Solucionario química raymond chang 10ma edición www elsolucionario org

CHAPTER 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 1.3 (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 1.12 (b) law (c) theory www.elsolucionario.org 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 (b) (g) Sn S (c) (h) (b) compound (b) (e) (d) (i) Cr Ar (c) (e) B Hg element (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 mL 5°C 9°F 5°C = 35°C 9°F 5°C (12 − 32)°F × = − 11°C 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 ⎞ ⎛ ? °F = ⎜ −273.15 °C × + 32°F = − 459.67°F °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) × 5°C 9°F CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE ? °C = (105 − 32)°F × (b) 5°C = 41°C 9°F Conversion from Celsius to Fahrenheit 9° F ⎞ ⎛ ? °F = ⎜ °C × + 32°F °C ⎟⎠ ⎝ 9° F ⎞ ⎛ ? °F = ⎜ −11.5 °C × + 32°F = 11.3 °F °C ⎟⎠ ⎝ (c) Conversion from Celsius to Fahrenheit 9° F ⎞ ⎛ + 32°F ? °F = ⎜ °C × 5°C ⎟⎠ ⎝ °F ⎞ ⎛ ? °F = ⎜ 6.3 × 103 °C × ⎟ + 32°F = 1.1 × 10 °F ° C ⎝ ⎠ (d) Conversion from Fahrenheit to Celsius ? °C = (°F − 32°F) × 5°C 9°F ? °C = (451 − 32)°F × 1.25 K = (°C + 273°C) 1K 1°C (a) K = 113°C + 273°C = 386 K (b) K = 37°C + 273°C = 3.10 × 10 K (c) K = 357°C + 273°C = 6.30 × 10 K (a) 1K 1°C °C = K − 273 = 77 K − 273 = −196°C (b) °C = 4.2 K − 273 = −269°C (c) °C = 601 K − 273 = 328°C 1.29 (a) 2.7 × 10 1.30 (a) 10 1.26 2 K = (°C + 273°C) −2 −8 (b) 3.56 × 10 10 −8 (c) 4.7764 × 10 (d) indicates that the decimal point must be moved two places to the left 1.52 × 10 (b) 5°C = 233°C 9°F −2 = 0.0152 indicates that the decimal point must be moved places to the left 7.78 × 10 −8 = 0.0000000778 9.6 × 10 −2 www.elsolucionario.org CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE 1.31 (a) (b) 1.32 −1 145.75 + (2.3 × 10 ) = 145.75 + 0.23 = 1.4598 × 10 79500 2.5 × 10 = 7.95 × 104 2.5 × 10 −3 = 3.2 × 102 −4 −3 −3 −3 (c) (7.0 × 10 ) − (8.0 × 10 ) = (7.0 × 10 ) − (0.80 × 10 ) = 6.2 × 10 (d) (1.0 × 10 ) × (9.9 × 10 ) = 9.9 × 10 (a) Addition using scientific notation 10 n Strategy: Let's express scientific notation as N × 10 When adding numbers using scientific notation, we must write each quantity with the same exponent, n We can then add the N parts of the numbers, keeping the exponent, n, the same Solution: Write each quantity with the same exponent, n n Let’s write 0.0095 in such a way that n = −3 We have decreased 10 by 10 , so we must increase N by 10 Move the decimal point places to the right 0.0095 = 9.5 × 10 −3 Add the N parts of the numbers, keeping the exponent, n, the same −3 9.5 × 10 −3 + 8.5 × 10 −3 18.0 × 10 The usual practice is to express N as a number between and 10 Since we must decrease N by a factor of 10 n to express N between and 10 (1.8), we must increase 10 by a factor of 10 The exponent, n, is increased by from −3 to −2 18.0 × 10 (b) −3 −2 = 1.8 × 10 Division using scientific notation n Strategy: Let's express scientific notation as N × 10 When dividing numbers using scientific notation, divide the N parts of the numbers in the usual way To come up with the correct exponent, n, we subtract the exponents Solution: Make sure that all numbers are expressed in scientific notation 653 = 6.53 × 10 Divide the N parts of the numbers in the usual way 6.53 ÷ 5.75 = 1.14 Subtract the exponents, n 1.14 × 10 (c) +2 − (−8) = 1.14 × 10 +2 + = 1.14 × 10 10 Subtraction using scientific notation n Strategy: Let's express scientific notation as N × 10 When subtracting numbers using scientific notation, we must write each quantity with the same exponent, n We can then subtract the N parts of the numbers, keeping the exponent, n, the same CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE Solution: Write each quantity with the same exponent, n Let’s write 850,000 in such a way that n = This means to move the decimal point five places to the left 850,000 = 8.5 × 10 Subtract the N parts of the numbers, keeping the exponent, n, the same 8.5 × 10 − 9.0 × 10 −0.5 × 10 The usual practice is to express N as a number between and 10 Since we must increase N by a factor of 10 n to express N between and 10 (5), we must decrease 10 by a factor of 10 The exponent, n, is decreased by from to −0.5 × 10 = −5 × 10 (d) Multiplication using scientific notation n Strategy: Let's express scientific notation as N × 10 When multiplying numbers using scientific notation, multiply the N parts of the numbers in the usual way To come up with the correct exponent, n, we add the exponents Solution: Multiply the N parts of the numbers in the usual way 3.6 × 3.6 = 13 Add the exponents, n 13 × 10 −4 + (+6) = 13 × 10 The usual practice is to express N as a number between and 10 Since we must decrease N by a factor of 10 n to express N between and 10 (1.3), we must increase 10 by a factor of 10 The exponent, n, is increased by from to 3 13 × 10 = 1.3 × 10 1.33 (a) (e) four three 1.34 (a) (e) one two or three 1.35 (a) 10.6 m 1.36 (a) Division (b) (f) two one (b) (f) (b) three one 0.79 g (c) (c) (g) five one (c) (g) three one or two 16.5 cm (d) (h) two, three, or four two (d) (d) four × 10 g/cm Strategy: The number of significant figures in the answer is determined by the original number having the smallest number of significant figures Solution: 7.310 km = 1.283 5.70 km The (bolded) is a nonsignificant digit because the original number 5.70 only has three significant digits Therefore, the answer has only three significant digits The correct answer rounded off to the correct number of significant figures is: 1.28 (Why are there no units?) www.elsolucionario.org CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE (b) Subtraction Strategy: The number of significant figures to the right of the decimal point in the answer is determined by the lowest number of digits to the right of the decimal point in any of the original numbers Solution: Writing both numbers in decimal notation, we have 0.00326 mg − 0.0000788 mg 0.0031812 mg The bolded numbers are nonsignificant digits because the number 0.00326 has five digits to the right of the decimal point Therefore, we carry five digits to the right of the decimal point in our answer The correct answer rounded off to the correct number of significant figures is: −3 0.00318 mg = 3.18 × 10 (c) mg Addition Strategy: The number of significant figures to the right of the decimal point in the answer is determined by the lowest number of digits to the right of the decimal point in any of the original numbers Solution: Writing both numbers with exponents = +7, we have 7 (0.402 × 10 dm) + (7.74 × 10 dm) = 8.14 × 10 dm Since 7.74 × 10 has only two digits to the right of the decimal point, two digits are carried to the right of the decimal point in the final answer (d) Subtraction, addition, and division Strategy: For subtraction and addition, the number of significant figures to the right of the decimal point in that part of the calculation is determined by the lowest number of digits to the right of the decimal point in any of the original numbers For the division part of the calculation, the number of significant figures in the answer is determined by the number having the smallest number of significant figures First, perform the subtraction and addition parts to the correct number of significant figures, and then perform the division Solution: (7.8 m − 0.34 m) 7.5 m = = 3.8 m /s (1.15 s + 0.82 s) 1.97 s 1.37 Calculating the mean for each set of date, we find: Student A: 87.6 mL Student B: 87.1 mL Student C: 87.8 mL From these calculations, we can conclude that the volume measurements made by Student B were the most accurate of the three students The precision in the measurements made by both students B and C are fairly high, while the measurements made by student A are less precise In summary: Student A: neither accurate nor precise Student B: both accurate and precise Student C: precise, but not accurate CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE 1.38 Calculating the mean for each set of date, we find: Tailor X: 31.5 in Tailor Y: 32.6 in Tailor Z: 32.1 in From these calculations, we can conclude that the seam measurements made by Tailor Z were the most accurate of the three tailors The precision in the measurements made by both tailors X and Z are fairly high, while the measurements made by tailor Y are less precise In summary: Tailor X: most precise Tailor Y: least accurate and least precise Tailor Z: most accurate 1.39 ? dm = 22.6 m × (b) ? kg = 25.4 mg × (c) ? L = 556 mL × (d) 1.40 dm = 226 dm 0.1 m (a) ? g cm 0.001 g kg × = 2.54 × 10−5 kg mg 1000 g × 10−3 L = 0.556 L mL = 1000 g ⎛ × 10−2 m ⎞ × ×⎜ ⎟ = 0.0106 g/cm ⎜ cm ⎟ kg m3 ⎝ ⎠ 10.6 kg (a) Strategy: The problem may be stated as ? mg = 242 lb A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g) This relationship will allow conversion from pounds to grams A metric conversion is then needed to convert −3 grams to milligrams (1 mg = × 10 g) Arrange the appropriate conversion factors so that pounds and grams cancel, and the unit milligrams is obtained in your answer Solution: The sequence of conversions is lb → grams → mg Using the following conversion factors, 453.6 g lb mg × 10−3 g we obtain the answer in one step: ? mg = 242 lb × 453.6 g mg × = 1.10 × 108 mg lb × 10−3 g Check: Does your answer seem reasonable? Should 242 lb be equivalent to 110 million mg? How many mg are in lb? There are 453,600 mg in lb www.elsolucionario.org CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE (b) Strategy: The problem may be stated as 3 ? m = 68.3 cm Recall that cm = × 10 −2 3 m We need to set up a conversion factor to convert from cm to m Solution: We need the following conversion factor so that centimeters cancel and we end up with meters × 10−2 m cm Since this conversion factor deals with length and we want volume, it must therefore be cubed to give ⎛ × 10−2 m ⎞ × 10−2 m × 10−2 m × 10−2 m × × = ⎜ ⎟ ⎜ cm ⎟ cm cm cm ⎝ ⎠ We can write ⎛ × 10−2 m ⎞ ? m = 68.3 cm × ⎜ ⎟ = 6.83 × 10−5 m ⎜ cm ⎟ ⎝ ⎠ 3 −6 Check: We know that cm = × 10 −6 −5 × 10 gives 6.83 × 10 3 m We started with 6.83 × 10 cm Multiplying this quantity by (c) Strategy: The problem may be stated as ? L = 7.2 m 3 In Chapter of the text, a conversion is given between liters and cm (1 L = 1000 cm ) If we can convert m −2 to cm , we can then convert to liters Recall that cm = × 10 m We need to set up two conversion 3 factors to convert from m to L Arrange the appropriate conversion factors so that m and cm cancel, and the unit liters is obtained in your answer Solution: The sequence of conversions is 3 m → cm → L Using the following conversion factors, ⎛ cm ⎞ ⎜ ⎟ ⎜ × 10−2 m ⎟ ⎝ ⎠ 1L 1000 cm3 the answer is obtained in one step: ⎛ cm ⎞ 1L = 7.2 × 103 L ? L = 7.2 m × ⎜ ⎟ × ⎜ × 10−2 m ⎟ 1000 cm3 ⎝ ⎠ 3 3 Check: From the above conversion factors you can show that m = × 10 L Therefore, m would equal × 10 L, which is close to the answer CHAPTER 1: CHEMISTRY THE STUDY OF CHANGE (d) Strategy: The problem may be stated as ? lb = 28.3 μg A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g) This relationship will allow conversion from grams to pounds If we can convert from μg to grams, we can then −6 convert from grams to pounds Recall that μg = × 10 g Arrange the appropriate conversion factors so that μg and grams cancel, and the unit pounds is obtained in your answer Solution: The sequence of conversions is μg → g → lb Using the following conversion factors, × 10−6 g μg lb 453.6 g we can write ? lb = 28.3 μg × × 10−6 g lb × = 6.24 × 10−8 lb μg 453.6 g Check: Does the answer seem reasonable? What number does the prefix μ represent? Should 28.3 μg be a very small mass? 1.41 1255 m mi 3600 s × × = 2808 mi/h 1s 1609 m 1h 1.42 Strategy: The problem may be stated as ? s = 365.24 days You should know conversion factors that will allow you to convert between days and hours, between hours and minutes, and between minutes and seconds Make sure to arrange the conversion factors so that days, hours, and minutes cancel, leaving units of seconds for the answer Solution: The sequence of conversions is days → hours → minutes → seconds Using the following conversion factors, 24 h day 60 1h 60 s we can write ? s = 365.24 day × 24 h 60 60 s × × = 3.1557 × 107 s day 1h Check: Does your answer seem reasonable? Should there be a very large number of seconds in year? 1.43 (93 × 106 mi) × 1.609 km 1000 m 1s × × × = 8.3 mi km 60 s 3.00 × 10 m 682 CHAPTER 24: ORGANIC CHEMISTRY 24.59 The structures are: H2 C (a) (b) H 2C H 3C CH3 C CH2 C H H 2C (c) H 3C OH H H H C C C C H H H H H CH2 (e) (d) Br Br alcohol (b) CH3 ether C CH3 aldehyde 24.61 Ethanol has a melting point of −117.3°C, a boiling point of +78.5°C, and is miscible with water Dimethyl ether has a melting point of −138.5°C, a boiling point of −25°C (it is a gas at room temperature), and dissolves in water to the extent of 37 volumes of gas to one volume of water 24.62 In Chapter 11, we found that salts with their electrostatic intermolecular attractions had low vapor pressures and thus high boiling points Ammonia and its derivatives (amines) are molecules with dipole−dipole attractions If the nitrogen has one direct N−H bond, the molecule will have hydrogen bonding Even so, these molecules will have much weaker intermolecular attractions than ionic species and hence higher vapor pressures Thus, if we could convert the neutral ammonia−type molecules into salts, their vapor pressures, and thus associated odors, would decrease Lemon juice contains acids which can react with ammonia−type (amine) molecules to form ammonium salts ⎯⎯ → NH4 + RNH2 + H + ⎯⎯ → RNH3 (e) amine (a) + (d) carboxylic acid 24.60 NH3 + H (c) C + 24.63 Cyclohexane readily undergoes halogenation; for example, its reaction with bromine can be monitored by seeing the red color of bromine fading Benzene does not react with halogens unless a catalyst is present 24.64 Marsh gas (methane, CH4); grain alcohol (ethanol, C2H5OH); wood alcohol (methanol, CH3OH); rubbing alcohol (isopropyl alcohol, (CH3)2CHOH); antifreeze (ethylene glycol, CH2OHCH2OH); mothballs (naphthalene, C10H8); vinegar (acetic acid, CH3COOH) 24.65 A mixture of cis and trans isomers would imply some sort of random addition mechanism in which one hydrogen atom at a time adds to the molecule The formation of pure cis or pure trans isomer indicates a more specific mechanism For example, a pure cis product suggests simultaneous addition of both hydrogen atoms in the form of a hydrogen molecule to one side of the alkyne In practice, the cis isomer is formed 24.66 The asymmetric carbons are shown by asterisks: (a) H H H H C *C C Cl H Cl H (c) (b) CH3 OH CH3 *C *C CH OH H H All of the carbon atoms in the ring are asymmetric Therefore there are five asymmetric carbon atoms CH3 www.elsolucionario.org CHAPTER 24: ORGANIC CHEMISTRY 24.67 (a) 683 Sulfuric acid ionizes as follows: + − → H (aq) + HSO4 (aq) H2SO4(aq) ⎯⎯ + − The cation (H ) and anion (HSO4 ) add to the double bond in propene according to Markovnikov’s rule: OSO3H CH3 + CH2 + H + HSO4 CH − CH3 C CH3 H Reaction of the intermediate with water yields isopropanol: OSO3H CH3 OH CH3 + H2O C CH3 H C CH3 + H2SO4 H Since sulfuric acid is regenerated, it plays the role of a catalyst (b) The other structure containing the −OH group is CH3−CH2−CH2−OH propanol 24.68 (c) From the structure of isopropanol shown above, we see that the molecule does not have an asymmetric carbon atom Therefore, isopropanol is achiral (d) Isopropanol is fairly volatile (b.p = 82.5°C), and the −OH group allows it to form hydrogen bonds with water molecules Thus, as it evaporates, it produces a cooling and soothing effect on the skin It is also less toxic than methanol and less expensive than ethanol The red bromine vapor absorbs photons of blue light and dissociates to form bromine atoms Br2 → 2Br• The bromine atoms collide with methane molecules and abstract hydrogen atoms Br• + CH4 → HBr + •CH3 The methyl radical then reacts with Br2, giving the observed product and regenerating a bromine atom to start the process over again: •CH3 + Br2 → CH3Br + Br• Br• + CH4 → HBr + •CH3 24.69 and so on From the molar mass of the alkene, we deduce that there can only be carbon atoms Therefore, the alkene is CH2=CH−CH3 (propene) The reactions are: H2C CH CH3 H2SO4 H2O OH H3C Markovnikov's rule CH CH3 K2Cr2O7 H+ O H3C C CH3 684 CHAPTER 24: ORGANIC CHEMISTRY 24.70 2−butanone is O H3C C CH2 CH3 CH2 CH3 Reduction with LiAlH4 produces 2-butanol OH H3C C H This molecule possesses an asymmetric carbon atom and should be chiral However, the reduction produces an equimolar d and l isomers; that is, a racemic mixture (see Section 22.4 of the text) Therefore, the optical rotation as measured in a polarimeter is zero 24.71 The structures of three alkenes that yield 2-methylbutane CH3 CH3CHCH2CH3 on hydrogenation are: CH3 CH3 H2C 24.72 C CH2 CH3 H3C C CH3 CH CH3 H3C CH CH CH2 To help determine the molecular formula of the alcohol, we can calculate the molar mass of the carboxylic acid, and then determine the molar mass of the alcohol from the molar mass of the acid Grams of carboxylic acid are given (4.46 g), so we need to determine the moles of acid to calculate its molar mass The number of moles in 50.0 mL of 2.27 M NaOH is 2.27 mol NaOH × 50.0 mL = 0.1135 mol NaOH 1000 mL soln The number of moles in 28.7 mL of 1.86 M HCl is 1.86 mol HCl × 28.7 mL = 0.05338 mol HCl 1000 mL soln The difference between the above two numbers is the number of moles of NaOH reacted with the carboxylic acid 0.1135 mol − 0.05338 mol = 0.06012 mol This is the number of moles present in 4.46 g of the carboxylic acid The molar mass is M = 4.46 g = 74.18 g/mol 0.06012 mol A carboxylic acid contains a −COOH group and an alcohol has an −OH group When the alcohol is oxidized to a carboxylic acid, the change is from −CH2OH to −COOH Therefore, the molar mass of the alcohol is 74.18 g − 16 g + (2)(1.008 g) = 60.2 g/mol www.elsolucionario.org CHAPTER 24: ORGANIC CHEMISTRY 685 With a molar mass of 60.2 g/mol for the alcohol, there can only be oxygen atom and carbon atoms in the molecule, so the formula must be C3H8O The alcohol has one of the following two molecular formulas OH CH3CH2CH2OH 24.73 H3C CH CH3 There are 18 structural isomers and 10 of them are chiral The asymmetric carbon atoms are marked with an asterisk C C C C C C OH OH OH C C C C *C C C C C C C C C C OH C C C C *C C OH C *C C C C *C C C *C C C C OH OH C C OH C C C C C C C C *C C C C C C C C OH C C C C OH C C C OH C C *C C C C *C C C OH C C C C C C C C OH C C C C C C C C *C OH C C OH C C C C C C C OH C OH C *C C C C C OH C C C 686 CHAPTER 24: ORGANIC CHEMISTRY 24.74 (a) Reaction between glycerol and carboxylic acid (formation of an ester) O CH2 O C R O (b) CH O C R' CH2 NaOH H2O O CH2 O C CH CH2 OH OH OH O + 3R Glycerol C − O Na + Fatty acid salts (soap) R'' A fat or oil (c) Molecules having more C=C bonds are harder to pack tightly together Consequently, the compound has a lower melting point (d) H2 gas with either a heterogeneous or homogeneous catalyst would be used See Section 13.6 of the text (e) Number of moles of Na2S2O3 reacted is: 20.6 mL × 0.142 mol Na 2S2 O3 1L × = 2.93 × 10−3 mol Na 2S2 O3 1000 mL 1L The mole ratio between I2 and Na2S2O3 is 1:2 The number of grams of I2 left over is: (2.93 × 10−3 mol Na 2S2 O3 ) × mol I2 253.8 g I2 × = 0.372 g I 2 mol Na 2S2 O3 mol I Number of grams of I2 reacted is: (43.8 − 0.372)g =  43.4 g I2 The iodine number is the number of grams of iodine that react with 100 g of corn oil iodine number = 43.4 g I × 100 g corn oil = 123 35.3 g corn oil www.elsolucionario.org CHAPTER 25 SYNTHETIC AND NATURAL ORGANIC POLYMERS Problem Categories Biological: 25.19, 25.20, 25.21, 25.35, 25.36, 25.39, 25.41, 25.43, 25.44 Conceptual: 25.22, 25.27, 25.28, 25.29, 25.30, 25.32, 25.34, 25.37, 25.38, 25.40, 25.42, 25.45 Descriptive: 25.7, 25.9, 25.10, 25.31, 25.33 Difficulty Level Easy: 25.9, 25.21, 25.22, 25.29, 25.31, 25.44 Medium: 25.7, 25.8, 25.10, 25.11, 25.12, 25.19, 25.20, 25.28, 25.30, 25.32, 25.33, 25.35, 25.36, 25.37, 25.38, 25.39, 25.40, 25.42, 25.46, 25.47 Difficult: 25.27, 25.34, 25.41, 25.43, 25.45, 25.48, 25.49, 25.50 25.7 The reaction is initiated be a radical, R• R• + CF2=CF2 → R−CF2−CF2• The product is also a radical, and the reaction continues R−CF2−CF2• + CF2=CF2 → R−CF2−CF2−CF2−CF2• 25.8 etc The repeating structural unit of the polymer is: H H H Cl C C C C H Cl H Cl n Does each carbon atom still obey the octet rule? 25.9 The general reaction is a condensation to form an amide O O R C OH + R' NH2 R C NH R' + H2O The polymer chain looks like: O O C C NH NH O O C C NH Note that both reactants are disubstituted benzene derivatives with the substituents in the para or 1,4 positions www.elsolucionario.org CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS 689 and NH2 CH2 CH2 CH2 CH2 O H O H2N CH C NH CH C OH glycine lysine 25.21 The structure of the polymer is: O C H N C H H 25.22 The rate increases in an expected manner from 10°C to 30°C and then drops rapidly The probable reason for this is the loss of catalytic activity of the enzyme because of denaturation at high temperature 25.27 There are two common structures for protein molecules, an α helix and a β−pleated sheet The α−helical structure is stabilized by intramolecular hydrogen bonds between the NH and CO groups of the main chain, giving rise to an overall rodlike shape The CO group of each amino acid is hydrogen-bonded to the NH group of the amino acid that is four residues away in the sequence In this manner all the main-chain CO and NH groups take part in hydrogen bonding The β−pleated structure is like a sheet rather than a rod The polypeptide chain is almost fully extended, and each chain forms many intermolecular hydrogen bonds with adjacent chains In general, then, the hydrogen bonding is responsible for the three dimensional geometry of the protein molecules In nucleic acids, the key to the double-helical structure is the formation of hydrogen bonds between bases in the two strands Although hydrogen bonds can form between any two bases, called base pairs, the most favorable couplings are between adenine and thymine and between cytosine and guanine More information concerning the importance of hydrogen bonding in biological systems is in Sections 25.3 and 25.4 of the text 25.28 Nucleic acids play an essential role in protein synthesis Compared to proteins, which are made of up to 20 different amino acids, the composition of nucleic acids is considerably simpler A DNA or RNA molecule contains only four types of building blocks: purines, pyrimidines, furanose sugars, and phosphate groups Nucleic acids have simpler, uniform structures because they are primarily used for protein synthesis, whereas proteins have many uses 25.29 When proteins are heated above body temperature they can lose some or all of their secondary and tertiary structure and become denatured The denatured proteins no longer exhibit normal biological activity 690 CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS 25.30 The sample that has the higher percentage of C−G base pairs has a higher melting point because C−G base pairs are held together by three hydrogen bonds The A−T base pair interaction is relatively weaker because it has only two hydrogen bonds Hydrogen bonds are represented by dashed lines in the structures below H δ+ H δ− O N H N N H guanine H δ− δ+ N H N N H δ− δ+ N N H H H O cytosine N H H H δ− N H N CH3 O δ+ N N δ− H δ+ H N N H H O thymine adenine 25.31 As is described in Section 25.3 of the text, acids denature enzymes The citric acid in lemon juice denatures the enzyme that catalyzes the oxidation so as to inhibit the oxidation (browning) 25.32 Leg muscles are active having a high metabolism, which requires a high concentration of myoglobin The high iron content from myoglobin makes the meat look dark after decomposition due to heating The breast meat is “white” because of a low myoglobin content 25.33 The cleavage reaction is: O O (CH2)4 C NH (CH2)6 NH C + H HOOC + + (CH2)4 COOH + H3N (CH2)6 NH3 25.34 Insects have blood that contains no hemoglobin Thus, they rely on simple diffusion to supply oxygen It is unlikely that a human-sized insect could obtain sufficient oxygen by diffusion alone to sustain its metabolic requirements 25.35 The best way to attack this type of problem is with a systematic approach Start with all the possible tripeptides with three lysines (one), then all possible tripeptides with two lysines and one alanine (three), one lysine and two alanines (three also −−Why the same number?), and finally three alanines (one) Lys−Lys−Lys Lys− Lys−Ala Lys−Ala− Lys Ala− Lys− Lys Lys−Ala−Ala Ala−Lys−Ala Ala−Ala−Lys Ala−Ala−Ala Any other possibilities? www.elsolucionario.org CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS 25.36 691 From the mass % Fe in hemoglobin, we can determine the mass of hemoglobin mass of Fe × 100% mass of compound (hemoglobin) % Fe = 0.34% = 55.85 g × 100% mass of hemoglobin minimum mass of hemoglobin = 1.6 × 10 g Hemoglobin must contain four Fe atoms per molecule for the actual molar mass to be four times the minimum value calculated 25.37 The main interaction between water molecules and the amino acid residues is that of hydrogen bonding In water the polar groups of the protein are on the exterior and the nonpolar groups are on the interior 25.38 The type of intermolecular attractions that occur are mostly attractions between nonpolar groups This type of intermolecular attraction is called a dispersion force 25.39 (a) deoxyribose and cytosine H NH2 H N OO P H O N O CH2 O H H O- H H OH H (b) ribose and uracil O H H N OO P O- H O N O CH2 O H H H H OH OH 25.40 This is as much a puzzle as it is a chemistry problem The puzzle involves breaking up a nine-link chain in various ways and trying to deduce the original chain sequence from the various pieces Examine the pieces and look for patterns Remember that depending on how the chain is cut, the same link (amino acid) can show up in more than one fragment 692 CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS Since there are only seven different amino acids represented in the fragments, at least one must appear more than once The nonapeptide is: Gly−Ala−Phe−Glu−His−Gly−Ala−Leu−Val Do you see where all the pieces come from? 25.41 pH = pKa + log [conjugate base] [acid] At pH = 1, −COOH = 2.3 + log [−COOH] [ −COO − ] −NH3 + [ −COO − ] [ −COOH] = 20 = 9.6 + log [− NH ] [− NH3+ ] [ − NH3+ ] = × 108 [− NH ] Therefore the predominant species is: + NH3 − CH2 − COOH At pH = 7, −COOH = 2.3 + log [−COO − ] [ −COOH] [−COO − ] = × 104 [−COOH] −NH3 + = 9.6 + log [− NH ] [− NH3+ ] [ − NH3+ ] = × 102 [− NH ] Predominant species: + NH3 − CH2 − COO − At pH = 12, −COOH 12 = 2.3 + log [−COO − ] [−COOH] [−COO − ] = × 109 [−COOH] www.elsolucionario.org CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS −NH3 + 12 = 9.6 + log [− NH ] [− NH3+ ] Predominant species: 693 [− NH ] [− NH3+ ] = 2.5 × 102 − NH2 − CH2 − COO 25.42 No, the milk would not be fit to drink Enzymes only act on one of two optical isomers of a compound 25.43 (a) The repeating unit in nylon 66 is O (CH2)4 C O N (CH2)6 H N C H and the molar mass of the unit is 226.3 g/mol Therefore, the number of repeating units (n) is n = 12000 g/mol = 53 226.3 g/mol (b) The most obvious feature is the presence of the amide group in the repeating unit Another important and related feature that makes the two types of polymers similar is the ability of the molecules to form intramolecular hydrogen bonds (c) We approach this question systematically First, there are three tripeptides made up of only one type of amino acid: Ala−Ala−Ala Gly−Gly−Gly Ser−Ser−Ser Next, there are eighteen tripeptides made up of two types of amino acids Ala−Ala−Ser Ala−Ser−Ala Ser−Ala−Ala Ser−Ser−Ala Ser−Ala−Ser Ala−Ser−Ser Gly−Gly−Ser Gly−Ser−Gly Ser−Gly−Gly Ser−Ser−Gly Ser−Gly−Ser Gly−Ser−Ser Ala−Ala−Gly Ala−Gly−Ala Gly−Ala−Ala Gly−Gly−Ala Gly−Ala−Gly Ala−Gly−Gly Finally, there are six different tripeptides from three different amino acids Ala−Gly−Ser Ser−Gly−Ala Ser−Ala−Gly Gly−Ala−Ser Ala−Ser−Gly Gly−Ser−Ala Thus, there are a total of twenty-seven ways to synthesize a tripeptide from three amino acids In silk, a basic six-residue unit repeats for long distances in the chain −Gly−Ser−Gly−Ala−Gly−Ala− The ability of living organisms to reproduce the correct sequence is truly remarkable It is also interesting to note that we can emulate the properties of silk with such a simple structure as nylon 694 CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS 25.44 We assume ΔG = 0, so that ΔG = ΔH − TΔS = ΔH − TΔS T = 25.45 ΔH 125 × 103 J/mol = = 315 K = 42°C 397 J/K ⋅ mol ΔS 2+ In deoxyhemoglobin, it is believed that the Fe ion has too large a radius to fit into the porphyrin ring (see 2+ Figure 25.15 of the text) When O2 binds to Fe , however, the ion shrinks somewhat so that it now fits into the plane of the ring As the ion slips into the ring, it pulls the histidine residue toward the ring and thereby sets off a sequence of structural changes from one subunit to another These structural changes occurring from one subunit to the next that cause deoxyhemoglobin crystals to shatter Myoglobin is only made up of one of the four subunits and thus does not have the structural changes from subunit to subunit described above Therefore, deoxymyoglobin crystals are unaffected by oxygen 25.46 O C CH3 O C O CH3 25.47 A DNA molecule has bases (A, C, G, T) A sequence of only two bases to define a particular amino acid has a total of or 16 possible combinations Because there are 20 different amino acids in proteins, we need a sequence of bases or = 64 combinations Because this number is greater than 20, some of the sequences are redundant; that is, they define the same amino acids 25.48 (a) The −COOH group is more acidic because it has a smaller pKa (b) We use the Henderson-Hasselbalch equation, Equation (16.4) of the text pH = pKa + log [conjugate base] [acid ] At pH = 1.0, −COOH 1.0 = 2.32 + log [− COOH] [− COO − ] −NH3 + [− COO − ] [− COOH ] = 21 1.0 = 9.62 + log [− NH ] + [− N H3 ] + [− N H3 ] = 4.2 × 10 [− NH ] Therefore the predominant species is: + CH(CH )2 − CH(NH ) − COOH www.elsolucionario.org CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS 695 At pH = 7.0, −COOH 7.0 = 2.32 + log [− COO − ] [− COOH ] [− COO − ] = 4.8 × 10 [− COOH ] −NH3 + 7.0 = 9.62 + log [− NH ] + [− N H3 ] + [− N H3 ] = 4.2 × 10 [− NH ] Predominant species: + CH(CH )2 − CH(NH ) − COO− At pH = 12.0, −COOH 12.0 = 2.32 + log [− COO − ] [− COOH ] [− COO − ] = 4.8 × 10 [− COOH ] −NH3 + 12.0 = 9.62 + log [− NH ] + [− N H3 ] [− NH ] + = 2.4 × 10 [− N H3 ] Predominant species: CH(CH )2 − CH(NH ) − COO − (c) 25.49 pI = pKa1 + pKa 2 = 2.32 + 9.62 = 5.97 ΔG° = ΔH° − TΔS° ⎡ J ⎞ kJ ⎤ ⎛ ΔG ° = 17 kJ/mol − ⎢(298 K) ⎜ 65 ⎥ ⎟× ⎝ K ⋅ mol ⎠ 1000 J ⎦ ⎣ ΔG° = −2 kJ/mol Since ΔG° < 0, the dimerization is favored at standard conditions and 25°C (298 K) As the temperature is lowered, ΔG° becomes less negative so that the dimerization is less favored At lower temperatures (T < 262 K), the reaction becomes spontaneous in the reverse direction and denaturation occurs For an enzyme to be cold labile, it must have ΔH° > and ΔS° > for folding to the native state so that below a certain temperature, the enthalpy term dominates, and denaturation occurs spontaneously 696 CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS 25.50 (a) All the sulfur atoms will have an octet of electrons and be sp hybridized (b) cysteine (c) Denaturation will lead to more disorder (more microstates) ΔS is positive To break a bond, energy must be supplied (endothermic) ΔH is positive Consider the equation ΔG = ΔH − TΔS This type of process with a positive ΔH and a positive ΔS is favored as the temperature is raised The TΔS term will become a larger negative number as the temperature is raised eventually leading to a negative ΔG (spontaneous) (d) If we assume that the probability of forming a disulfide bond between any two cysteine residues is the same, then, statistically, the total number of structurally different isomers formed from eight cysteine residues is given by × × = 105 Note that the first cysteine residue has seven choices in forming an S−S bond, the next cysteine residue has only five choices, and so on This relationship can be generalized to (N − 1)(N − 3)(N − 5) ⋅ ⋅ ⋅ 1, where N is the total (even) number of cysteine residues present The observed activity of the mixture—the "scrambled protein"—is less than 1% of that of the native enzyme (1/105 < 0.01) This finding is consistent with the fact that only one out of every 105 possible structures corresponds to the original state (e) Oxidation causes sulfur atoms in two molecules to link, similar to the cross-linking described in the problem The new compound formed has less odor compared to the compound secreted by the skunk ... Chemical change Photosynthesis changes water, carbon dioxide, etc., into complex organic matter (e) Physical change The salt can be recovered unchanged by evaporation 1.12 (b) law (c) theory www. elsolucionario. org. .. 22) (c) Elements: (b) and (d) Compounds: (a) and (c) (a) Chemical change: (b) and (c) Physical change: (d) (a) www. elsolucionario. org CHAPTER ATOMS, MOLECULES, AND IONS Problem Categories Conceptual:... Chromium, Iodine www. elsolucionario. org CHAPTER 2: ATOMS, MOLECULES, AND IONS 2.96 33 The change in energy is equal to the energy released We call this ΔE Similarly, Δm is the change in mass E