all chapter Chemistry and chemical reactivity Kotz , Treichel and Townsend 9th edition solution manual pdf PDF

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Authors: John C. Kotz , Paul M. Treichel , John Townsend , David Treichel
Published: Cengage Learning 2014
Edition: 9th
Pages: 484
Type: pdf
Size: 11MB
Content: 9th edition solutions manual
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https://gioumeh.com/product/chemistry-chemical-reactivity-solutions/ Let’s Review

FOLFNKHUHWRGRZQORDG Let’s Review

The Tools of Quantitative Chemistry INSTRUCTOR’S NOTES This section reviews a wide range of mathematical topics that will be needed by the student in the course, so mastery of this material should be stressed. The need for accurate measurement of physical quantities and the correct recording of this information can be characterized as necessary for reasons of safety using medical (drug dosage) and engineering examples (material composition and properties). Mistakes in this area can mean the difference between life and death. A pre-quiz selected from the Study Questions at the end of this section could help establish what degree of review is needed for your students.

SUGGESTED DEMONSTRATIONS 1.

Units of Measurement 

Add a few drop of bromcresol green to a 2-L flask before the lecture. On filling with water during lecture, the water becomes blue. When the flask is almost full, it is topped off with dilute HCl, and the solution turns yellow. Next, the water is poured into an ordinary 1-L flask or beaker that already contains some dilute base, and the solution turns blue again. This is then poured into a graduated cylinder containing some phenolphthalein. We use this sequence to comment on the relative accuracies of the different types of glassware. We take some containers such as soda cans so students can connect metric units with familiar objects.



A weighed piece of fruit or some other solid gives some meaning to mass expressed in grams.



Earley, C. W. “A Simple Demonstration for Introducing the Metric System to Introductory Chemistry Classes,” Journal of Chemical Education 1999, 76, 1215.

2.

Demonstrations Using Significant Figures 

Kirksey, H. G. “Significant Figures: A Classroom Demonstration,” Journal of Chemical Education 1992, 69, 497.



Abel, K. B.; Hemmerlin, W. M. “Significant Figures,” Journal of Chemical Education 1990, 67, 213.

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9

https://gioumeh.com/product/chemistry-chemical-reactivity-solutions/ The Tools of Quantitative Chemistry SOLUTIONS TO STUDY QUESTIONS FOLFNKHUHWRGRZQORDG 1K = 298 K 1 C

LR.1

(25 °C + 273.15 °C)

LR.2

(5.5  103 °C + 273.15 °C)

LR.3

(a) 289 K

(b) 97 °C

(c) 310 K

LR.4

(a) –196 °C

(b) 336 K

(c) 1177 °C

LR.5

42.195 km ·

1000 m = 4.2195  104 m 1 km

42.195 km ·

0.62137 miles = 26.219 miles 1 km

LR.6

LR.7

19 cm ·

10 mm = 190 mm 1 cm

19 cm ·

1m = 0.19 m 100 cm

1K = 5.8  103 K 1 C

2.5 cm  2.1 cm = 5.3 cm2 2

 1m  –4 2 5.3 cm2 ·   = 5.3  10 m 100 cm   2

LR.8

11.8 cm  2 Area =  r2 =    = 109 cm 2   2

 1m  –2 2 109 cm 2    = 1.09  10 m  100 cm 

LR.9

250. mL ·

1 cm3 = 250. cm3 1 mL

250. mL ·

1L = 0.250 L 103 mL

250. mL ·

250. mL ·

10

3

1 cm3

 1m  –4 3 ·   = 2.50  10 m 1 mL  100 cm  1L

103 mL

·

1 dm3 1L

= 0.250 dm3

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LR.10

1.5 L · 1.5 L ·

103 mL 1L 103 mL 1L

FOLFNKHUHWRGRZQORDG

= 1.5  103 mL ·

1 cm3

= 1.5  103 cm3

1 mL 3

1 dm  3 1.5  103 cm3 ·   = 1.5 dm  10 cm 

LR.11

2.52 kg ·

103 g = 2.52  103 g 1 kg

LR.12

2.265 g ·

1 kg = 2.265  10–3 kg 103 g

2.265 g ·

103 mg = 2.265  103 mg 1g

LR.13

500. mL ·

LR.14

2.365 g 

LR.15

density =

1 cm3 1 mL

·

1.11 g 1 cm3

= 555 g

1 cm3 = 0.225 cm3 10.5 g 2.361 g = 7.14 g/cm3 (2.35 cm  1.34 cm  0.105 cm)

The metal is (c) Zinc LR.16

600 g H2O ·

600 g Pb ·

1 cm3 = 600 cm3 0.995 g

1 cm 3 = 50 cm3 11.35 g

Given equal masses the less dense substance (water) will have a greater volume.

10 3 cal 4.184 J · = 5.0  106 J 1 Cal 1 cal

LR.17

1200 Cal ·

LR.18

1670 kJ ·

1 Cal 10 3 J 1 cal · = 399 Cal · 4.184 J 103 cal 1 kJ

LR.19

170 kcal ·

1 kJ 103 cal 4.184 J · 3 = 710 kJ · 10 J 1 cal 1 kcal

The food product with 170 kcal per serving has a greater energy content per serving.

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11

https://gioumeh.com/product/chemistry-chemical-reactivity-solutions/ The Tools of Quantitative Chemistry LR.20

130 Calories × 4.184 J/cal × 1000 cal/Calorie = 5.4 × 105 J = 5.4 × 102 kJ. The juice provides more total FOLFNKHUHWRGRZQORDG energy. 5.4 × 102 kJ/335 mL = 1.6 kJ/mL 630 kJ/295 mL = 2.1 kJ/mL. The juice provides more energy per milliliter.

LR.21

(a) Method A average = Method B average =

2.2 + 2.3 + 2.7 + 2.4 = 2.4 g/cm3 4 2.703 + 2.701 + 2.705 + 5.881 = 3.480 g/cm3 4

For method B the reading of 5.811 can be excluded because it is more than twice as large as all other readings. Using only the first three readings, average = 2.703 g/cm3. (b) Method A: Percent error = Method B: Percent error =

2.4 – 2.702  100 % = 10 % 2.702 2.703 – 2.702  100 % = 0.04 % (omitting data point) 2.702

(c) Method A: Standard deviation =

Method B: Standard deviation =

0.22 + 0.12 + 0.32 + 02 = 0.2 g/cm3 3 02 + 0.002 2 + 0.002 2 = 0.002 g/cm3 (omitting data point) 2

(d) Before excluding a data point for B, method A is more accurate and more precise. After excluding the fourth data point for B, this method gives a more accurate and more precise result. LR.22

(a) Student A: Average = 135 °C

Percent error =

135 – 135  100 % = 0 % 135

Student B: Average = 138 °C

Percent error =

138 – 135  100 % = 2 % 135

(b) Student B is more precise; Student A is more accurate LR.23

LR.24

LR.25

(a) 5.4  10–2 g (2 significant figures)

(c) 7.92  10–4 g (3 significant figures)

(b) 5.462  103 g (4 significant figures)

(d) 1.6  103 mL (2 significant figures)

(a) 1623 (4 significant figures)

(c) 0.0632 (3 significant figures)

(b) 0.000257 (3 significant figures)

(d) 3404 (4 significant figures)

(a) 9.44  10–3

(c) 11.9

3

LR.26

12

(b) 5.694  10

(d) 0.122

(a) 2.44  108

(c) 0.133

(b) 4.85  10–2

(d) 0.0286

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FOLFNKHUHWRGRZQORDG

LR.27

7 6 5 4 3 2 1 0

y = 0.1637x + 0.0958 0

20

40

Number of kernels

slope = 0.1637 g/kernel

The slope represents the average mass of a kernel.

Using the straight-line equation, mass = (0.1637 g/kernel)(20 kernels) + 0.0958 g = 3.370 g 20.88 g = (0.1637 g/kernel)(number of kernels) + 0.0958 g number of kernels = 127 LR.28

(a) 0.21 (b) 5.6 (c) slope =

5.6 – 4.0 = 18; intercept = 0.20 0.30 – 0.21

(d) y = 18x + 0.20 = (18)(1.0) + 0.20 = 18 LR.29

(a) slope =

20.00 – 4.00 = –4.00, intercept = 20.00, y = –4.00x + 20.00 0.00 – 4.00

(b) y = –4.00x + 20.00 = (–4.00)(6.0) + 20.00 = –4.00 LR.30

(a) 7 6 5 4 3 2 1 0 0.E+00

y = 4E-05x - 0.0416 1.E+05

2.E+05

1/speed

(b) y = 4.0  10–5x – 0.0416 y-intercept = –0.0416 LR.31

C = 0.0823

LR.32

n = 1.63

LR.33

T = 295

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13

https://gioumeh.com/product/chemistry-chemical-reactivity-solutions/ The Tools of Quantitative Chemistry LR.34

n = 1.5 (the answer is 2 to 1 significant figure) FOLFNKHUHWRGRZQORDG

LR.35

1.97 Å · 1.97 Å ·

LR.36

10–10 m 1Å

1 nm 10–9 m

= 0.197 nm

10–10 m 1 pm · –12 = 197 pm 1Å 10 m

0.154 nm ·

10Ğ9 m = 1.54  10–10 m 1 nm

1.54  10–10 m · 1.54  10–10 m ·

LR.37

·

(a) 7.5 m ·

1 pm 10Ğ12 m 1Å 10–10 m

= 154 pm = 1.54 Å

10 –6 m = 7.5  10–6 m 1 m

(b) 7.5  10–6 m · (c) 7.5  10–6 m ·

LR.38

1.53 g cisplatin ·

LR.39

0.50 mL ·

LR.40

7.6 g ·

1 nm

= 7.5  103 nm

10–9 m 1 pm 10–12 m

= 7.5  106 pm

65.0 g Pt 100.0 g cisplatin

= 0.995 g Pt

10. g procaine hydrochloride 103 mg · = 50. mg procaine hydrochloride 100 mL solution 1g

1 cm3 = 2.8 cm3 2.698 g

cube edge =

3

3

2.8 cm = 1.4 cm

LR.41

95.2 g 1 mL  = 2.5 g/cm3 (99 – 61) mL 1 cm3

LR.42

18.82 g 1 mL  = 2.8 g/cm3 (15.3 – 8.5) mL 1 cm 3 The white solid’s density matches that of (c) KBr.

LR.43

(a) Volume = (0.563 nm)3 = 0.178 nm3 3

3

10–9 m   100 cm  –22 3 0.178 nm3 ·   ·   = 1.78  10 cm 1 m 1 nm    

14

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(b) (c)

2.17 g 1 cm 3

3 FOLFNKHUHWRGRZQORDG · 1.78  10–22 cm = 3.87  10–22 g

3.86  10–22 g = 9.68  10–23 g/formula unit 4 NaCl formula units

1 cm3 0.200 g = 0.0854 cm3 · 3.513 g 1 carat

LR.44

1.50 carat ·

LR.45

The normally accepted value for a human temperature is 98.6 °F. On the Celsius scale, this corresponds to 37 °C. Gallium has a melting point of 29.8 °C, so the solid should melt in your hand.

LR.46

The lab partner is on the right track in estimating the density to lie between that at 15 °C and that at 25 °C. If one assumes is it halfway between those densities, a better estimate would be 0.99810 g/cm3. A more accurate analysis, however, shows that the decrease in density with increasing temperature is not linear. A more careful analysis would lead to a value of perhaps 0.99820 g/cm3 (not far from the actual value of 0.99840 g/cm3)

LR.47

(a)

(0.125 – 0.106) g · 100% = 15% of mass lost on popping 0.125 g

(b) 1 lb popcorn ·

LR.48

12 oz. ·

1 kernel 453.6 g · = 3630 kernels 1 lb 0.125 g

28.4 g 1 cm3 · = 126 cm3 1 oz. 2.70 g 2

2

 12 in   2.54 cm  4 2 75 ft2     = 7.0  10 cm  1 ft   1 in 

thickness =

LR.49

volume 126 cm3 = = 1.8  10–3 cm = 1.8  10–2 mm area 7.0  104 cm2

150,000 people ·

1 kg 660 L water 365 days 103 mL 1 cm3 1.00 g · · · · · 1 year 1L 1 mL 1 cm3 103 g 1 day

·

1 kg fluoride 6

10 kg water

·

100.0 kg NaF = 8.0  104 kg NaF/year 45.0 kg fluoride

2

LR.50

0.5 acre ·

1.0  104 m2  100 cm  7 2   = 2  10 cm 2.47 acres  1 m 

thickness =

volume 5 cm3 = = 2  10–7 cm area 2  107 cm2

This is likely related to the “length” of oil molecules. LR.51

500. mL ·

1 cm3 1.285 g 38.08 g sulfuric acid · · = 245 g sulfuric acid 1 mL 100.00 g solution 1 cm3

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15

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LR.52

279 kg ·

103 g 1 cm3 FOLFNKHUHWRGRZQORDG · = 1.45  104 cm3 1 kg 19.3 g

volume = (area)(thickness) 1 cm   1.45  104 cm3 = (area)  0.0015 mm ·  10 mm   2

 1m  3 2 area = 9.6  107 cm2 ·   = 9.6  10 m  100 cm 

LR.53

(a) At 25 °C: 250. mL · 249 g ice ·

1 cm3 0.997 g · = 249 g water (= 249 g ice when cooled) 3 1 mL 1 cm

1 cm3 1 mL · = 272 mL ice 3 0.917 g 1 cm

(b) The ice cannot be contained in the can. 3

LR.54

3

3

 12 in   2.54 cm   1 m  3 (a) volume = (length)(width)(height) = (18 ft)(15 ft)(8.5 ft)       = 65 m 1 ft 1 in 100 cm      

65 m3 ·

1L = 6.5  104 L 3 10 m –3

(b) 6.5  104 L ·

78 kg ·

1.2 g 1 kg = 78 kg · 3 1L 10 g

103 g 1 lb = 170 lb · 454 g 1 kg 3

LR.55

1 cm   9.40 mm 3 volume = (4/3) r3 = (4/3)   ·  = 0.435 cm 2 10 mm  

density =

LR.56

3.475 g 3

0.435 cm

(a) density =

= 7.99 g/cm3

16.08 g – 12.20 g 1 mL · = 1.11 g/cm3 3.50 mL 1 cm3

The liquid is ethylene glycol.

(b) The calculated density would be 1.1 g/cm3. While this value still suggests that the liquid is ethylene glycol, it is close to the value for acetic acid. Further testing should be done on the liquid. LR.57

(a) density =

74.122 g 1 mL · = 8.7 g/cm3 36.7 mL – 28.2 mL 1 cm 3

(b) The metal is probably cadmium, but the calculated density is close to that of cobalt, nickel, and copper. Further testing should be done on the metal.

16

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LR.58

(a) density =

3.2745 g FOLFNKHUHWRGRZQORDG = 0.65 g/mL 5.0 mL

(b) No, all five hydrocarbon density values fall within the range of possible values for the liquid. (c) No, all five hydrocarbon density values fall within the range of possible values for the liquid. (d) Using a volume of (4.93 ± 0.01) mL, calculated maximum and minimum density values are 0.666 g/mL and 0.663 g/mL. The liquid is 2-methylpentane. LR.59

3.416 g – 3.263 g = 0.153 g Hg volume of Hg = 0.153 g Hg ·

1 cm 3 = 0.0113 cm3 13.546 g

1 cm   0.0113 cm3 = (radius)2 16.75 mm · 10 mm  radius = 0.0463 cm diameter = 2  radius = 0.0927 cm LR.60

57 kg ·

103 g 1 cm 3 · = 6.4  103 cm3 copper 1 kg 8.96 g

length of wire =

LR.61

volume 6.4  103 cm3 = 9.0  103 cm = 2 2 ( )(radius) ( )(0.950/2 cm)

9.0  103 cm ·

1m = 90. m 100 cm

(a) 0.1206 g ·

1 Cu atom = 1.143  1021 Cu atoms 1.055  10–22 g

cube volume = (0.236 cm)3 = 0.0131 cm3 3

 1m 100 cm  –24 3 · Cu atom volume = 4/3() 128 pm · 12  = 8.78  10 cm 10 pm 1m   Fraction of occupied space =

(1.143  10 21 Cu atoms)(8.78  10 –24 cm 3/Cu atom) = 0.766 3 0.0131 cm

The empty space is due to the inability of spherical atoms to completely fill a given volume. 3

 1m 100 cm  8.960 g · = 4.232  10–22 g (b) Mass of repeating unit = 361.47 pm · 12  · 3 10 pm 1 m 1 cm   4.232  10–22 g ·

1 Cu atom = 4 Cu atoms 1.055  10–22 g

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17

https://gioumeh.com/product/chemistry-chemical-reactivity-solutions/ The Tools of Quantitative Chemistry

LR.62

1.77 lb 453.6 g 1 kg FOLFNKHUHWRGRZQORDG · · 3 = 0.803 kg/L (the correct conversion factor) 1L 1 lb 10 g

0.803 kg · 7682 L = 6170 kg 1L 22,300 kg – 6170 kg = 16,130 kg additional fuel needed 16,130 kg ·

LR.63

1L = 20,100 L fuel needed 0.803 kg

37.5 g (20.2 – 6.9) mL



1 mL 1 cm3

= 2.82 g/cm3

The sample’s density matches that of aluminum. LR.64

23.5 g 1 mL  = 5.0 g/cm3 (52.2 – 47.5) mL 1 cm 3

The sample's density matches that of fool's gold. LR.65

1.2 1 0.8 0.6 0.4

y = 248.4x + 0.0022

0.2 0 0

0.002

0.004

0.006

Concentration (g/L) slope = 248.4; y-intercept = 0.0022 0.635 = (248.4)(concentration) + 0.0022 concentration = 2.55  10–3 g/L

2.55  10–3 g 1L 10 3 mg = 2.55  10–3 mg/mL · 3 · L 10 mL 1g LR.66

5

y = 2.0925x + 0.2567

4 3 2 1 0 0

0.5

1

1.5

2

% isooctane

18

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https://gioumeh.com/product/chemistry-chemical-reactivity-solutions/ Let’s Review 2.75 = 2.0925(% isooctane)FOLFNKHUHWRGRZQORDG + 0.2567 % isooctane = 1.19% LR.67

Average = (5.22 % + 5.28 % + 5.22 % + 5.30 % + 5.19 % + 5.23 % + 5.33 % + 5.26 % + 5.15 % + 5.22 %)/10 = 5.24 % Standard deviation = [(0.022 + 0.042 + 0.022 + 0.062 + 0.052 +0.012 + 0.092 + 0.022 + 0.092 + 0.022)/10]1/2 = 0.05% Seven points fall within 5.19 < x < 5.29.

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