Chemistry and the Scientific Method

CHAPTER. Chemistry and the Scientific Method -. Write.720 0 3 as 72.0 0 5 and add it to 2.26 0 5 to obtain 74.27 0 5 or.7427 0 3. -2. Consulting Table.2 we have that (a) 400 nm = 400 0 9 m = 4.00 0 7 m (b) 20 ps = 20 0 2 s = 2.0 0 s -3. (a) Using Equation.2, Using Equation., t(in C) = (b) Let x = t(in C) = t(in F), then -4. Using Equation.3, d = m V = -5. We must convert kilometers to meters and hours to seconds ( ) 5 ( 90 32.0) = 68 C 9 T(in K) = 68 + 273.5 = 205 K x = 5 5x (x 32.0) = 9 9 7.8 9x = 5x 60 4x = 60 x = 40 22.55 g.667 ml = 3.53 g ml = 3.53 g cm 3 950 km = 950 0 3 m = 9.50 0 5 m hr= 60 min = 3600 s

2 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly Thus, the velocity is v = 9.50 05 m 3600 s = 263.9 m s The kinetic energy is E K = 2 mv2 = ( ) (3.22 0 5 kg)(263.9 m s ) 2 2 =.2 0 0 kg m 2 s 2 =.2 0 0 J =.2 0 7 kj -6. We use Equation.5 to calculate the potential energy E p = mgh = ( kg)(9.8 m s 2 )(200 m) =.96 0 3 J =.96 kj This energy is converted into kinetic energy as it falls. -7. The energy expended is given by ( kw = kj s ) ( ) 60 s E = (7.5 kj s )(55 min) min ( ) hr = 24 750 kj = 24 750 kw s 3600 s At a cost of $0.0 per kw h, we have $0.69 to two significant figures. -8. The average of the measurements is given by = 6.875 kw h ave = (0.203 + 0.209 + 0.99) μm 3 = 0.204 μm The percentage error is given by (Equation.6) percentage error = = (average value) (true value) 00 true value 0.204 μm 0.200 μm 00 = 2% 0.200 μm See the next Section for a discussion of significant figures. -9. (a) 30 000 000; the last six zeros are not significant figures (b) 30006; all five digits are significant (c) 0.002 9060; the last five digits are significant; the leading zeros are not significant figures (d) 2 is exact because it is a counting number 8.34 298.5-0. y = = 0.02569 96 485.3 The answer should be given to only four significant figures. -. y = 7.2960 32.0 = 0.8978 32.0 = 3.2 8.9000 University Science Books, 20. All rights reserved. www.uscibooks.com

Chapter : Chemistry and the Scientific Method 3-2. The percentage error is given by (Equation.6) (average value) (true value) percentage error = 00 true value 3.03 2.70 = 00 = 0.33 00 = 2% 2.70 2.70-3. Using the conversions given in Appendix B, we have ( ) mile (72 km) = 45 mile.6093 km ( )( ).0936 yd 3ft (260 m) = 430 ft m yd -4. Using Table.2, the time to transfer the data is ( ) (.2 0 9 ) bytes s (25 photos) = 670 s = min photo 45 0 6 bytes -5. Express each piece of data as Charge/0 5 C as in the following table: -6. The appropriate table is Charge/0 5 C 7.05 3.24 0.986 ν/m s t 2 /s 2 0 0 6.0 64 4.0 44 9.0 256 6.0 A plot of ν/m s against t 2 /s 2 is a straight line as shown below. 250 v/m s 200 50 00 50 2 4 6 8 0 2 4 6 t 2 /s 2

CHAPTER 2. Atoms and Molecules 2-. We have that mass of the element mass % of an element = 00 mass of sample Thus, (mass % of an element)(mass of sample) mass of the element = 00 The mass of each element is given by ( ) 38.67 mass of K = (5.650 g) = 2.85 g 00 ( ) 3.86 mass of N = (5.650 g) = 0.783 g 00 mass of O = 5.650 g 2.85 g 0.783 g = 2.682 g 2-2. The atomic mass of O is 5.9994. The atomic mass of C based on O as exactly 6 would be ( ) 6 atomic mass C = (2.007) = 2.02 5.9994 The difference is not significant to five significant figures. 2-3. See the text. 2-4. See the text. 2-5. We will use the atomic masses of bromine and fluorine to five significant figures. The molecular mass of BrF 5 is molecular mass = ()(79.904) + (5)(8.998) = 74.894 mass % Br = 79.904 00 = 45.687% 74.894 mass % F = 5 8.998 00 = 54.33% 74.894 4

Chapter 2: Atoms and Molecules 5 2-6. Assume that both the nucleus and the atom are spheres. The fraction of the volume of the atom that is occupied by the nucleus is given by We are given that fraction = volume of the nucleus volume of the atom = 4 3 πr 3 nuc 4 3 πr 3 atom = r 3 nuc r 3 atom = ( ) 3 rnuc r atom Thus, the fraction is given by r nuc = d nuc 2 = ( 0 5 )d atom 2 fraction = = 0 5 r atom ( ) 0 5 3 r atom = 0 5 r atom The percentage of the atom occupied by the nucleus is 0 3 %. 2-7. The atomic number of phosphorus is 5, so there are 5 protons in its nucleus and 5 electrons in the atom. The number of neutrons in the nucleus is 32 5 = 7. 2-8. Let x be the percentage of lithium-7. Thus, the percentage of lithium-6 is 00 x. Then we have ( ) ( ) 00 x x 6.94 = (6.05 223) + (7.06 0040) 00 00 Multiplying both sides of the equation by 00 gives 694. = (00 x)(6.05 223) + (x)(7.06 0040) Solving for x gives x = 92.5%. The natural abundance of lithium-7 is 92.5% and that of lithium-6 is 00% 92.5% = 7.5%. 2-9. The atomic number of oxygen is 8. The number of electrons in the oxide ion is 8 + 2 = 0 electrons. Examples of cations that are isoelectronic to O 2 are F,Na +,Mg 2+, and Al 3+. 2-0. (a) Silicon has an atomic number of 4. Therefore, there are 4 protons, 28 4 = 4 neutrons, and 4 + 4 = 8 electrons in 28 4 Si4. (b) Tungsten has an atomic number of 74. Therefore, there are 74 protons, 86 74 = 2 neutrons, and 74 5 = 69 electrons in 86 74 W+5. (c) Uranium has an atomic number of 92. Therefore, there are 92 protons, 235 92 = 43 neutrons, and 92 electrons in 235 92 U. (d) Iron has an atomic number of 26. Therefore, there are 26 protons, 58 26 = 32 neutrons, and 26 3 = 23 electrons in 58 26 Fe+3.

CHAPTER 3. The Periodic Table and Chemical Periodicity 3-. Step is to write the following equation P + O 2 P 4 O 0 (not balanced) Using Step 2, there is one phosphorus atom on the left and four phosphorus atoms on the right, so we must add a 4 in front of P. Using Steps 3 and 4, there are 0 oxygen atoms on the right side, and so we add a 5 in front of O 2 to obtain 4P + 5O 2 P 4 O 0 Finally adding the states of the reactants and product, we have 4P(s) + 5O 2 (g) P 4 O 0 (s) 3-2. By analogy with the reaction of calcium with chlorine, so we predict (correctly) that Sr(s) + Cl 2 (g) SrCl 2 (s) 3-3. Iodine is in the same group as chlorine, so predict CI 4 (s) and NI 3 (s). 3-4. Rubidium and sodium are in the same group. Using the reaction between sodium and water as an analogy, we write 3-5. See the text. 3-6. See the text. 2Rb(s) + 2H 2 O(l) 2 RbOH(aq) + H 2 (g) 6

CHAPTER 4. Early Quantum Theory 4-. The energy required to produce B +3 from a boron atom is the sum of the first three ionization energies of a boron atom. 4-2. We have the following figure: energy = (.33 + 4.04 + 6.08) aj =.45 aj 5 ln(i a /aj) 4 3 2 2 3 4 5 6 7 8 9 n 4-3. We use Equation 4.. λ = c ν = 2.9979 08 m s.50 0 8 Hz = 2.00 0 0 m = 0.200 nm = 2.9979 08 m s.50 0 8 s This wavelength corresponds to an X-ray. 4-4. The slope of the plot of E k against ν gives the value of h. The frequencies corresponding to 0.6 aj and 0.2 aj are 4 0 4 s and 8 0 4 s, respectively. Thus, to one significant figure. slope = h = (E k) 2 (E k ) ν 2 ν = 6 0 9 J 2 0 9 J 4 0 4 s 8 0 4 s = 6.7 0 34 J s = 7 0 34 J s 7

8 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly 4-5. We first use Equation 4. The energy is given by Equation 4.3 4-6. We use Equation 4.7 λ = h mv = ν = c λ = 2.9979 08 m s 656.3 0 9 m = 4.568 04 s E = hν = (6.626 0 34 J s)(4.568 0 4 s ) = 3.027 0 9 J = 0.3027 aj = 2.69 0 3 m 6.626 0 34 J s (.6093 km ( ) kg (45.9 g) (20 mi h ) 000 g mi )( ) hr 3600 s A wavelength of 2.69 0 3 m is totally negligible relative to the diameter of a golf ball. 4-7. Using Equation 4., we have ( λ = (.097 07 m ) 3 ) = 7.8009 0 5 m 2 5 2 λ = 7.8009 0 5 m =.282 0 6 m = 282 nm 4-8. We use Equation 4. and the fact that n f = 2 for the Balmer series ( 656.3 0 9 m = (.097 07 m ) 4 ) n 2 0.39 = 4 n 2 or n 2 = 9orn = 3. The state n = 3 is called the second excited state because it is two levels above the ground state of hydrogen (n = ). 4-9. We use Equation 4.5 with E = 2.8 aj (from Example 4-9) λ = hc E = (6.626 0 34 J s)(2.9979 0 8 m s ) 2.8 0 8 J = 9. 0 8 m = 9. nm University Science Books, 20. All rights reserved. www.uscibooks.com

CHAPTER 5. Quantum Theory and Atomic Structure 5-. We use Equation 5. p The uncertainty in the speed is The speed is h 4π x = 6.626 0 34 J s 4π( 0 3 m) = 5.3 0 32 kg m s v = p m = 5.3 0 32 kg m s 45.9 0 3 kg 0 30 m s ( )( )( ) 200 km 000 m hr v = = 60 m s hr km 3600 s Thus, the percentage uncertainty in the speed is 5-2. See the text. 5-3. See the text. 5-4. See the text. 5-5. See the text. 5-6. See the text. v = 0 30 m s 60 m s 00 = 2 0 30 % 5-7. The ground-state electron configuration of an O 2 ion (0 electrons) is s 2 2s 2 2p 6. The first excited state occurs when one of the 2p electrons is promoted to the 3s state, giving s 2 2s 2 2p 5 3s. 5-8. (a) A neon atom, with its ground-state electron configuration [He]2s 2 2p 6, has eight valence electrons. (b) The ground-state configuration of Al 3+ is simply [Ne] in the notation of Table 5.6. Being a cation, we say that Al 3+ has no valence electrons. 9

0 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly (c) The ground-state configuration of Mg 2+ is simply [Ne] in the notation of Table 5.6. Being a cation, we say that Mg 2+ has no valence electrons. (d) A phosphorus atom, with its ground-state electron configuration [Ne]3s 2 3p 3, has five valence electrons. (e) The ground-state configuration of Cl is [Ne]3s 2 3p 6 or [Ar] in the notation of Table 5.6. Being an anion, we say that Cl has eight valence electrons. University Science Books, 20. All rights reserved. www.uscibooks.com

CHAPTER 6. Ionic Bonds and Compounds 6-. Aluminum belongs to Group 3. An aluminum atom has three more electrons than a neon atom, or [Ne]3s 2 3p, and so it achieves a ground-state electron configuration of [Ne] by losing three electrons to give Al 3+. 6-2. The ground-state electron configuration of a selenium atom is [Ar]3s 2 3p 4. It can achieve a krypton-like electron configuration by adding two electrons, giving Se 2. The ground-state electron configuration of a nitrogen atom is [He]2s 2 2p 3. It can achieve a neon-like electron configuration by adding three electrons, giving N 3. The ground-state electron configuration of a phosphorus atom is [Ne]3s 2 3p 3. It can achieve an argon-like electron configuration by adding three electrons, giving P 3. 6-3. See the text. 6-4. See the text. Both Mg 2+ and N 3 have a [Ne] electron configuration (s 2 2s 2 2p 6 ). To balance the total ionic charges, the formula of magnesium nitride is Mg 3 N 2, which is a solid at room temperature, as are all ionic compounds. 6-5. An indium atom has a ground-state electron configuration of [Kr]5s 2 4d 0 5p. It can achieve an 8-electron configuration by losing its two 5s electrons and its one 5p electron to become [Ar]4s 2 4p 6 4d 0 or In 3+. 6-6. See the text. 6-7. The ground-state electron configuration of a Pd 2+ ion is regular: [Kr]4d 8. 6-8. The ground-state electron configuration of these ions are as follows: Cr(II) : [Ar]3d 4 Cr(III) : [Ar]3d 3 Cr(VI) : [Ar] 6-9. Aluminum, magnesium, and sodium are in the same row of the periodic table (3rd row) and the ions are isoelectronic, and so the order of their sizes goes as Al 3+ < Mg 2+ < Na +. Bromine and iodine are in higher rows, and so we have Al 3+ < Mg 2+ < Na + < Br < I.

2 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly 6-0. As in Example 6-0, we describe the reaction in three steps Cs(g) Cs + (g) + e Cl(g) + e Cl (g) Cs + (g) + Cl (g) Cs + Cl E = I = 0.624 aj E 2 = EA = 0.580 aj ( ) ()( ) E 3 = (23 aj pm) = 0.664 aj 67 + 8 where we have used values from Tables 6.4 and 6.5. The total is E rxn = E + E 2 + E 3 = 0.620 aj University Science Books, 20. All rights reserved. www.uscibooks.com