟ (215.0ºF − 32ºF) = 101.7ºC Thus, rate of change in ºC per second = (101.7D C − 26.7o C) 1 min × 1 min 60 s = 1.25ºC/s 1.104 See Section 1.2. Maybe something like “similar substances may have very similar physical properties and chemical reactivity that may appear identical within the accura cy and precision of our measuring devices. However, upon careful analysis, 2 different substances must have some measurable differences in their properties. 1.106 See Section. 1.4. (a) ? height in mm = (9.2 × 1012) × (0.166 mm) = 1.53 × 1012 mm ? height in km = 1.53 × 1012 × 61 km 1 10 mm× = 1.5 × 106 km (b) ? mass in g = (9.2 × 1012) × (1.01 g) = 9.3 × 1012 g 9Chapter 2: Atoms, Mo lecules, and Ions 2.2 See Section 2.1. (a) Dalton’s fourth postulate explains the Law of Conservation of Mass. (b) Dalton’s third postulate explains the Law of Constant Composition. 2.4 See Section 2.2. The masses of protons and neutrons are approximately equal and approximately 1836 times greater than the mass of an electron. The protons and elect rons have equal but opposite charges, and neutrons have no charge. On a relative scale, we say the masses of protons and neutrons are each one and the electron has a mass of zero. Similarly, on a relative scale, we say the charge on a proton is +1 and the charge on an electron is −1. 2.6 See Section 2.2. To explain the deflection of alpha particles, Ruther ford proposed that the positive charge and nearly all of the mass of atom are in the central core, with the electrons at a relatively large distance from this core. 2.8 See Section 2.2. The protons and neutrons are located in a central core that is called the nucleus. The electrons surround the nucleus and occupy most of the volume occupied by atoms. 2.10 See Section 2.3. (a) Atomic number is the number of protons in the nucleus of an atom. (b) Mass number is the total number of protons an d neutrons in an atom. Hence, the number of neutrons is equal to the mass number minus the atomic number. (c) The chemical symbol of an element does not tell us directly the number of particles of any kind in atoms of that element. However, the chemical symbol is always accompanied by the atomic number of the element in the periodic table, and the atomic number is the number of protons in the nucleus of an atom. 2.12 See Section 2.5. 4 P represents 4 atoms of phosphorus, whereas P 4 represents a molecule composed of 4 atoms of phosphorus. 2.14 See Section 2.5 and Figure 2.11. Each molecule of methane, CH 4, is composed of one carbon atom and four hydrogen atoms. 2.16 See Sections 2.5, 2.6. Carbon monoxide, CO, exists as molecules composed of one carbon atom and one oxygen atom each. Cesium bromide, CsBr, exists as Cs+ ions and Br− ions present in a one-to-one ratio. 10 2.18 See Section 2.8. In naming acids corresponding to anions, if the name of the anion ends in ate, the ate ending is changed to ic and the word acid is added. If the name of the anion ends in ite, the ite ending is changed to ous and the word acid is added. 2.20 See Section 2.8. A Roman numeral in parentheses represnets the posi tive charge on transition metal ions. The correct name for CrCl 3 is chromium(III) chloride. 2.22 See Section 2.7. Ionic compounds are generally hard crystalline solids which do not conduct electricity and have high melting points. However, ionic compounds do conduct electricity when melted or when dissolved in water. Molecular compounds consisting of small molecules ge nerally exist as low-melting solids, liquids or gases at room temperature and generally do not conduct electricity. 2.24 See Section 2.7. The chemist should conclude the solid is an ionic compound because ionic compounds exist as crystalline solids having high melting points and conduct electricity when dissolved in water. 2.26 See Section 2.9. In a molten (liquid) ionic compound, the ions present are free to move about and can carry an electrical current. Molten molecular compounds are po or conductors of electricity because they would contain very little (if any) charged particles. 2.28 See Section 2.5. Many responses are equally valid here. Below are a few common examples. These lists are not comprehensive; many other answers are also right. Th e periodic table on the inside cover of your text is color coded to indicate metals, non-metals and metalloids. (a) Common metallic elements: iron, Fe; gold, Au; lead, Pb; copper, Cu; aluminum, Al (b) Common non-metallic elements: carbon, C; hydrogen, H; oxygen, O; nitrogen, N (c) Metalloids: boron, B; silicon, Si; germanium, Ge; arsenic, As; antimony, Sb; tellurium, Te (d) Elements that are diatomic molecules: nitrogen, N 2; oxygen, O 2; hydrogen, H 2; fluorine, F 2; chlorine, Cl 2; bromine, Br 2; iodine, I 2 2.30 See Section 2.3. (a)59 28Ni (b) 184 74W 11 2.32 See Section 2.3. (a) 11 5B (b) 55 25Mn (c) 28 14Si 2.34 See Section 2.3 and Examples 2.1, 2.2, 2.3. (a) 32 16S : 16 protons, 16 neutrons, 16 electrons (b) 24 12Mg : 12 protons, 12 neutrons, 12 electrons (c) 37 17Cl : 17 protons, 20 neutrons, 17 electrons 2.36 See Section 2.3. Since the charge of the ion is –2, the number of electrons in this ion is two more than the number of protons. e- = p + 2 It is given in the problem that the number of neut rons is 20 more than the number of electrons. Therefore, there are 22 more neutrons than protons. n = e- + 20 = ( p + 2 ) + 20 = p + 22 It is also given that the ion has a mass number of 126. Therefore, the sum of the protons and neutrons is equal to 126. p + n = 126 Since n = p + 22, this information can be used to solve for the number of protons. p + ( p + 22 ) = 126 2p = 104 p = 52, therefore the element must be tellerium, Te. n = p + 22 = 52 + 22 = 74, therefore there must be 74 neutrons in this ion. e- = p + 2 = 52 + 2 = 54, therefore there must be 54 electrons. The symbol for this ion is therefore 126 2- 52Te 2.38 See Section 2.3 and Example 2.2. (a) 92 4Be+ (b) 72 2+ 32Ge (c) 79 35Br− 12 2.40 See Section 2.3 and Examples 2.1, 2.2. (a) 51 3+ 23V (b) 127 53I− (c) 102 3+ 44Ru (d) 31 15P 2.42 See Section 2.3. 19 9F 0 9 10 9 31 15P 0 15 16 15 57 27Co3+ +3 27 30 24 32 16S2- - 2 16 16 18 2.44 See Section 2.4 and Example 2.5. Multiply the mass of each isotope by its decimal fr action, and add these numbers together to obtain the atomic mass of the element. Atomic mass = (0.6917)(62.9396 u) + (0.3083)(64.9278 u) = 63.55 u The element is copper, Cu. 2.46 See Section 2.4 and Example 2.5. Multiply the mass of each isotope by its decimal fr action, and add these numbers together to obtain the atomic mass of the element. Atomic mass of In = (0.957)(114.904 u) + (0.043)(112.904 u) = 115 u 2.48 See Section 2.4 and Example 2.5. (a) Multiply the mass of each isotope by its decima l fraction, and add these numbers together to obtain the atomic mass of the element. Atomic mass = (0.922)(27.977 u) + (0.0467)(28.976 u) + (0.0310)(29.974 u) = 28.1 μ The element is silicon, Si. (b) 28Si, 29Si, and 30Si 2.50 See Section 2.4 and Example 2.5. Let y be the percentage of atoms having a mass of 64.928 u and 100.0 44.58⎛⎞ ⎜⎟⎝⎠ y be the percentage of atoms having a mass of 62.940 u giving: y + 100.0 44.58⎛⎞ ⎜⎟⎝⎠ y = 100.100%, 3.243 y = 100.00%, and y = 30.83%. 13Atomic mass = (0.3083)(64.928 u) + (0.6917)(62.940 u) = 63.55 u The element is copper, Cu. 2.52 See Section 2.4 and Example 2.5 . (a) 79Br and 81Br (b) Let y be the percentage of the atoms having a mass of 78.918 u and (100.00-y) be the percentage of the atoms having a mass of 80.916 u giving: y 100.00⎛⎞ ⎜⎟⎝⎠ (78.918 u) + 100.00 y 100.00− ⎛⎞ ⎜⎟⎝⎠(80.916 u) = 79.904 u (y)(78.918 u) + (100.00-y)(80.916 u) = (100.00)(79.904 u) 1.998y = 101.2 and y = 50.65 Hence, %79Br = 50.65% and %81Br = 100.00% − 50.65% = 49.35%. 2.54 See Section 2.5 and Figure 2.8. (a) Lead, Pb (b) Barium, Ba (c) Lanthanum, La (d) Bismuth, Bi 2.56 See Section 2.5 and Figure 2.8. (a) He, Ne, Kr, Xe or Rn (b) P, As, Sb or Bi (c) Fe, Ru or Hs (d) Cr, Mo or Sg 2.58 See Section 2.5 and Figure 2.8. (a) Ba, representative (b) Mo, transition (c) F, representative (d) Hf, transition 2.60 See Section 2.5 and Figure 2.8. (a) Br, representative (b) Pt, transition (c ) Ru, transition (d) U, inner transition 2.62 See Section 2.5 and Figure 2.8. (a) Mg, magnesium (b) Kr, krypton (c) Fr, francium (d) I, iodine 2.64 See Section 2.5. There are five elements in Group 4A of the peri odic table. They are non-metal: carbon (C), metalloids: silicon (Si) and germanium (G e), and metals: tin (Sn) and lead (Pb). 2.66 See Section 2.6 and Table 2.1. S and Se should have similar properties, since both are members of Group VIA 2.68 See Section 2.5 and Figure 2.8. Cl and I should have similar properties, since both are members of Group VIIA. 2.70 See Section 2.8. 14 (a) C 2H2 (b) PCl 3 15 2.72 See Section 2.8. 2.74 See Section 2.5 and Example 2.7. (a) Molecular mass of P 4O10: 4[P] × 30.97 =123.88 10[O] × 16.00 = 160.00 283.88 u (b) Molecular mass of C 6H7N: 6[C] × 12.01 = 72.06 7[H] × 1.01 =7.07 1[N] × 14.01 =14.01 93.14 u (c) Molecular mass of H 3PO 4: 3[H] × 1.01 = 3.03 1[P] × 30.97 = 3097 4[O] × 16.00 =64.00 98.00 u 2.76 See Section 2.5 and Example 2.7. Molecular mass of B 10H14: 10[B] × 10.81 = 108.10 14[H] × 1.01 = 14.14 122.24 u 2.78 See Section 2.6 and Table 2.1. (a) K+ (b) Br− (c) Ba2+ (d) S2− 2.80 See Section 2.6 and Example 2.8. (a) LiI (b) Cs 2O (c) YCl 3 2.82 See Section 2.6, Table 2.1, and Example 2.8. (a) MgF 2 (b) Na 2O (c) Sc 2Se3 (d) Ba 3N2 2.84 See Table 2.2. (a) chromate, CrO2 4− (b) carbonate, CO2 3− (c) sulfate, SO24− 2.86 See Table 2.2. (a) perchlorate, ClO 4− (b) sulfite, SO2 3− (c) hydrogen carbonate HCO 3− 16 2.88 See Section 2.6, Tables 2.1, 2.2, and Example 2.9. (a) sodium nitrate, NaNO 3 (b) beryllium hydroxide, Be(OH) 2 (c) ammonium acetate, NH 4CH 3CO 2 (d) potassium sulfite, K 2SO 3 2.90 See Section 2.7. The symbol and charge for the following ions: (a) barium ion Ba 2+ (b) perchlorate ion ClO 4- (c) cobalt(II) ion Co2+ (d) sulfate ion SO 42- 2.92 See Section 2.6 and Example 2.10. (a) Formula mass of NaOH: 1[Na] × 22.99 = 22.99 1[O] × 16.00 = 16.00 1[H] × 1.01 = 1.01 40.00 u (b) Formula mass of K 2CO 3: 2[K] × 39.10 = 78.20 1[C] × 12.01 = 12.01 3[O] × 16.00 = 48.00 138.2 1 u (c) Formula mass of Ca 3(PO 4)2: 3[Ca] × 40.08 = 120.24 2[P] × 30.97 = 61.94 8[O] × 16.00 = 128.00 310.18 u 2.94 See Section 2.8 and Tables 2.1, 2.2, 2.3. (a) NH 4Br, ammonium bromide (b) BaCl 2, barium chloride (c) K 2O, potassium oxide (d) Sr(NO 3)2, strontium nitrate 2.96 See Section 2.8, Tables 2.2, 2.3, 2.4, and Example 2.13. (a) RhBr 2, rhodium(II) bromide (b) CuCN, copper(I) cyanide c) V(NO 3)3, vanadium(III) nitrate 2.98 See Section 2.8, Tables 2.1, 2.2, 2.3, 2.4, and Examples 2.12, 2.13. (a) calcium nitride, Ca 3N2 (b) chromiun(III) perchlorate, Cr(ClO 4)3 (c) tin(II) fluoride, SnF 2 17(d) potassium permanganate, KMnO 4 18 2.100 See Section 2.8 and Tables 2.2, 2.3. (a) cyanide ion: HCN, hydrocyanic acid (b) nitrate ion: HNO 3, nitric acid (c) phosphate ion: H 3PO 4, phosphoric acid 2.102 See Section 2.8 and Tables 2.2, 2.3. (a) H 2CO 3 is related to the carbonate ion, CO2 3−, and is carbonic acid. (b) HBr is related to the bromide ion, Br−, and is hydrobromic acid. (c) HNO 2 is related to the nitrite ion, NO 2−, and is nitrous acid. 2.104 See Section 2.8 and Tables 2.3 and 2.4. MnO, manganese(II) oxide 2.106 See Section 2.8 and Table 2.5. (a) sulfur difluoride, SF 2 (b) silicon tetrachloride, SiCl 4 (c) gallium trichloride, GaCl 3 (d) dinitrogen trioxide, N 2O3 2.108 See Section 2.8, Table 2.5, and Example 2.15. (a) HI, hydrogen iodide (b) NF 3, nitrogen trifluoride (c) SO 2, sulfur dioxide (d) N 2Cl4, dinitrogen tetrachloride 2.110 See Section 2.8. (a) cyclohexane (b) 2-bromopentane 2.112 See Section 2.8. (a) 2-bromopentane (b) 2-iodobutane 2.114 See Section 2.6 and 2.7. Na2CO 3 will dissolve in water because is an ionic compound. Cl 2 will be a gas because it is a molecular compound with a relatively low molecular mass. 2.116 See Section 2.7. Beaker B. Lithium sulfide has the formula Li 2S. 2 molecules Li 2S dissociate in water to form 4 Li+ and 2S2-. 19 2.118 See Section 2.7. (a) Solution contains Al3+ and Br− ions and conducts electricity. (b) Solution contains C 2H4(OH) 2 molecules and does not conduct electricity. (c) Solution contains Ca2+ and NO 3− ions and conducts electricity. (d) Solution contains NH 4+ and SO2 4− ions and conducts electricity. (e) Solution contains K+ and Cr 2O2 7− ions and conducts electricity. 2.120 See Section 2.8, Tables 2.2, 2.4, and Example 2.9. iron(III) sulfate, Fe 2(SO 4)3 2.122 See Section 2.8, Tables 2.2, 2.4, and Example 2.9. Potassium nitrate KNO 3 and ammonium carbonate (NH 4)2CO 3 2.124 See Section 2.3 (a) 39 18Ar (must be Ar to have a mass number for an isotope that is close to its atomic mass) (b) 40 20 Ca2+ (must be Ca, since the mass number of the isotope is close to the atomic mass of Ca but not that of any other alkaline earth metal) 2.126 See Section 2.8, Tables 2.3, 2.4, 2.5, and Examples 2.12, 2.14. (a) calcium phosphate, Ca 3(PO 4)2, ionic (b) germanium dioxide, GeO 2, ionic (c) iron(III) sulfate, Fe 2(SO 4)3, ionic (d) phosphorous tribromide, PBr 3, molecular 2.128 See Section 2.3 and Example 2.1. 239 94Pu: 94 protons, 145 neutrons, 94 electrons 2.130 See Section 2.3 and Figure 2.7. 1 1 H (protium): 1 proton, 1 electron 2 1 H (deuterium): 1 proton, 1 neutron, 1 electron 3 1 H (tritium): 1 proton, 2 neutron, 1 electron 2.132 See Section 2.8, Tables 2.1, 2.2, 2.3, 2.4, and Examples 2.12, 2.14. (a) sodium selenide, Na 2Se (b) nickel (II) bromide, NiBr 2 (c) dinitrogen pentoxide, N 2O5 (d) copper(II) sulfate, CuSO 4 (e) ammonium sulfite, (NH 4)2SO 3 20 2.134 See Section 2.3 and 2.5. (a) 78 38Sr, strontium (b) Group 2, alkaline earth (c) 2+ (d) Mg, magnesium 2.136 See Section 2.4 and Example 2.5. The accepted mass of carbon is 12.011 u 12.011 = 0.98938 x 12.00 + 0.01062 x (isotopic mass of 13C) Isotopic mass of 13C = 13.036 u 21Chapter 3: Stoichiometry I 3.2 See Section 3.1. 1. Assume the equation contains one formula unit of the most complicated substance, and bring the atoms of this substance into balance by adjusting the coefficients of the substances on the other side of the equation. 2. Continue by adjusting the coefficients of the reactants and products until the same numbers of each type of atom appear on both sides of the equation and the coefficients are given in terms of whole numbers. 3. Check to make sure your final equation contains the same number s of atoms of each type on both sides. 3.4 See Section. 3.1 3.6 See Section 3.2. The number of objects in one mole has been determined experimentally to be 6.022 × 1023 (when expressed to four significant figures) and is known as Avogadro’s number. 3.8 See Section 3.2 and Table 3.2. The units for atomic, molecular and formula masses are atomic mass units ( μ), and the units for molar masses are grams per mole (g/mol). 3.10 See Section 3.2. moles Avogadro’s number atoms where the number of moles is multiplied by 6.022 × 1023 atoms 1 mol . 3.12 See Section 3.3, Figure 3.4, and Question 3.9. The masses of C and H are obtained as outlined in 3. 9. These masses are subtracted from the total mass of sample burned to obtain the mass of O in the sample. The percentage O by mass is calculated using (mass of O/mass of sample) × 100%. 3.14 See Section 3.4. One mol of N 2 reacts with two mol H 2 to form one mole N 2H4. H2 O2 H2O2 22 3.16 See Section 3.4. “The reaction was carried out with the reactants pres ent in stoichiometric amounts,” means the reactants were present in the exact amounts n ecessary to react with one another and none was present in excess. 3.18 See Section 3.4 The statement, “C 2H4 is the limiting reactant and oxygen is pr esent in excess in a combustion reaction,” means the amount of carbon dioxide and water that can be produced by the reaction is limited by the amount of C 2H4 present. It also means the C 2H4 is completely consumed and an excess of oxygen remains when the reaction is complete. 3.20 See Section. 3.1 (a) 2SO 2 + O 2 Æ 2SO 3 (b) sulfur trioxide 3.22 See Section 3.1 and Examples 3.1, 3.2. (a) Unbalanced: Mg 3N2 + H 2O → NH 3 + Mg(OH) 2 Start with Mg 3N2. Step 1: Mg 3N2 + H 2O → 2NH 3 + 3Mg(OH) 2 Balances Mg, N. Step 2: Mg 3N2 + 6H2O → 2NH 3 + 3Mg(OH) 2 Balances H,O. (b) Unbalanced: Fe + O 2 → Fe 2O3 Start with Fe. Step 1: 2 Fe + O 2 → Fe 2O3 Balances Fe. Step 2: 2Fe + 3 2−O2 → Fe 2O3 Balances O. Step 3: 4 Fe + 3 O2 → 2Fe2O3 Gives whole number coefficients. (c) Unbalanced: Zn + H 3PO 4 → H2 + Zn 3(PO 4)2 Start with H 3PO 4. Step 1: Zn + 2 H3PO 4 → H2 + Zn 3(PO 4)2 Balances PO 4 units. Step 2: 3 Zn + 2H 3PO 4 → H2 + Zn 3(PO 4)2 Balances Zn. Step 3: 3Zn + 2H 3PO 4 → 3H 2 + Zn 3(PO 4)2 Balances H. 3.24 See Section. 3.1 (a)? Al(s) + ? O 2(g) ? Al2O3(s) Select order: Al then O 2 Al(s) + ? O 2(g) 1 Al2O3(s) 2 Al 2 Al(s) + 3 2 O2(g) 1 Al2O3(s) 3 O 4 Al(s) + 3 O 2(g) 2 Al2O3(s) (b)? N 2(g) + ? H 2(g) ? NH 3(g) Select order: H then N ? N 2(g) + 3 H 2(g) 2 NH 3(g) 6 H 1 N 2(g) + 3 H 2(g) 2 NH 3(g) 2 N 23 (c)? C 6H6(l) + ? O 2(g) ? H2O(l) + ? CO 2(g) Select order: C, H then O C 6H6(l) + ? O 2(g) ? H2O(l) + 6 CO 2(g) 6 C C 6H6(l) + ? O 2(g) 3 H2O(l) + 6 CO 2(g) 6 H C 6H6(l) + 15 2 O2(g) 3 H2O(l) + 6 CO 2(g) 15 O 2 C 6H6(l) + 15 O 2(g) 6 H2O(l) + 12 CO 2(g) 3.26 See Section. 3.1. (a) HF (g) 2HO⎯⎯⎯→H+ (aq) + F- (aq) (b) Li 2SO 4 (s) 2HO⎯⎯⎯→2 Li+ (aq) + SO 42- (aq) 3.28 See Section. 3.1. (a) HCl (g) 2HO⎯⎯⎯→ H+ (aq) + Cl- (aq) (b) Ba(NO 3)2 (s) 2HO⎯⎯⎯→ Ba2+ (aq) + 2NO 3- (aq) 3.30 See Section 3.1 and Example 3.3. (a) Unbalanced: Mg(OH) 2 + HF → MgF 2 + H 2O Start with MgF 2. Step 1: Mg(OH) 2 + 2HF → MgF 2 + H 2O Balances F. Step 2: Mg(OH) 2 + 2HF → MgF 2 + 2H2O Balances H, O. (b) Unbalanced: HCl + NaOH → H2O + NaCl Start with NaCl. Step 1: HCl + NaOH → H2O + NaCl Balances H, O, Na, Cl (c) Unbalanced: H 2SO 4 + Sr(OH) 2 → H2O + SrSO 4 Start with H 2O. Step 1: H 2SO 4 + Sr(OH) 2 → 2H2O + SrSO 4 Balances H, O. 3.32 See Section. 3.1. (a) Unbalanced: ? C 4H10(g) + ? O 2(g) ? CO 2(g) + ? H 2O(g) Balance C: C 4H10(g) + ? O 2(g) 4 CO 2(g) + ? H 2O(g) 4 C’s Balance H: C 4H10(g) + ? O 2(g) 4 CO 2(g) + 5 H 2O(g) 10 H’s Balance O: C 4H10(g) + 13 2 O2(g) 4 CO 2(g) + 5 H 2O(g) 13 O’s Multiply by 2: 2 C 4H10(g) + 13 O 2(g) 8 CO 2(g) + 10 H 2O(g) (b) Unbalanced: ? C 6H12O6(s) + ? O 2(g) ? CO 2(g) + ? H 2O(g) Balance C: C 6H12O6(s) + ? O 2(g) 6 CO 2(g) + ? H 2O(g)6 C’s Balance H: C 6H12O6(s) + ? O 2(g) 6 CO 2(g) + 6 H 2O(g) 12 H’s Balance O: C 6H12O6(s) + 6 O 2(g) 6 CO 2(g) + 6 H 2O(g) 18 O’s 24 (c) Unbalanced: ? C 4H8O(A) + ? O 2(g) ? CO 2(g) + ? H 2O(g) Balance C: C 4H8O(A) + ? O 2(g) 4 CO 2(g) + ? H 2O(g) 4 C’s Balance H: C 4H8O(A) + ? O 2(g) 4 CO 2(g) + 4 H 2O(g) 8 H’s Balance O: C 4H8O(A) + 11 2 O2(g) 4 CO 2(g) + 4 H 2O(g) 12 O’s Multiply by 2: 2 C 4H8O(A) + 11 O 2(g) 8 CO 2(g) + 8 H 2O(g) 3.34 See Section 3.1 and Examples 3.3, 3.4. (a) Unbalanced: C 8H8 + O 2 → CO 2 + H 2O Start with C 8H8. Step 1: C 8H8 + O 2 → 8CO 2 + 4H2O Balances C, H. Step 2: C 8H8 + 10O 2 → 8CO 2 + 4H 2O Balances O. (b) Unbalanced: HCl + KOH → H2O + KCl Start with KCl. Step 1: HCl + KOH → H2O + KCl Balances H, O, K, Cl. 3.36 See Section 3.1 and Example 3.4. Unbalanced: CH 3OC(CH 3)3 + O 2 → CO 2 + H 2O Start with CH 3OC(CH 3)3. Step 1: CH 3OC(CH 3)3 + O 2 → 5CO 2 + 6H2O Balance C, H. Step 2: CH 3OC(CH 3)3 + 15 2−O2 → 5CO 2 + 6H 2O Balances O. Step 3: 2 CH 3OC(CH 3)3 + 15 O2 → 10CO 2 + 12 H2O Gives whole number coefficients. 3.38 See Section 3.1 and Example 3.3. Unbalanced: H 3PO 3 → H3PO 4 + PH 3 Start with H 3PO 3 and H 3PO 4. Step 1: 4 H3PO 3 → 3H3PO 4 + PH 3 Balances O, P, H. 3.40 See Section 3.1 and Examples 3.1, 3.2. Unbalanced: UO 2 + CCl 4 → UCl 4 + COCl 2 Start with CCl 4. Step 1: UO 2 + 2CCl 4 → UCl 4 + 2COCl 2 Balances Cl, C, O. 3.42 See Section. 3.1. (a) NH 4Cl: Assume hydrogen has an an oxidation number of +1 and chlorine -1; therefore N has an oxidation number of -3. (b) N 2O: Assume oxygen has an oxidation number of -2; therefore each nitrogen has an oxidation state of +1. (c) Ag: All elements in their elemental state have an oxidation number of zero 25(d) AuI 3: Assume each iodine has an oxidation number of -1; therefore gold must have an oxidation number of +3. 3.44 See Section. 3.1. Na2O2: Assume that each sodium has an oxidation number of +1; therefore each oxygen must have an oxidation number of -1. 3.46 See Section 3.1 and Example 3.5. Unbalanced: Fe + O 2 → Fe 2O3 Start with O 2. Step 1: Fe + 3 O2 → 2Fe2O3 Balances O. Step 2: 4 Fe + 3O 2 → 2Fe 2O3 Balances Fe. Assigning an ON of -2 to oxygen in Fe 2O3 gives 4Fe + 3O 2 → 2FeO 3 ON: 0 0 +3,-2 The ON of the Fe atoms increase, and the ON of the O atoms decrease. The Fe metal is oxidized, and the O atoms in O 2 are reduced. 3.48 See Section 3.1 and Example 3.5. Unbalanced: P 4(s) + O 2(g) → P4O10(s) Start with O 2. Step 1: P 4(s) + 5 O2(g) → P4O10(s) Balances O. Assigning an ON of −2 to O in P 4O10 gives P 4(s) + 5O 2(g) → P4O10(s) ON 0 0 +5, −2 The ON of the P atoms increases, and the ON of the O atoms decreases. The P atoms in P 4 are oxidized, and the O atoms in O 2 are reduced. 3.50 See Section 3.1 and Example 3.5. Unbalanced: MnO 2 + HCl → MnCl 2 + Cl 2 + H 2O Start with HCl. Step 1: MnO 2 + 4HCl → MnCl 2 + Cl 2 + H 2O Balances Cl. Step 2: MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2O Balances H, O. Assigning an ON of −2 to O in MnO 2 and H 2O, an ON of +1 to H in HCl and H 2O and an ON of −1 to Cl in HCl and MnCl 2 gives MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2O ON: +4, −2 +1, −1 +2, −1 0 +1, −2 The ON of some of the Cl atoms in HCl incr eases, and the ON of the Mn atoms decreases. The Cl atoms in HCl that undergo an increase in ON are oxidized, and the Mn atoms in MnO 2 are reduced. 3.52 See Section 3.2 and Example 3.6. (a) ? atoms Xe = 0.0778 mol Xe × 236.022 10 atoms Xe 1 mol Xe× = 4.69 × 1023 atoms Xe (b) ? atoms K = 1.45 mol K × 236.022 10 atoms K 1 mol K× × 8.73 × 1023 atoms K (c) ? atoms Ti = 55.8 mol Ti × 236.022 10 atoms Ti 1 mol Ti× = 3.36 × 1025 atoms Ti