Chemistry 13th edition by Raymond Chang PDF

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textbook for general chemistry 1 in winter session 2021 / 2022...

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STUDENT SOLUTION MANUAL

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CHAPTER 1 CHEMISTRY: THE STUDY OF CHANGE Problem Categories Biological: 1.24, 1.48, 1.69, 1.70, 1.78, 1.84, 1.93, 1.95, 1.96, 1.97, 1.105. Conceptual: 1.3, 1.4, 1.11, 1.12, 1.15, 1.16, 1.54, 1.62, 1.89, 1.101, 1.103. Environmental: 1.70, 1.87, 1.89, 1.92, 1.98. Industrial: 1.51, 1.55, 1.72, 1.81, 1.91. Difficulty Level Easy: 1.3, 1.11, 1.13, 1.14, 1.15, 1.21, 1.22, 1.23, 1.24, 1.25, 1.26, 1.29, 1.30, 1.31, 1.32, 1.33, 1.34, 1.54, 1.55, 1.63, 1.64, 1.77, 1.80, 1.84, 1.89, 1.91. Medium: 1.4, 1.12, 1.16, 1.35, 1.36, 1.37, 1.38, 1.39, 1.40, 1.41, 1.42, 1.43, 1.44, 1.45, 1.46, 1.47, 1.48, 1.49, 1.50, 1.51, 1.52, 1.53, 1.56, 1.57, 1.59, 1.60, 1.61, 1.62, 1.70, 1.71, 1.72, 1.73, 1.74, 1.75, 1.76, 1.78, 1.79, 1.81, 1.82, 1.83, 1.85, 1.94, 1.95, 1.96, 1.97, 1.98. Difficult: 1.58, 1.65, 1.66, 1.67, 1.68, 1.69, 1.86, 1.87, 1.88, 1.90, 1.92, 1.93, 1.99, 1.100, 1.101, 1.102, 1.103, 1.104, 1.105, 1.106. (a)

Quantitative. This statement clearly involves a measurable distance.

(b)

Qualitative. This is a value judgment. There is no numerical scale of measurement for artistic excellence.

(c)

Qualitative. If the numerical values for the densities of ice and water were given, it would be a quantitative statement.

(d)

Qualitative. Another value judgment.

(e)

Qualitative. Even though numbers are involved, they are not the result of measurement.

1.4

(a)

hypothesis

1.11

(a)

Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity are changed.

(b)

Chemical property. The fertilizer is consumed by the growing plants; it is turned into vegetable matter (different composition).

(c)

Physical property. The measurement of the boiling point of water does not change its identity or composition.

(d)

Physical property. The measurement of the densities of lead and aluminum does not change their composition.

(e)

Chemical property. When uranium undergoes nuclear decay, the products are chemically different substances.

(a)

Physical change. The helium isn't changed in any way by leaking out of the balloon.

(b)

Chemical change in the battery.

(c)

Physical change. The orange juice concentrate can be regenerated by evaporation of the water.

(d)

Chemical change. Photosynthesis changes water, carbon dioxide, etc., into complex organic matter.

(e)

Physical change. The salt can be recovered unchanged by evaporation.

1.3

1.12

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(b)

law

(c)

theory

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CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE

1.13

Li, lithium; F, fluorine; P, phosphorus; Cu, copper; As, arsenic; Zn, zinc; Cl, chlorine; Pt, platinum; Mg, magnesium; U, uranium; Al, aluminum; Si, silicon; Ne, neon.

1.14

(a) (f)

K Pu

1.15

(a)

element

1.16

(a) (d) (g)

homogeneous mixture homogeneous mixture heterogeneous mixture

1.21

density =

1.22

Strategy: We are given the density and volume of a liquid and asked to calculate the mass of the liquid. Rearrange the density equation, Equation (1.1) of the text, to solve for mass.

Sn S

(b) (g)

(b)

(c) (h)

Cr Ar

(d) B (i) Hg

compound (b) (e)

(c)

element

(e)

(d)

element heterogeneous mixture

(c) (f)

Ba

compound compound homogeneous mixture

mass 586 g = = 3.12 g/mL volume 188 mL

density =

mass volume

Solution: mass = density × volume mass of ethanol =

1.23

? °C = (° F − 32° F) ×

0.798 g × 17.4 mL = 13.9 g 1 mL

5° C 9° F

5 °C = 35°C 9° F 5 °C = − 11° C (12 − 32)° F × 9° F 5°C (102 − 32)° F × = 39°C 9° F 5°C (1852 − 32)° F × = 1011 °C 9° F 9 °F ⎞ ⎛ ⎜ ° C × 5 °C ⎟ + 32° F ⎝ ⎠

(a)

? °C = (95 − 32)° F ×

(b)

? °C =

(c)

? °C =

(d)

? °C =

(e)

? °F =

9°F ⎞ ⎛ + 32°F = − 459.67° F ? ° F = ⎜ −273.15 °C × 5°C ⎟⎠ ⎝

1.24

Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between Celsius and Fahrenheit given in Section 1.7 of the text. Substitute the temperature values given in the problem into the appropriate equation. (a)

Conversion from Fahrenheit to Celsius. ? °C = (° F − 32° F) ×

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5°C 9° F

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99

4.100

First, the gases could be tested to see if they supported combustion. O2 would support combustion, CO2 would not. Second, if CO2 is bubbled through a solution of calcium hydroxide [Ca(OH)2], a white precipitate of CaCO3 forms. No reaction occurs when O2 is bubbled through a calcium hydroxide solution.

4.101

Choice (d), 0.20 M Mg(NO3)2, should be the best conductor of electricity; the total ion concentration in this solution is 0.60 M. The total ion concentrations for solutions (a) and (c) are 0.40 M and 0.50 M, respectively. We can rule out choice (b), because acetic acid is a weak electrolyte.

4.102

Starting with a balanced chemical equation:

→ MgCl2(aq) + H2(g) Mg(s) + 2HCl(aq) ⎯⎯ From the mass of Mg, you can calculate moles of Mg. Then, using the mole ratio from the balanced equation above, you can calculate moles of HCl reacted. 4.47 g Mg ×

1 mol Mg 2 mol HCl × = 0.3677 mol HCl reacted 24.31 g Mg 1 mol Mg

Next we can calculate the number of moles of HCl in the original solution. 2.00 mol HCl × (5.00 × 10 2 mL) = 1.00 mol HCl 1000 mL soln

Moles HCl remaining = 1.00 mol − 0.3677 mol = 0.6323 mol HCl conc. of HCl after reaction =

4.103

mol HCl 0.6323 mol HCl = = 1.26 mol/L = 1.26 M L soln 0.500 L

The balanced equation for the displacement reaction is: Zn(s) + CuSO4(aq) ⎯⎯ → ZnSO4(aq) + Cu(s) The moles of CuSO4 that react with 7.89 g of zinc are: 7.89 g Zn ×

1 mol Zn 1 mol CuSO 4 × = 0.1207 mol CuSO 4 65.39 g Zn 1 mol Zn

The volume of the 0.156 M CuSO4 solution needed to react with 7.89 g Zn is: L of soln =

mole solute 0.1207 mol CuSO4 = = 0.774 L = 774 mL 0.156 mol/L M 2+

Would you expect Zn to displace Cu

4.104

from solution, as shown in the equation?

The balanced equation is: 2HCl(aq) + Na2CO3(s) ⎯⎯ → CO2(g) + H2O(l) + 2NaCl(aq) The mole ratio from the balanced equation is 2 moles HCl : 1 mole Na2CO3. The moles of HCl needed to react with 0.256 g of Na2CO3 are: 0.256 g Na 2CO3 ×

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1 mol Na 2CO 3 2 mol HCl × = 4.831 × 10−3 mol HCl 105.99 g Na 2CO 3 1 mol Na 2CO 3

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CHAPTER 7 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS Problem Categories Biological: 7.110, 7.122, 7.125, 7.132, 7.140. Conceptual: 7.25, 7.26, 7.27, 7.28, 7.59, 7.60, 7.68, 7.94, 7.98, 7.99, 7.101, 7.112, 7.116, 7.140, 7.142. Descriptive: 7.127. Environmental: 7.109, 7.125. Difficulty Level Easy: 7.7, 7.8, 7.9, 7.10, 7.11, 7.12, 7.15, 7.16, 7.19, 7.20, 7.26, 7.28, 7.30, 7.33, 7.39, 7.40, 7.41, 7.42, 7.55, 7.56, 7.60, 7.63, 7.65, 7.67, 7.69, 7.70, 7.75, 7.76, 7.77, 7.89, 7.90, 7.101, 7.115. Medium: 7.17, 7.18, 7.21, 7.22, 7.25, 7.27, 7.29, 7.31, 7.32, 7.34, 7.57, 7.58, 7.59, 7.61, 7.62, 7.64, 7.66, 7.78, 7.87, 7.88, 7.91, 7.92, 7.93, 7.94, 7.95, 7.96, 7.97, 7.98, 7.100, 7.107, 7.108, 7.110, 7.113, 7.114, 7.116, 7.117, 7.118, 7.119, 7.120, 7.121, 7.124, 7.128, 7.129, 7.130, 7.132, 7.138, 7.140, 7.142, 7.143. Difficult: 7.68, 7.99, 7.102, 7.103, 7.104, 7.105, 7.106, 7.109, 7.111, 7.112, 7.122, 7.123, 7.125, 7.126, 7.127, 7.129, 7.131, 7.133, 7.134, 7.135, 7.136, 7.137, 7.139, 7.141.

7.7

7.8

(a)

λ =

3.00 × 108 m/s c = = 3.5 × 10− 6 m = 3.5 × 103 nm ν 8.6 × 1013 /s

(b)

ν =

3.00 × 108 m/s c 14 = = 5.30 × 1014 /s = 5.30 × 10 Hz −9 λ × 566 10 m

(a) Strategy: We are given the wavelength of an electromagnetic wave and asked to calculate the frequency. Rearranging Equation (7.1) of the text and replacing u with c (the speed of light) gives: ν =

c λ

Solution: Because the speed of light is given in meters per second, it is convenient to first convert −9 wavelength to units of meters. Recall that 1 nm = 1 × 10 m (see Table 1.3 of the text). We write: −

456 nm ×

1 × 10 9 m − − = 456 × 10 9 m = 4.56 × 10 7 m 1 nm 8

Substituting in the wavelength and the speed of light (3.00 × 10 m/s), the frequency is:

m 3.00 × 108 c s = 6.58 × 1014 s− 1 or 6.58 × 1014 Hz ν = = − λ 4.56 × 10 7 m 14

Check: The answer shows that 6.58 × 10 waves pass a fixed point every second. This very high frequency is in accordance with the very high speed of light.

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499

Thus the pH range varies from 2.51 to 4.41 as the [HIn] varies from 90% to 10%. 16.89

Referring to Figure 16.5, at the half-equivalence point, [weak acid] = [conjugate base]. Using the HendersonHasselbalch equation: [conjugate base] pH = pKa + log [acid] so, pH = pKa

16.90

First, calculate the pH of the 2.00 M weak acid (HNO2) solution before any NaOH is added. −

+ HNO2(aq) U H (aq) + NO2 (aq)

Initial (M): 2.00 Change (M): −x Equilibrium (M): 2.00 − x Ka =

0 +x x

0 +x x

[H + ][NO 2− ] [HNO 2 ]

− 4.5 × 10 4 =

x2 x2 ≈ 2.00 − x 2.00

+

x = [H ] = 0.030 M pH = −log(0.030) = 1.52 Since the pH after the addition is 1.5 pH units greater, the new pH = 1.52 + 1.50 = 3.02. +

From this new pH, we can calculate the [H ] in solution. +

−pH

[H ] = 10

−3.02

= 10

−4

= 9.55 × 10

M

When the NaOH is added, we dilute our original 2.00 M HNO2 solution to: MiVi = MfVf (2.00 M)(400 mL) = Mf(600 mL) Mf = 1.33 M Since we have not reached the equivalence point, we have a buffer solution. The reaction between HNO2 and NaOH is:

→ NaNO2(aq) + H2O(l) HNO2(aq) + NaOH(aq) ⎯⎯ Since the mole ratio between HNO2 and NaOH is 1:1, the decrease in [HNO2] is the same as the decrease in [NaOH]. We can calculate the decrease in [HNO2] by setting up the weak acid equilibrium. From the pH of the + −4 solution, we know that the [H ] at equilibrium is 9.55 × 10 M. +

HNO2(aq) U H (aq) + Initial (M): Change (M): Equilibrium (M):

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1.33 −x 1.33 − x

0

−

NO2 (aq) 0 +x

−4

9.55 × 10

x

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CHAPTER 22: TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS

22.75

(a)

The second equation should have a larger ΔS° because more molecules appear on the products side compared to the reactants side.

(b)

Consider the equation, ΔG° = ΔH° − TΔS°. The ΔH° term for both reactions should be approximately the same because the C−N bond strength is approximately the same in both complexes. Therefore, ΔG° will be predominantly dependent on the −TΔS° term. Next, consider Equation (18.14) of the text.

ΔG° = −RT ln K A more negative ΔG° will result in a larger value of K. Therefore, the second equation will have a larger K because it has a larger ΔS°, and hence a more negative ΔG°. This exercise shows the effect of bi- and polydentate ligands on the position of equilibrium.

22.76

3+

2+

(a)

Cu would not be stable in solution because it can be easily reduced to the more stable Cu . From Figure 3+ 22.3 of the text, we see that the 3rd ionization energy is quite high, about 3500 kJ/mol. Therefore, Cu has a great tendency to accept an electron.

(b)

Potassium hexafluorocuprate(III). Cu is 3d . CuF6 has an octahedral geometry. According to 3− Figure 22.17 of the text, CuF6 should be paramagnetic, containing two unpaired electrons. (Because it is 8 3d , it does not matter whether the ligand is a strong or weak-field ligand. See Figure 22.22 of the text.)

(c)

We refer to Figure 22.24 of the text. The splitting pattern is such that all the square-planar complexes of 3+ Cu should be diamagnetic.

3+

8

3−

Answers to Review of Concepts Section 22.1 (p. 956)

Section 22.3 (p. 961)

Section 22.5 (p. 971)

CrCl3 · 6H2O: This is a hydrate compound (see Section 2.7 of the text). The water molecules are associated with the CrCl3 unit. [Cr(H2O)6]Cl3: This is a coordination 3+ compound. The water molecules are ligands that are bonded to the Cr ion. 3+ The yellow color of CrY6 means that the compound absorbs in the blue-violet region, which has a larger ligand-field splitting (see Figure 22.18 of the text). Thus, Y has a stronger field strength.

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