Solution manual chemistry k c timberlake 2nd edition

www.elsolucionario.net 47374_01_p1-2.qxd 2/9/07 7:41 AM Page www.elsolucionario.net Chemistry in Our Lives 1.1 a Chemistry is the science of the composition and properties of matter b A chemist is a scientist who studies the composition and changes of matter c A chemical is a substance that is used in or produced by a chemical process 1.2 Your friends may give a variety of definitions, most of which will probably not agree with the dictionary definitions 1.3 Many chemicals are listed on a vitamin bottle such as: vitamin A, vitamin B3, vitamin B12, vitamin C, folic acid, etc 1.4 Many chemicals are listed on a cereal box such as: vitamin A, vitamin B6, vitamin B12, vitamin C, folic acid, sugar, salt, iron, etc 1.5 Typical items found in a medicine cabinet and chemicals they contain: Antacid tablets: calcium carbonate, cellulose, starch, stearic acid, silicon dioxide Mouthwash: water, alcohol, glycerol, sodium benzoate, benzoic acid Cough suppressant: menthol, beta-carotene, sucrose, glucose 1.6 Typical items found in cleansers are: water, ammonia, sodium silicate, and sodium phosphate 1.7 No All of these ingredients are chemicals 1.8 No All of these ingredients are chemicals 1.9 An advantage of a pesticide is that it protects crops from damage by various insects Some disadvantages are that a pesticide can destroy beneficial insects, be retained in a crop that is eventually eaten by animals or humans, or pollute ground water 1.10 Some advantages of eating sugar is that it gives energy and tastes sweet Some disadvantages are that sugar can lead to obesity and tooth decay 1.11 a b c d A hypothesis proposes a possible explanation for a natural phenomenon An experiment is a procedure that tests the validity of a hypothesis A theory is a hypothesis that has been validated many times by many scientists An observation is a description or measurement of a natural phenomenon 1.12 a b c d hypothesis observation experiment theory 1.13 (1) (2) (3) (4) (5) (6) observation hypothesis experiment observation observation theory 1.14 (1) observation (2) hypothesis (3) experiment 47374_01_p1-2.qxd 2/9/07 7:41 AM Page www.elsolucionario.net Chapter (4) experiment (5) observation (6) hypothesis or theory 1.15 There are several things a student can to be successful in chemistry, including forming a study group, going to lecture, working sample problems and study checks, working problems and checking answers, reading the assignment ahead of class, going to the instructor’s office hours, and keeping a problem notebook 1.16 There are many things that make it difficult to learn chemistry, including not going to lecture regularly, not working problems and study checks, not reading the assignment ahead of class, not going to the instructor’s office hours, and waiting until the night before an exam to study 1.17 a Form a study group c Visit the professor during office hours e Become an active learner 1.18 a, c, d, and e 1.19 Yes Sherlock’s investigation includes observations (gathering data), formulating a hypothesis, testing the hypothesis, and modifying it until the hypothesis is validated 1.20 Sherlock meant that you should not propose a theory until you have data from experiments and observations 1.21 a Determination of a melting point with a thermometer is an observation b Describing a reason for the extinction of dinosaurs is a hypothesis or theory c Measuring the speed of a race is an observation 1.22 a observation b observation c hypothesis or theory 1.23 A hypothesis, which is a possible explanation for an observation, can be tested with experiments 1.24 Experimentation is used to test and verify a hypothesis 1.25 b Another hypothesis needs to be written when experimental results not support the previous hypothesis c More experiments are needed for a new hypothesis 1.26 b Many experiments by many scientists validate the hypothesis 1.27 A successful study plan would include: b Working the sample problems throughout the chapter c Going to the instructor’s office during office hours 1.28 A successful study plan would include: b Forming a study group and discussing problems with others c Working problems in a notebook for reference before exams 1.29 a b c d (1) (2) (3) (2) observation hypothesis experiment hypothesis 1.30 a b c d (1) (2) (1) (1) observation hypothesis observation observation 47374_02_p3-14.qxd 2/9/07 7:45 AM Page www.elsolucionario.net Measurements 2.1 a meter, length d second, time b gram, mass e Celsius, temperature c liter, volume 2.2 a liter, volume d gram, mass b meter, length e kelvin, temperature c kilogram, mass 2.3 a meter; both d second, both b kilogram, both e Celsius, metric c foot, neither 2.4 a cubic meter, both d liter, metric b kelvin, SI e gram, metric c Fahrenheit; neither 2.5 a gram; metric d meter; both b liter; metric e second; both c Celsius; metric 2.6 a kelvin, SI d meter, both b kilogram, both e cubic meter, both c liter, metric 2.7 a b c d e f 2.8 a 1.8 ϫ 108 g e 2.4 ϫ 10Ϫ2 s 2.9 a b c d Move the decimal point left four places to give 5.5 ϫ 104 m Move the decimal point left two places to give 4.8 ϫ 102 g Move the decimal point right six places to give ϫ 10Ϫ6 cm Move the decimal point right four places to give 1.4 ϫ 10Ϫ4 s Move the decimal point right three places to give 7.85 ϫ 10Ϫ3 L Move the decimal point left six places to give 6.7 ϫ 105 kg c 7.5 ϫ 105 g d 1.5 ϫ 10Ϫ1 m The value 7.2 ϫ 103, which is also 72 ϫ 102, is greater than 8.2 ϫ 102 The value 3.2 ϫ 10Ϫ2, which is also 320 ϫ 10Ϫ4, is greater than 4.5 ϫ 10Ϫ4 The value ϫ 104 or 10 000 is greater than ϫ 10Ϫ4, or 0.0001 The value 6.8 ϫ 10Ϫ2 or 0.068 is greater than 0.00052 2.10 a 5.5 ϫ 10Ϫ9 2.11 a b c d b ϫ 10Ϫ5 m f 1.5 ϫ 103 m3 b 3.4 ϫ 102 c ϫ 10Ϫ8 d ϫ 10Ϫ10 The standard number is 1.2 times the power of 104, or 10 000, which gives 12 000 The standard number is 8.25 times the power of 10Ϫ2, or 0.01, which gives 0.0825 The standard number is times the power of 106, or 000 000, which gives 000 000 The standard number is times the power of 10Ϫ3, or 0.001, which gives 0.005 2.12 a 0.000 036 b 87 500 c 0.03 d 212 000 2.13 a The estimated digit is the last digit reported in a measurement In 8.6 m, the in the first decimal (tenths) place was estimated and has some uncertainty b The estimated digit is the in the second decimal (hundredths) place c The estimated digit would be the in the first decimal (tenths) place 2.14 a The estimated digit is the last digit reported in a measurement In 125.04 g, the in the hundredths place is estimated and has some uncertainty b The estimated digit is the in the third decimal (thousandths) place c The estimated digit would be the in the first decimal (tenths) place 47374_02_p3-14.qxd 2/9/07 7:45 AM Page www.elsolucionario.net Chapter 2.15 Measured numbers are obtained using some kind of measuring tool Exact numbers are numbers obtained by counting or from a definition in the metric or the U.S measuring systems a measured b exact c exact d measured 2.16 a exact b measured c measured d measured 2.17 Measured numbers are obtained using some kind of measuring tool Exact numbers are numbers obtained by counting or from a definition in the metric or the U.S measuring systems a The value oz of meat is obtained by measurement, whereas hamburgers is a counted/ exact number b None; both table and chairs are counted/exact numbers c Both 0.75 lb and 350 g are obtained by measurements d None; the values in a definition are exact numbers 2.18 a pizzas 2.19 a b c d e c onions d cars Zeros preceding significant digits are not significant Zeros between significant digits are significant Zeros after significant digits in a decimal number are significant Zeros in the coefficient of a number written in scientific notation are significant Zeros in a number with no decimal point are considered as placeholders only and not significant 2.20 a significant e significant 2.21 a b c d e f b nickels b significant c not significant d not significant All five numbers are significant figures Only the two nonzero numbers are significant; the preceding zeros are placeholders Only the two nonzero numbers are significant; the zeros that follow are placeholders All three numbers in the coefficient of a number written in scientific notation are significant All four numbers including the last zero in a decimal number are significant All three numbers including the zeros that follow a nonzero digit in a decimal number are significant 2.22 a SF e SF b SF f SF c SF d SF 2.23 Both measurements in c have significant figures and both measurements in d have significant figures 2.24 In a and b both pairs have three significant figures In d both pairs have two significant figures 2.25 a 000 is the same as ϫ 1000, which is written in scientific notation as ϫ 103 b 30 000 is the same as ϫ 10 000, which is written in scientific notation as ϫ 104 c 100 000 is the same as ϫ 100 000, which is written in scientific notation as ϫ 105 d 0.000 25 is the same as 2.5 ϫ , which is written in scientific notation as 2.5 ϫ 10Ϫ4 10 000 2.26 a 5.1 ϫ 106 g b 2.6 ϫ 104 s c 4.0 ϫ 104 m d 8.2 ϫ 10Ϫ4 kg 2.27 Calculators carry out mathematical computations and display without regard to significant figures Our task is to round the calculator’s answer to the number of significant figures or digits allowed by the values of the original data 2.28 The number in the calculator display does not show the correct number of significant figures Thus, a significant zero must be added 2.29 To round a number, determine how many significant figures are kept and drop all remaining digits There is no change in the retained figures if the first digit dropped is to However, if the first digit dropped is to 9, raise the last retained digit by 47374_02_p3-14.qxd 2/9/07 7:45 AM Page www.elsolucionario.net Measurements To round 1.854, drop the and keep 1.85 To round 184.2038, drop 2038 and keep 184 To round 0.004 738 265, drop 8265 and increase the retained digits by 1, or 0.004 74 To round 8807 to three significant figures, drop and increase the retained digits to 8810, keeping a zero in the ones place as a placeholder In scientific notation: 8.81 ϫ 103 e To round 1.832 149, drop 2149 and keep 1.83 as the rounded value a b c d 2.30 a 1.9 2.31 a b c d b 180 c 0.0047 d 8800 e 1.8 Drop 55 and increase the last digit by 1, which gives 56.9 m Drop 25, and keep remaining digits as 0.00228 g Drop 27, keep remaining digits, and add two zeros as placeholders, 11 500 s (1.15 ϫ 104 s) Add a significant zero to give three significant figures, 8.10 L 2.32 a 3.3 m b 1.9 ϫ 102 g c 0.0023 m d 2.0 L 2.33 a Because the value of 0.034 has SFs, the answer 1.6 can have only SFs b The measurement has SF, which allows SF in the answer (0.01) c The measurement 1.25 has SFs, which allows SFs in the answer (27.6): 34.56 ϭ 27.6 1.25 d The measurement 25 has SFs, which allows SFs in the answer (3.5): (0.2465)(25) ϭ 3.5 1.78 e The measurement 2.8 ϫ 104 has SFs, which allows SFs in the answer (0.14): (2.8 ϫ 104)(5.05 ϫ 10Ϫ6) ϭ 0.14 (1.4 ϫ 10Ϫ1) f The measurement ϫ 103 has SF, which allows SF in the answer (0.8): (3.45 ϫ 10Ϫ2)(1.8 ϫ 105) ϭ ϫ 10Ϫ1 (0.8) (8 ϫ 103) 2.34 a ϫ 104 d 0.0055 b 0.005 e ϫ 106 c 15 f 8.58 2.35 The answer of addition/subtraction problems has the same number of places as the measurement with the largest place a 45.48 cm decimal places + 8.057 cm decimal places 53.54 cm decimal places b 23.45 g 104.1 g + 0.025 g 127.6 g decimal places decimal place decimal places decimal place c 145.675 mL Ϫ 24.2 mL 121.5 mL decimal places decimal place decimal place d 1.08 L Ϫ 0.585 L 0.50 L decimal places decimal places decimal places 2.36 a b c d 5.08 g ϩ 25.1 g ϭ 30.2 g 85.66 cm ϩ 104.10 cm ϩ 0.025 cm ϭ 189.79 cm 24.568 mL Ϫ 14.25 mL ϭ 10.32 mL 0.2654 L Ϫ 0.2585 L ϭ 0.0069 L 47374_02_p3-14.qxd 2/9/07 7:45 AM Page www.elsolucionario.net Chapter 2.37 The km/h markings indicate how many kilometers (how much distance) will be traversed in hour’s time if the speed is held constant The mph markings indicate the same distance traversed but measured in miles during the hour of travel 2.38 On the speedometer, 80 kph is about 50 mph You are not exceeding the 55 mph speed limit if your speedometer reads 80 km/hr (kph) 2.39 Because the prefix kilo means one thousand times, a kilogram is equal to 1000 grams 2.40 Because the prefix centi means one hundredth, a centimeter is one hundredth of a meter 2.41 a mg e ␮L b dL f ns c km 2.42 a centimeter d gigameter b kilogram e microgram c deciliter f picogram 2.43 a 0.01 d 0.1 b 1000 e 000 000 c 0.001 f 10Ϫ9 2.44 a decigram d centigram b microgram e milligram c kilogram f picogram 2.45 a 100 cm b 1000 m c 0.001 m d 1000 mL 2.46 a kg ϭ 1000 g b mL ϭ 0.001 L c g ϭ 0.001 kg d g ϭ 1000 mg 2.47 a b c d d kg A kilogram, which is 1000 g, is larger than a milligram (0.001 g) A milliliter, which is 10Ϫ3 L, is larger than a microliter (10Ϫ6 L) A km, which is 103 (1000) m, is larger than a cm (10Ϫ2 m or 0.01 m) A kL, which is 103 (1000) L, is larger than a dL (10Ϫ1 L or 0.1 L) 2.48 a mg c ␮m b mm 2.49 Because a conversion factor can be inverted to give a second factor d mL 1m 100 cm and 100 cm 1m 2.50 Verify that the units cancel when the conversion factors are applied 2.51 The numerator and denominator are from the equality: kg ϭ 1000 g 2.52 m ϭ 100 cm 2.53 a yd ϭ ft, b L ϭ 1000 mL, c ϭ 60 s, d dL ϭ 100 mL, 2.54 a gal ϭ qt, b m ϭ 1000 mm, c week ϭ days, d $1 ϭ quarters, yd ft 1L 1000 mL 60 s dL 100 mL gal qt 1m 1000 mm week days $1 quarters and and and and and and and and ft yd 1000 mL 1L 60 s 100 mL dL qt gal 1000 1m days week quarters $1 47374_02_p3-14.qxd 2/9/07 7:45 AM Page www.elsolucionario.net Measurements 2.55 The equalities between the metric prefixes can be written as two conversion factors 1m 100 cm a m ϭ 100 cm, and 100 cm 1m 1g 1000 mg b g ϭ 1000 mg, and 1000 mg 1g 1L 1000 mL c L ϭ 1000 mL, and 1000 mL 1L kg 10 mg d kg ϭ 106 mg, and kg 106 mg e (1 m)3 ϭ (100 cm)3, 2.56 a in ϭ 2.54 cm, b kg ϭ 2.205 lb, c lb ϭ 453.6 g, d 1.057 qt ϭ L, e in.2 ϭ (2.54)2 cm2, (100 cm)3 (1 m)3 in 2.54 cm kg 2.205 lb lb 453.6 g 1.057 qt 1L (2.54)2 cm2 in.2 and and and and and and (1 m)3 (100 cm)3 2.54 cm in 2.205 lb kg 453.6 g lb 1L 1.057 qt in.2 (2.54)2 cm2 2.57 An equality stated in a problem can be written as two conversion factors, which are true only for that problem 3.5 m 1s a 3.5 m ϭ s, and 1s 3.5 m 0.65 g mL b mL ϭ 0.65 g, and mL 0.65 g 46.0 km 1.0 gal c 1.0 gal ϭ 46.0 km, and 1.0 gal 46.0 km d Percent means parts silver per 100 parts sterling silver Using grams (g) as the mass unit, 100 g sterling ϭ 93 g silver, e ppb indicates ␮g/kg, 2.58 a 32 mi ϭ gal, b 20 drops ϭ mL, c ppm indicates mg/kg, d 58 g gold ϭ 100 g jewelry, e $3.19 ϭ gal, 93 g silver 100 g sterling 29 ␮g kg 32 mi gal 20 drops mL 32 mg kg 58 g gold 100 g jewelry $3.19 gal and and and and and and and 100 g sterling 93 g silver kg 29 ␮g gal 32 mi mL 20 drops kg 32 mg 100 g jewelry 58 g gold gal $3.19 47374_02_p3-14.qxd 2/9/07 7:45 AM Page www.elsolucionario.net Chapter 2.59 When using a conversion factor, you are trying to cancel existing units and arrive at a new (desired) unit The conversion factor must be set up to give unit cancellation 2.60 The new (desired) unit should be in the numerator of the conversion factor 2.61 a Plan: cm : m 1m ϭ 1.75 m 175 cm ϫ 100 cm b Plan: mL : L 1L ϭ 5.5 L 5500 mL ϫ 1000 mL c Plan: kg : g 1000 g 0.0055 kg ϫ ϭ 5.5 g kg d Plan: cm3 : m3 m3 350 cm3 ϫ ϭ 3.5 ϫ 10Ϫ4 m3 (100)3 cm3 2.62 a Plan: mg : g 1g ϭ 0.8 g 1000 mg b Plan: dL : mL 100 mL ϭ 85 mL 0.85 dL ϫ dL c Plan: mg : g 1g 2840 mg ϫ ϭ 2.84 g 1000 mg d Plan: m2 : km2 km2 150 000 m2 ϫ ϭ 0.15 km2 (1000)2 m2 800 mg ϫ 2.63 a Plan: qt : mL 1L 1000 mL ϫ ϭ 710 mL 0.750 qt ϫ 1.057 qt 1L b Plan: stone : lb : kg 14 lb kg ϫ ϭ 74.9 kg 11.8 stones ϫ stone 2.205 lb c Plan: in : cm : mm 2.54 cm 10 mm ϫ ϭ 495 mm 19.5 in ϫ in cm d Plan: ␮m : m : cm : in 1m 100 cm in 0.50 ␮m ϫ ϫ ϫ ϭ 2.0 ϫ 10Ϫ5 in 10 ␮m 1m 2.54 cm lb 453.6 g ϫ ϭ 110 g 16 oz lb qt 1L 1000 mL b 5.0 pt ϫ ϫ ϫ ϭ 2400 mL pt 1.057 qt 1L km c 120 000 mi ϫ ϭ 190 000 km 0.6214 mi 2.64 a 4.0 oz ϫ 47374_02_p3-14.qxd 2/9/07 7:45 AM Page www.elsolucionario.net Measurements 1.057 qt gal ϫ ϭ 12.2 gal 1L qt 18.5 gal Ϫ 12.2 gal ϭ 6.3 gal d 46.0 L ϫ 2.65 a Plan: ft : in : cm : m 12 in 2.54 cm 1m 78.0 ft ϫ ϫ ϫ ϭ 23.8 m (length) ft in 100 cm b Plan: ft : in : cm : m : m2 12 in 2.54 cm 1m 27.0 ft ϫ ϫ ϫ ϭ 8.23 m (width) 1ft in 100 cm Area ϭ 23.8 m ϫ 8.23 m ϭ 196 m2 c Plan: m : km : hr : : s km hr 60 60 s ϫ ϫ ϫ ϭ 0.463 s 23.8 m ϫ 1000 m 185 km hr d Plan: m2 : cm2 : in.2 : ft2 : gal : qt : L (100 cm)2 (1 in.)2 (1 ft)2 gal qt 1L 196 m2 ϫ ϫ ϫ ϫ ϫ ϫ ϭ 53.2 L 2 2 1m (2.54 cm) (12 in.) 150 ft gal 1.057 qt 2.66 a 91.4 m b 41 m c 4500 m2 d 2.7 s 2.67 Each of the following require a percent factor from the problem information a Plan: g crust : g oxygen (percent equality: 100.0 g crust ϭ 46.7 g oxygen) 46.7 g oxygen ϭ 152 g oxygen 325 g crust ϫ 100.0 g crust b Plan: g crust : g magnesium (percent equality: 100.0 g crust ϭ 2.1 g magnesium) 2.1 g magnesium ϭ 0.026 g magnesium 1.25 g crust ϫ 100.0 g crust c Plan: oz : lb : g : g nitrogen (percent equality: 100.0 g fertilizer ϭ 15 g nitrogen) lb 453.6 g 15 g nitrogen ϫ ϫ ϭ 43 g nitrogen 10.0 oz fertilizer ϫ 16 oz lb 100.0 g fertilizer d Plan: kg pecans : kg choc bars : lb (percent equality: 100.0 kg bars ϭ 22.0 kg pecans) 100 kg choc bars 2.205 lb 5.0 kg pecans ϫ ϫ ϭ 50 lb of chocolate bars 22.0 kg pecans kg 2.68 a 0.045 kg b 2530 g c 29 g fiber/cake d 4.0 oz 2.69 Because the density of aluminum is 2.70 g/cm3, silver is 10.5 g/cm3, and lead is 11.3 g/cm3, we can identify the unknown metal by calculating its density as follows: 217 g metal ϭ 11.3 g/cm3 The metal is lead 19.2 cm3 metal 2.70 The volume of a cube, 2.0 cm on each edge, is calculated as follows: (2.0 cm)3 ϫ mL/1 cm ϭ 8.0 mL A cube will displace its volume when submerged in water, so the final volume reading in each graduated cylinder is: 40.0 mL water ϩ 8.0 mL metal ϭ 48.0 mL total volume 2.71 Density is the mass of a substance divided by its volume The densities of solids and liquids are usually stated in g/mL or g/cm3 Mass (grams) Density ϭ Volume (mL) 47374_17_p189-200.qxd 2/9/07 10:19 AM Page 197 www.elsolucionario.net Organic Chemistry 17.81 a carboxylic acid d amine b alkene e aldehyde c ester 17.82 a amine d alkyne b ketone e ether c alcohol 17.83 a alcohol d alkane b alkene e carboxylic acid c aldehyde f amine 17.84 a tetrahedral d alkyne b ester e functional group c ether f ketone 17.85 17.86 F F F F F F F F C C C C C C C C F H F Cl F H F Cl F H F Cl F H F Cl C C C C C C C C H H H H 17.87 a aromatic, aldehyde H H H H b aromatic, aldehyde, alkene 17.88 a phenol, ether c ketone b phenol, ether, aldehyde 17.89 carboxylic acid, aromatic, amine, amide, ester 17.90 aromatic, amide, ether CH2 17.91 a CH3 c Cl CH2 CH2 CH CH2 CH3 CH2 CH C CH2 CH2 CH3 CH2 CH2 b CH3 CH3 CH3 CH CH CH2 CH3 CH3 Cl 17.92 a 2, 2-dimethylbutane b 1-chloroethane c 2-bromo-4-ethylhexane 17.93 a This compound contains a five-carbon chain with a double bond between carbon and carbon and a methyl group attached to carbon The IUPAC name is 2-methyl-1-pentene b This compound contains a four-carbon chain with a triple bond between carbon and carbon The IUPAC name is 1-butyne c This compound contains a five-carbon chain with a double bond between carbon and carbon The IUPAC name is 2-pentene 17.94 a CH3 9C #C 9CH2 CH3 b CH3 9CH "CH9 CH2 CH2 CH2 CH3 c CH3 9CH2 9CH" CH9 CH2 CH3 197 47374_17_p189-200.qxd 2/9/07 10:19 AM Page 198 www.elsolucionario.net Chapter 17 17.95 a toluene b 2-chlorotoluene c 4-ethyltoluene CH3 CH2 CH3 CH3 Cl 17.96 a b c CH3 Cl 17.97 a In a primary (1Њ) alcohol, there is one carbon group attached to the carbon with the OH b In a secondary (2Њ) alcohol, there are two carbon groups attached to the carbon with the OH c In a primary (1Њ) alcohol, there is one carbon group attached to the carbon with the OH 17.98 a In a primary (1Њ) alcohol, there is one carbon group attached to the carbon with the OH b In a tertiary (3Њ) alcohol, there are three carbon groups attached to the carbon with the OH c In a primary (1Њ) alcohol, there is one carbon group attached to the carbon with the OH 17.99 a OH Cl CH3 OH b CH3 CH CH2 CH CH3 O c CH3 CH2 C CH2 CH3 OH 17.100 a CH3 CH2 CH CH2 CH3 OH b CH3 CH CH2 CH2 CH3 c CH3 O9CH2 9CH2 9CH3 O 17.101 a CH3 CH2 C H O b CH3 CH2 C CH3 CH3 O 17.102 a CH3 CH C CH3 O b CH3 CH2 CH2 C CH3 17.103 a 4-chloro-3-hydroxybenzaldehyde c 2-chloro-3-pentanone 198 b 3-chloropropanal 47374_17_p189-200.qxd 2/9/07 10:19 AM Page 199 www.elsolucionario.net Organic Chemistry 17.104 a butanone; ethyl methyl ketone c 3-hydroxy-4-methylpentanal b 3, 5-dichlorobenzaldehyde 17.105 a 4-Chlorobenzaldehyde is benzene with an aldehyde group and a chlorine atom on carbon CHO Cl b 3-Chloropropionaldehyde is a three-carbon aldehyde with a chlorine atom located two carbons from the carbonyl group O Cl CH2 CH2 C H c In ethyl methyl ketone, the ketone carbon is attached to a methyl and an ethyl group O CH3 C CH2 CH3 O 17.106 a CH3 b CH3 CH2 CH2 C H Cl O CH C CH3 c CH3 CH H CH3 O CH CH2 CH2 C H 17.107 Primary alcohols oxidize to aldehydes Secondary alcohols oxidize to ketones O a CH3 CH2 C H O b CH3 C CH2 CH3 CH2 O c CH3 CH2 CH2 H C 17.108 Primary alcohols oxidize to aldehydes Secondary alcohols oxidize to ketones O O O a CH3 CH2 C CH H b CH3 C CH2 CH3 O CH3 c CH3 C CH2 C H 199 47374_17_p189-200.qxd 2/9/07 1:10 PM Page 200 www.elsolucionario.net Chapter 17 17.109 a 3-methylbutanoic acid b ethyl benzoate c ethyl propanoate; ethyl propionate 17.110 a 4-methylpentanoic acid c methylbenzoate b 3,5-dichlorobenzoic acid 17.111 a CH3 9CH2 9NH2 CH2 b CH3 NH 9CH3 c CH3 N CH2 CH3 CH2 17.112 a methanamide CH3 b propanamide c ethanamide 17.113 aromatic ring, ketone, amine, carboxylic acid 17.114 aromatic rings, amine, carboxylic acid 17.115 a b c d e alcohol; 1-propanol amine; methyl propyl amine ketone; 3-pentanone; diethylketone alkene; 4-methyl-2-pentene carboxylic acid; propanoic acid; propionic acid Cl Cl CH3 CH3 CH 17.116 a CH3 CH2 CH CH3 b c CH3 9CH2 9NH 9CH3 O d CH3 C CH3 17.117 Alcohols with four carbon atoms in a continuous chain OH CH2 CH2 CH3 CH2 OH CH3 CH CH2 CH3 Branched alcohols with three carbon atoms in the longest chain CH3 OH CH3 CH CH2 OH CH3 C CH3 CH3 Ethers CH3 CH2 CH2 O CH3 CH3 CH2 O CH2 CH3 CH3 CH3 CH O CH3 Heat 17.118 a C5H12 ϩ 8O2 9: 5CO2 ϩ 6H2O Ni b CH3 9CH "CH9CH3 ϩ H2 9: CH3 CH2 9CH2 9CH3 OH CH3 [O] c CH3 200 CH CH3 CH3 CH O CH3 47374_18_p201-211.qxd 2/9/07 11:03 AM Page 201 www.elsolucionario.net 18 Biochemistry 18.1 Hydroxyl ( 9OH) groups and a carbonyl (C "O) are found in all monosaccharides 18.2 An aldose is a monosaccharide with an aldehyde group; a ketose is a monosaccharide with a ketone group 18.3 A ketopentose has five carbon atoms with a ketone functional group and hydroxyl groups on the other carbon atoms 18.4 An aldohexose has six carbon atoms, several hydroxyl groups, and an aldehyde group 18.5 a b c d e This monosaccharide is a ketose; it has a carbonyl between two carbon atoms This monosaccharide is an aldose; it has a CHO, an aldehyde group This monosaccharide is a ketose; it has a carbonyl between two carbon atoms This monosaccharide is an aldose; it has a CHO, an aldehyde group This monosaccharide is an aldose; it has a CHO, an aldehyde group 18.6 a hexose b pentose c triose d pentose e hexose 18.7 In galactose, the hydroxyl on carbon four extends to the left, whereas in glucose it goes to the right 18.8 Glucose is an aldehyde with the carbonyl group on carbon 1; fructose is a ketone with the carbonyl group on carbon 18.9 In the cyclic structure of glucose, there are five carbon atoms and an oxygen atom in the ring 18.10 The cyclic structure of fructose is a five-atom ring of four carbon atoms and an oxygen atom The five-atom ring is a result of the hydroxyl group on carbon reacting with the ketone group on carbon 18.11 a This is the ␣ form because the OH on carbon is down b This is the ␣ form because the OH on carbon is down 18.12 a This is the ␤ form b This is the ␣ form 18.13 a When lactose is hydrolyzed, galactose and glucose are produced The glycosidic bond is a ␤-1,4 bond because the ether bond is up from carbon of the galactose, which is on the left in the drawing, to carbon of the glucose on the right ␤-Lactose is the name of this disaccharide because the free hydroxyl is up b When this disaccharide is hydrolyzed, two molecules of glucose are produced The glycosidic bond is an ␣-1,4 bond because the ether bond is down from the carbon of the glucose on the left to the carbon of the glucose on the right ␣-Maltose is the name of this disaccharide because the free hydroxyl is down 18.14 a The disaccharide is ␤-maltose, which has an ␣-1,4-glycosidic bond between two glucose monosaccharide units b The disaccharide is sucrose, which has an ␣-1,␤-2-glycosidic bond between the ␣-glucose and ␤-fructose monosaccharide units 18.15 a Another name for table sugar is sucrose b Lactose is the disaccharide found in milk and milk products 201 47374_18_p201-211.qxd 2/9/07 11:03 AM Page 202 www.elsolucionario.net Chapter 18 c Maltose is also called malt sugar d When lactose is hydrolyzed, the products are the monosaccharides galactose and glucose 18.16 a maltose b maltose c lactose d sucrose 18.17 a Cellulose is not digestible by humans because we not have the enzymes necessary to break the ␤-1,4-glycosidic bonds in cellulose b Amylose and amylopectin are the storage forms of carbohydrates in plants c Amylose is the polysaccharide that contains only ␣-1,4 glycosidic bonds d Glycogen contains many ␣-1,4 and ␣-1,6 bonds and is the most highly branched polysaccharide 18.18 a glycogen 18.19 a b c d b cellulose c amylopectin, glycogen d amylose, amylopectin Lauric acid has only carbon-carbon single bonds; it is saturated Linolenic acid has three carbon-carbon double bonds; it is unsaturated Palmitoleic acid has one carbon-carbon double bond: it is unsaturated Stearic acid has only carbon-carbon single bonds; it is saturated 18.20 a unsaturated b saturated c saturated d unsaturated 18.21 Tripalmitin has three palmitic acids (16 carbon saturated fatty acid) forming ester bonds with glycerol O CH2 HC O O O (CH2)14CH3 C (CH2)14 C CH3 O CH2 O C 18.22 (CH2)14 CH3 O CH2 HC O O O C C (CH2)7CH (CH2)7CH CH(CH2)7 CH(CH2)7 CH3 CH3 O O CH2 C Triolein (CH2)7CH CH(CH2)7 CH3 18.23 Safflower oil contains fatty acids with two or three double bonds; olive oil contains a large amount of oleic acid, which has a single (monounsaturated) double bond 18.24 Olive oil contains more unsaturated fatty acids than butter fat 18.25 a Partial hydrogenation means that some of the double bonds in the unsaturated fatty acids have been converted to single bonds b Because the margarine now has mostly saturated fatty acids that can interact more strongly, it will be a solid 18.26 O CH2 HC O C O O C O CH2 202 O C (CH2)16 (CH2)16 (CH2)16 CH3 CH3 CH3 47374_18_p201-211.qxd 2/9/07 11:03 AM Page 203 www.elsolucionario.net Biochemistry 18.27 18.28 Hydroxyl groups ( 9OH), ketone (C " O), and methyl groups ( CH3) 18.29 All amino acids contain a carboxylic acid group and an amino group on the alpha carbon 18.30 The side chain in leucine is a hydrocarbon group that is nonpolar In serine, the side group contains an OH group, which makes it polar CH3 + 18.31 a H3N ϩ CH3 O CH C O– HO + b H3N CH O CH C O– CH2 O c H3N9CH9C9OϪ OH CH2 O + 18.32 a H3N CH O– C CH3 CH CH3 CH2 HCOH O + b H3N ϩ CH C O– CH2 O c H3N9CH9C9OϪ 18.33 a Alanine is hydrophobic (nonpolar) because it has a methyl (hydrocarbon) side group b Threonine is hydrophilic (polar) because it has a polar OH side group c Phenylalanine is hydrophobic (nonpolar) with a nonpolar benzene ring in its side group 18.34 a hydrophilic (polar) b hydrophobic (nonpolar) c hydrophilic (polar) 18.35 The abbreviations of most amino acids are derived from the first three letters in the name a alanine b valine c lysine d cysteine 18.36 a tryptophan b methionine c proline d glycine 203 47374_18_p201-211.qxd 2/9/07 11:03 AM Page 204 www.elsolucionario.net Chapter 18 18.37 In a peptide, the amino acids are joined by peptide bonds (amide bonds) The first amino acid has a free amine group and the last one has a free carboxyl group SH CH3 O OH CH O + a H3N CH C NH C CH CH2 O ϩ CH2 O b H3N9 CH9 C9 NH9CH 9C OϪ O– CH3 O CH3 O H3C CH O CH NH CH C + c H3N CH2 C NH C O– H 18.38 CH3 N S O CH2 C CH2 O CH2 O O– + a H3N CH C NH CH C CH2 ϩ CH3 O S C CH2 CH2 NH3 CH2 O CH2 O (CH2)4 O CH C NH CH CH2 O b H3N9 CH9 C9NH9CH9 C OϪ O– O– + + c H3N CH3 O C NH CH C O– 18.39 The possible primary structures of a tripeptide of one valine and two serines are Val-Ser-Ser, Ser-Val-Ser, and Ser-Ser-Val 18.40 The three types of secondary structures are ␣ helix, ␤-pleated sheet, and the triple helix 18.41 In an ␣ helix, hydrogen bonds form between the carbonyl oxygen atoms and the hydrogen atom in the amine groups in different turns of the helix In a ␤-pleated sheet, the hydrogen bonds occur between two adjacent protein chains or between different parts of a long protein 18.42 In a ␤-pleated sheet, there are hydrogen bonds between the polypeptide chains that lie side by side in a sheet In a triple helix, there are three polypeptide chains that are woven together like a rope 18.43 a The two cysteine residues have 9SH groups, which react to form a disulfide bond b Serine has a polar 9OH group that can form a hydrogen bond with the carboxyl group of aspartic acid c Two leucine residues that are nonpolar would have a hydrophobic interaction 18.44 a Nonpolar side groups would be found in the center of the tertiary structure away from the water b Polar groups would be on the outside of the globular shape c Myoglobin is a globular protein that is compact and spherical in shape with polar side groups on the surface to make it soluble in water Silk and wool are fibrous proteins of pleated sheets with many nonpolar side groups that are not soluble in water 204 47374_18_p201-211.qxd 2/9/07 11:03 AM Page 205 www.elsolucionario.net Biochemistry 18.45 a The side group of cysteine with the SH group can form disulfide cross-links b Leucine and valine will be found on the inside of the protein because they have nonpolar side groups and are hydrophobic c The polar cysteine and aspartic acid would be on the outside of the protein to react with water d The order of the amino acid (the primary structure) provides the side chains that determine the tertiary structure of the protein 18.46 a Disulfide bonds join different sections of the protein chain to give a three-dimensional shape Disulfide bonds are important in the tertiary and quaternary structures b Peptide bonds join the amino acid building blocks in the primary structure c Hydrogen bonds that hold protein chains together are found in the secondary structures of ␤pleated sheets of fibrous proteins and in the triple helices of collagen d In the secondary structure of ␣ helices, hydrogen bonding occurs between the carbonyl oxygen and nitrogen atom in the amide bonds e The hydrophobic side chains pull a protein chain into a tertiary structure f A triple helix is a secondary structure g The combination of two or more protein units is a quaternary structure 18.47 a An enzyme has a tertiary structure that recognizes the substrate b The combination of the enzyme and substrate is the enzyme-substrate complex c The substrate has a structure that complements the structure of the enzyme 18.48 a The active site (1) on an enzyme is where catalytic activity occurs b In the induced-fit model (3), the active site adjusts to the substrate shape c In the lock-and-key model (2), the active site is considered to have a rigid shape that fits only a substrate with that geometry 18.49 a The equation for an enzyme-catalyzed reaction is E ϩ S EF ES : E ϩ P E ϭ enzyme, S ϭ substrate, ES ϭ enzyme-substrate complex, P ϭ products b The active site is a region or pocket within the tertiary structure of an enzyme that accepts the substrate, aligns the substrate for reaction, and catalyzes the reaction 18.50 a An enzyme speeds up the reaction of substrate because an enzyme lowers the activation energy for the reaction of that substrate b When the enzyme releases product, the enzyme is available to catalyze the reaction of more substrate 18.51 DNA contains two purines, adenine (A) and guanine (G), and two pyrimidines, cytosine (C) and thymine (T) RNA contains the same bases, except thymine (T) is replaced by the pyrimidine uracil (U) a DNA b both DNA and RNA 18.52 a Present in both DNA and RNA b Adenine is present in both DNA and RNA 18.53 The two DNA strands are held together by hydrogen bonds between the nitrogen bases in each strand 18.54 In complementary base pairing, there are three hydrogen bonds between G and C (or C G), and two hydrogen bonds between A and T (or T9 A) There are no other combinations of bases 18.55 a Because T pairs with A, if one strand of DNA has the sequence AAAAAA, the second strand would be TTTTTT b Because C pairs with G, if one strand of DNA has the sequence GGGGGG, the second strand would be CCCCCC 205 47374_18_p201-211.qxd 2/9/07 11:03 AM Page 206 www.elsolucionario.net Chapter 18 c Because T pairs with A and C pairs with G, if one strand of DNA has the sequence AGTCCAGGT, the second strand would be TCAGGTCCA d Because T pairs with A and C pairs with G, if one strand of DNA has the sequence CTGTATACGTTA, the second strand would be GACATATGCAAT 18.56 a AAAAAA c TACCGT b GGGGGGGGG d TATACGCGATTT 18.57 Once the two DNA strands separate, DNA polymerase begins to pair each base along each strand with its complementary base A is paired with T and C is paired with G Then each base is joined to the new, growing DNA strand 18.58 DNA polymerase catalyzes the formation of phosphodiester bonds between the nucleotides 18.59 The three types of RNA are the messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA) 18.60 Ribosomal RNA (rRNA) is found in the ribosomes, which are the sites for protein synthesis Transfer RNA (tRNA) brings specific amino acids to the ribosomes for protein synthesis Messenger RNA (mRNA) carries the information needed for protein synthesis from the DNA in the nucleus to the ribosomes 18.61 In transcription, the sequence of nucleotides on a DNA template (one strand) is used to produce the base sequences of a messenger RNA The DNA unwinds, and one strand is copied as complementary bases are placed in the mRNA molecule In RNA, U (uracil) is paired with A in DNA 18.62 In RNA, U, A, C, and G complement A, T, C, and G in DNA 18.63 In mRNA, C, G, and A pair with G, C, and T in DNA However, in mRNA, U will pair with A in DNA The strand of mRNA would have the following sequence: GGC9 UUC 9CAA 9GUG 18.64 A9U 9G9C 9C G9U9U 9C 9G9 A 9U 18.65 A codon is a sequence of three bases (triplet) in mRNA that code for a specific amino acid in a protein 18.66 Protein synthesis takes place at the ribosomes in the cytoplasm 18.67 a Melezitose is a trisaccharide b Melezitose contains two glucose molecules and a fructose molecule 18.68 a sucrose c lactose b cellulose d amylose, amylopectin, and cellulose in the peel 18.69 O CH2 O C (CH2)14 CH O H C O C (CH2)14 CH3 O CH2 206 O C (CH2)14 CH3 Glyceryl tripalmitate 47374_18_p201-211.qxd 2/9/07 11:03 AM Page 207 www.elsolucionario.net Biochemistry 18.70 a b c d polyunsaturated; omega-6 fatty acid polyunsaturated; omega-3 fatty acid saturated fatty acid monounsaturated fatty acid 18.71 a O CH2 O C (CH2)7 CH2 CH CH CH CH (CH2)4 CH3 O H C O C (CH2)7 CH3 (CH2)7 CH CH O O CH2 C (CH2)7 CH CH CH2 CH CH (CH2)4 CH3 (CH2)7 CH CH CH2 CH CH (CH2)4 CH3 O CH2 O C O H C O CH (CH2)7 C CH CH2 CH CH (CH2)4 CH3 O O CH2 C (CH2)7 CH (CH2)7 CH (CH2)7 CH CH3 O b CH2 C O CH2 CH CH CH (CH2)4 CH3 O H C O C (CH2)7 CH CH (CH2)7 + 3H2 CH3 Pt O CH2 O C (CH2)7 CH CH CH2 CH CH (CH2)4 CH3 O CH2 CH CH2 18.72 a yes d yes 18.73 a b c d b yes e no O O O C O (CH2)16 C (CH2)16 O C (CH2)16 CH3 CH3 CH3 c no f yes asparagine and serine; hydrogen bond (3) aspartic acid and lysine; salt bridge (5) two cysteines; disulfide bond (4) valine and alanine; hydrophobic attraction (1) 207 47374_18_p201-211.qxd 2/9/07 12:57 PM Page 208 www.elsolucionario.net Chapter 18 18.74 a A G G T C G C C T Parent strand T C C A G C G G A Template strand b A G G U C G C C U 18.75 They differ only at carbon The 9OH in glucose is on the right side, and in galactose it is on the left side 18.76 The carbonyl carbon in glucose is carbon (aldehyde), whereas in fructose the carbonyl group is on carbon (ketone) 18.77 CH2OH O CH2OH O OH OH OH OH OH OH a-gulose 18.78 a B 18.79 OH OH b-gulose b A CH2OH O c C O CH2 OH OH O OH OH OH OH OH 18.80 Hibernating animals in cold climates have more unsaturated triacylglycerols because unsaturated triacylglycerols will not solidify at cold temperatures 18.81 + H 3N CH O H C N CH2OH CH (CH2)4 + NH3 O H C N O CH CH2 C O 208 C O– O– 47374_18_p201-211.qxd 2/9/07 11:03 AM Page 209 www.elsolucionario.net Biochemistry 18.82 CH3 CH CH3 CH3 + H3N CH3 CH O H CH3 O H CH2 O CH C N CH C N CH C O– 18.83 a thymine and deoxyribose c cytosine and ribose b adenine and ribose d guanine and deoxyribose 18.84 a cytosine, ribose c guanine, deoxyribose b adenine, deoxyribose d uracil, ribose 18.85 a C9T9 G9A9A T9C C G b A9C G9 T9T9 T9 G 9A9 T9 C9 G9T c T9A 9G9C9T9 A9G 9C 9T9 A9 G9 C 18.86 a A9A9T9G C9 C 9T9 G9G9 C G b T9A9T9C9 G G9G 9A9 A9T9 G9A 9C 9C c C 9C9G9G9 A9 T9G9 G 9A9 A9 T9 T9 G9C 9T9 G9C 18.87 a mRNA b mRNA 18.88 a ribosomal RNA (rRNA) b transfer RNA (tRNA) 18.89 galactose, glucose, and fructose 18.90 O O CH2 CH O O C O (CH2)7CH C (CH2)7CH O CH2 O C(CH2)7CH Triolein CH(CH2)7 CH(CH2)7 CH(CH2)7 CH3 CH3 + 3H CH3 CH2 Pt O (CH2)16 CH3 C (CH2)16 O CH3 C O CH O CH2 O C (CH2)16 Tristearin CH3 The hydrogenation of the oleic acids converts them into stearic acids, which makes the product tristearin One mol triolein requires mol H2 or 6.0 g H2 to completely hydrogenate the fat At STP, mol ϫ 22.4 L/mol or 67.2 L hydrogen is needed for the reaction 18.91 a The secondary structure of a protein depends on hydrogen bonds to form a helix or a pleated sheet The tertiary structure is determined by the interaction of side chains and determines the three-dimensional structure of the protein b Nonessential amino acids are synthesized by the body, but essential amino acids must be supplied by the diet c Polar amino acids have hydrophilic side groups, whereas nonpolar amino acids have hydrophobic side groups d Dipeptides contain two amino acids, whereas tripeptides contain three e An ionic bond is an interaction between a basic and acidic side group; a disulfide bond links the sulfides of two cysteines f The ␣ helix is the secondary shape like a spiral staircase or corkscrew The ␤-pleated sheet is a secondary structure that is formed by many proteins side by side g The tertiary structure of a protein is its three-dimensional structure In the quaternary structure, two or more peptide subunits are grouped 209 47374_18_p201-211.qxd 2/9/07 11:03 AM Page 210 www.elsolucionario.net Chapter 18 18.92 a hydrogen bonding b hydrophobic c salt bridges (ionic) 18.93 Because A bonds with T, T is also 28% Because A and T ϭ 56%, there is 44% for the other nucleotides, or 22% G and 22% C 18.94 Adenine and thymine can only form two hydrogen bonds, whereas guanine and cytosine form three hydrogen bonds Each base pair has the same distance between the two DNA strands Other combinations of base pairs would not maintain an equal distance along the DNA polymer Answers to Combining Ideas Chapters 17 and 18 CI 35 a aromatic, ether, alcohol, ketone b aromatic, ether, alkene, ketone CI 36 a aromatic, alcohol (phenol), amine b aromatic, ether, amide CI 37 a 2C2H2 ϩ 5O2 : 4CO2 ϩ 2H2O C2H2 26.0 g/mol mol C2H2 mol O2 32.0 g O2 ϫ ϫ ϭ 30 g O2 b 8.5 L C2H2 ϫ 22.4 L C2H2 mol C2H2 mol O2 mol C2H2 mol CO2 22.4 L CO2 ϫ ϫ ϭ 51.6 L CO2 c 30.0 g C2H2 ϫ 26.04 g C2H2 mol C2H2 mol CO2 CH3 O CI 38 CH3 CH C H 2-methylpropanal 2.7 g O 88.15 g MTBE ϫ ϭ 15 g MTBE 100 g gasoline 16.00 g O 1000 mL 0.740 g fuel 15 g MTBE mL MTBE 1L b L fuel ϫ ϫ ϫ ϫ ϫ ϭ L fuel mL 100 g fuel 0.740 g MTBE 1000 mL 0.15 L MTBE c 2C5H12O ϩ 15O2 : 10CO2 ϩ 12H2O MTBE 1000 mL MTBE 0.740 g MTBE mol MTBE d 1.00 L MTBE ϫ ϫ ϫ L MTBE mL MTBE 88.15 g MTBE 15 mol O2 22.4 L O2 100 L air ϫ ϫ ϫ ϭ 6700 L (6.7 ϫ 103 L air) mol MTBE mol O2 21 L O2 CI 39 a 100 g gasoline ϫ CI 40 ϫ 1014 molecules 210 47374_18_p201-211.qxd 2/9/07 11:03 AM Page 211 www.elsolucionario.net Biochemistry CI 41 a Adding NaOH would hydrolyze the tristearate lipid, breaking it up to wash down the drain b O O CH2 C (CH2)16 CH3 O O C H O CH2 CH2 C (CH2)16 O C (CH2)16 CH3 + NaOH CH3 OH O H OH C CH2 Glycerol + Na+ –O C (CH2)16 CH3 OH salts of stearic acid CI 42 a Glyceryl triolein ϭ C57H104O6 O CH2 CH CH2 O O O C O (CH2)7 C (CH2)7 O C (CH2)7 CH CH CH CH CH CH (CH2)7 (CH2)7 (CH2)7 CH3 CH3 ϩ 3H2 (to saturate) or NaOH (to saponify) CH3 mol triolein mol H2 22.4 L ϫ ϫ ϭ 7.60 L H2 884.0 g triolein mol triolein mol H2 mol triolein mol NaOH 100 mL NaOH c 100 g triolein ϫ ϫ ϫ 884.0 g triolein mol triolein 6.00 mol NaOH ϭ 56.6 mL NaOH b 100 g triolein ϫ 211 ... 260 kcal) Total 520 kcal french fries (3 g ϩ 29 g) ϫ kcal/g ϭ 130 kcal ϩ (11 ϫ ϭ 100 kcal) Total 230 kcal chocolate shake (11 g ϩ 60 g) ϫ kcal/g ϭ 280 kcal ϩ (9 ϫ ϭ 80 kcal) Total 360 kcal kcal... J/g ? ?C kJ ϭ b 121? ?C c 2.02? ?C d 57.4? ?C 4.184 J kJ kcal ϫ 10.5? ?C ϫ ϭ 22.2 kJ ϫ ϭ 5.30 kcal g ? ?C 1000 J 4.184 kJ 4.184 J kJ kcal b 4980 g ϫ ϫ 42? ?C ϫ ϭ 871 kJ ϫ ϭ 208 kcal g ? ?C 1000 J 4.184 kJ 4.184... www.elsolucionario.net Chapter 5.13 a Potassium loses eϪ, and chlorine gains eϪ K + – → K+ + Cl → KCl Cl b Calcium loses eϪ, and two chlorine atoms each gain eϪ → Ca2+ + Ca + Cl + Cl – → CaCl2 Cl c Each