Chapter 2 Answers to Review Problems PDF

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Solutions Manual to accompany

Chemistry 2nd Edition By Blackman, Bottle, Schmid, Mocerino & Wille

Prepared by John Hill

© John Wiley & Sons Australia, Ltd 2012

Solutions Manual to accompany Chemistry 2e by Blackman et al

Chapter 2 - The language of chemistry Practice Exercises /2 mu2 will have units of mass  (velocity)2 = kg·(m s–1)2 = kg·m2 s–2. (Note that the numbers and factors, such as ½, are unit-less).

2.1

1

2.2

1.

(a) (b) (c)

g m ns

2.

(a) (b) (c)

110–9 110–2 110–12

3.

(a) (b) (c)

cg Mm s

1.

Since 1 x 10 -5 bar = 1 pascal 0.3 0.3 bar = = 3 x 104 pascal 1 x 10-5

2.

Since 7.50 x 10-3 mm Hg = 1 pascal 1 451 mm Hg = x 451 = 3.38 x 10 6 pascal -3 7.50 x 10

3.

Since 9.87 x 10-6 atmosphere = 1 pascal 1 3.81 atmosphere = x 3.81 = 3.86 x 10 5 pascal. 9.87 x 10-6

(a)

molar mass M =

2.3

2.4

m mass m so M  n amount n

Since pV = nRT (b)

(c)

R=

Since E =

pV Pa m3 kg m1 s2  m3  J mol 1 K 1 so R = = nT mol K mol K

hc



,c =

E Jm 1 so c = = ms h Js

© John Wiley & Sons Australia, Ltd 2012 2.2

Chapter 2: The language of chemistry

2.5

Writing the numbers in scientific notation helps us determine the number of significant figures. (a) 1.000405 = 1.000405  100 (7 significant figures) (b) 0.001000 = 1.000  10–3 (4 significant figures) (c) 1000010.0 = 1.0000100  106 (8 significant figures)

2.6

(a) (b) (c) (d) (e) (f)

37.28 6.75 53.3 75.00 1.08 127

2.7

(a) (b) (c) (d) (e) (f)

42.0 g 30.0 mL 54.155 g 11.3 g 0.857 g mL–1 8.3 m3

2.8

% uncertainty in mass of NaCI is % undertainty in volume is

0.001 100 1.0% 0.1

1 = 0.01% 100

Uncertainty in concentration is 1.00 + 0.01 = 1.01% Concentration of prepared solution is 0.1g per 100mL = 1.0gL-1

2.9

The data set from Worker C has the best precision (all data are very close to the same value). The data set from Worker A has the best accuracy (all data are very close to the correct value of 10.000 g).

2.10

NH4Br NH4ClO4 (NH4)2CO3 (NH4)3PO4 KBr KClO4 K2CO3 K3PO4 CaBr2 Ca(ClO4)2 CaCO3 Ca3(PO4)2

© John Wiley & Sons Australia, Ltd 2012 2.3

Solutions Manual to accompany Chemistry 2e by Blackman et al

AlBr3 Al(ClO4)3 Al2(CO3)3 AlPO4 2.11

The structural formula is: H

H

H

H

C

N

H C

C

C

C

O

H

C C

H O

C

H

H

The chemical formula is C8H9NO2. 2.12

(a) (b) (c)

silicon tetrabromide sulfur trioxide chlorine trifluoride

2.13

CH3CH2CH2CH2OH CH3CH2CH(OH)CH3 (CH3)2CHCH2OH (CH3)3COH

2.14

CH3CH2COCH2CH3 CH3COCH2CH2CH3 (CH3)2CHCOCH3

2.15

CH3CH2CH2COOH (CH3)2CHCOOH

2.16

To name these, first identify the longest carbon chain. (a) The longest carbon chain has 8 carbons; therefore this is named as a substituted octane. The name is 5-isopropyl-2-methyloctane (b) The longest carbon chain has 8 carbons; therefore this is named as a substituted octane. The name is 4-isopropyl-4-propyloctane.

2.17

(a) (b)

(primary) (secondary) (primary) (tertiary)

The two structural formulae represent constitutional isomers. The two structural formulae represent the same compound.

© John Wiley & Sons Australia, Ltd 2012 2.4

Chapter 2: The language of chemistry

2.18

To do questions of this type, start by drawing the isomer with the longest chain of C atoms (in this case, 7). Then successively decrease the longest chain by one carbon atom and draw all the possible isomers for that longest chain length. In this case, the isomers are as follows:

Chain length 7

Chain length 6

Chain length 5

Chain length 4

2.19

(a) (b)

propanone pentanal

© John Wiley & Sons Australia, Ltd 2012 2.5

Solutions Manual to accompany Chemistry 2e by Blackman et al

Review questions 2.1

Measurements involve a comparison relative to a standard. The unit gives the number scientific meaning.

2.2

Système International

2.3

kilogram

2.4

kilogram

2.5

force = mass × acceleration. Therefore the units are kg × m s–2 = kg m s–2

2.6

(a) (b) (c) (d) (e) (f) (g)

10–2 10–3 103 10–6 10–9 10–12 106

2.7

(a) (b) (c) (d) (e) (f) (g)

c m k µ n p M

2.8

(a ) (b) (c)

centimetres or millimetre millilitre gram

2.9

The figures that are significant figures in a quantity are those that are known (measured) with certainty plus the last figure, which contains some uncertainty.

2.10

The accuracy of a measured value is the closeness 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.

2.11

The maximum possible concentration of the toxin is 1.499 ug L-1 which reduces to 1.50 ug L-1, correct to 3 significant figures.

© John Wiley & Sons Australia, Ltd 2012 2.6

Chapter 2: The language of chemistry

2.12

To convert 250 s to hours, multiply 250 s by multiply 3.84 h by

1h . To convert 3.84 h to seconds, 3600 s

3600 s . 1h

2.13

1. All bonds except C—H bonds are shown as lines. 2. C—H bonds and H atoms attached to carbon are not shown in the line structure. 3. Single bonds are shown as one line; double bonds are shown as two lines; triple bonds are shown as three lines. 4. Carbon atoms are not labelled. All other atoms are labelled with their elemental symbols.

2.14

Isomers have the same chemical formulae but different chemical structures.

2.15

(a) (b) (c)

2.16

(a)

C18H36O2 SiCl4 C2Cl3F3

H H

C

H

C H

H C

C

C

H

H

H

(b) H H

C

H

C H

C C

C

C H

H

C C

H

H

(c)

© John Wiley & Sons Australia, Ltd 2012 2.7

Solutions Manual to accompany Chemistry 2e by Blackman et al

H H

O

C

H

C C

C

H

O

H

H C

H

2.17

H

(a)

(b)

(c)

(d)

(e)

© John Wiley & Sons Australia, Ltd 2012 2.8

Chapter 2: The language of chemistry

(f)

2.18

(a) (b) (c)

C10H22 (CH3)2CHCH(CH(CH3)2)CH2CH2CH3 C8H18 (CH3)3CC(CH3)3 C11H24 CH3CH2CH2CH(C(CH3)3)CH2CH2CH3

2.19

(a)

CH3(CH2)4CH(CH3)2

(b) (CH2)2CH3 H C(CH2)2CH3 (CH2)2CH3

(c) (CH2)2CH3 CH3 C(CH2)4CH3 (CH2)2CH3

2.20

(a) (b) (c) (d) (e) (f)

ClF SeO3 HBr SiCl4 SO2 H 2O 2

2.21

(a) (b) (c) (d) (e) (f) (g) (h)

chlorine trifluoride hydrogen selenide chlorine dioxide antimony trichloride phosphorus pentachloride dinitrogen pentoxide dinitrogen tetrachloride ammonia (trivial name)

2.22

(a)

© John Wiley & Sons Australia, Ltd 2012 2.9

Solutions Manual to accompany Chemistry 2e by Blackman et al

H

H

H

H

H

C

C

C

C

H

H

H

H

H

H

H

H

H

C

C

C

C

C

H

H

CH3

H

H

H

(b)

H

H

(c)

H

2.23

H

H

H

H

H

H

C

C

C

C

C

C

H

CH3

CH3

H

H

H

H

Unlike the alcohol function group, the aldehyde and carboxylic acid functional groups can only be situated at the end of a carbon chain and therefore only be attached to a single C atom.

© John Wiley & Sons Australia, Ltd 2012 2.10

Chapter 2: The language of chemistry

Review problems 2.24

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l)

0.01 m 1000 m 11012 pm 0.1 m 0.001 kg 0.01 g 1  10–9 m 1  10–6 g 1000 g 1  106 g 1 10–3 g 0.1 g

2.25

To convert oC to K, add 273. To convert K to oC, subtract 273. (a) 333 K (b) 243 K (c) 0 oC (d) 26 oC (e) 313 K

2.26

–162 oC

2.27

–269 C

2.28

(a) (b) (c) (d) (e) (f)

4 5 4 2 4 2

2.29

(a) (b) (c) (d) (e) (f)

3 6 1 5 1 5

2.30

(a) (b) (c) (d) (e)

5 5 2 5 4

© John Wiley & Sons Australia, Ltd 2012 2.11

Solutions Manual to accompany Chemistry 2e by Blackman et al

2.31

(a) (b) (c) (d) (e)

4 4 4 2 3

2.32

(a) (b) (c) (d) (e)

0.72 m2 84.24 kg – 0.465 g cm–3 (dividing a number with 4 sig. figs by one with 3 sig. figs) 19.42 g mL–1 858.0 cm2

2.33

(a) (b) (c) (d) (e)

2.06 g mL–1 4.02 mL 12.4 g mL–1 0.276 g mL–1 0.0006 m s–1

2.34

(a) (b) (c) (d) (e) (f)

4.34 × 103 3.20 × 107 3.29 × 10−3 4.20 × 104 8.00 × 10−6 3.24 × 105

2.35

(a) (b) (c) (d) (e) (f)

4.89 × 102 3.75 × 10–3 8.23 × 104 1.225 × 10–2 2.43 × 100 2.732 × 104

2.36

(a) (b) (c) (d) (e) (f)

310 000 0.000 004 35 3 900 0.000 000 000 004 4 0.000,003 56 880,000,000

2.37

(a) (b) (c) (d) (e) (f)

0.000 000 052 7 712 000 0.000 000 043 5 0.023 5 40 000 000 3720

© John Wiley & Sons Australia, Ltd 2012 2.12

Chapter 2: The language of chemistry

2.38

(a) (b) (c) (d) (e)

4.0 × 107 (derived from 4.0 × 107 − 0.021 × 107) 3.0 × 10−2 4.4 × 1012 2.5 2.3 × 1018

2.39

(a) (b) (c) (d) (e)

6.3 × 109 2.0 × 1018 5.5 × 10–9 1.1 × 105 2.55 × 10–2

2.40

When converting from a smaller unit to a larger unit, the numerical value obtained is less than the original. Conversely, converting from a larger to a smaller unit, the numerical value obtained is greater than the original. (a)

32.0 dm = 32.0  10–1 m 1 km = 1000 m therefore 32.0 dm =

(b)

8.2 mg = 8.2  10–3 g 1 g = 1  10–6 g

8.2 10 3 g = 8.2  103 g 1 106

therefore 8.2 mg = (c)

75.3 mg = 75.3  10–3 g 1 kg = 1000 g therefore 75.3 mg =

75.3 10 3 kg = 7.53  10–5 kg 1000

(d)

137.5 mL = 137.5  10–3 L = 0.1375 L

(e)

0.025 L = 0.025  103 mL = 25 mL

(f)

342 pm = 342  10–12 m 1 dm = 1  10–1 m therefore 342 pm =

2.41

32.0 10 1 km = 3.20  10–3 km 1000

(a)

92 dL = 92  10–1 L 1 mL = 1  10–3 L therefore 92 dL =

(b)

342  10 12 dm = 3.42  10–9 dm 1 10 1

92 101 mL = 9200 mL 1 103

22 ng = 22  10–9 g 1 g = 1  10–6 g

© John Wiley & Sons Australia, Ltd 2012 2.13

Solutions Manual to accompany Chemistry 2e by Blackman et al

therefore 22 ng = (c)

22 10 9 g = 2.2  10–2 g 1 106

83 pL = 83  10–12 L 1 nL = 1  10–9 L therefore 83 pL =

83 1012 nL = 8.3  10–2 nL 9 1 10

(d)

230 km = 230  103 m = 2.30  105 m

(e)

87.3 cm = 87.3  10–2 m 1 km = 1000 m therefore

(f)

87.3 10 2 km = 8.73  10–4 km 1000

238 mm = 238  10–3 m 1 nm = 1  10–9 m therefore 238 mm =

2.42

(a)

230 km = 230  103 m 1 cm = 1  10–2 m therefore 230 km =

(b)

423 10 3 8 3 mg = 4.23  10 mg 1 10

423 kg = 423  103 g 1 Mg = 1  106 g

423 10 3 Mg = 4.23  10–1 Mg 6 1 10

therefore 423 kg = (d)

230  10 3 cm = 2.30  107 cm 2  1 10

423 kg = 423  103 g 1 mg = 1  10–3 g therefore 423 kg =

(c)

238 10 3 nm = 2.38  108 nm 1 10 9

430 L = 430  10–6 L 1 mL = 1  10–3 L

430  10 6 mL = 4.30  10–1 mL therefore 430 L = 3 1 10 (e)

27 ng = 27  10–9 g 1 kg = 1000 g therefore 27 ng =

(f)

27 10 9 kg = 2.7  10–11 kg 1000

730 nL = 730  10–9 L 1 kL = 1000 L © John Wiley & Sons Australia, Ltd 2012 2.14

Chapter 2: The language of chemistry

730 10 9 kL = 7.3  10–10 kL 1000

therefore 730nL =

2.43

(a)

183 nm = 183  10–9 m 1 cm = 1  10–2 m therefore 183 nm =

183 10 9 cm = 1.83  10–5 cm 2  1 10

(b)

3.55 g = 3.55  101 dg

(c)

6.22 km = 6.22  103 m 1 nm = 1  10–9 m

6.22 10 3 12 therefore 6.22 km = 9 nm = 6.22  10 nm 1 10 (d)

33 dm = 33  10–1 m 1 mm = 1  10–3 m therefore 33 dm =

(e)

33 10 1 mm = 3.3  103 mm 3 1 10

0.55 dm = 0.55  10–1 m 1 km = 1000 m therefore 0.55 dm =

(f)

0.55 10 1 km = 5.5  10–5 km 1000

53.8 ng = 53.8  10–9 g 1 pg = 1  10–12 g therefore 53.8 ng =

2.44

(a) (b) (c) (d)

alkene alcohol carboxylic acid alcohol

2.45

(a) (b) (c)

alkyne aldehyde ketone

2.46

(a) (b) (c) (d) (e) (f)

identical identical unrelated isomers identical isomers

53.8 10 9 pg = 5.38  104 pg 12 1 10

© John Wiley & Sons Australia, Ltd 2012 2.15

Solutions Manual to accompany Chemistry 2e by Blackman et al

2.47

(a) (b) (c) (d) (e)

identical isomers isomers identical identical

2.48

(a) (b) (c) (d)

pentane 2-methylpentane 2,4-dimethylhexane 2,4-dimethylhexane

2.49

There are only 5 possible saturated alcohols with 3 or fewer C atoms per molecule: H

H

C

OH

H

H

H

H

C

C

H

H

OH

H

H

H

H

C

C

C

H

H

H

OH

H H

H

H

H

C H

C

C

C

OH C

H C

H

H

H

OH

H

H

2.50

H

2.51

There are only 3 possible structures: H

H

H

H

C

C

C

C

H

H

H

H

OH

H

H

H

H

C

C

C

H

H CH2OH

H

H

H

H

H

H

C

C

C

C

H

H

OH

H

Determine the chemical formula from a ball-and-stick model by counting balls of each colour and consulting the colour code for the elements (appendix J): (a) CH4 (b) C2H4 (c) C2H6O (d) HBr (e) PCl3 (f) CH4N2O (g) C2H5I

© John Wiley & Sons Australia, Ltd 2012 2.16

H

Chapter 2: The language of chemistry

2.52

Determine the chemical formula from a ball-and-stick model by counting balls of each colour and consulting the colour code for the elements (appendix J): (a) NH3 (b) C2H6 (c) CH4O (d) I2 (e) HCN (f) C2H6SO (g) C3H6O

2.53

Structural formulae are derived from ball-and-stick models by replacing the ‘sticks’ with lines and the ‘balls’ with elemental symbols:

(a)

(b) H

H

C

H

H

C

(c)

(d) H C

C

H

H

H

H

H

H

Br

H O

H

C

H

H

(e)

(f) O

H Cl Cl

P

N Cl

C

N

H

(g)

H

H

H

(h) H

H

C

C

H

H

I

H H

N

© John Wiley & Sons Australia, Ltd 2012 2.17

H

Solutions Manual to accompany Chemistry 2e by Blackman et al

(i)

H

(k)

(j) H

H

C

C

H

H

I

I

H H

H

H

(l)

H

C

N

H

O

H

C

C

C

(n) H

O

H

C

S

C

H

2.54

O

H

(m)

H

C

H

H

H

H

H

H

To convert a structural formula into a line structure, remove all hydrogen atoms bonded to carbon atoms and remove all C atom labels.

(a) (b) H N

(c) O

(d)

© John Wiley & Sons Australia, Ltd 2012 2.18

Chapter 2: The language of chemistry

(e) H O

(f) O

O

2.55

To convert a structural formula into a line structure, remove all hydrogen atoms bonded to carbon atoms and remove all C atom labels. (a) O H O

(b) HO

(c) H

H N

(d) O O

(e)

(f)

© John Wiley & Sons Australia, ...