Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition

Name: Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition
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Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition condenses textbook information into an easy-to-follow study resource.

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1
Chapter 1: Matter, Energy, and Measurement
QC 1.1 Multiplication: (a) 4.69  105 (b) 2.8  10-15
Division: (a) 1.94  1018 (b) 1.37  105
QC 1.2 (a) o F = 9
5
o C + 32 = 9
5 64.0 o C + 32 = 147 o F
(b) o C = 5
9 ( o F - 32) = 5
9 (47 o F - 32) = 8.3o C
QC 1.3 8.55 mi 1.609 km
1 mi



 = 13.8 km
QC 1.4 332 m
s
1 km
1000 m





 1 mi
1.609 km




60 s
1 min





 60 min
1 hr





 = 743 mi/hr
QC 1.5
50. mLsol
hr
1.5 g antibiotic
1000. mL sol





 1000 mg antibiotic
1 g antibiotic





 1 hr
60 min





 = 1.3 mg antibiotic/min
QC 1.6 Mass of Ti = 17.3 mL 4.54 g Ti
1 mL



 = 78.5 g Ti
QC 1.7 d = m/V = 56.8 g
23.4 mL = 2.43 g/mL
QC 1.8 The specific gravity of a substance is the density of this substance divided by the
density of water, which is 1.000 g/mL.
Specific gravity = 1.016 = d
1.000 g/mL
d = 1.016 g/mL
1.2 A major reason for the increase of the average life expectancy of humans in the last 80
years has been great progresses in medical science. Diseases that were once fatal have
either been eradicated or cures developed. The causes and cures for many major diseases
are now better understood and treatments more effective.

Chapter 1: Matter, Energy, and Measurement
2
1.4 (a) Chemical change: burning gasoline is converted to carbon dioxide and water.
(b) Physical change: ice forming from liquid water is still H2O, just a different state of
matter.
(c) Physical change: boiling oil is still oil, just a different state of matter.
(d) Physical change: melting lead remains lead, just a different state of matter.
(e) Chemical change: elemental Fe has been converted to rust, Fe2O3.
(f) Chemical change: nitrogen and hydrogen converted ammonia, NH3, involves a change
in chemical composition.
(g) Chemical change: the chemical components in food are converted to energy, carbon
dioxide, and water thus changing chemical composition.
1.6 (a) 403,000 (b) 3,200 (c) 0.0000713 (d) 0.000000000555
1.8 (a) 2.10  1013 (b) 1.07  104 (c) 2.93  10-12 (d) 8.4  101 (e) 1.36  10-26
1.10 (a) 7.74  103 (b) 8.808  10-2 (c) 1.3022  102
1.12 3.25  10-10
1.14 (a) 3 (b) 2 (c) 1 (d) 4 (e) 5
1.16 (a) 2.5  104 (b) 4.1 (c) 15.5
1.18 (a) 10963.1 (b) 244 (c) 172.34
1.20 (a) 1 kg = 1000 g (b) 1 mg = 0.001 g
1.22 (a) 100 cm (b) 230 mL (c) 75 kg (d) 15 mL
(e) 50 mg (f) 100 mm (g) 8 g
1.24 Temperature conversions: o C = 5
9 ( o F - 32) and K = 273 + o C
(a) 5
9 (320 o F - 32) = 160 o C and 273 + 160 o C = 433 K
(b) 5
9 (212 o F - 32) = 100 o C and 273 + 100 o C = 373 K
(c) 5
9 (0 o F - 32) = -18 o C and 273 + (-18) o C = 255 K
(d) 5
9 (-250 o F - 32) = -157 o C and 273 + (-157) o C = 116 K

Chapter 1: Matter, Energy, and Measurement
3
1.26 Unit conversions:
(a) 42.6 kg 2.205 lb
1 kg





 = 93.9 lb (b) 1.62 lb 453.6 g
1 lb



 = 735 g
(c) 34 in 2.54 cm
1 in



 = 86 cm (d) 37.2 km 1 mi
1.609 km



 = 23.1 mi
(e) 2.73 gal 3.785 L
1 gal





 = 10.3 L (f) 62 g 1 oz
28.35 g





 = 2.2 oz
(g) 33.61 qt 1 L
1.057 qt





 = 31.80 L (h) 43.7 L 1 gal
3.785 L



 = 11.5 gal
(i) 1.1 mi 1.609 km
1 mi



 = 1.8 km (j) 34.9 mL 1 fl oz
29.57 mL



 = 1.18 fl oz
1.28 First, convert one of the bottle's units to the other.
9.5 fl. oz . 29.57 mL
1 fl. oz.





 1 cc
1 mL



 = 2.8  10 2 cc, which is less than 300. cc bottle
1.30 The key here is to first convert speed from mph to kph. Yes, you would reach Ottawa
within an hour.
Speed = 75 mi
1 hr
1.609 km
1 mi



 = 120 kph
Time to reach Ottawa = 80 km 1 hr
120 km



 = 0.7 hr (rounded to one significant figure)
1.32 Car efficiency = 25.00 mi
gal





 1.609 km
1 mi




1 gal
3.785 L





 = 10.63 km/L
1.34 186 lb 1 kg
2.205 lb





 2.0 mg drug
1 kg





 1 cc IV sol
10. mg drug





 1 mL IV sol
1 cc IV sol



 = 17 mL IV sol
1.36 36 lb child 1 kg
2.205 lb





 20. mg Velosef/day
1 kg





 = 3.3  10 2 mg/day
36 lb child 1 kg
2.205 lb





 20. mg Velosef /day
1 kg





 1 mL sol
208 mg Velosf





 = 1.6 mL sol
day

Chapter 1: Matter, Energy, and Measurement
4
1.38 750 mg lidocaine 150 mL IV sol
500. mg lidocaine





 1 min
5 mL IV sol





 = 5  101 min
The answer is rounded to one significant figure because both 1 min and 5 mL only contain
one significant figure.
1.40 Vol of pantoprazole IV = 40. mg PAN 1 mL IV sol
0.4 mg PAN





 = 1  10 2 mL PAN IV solution
Vol of MgSO 4 IV = 5 g MgSO 4
1 mL IV sol
0.02 g MgSO 4







 = 3  10 2 mL MgSO 4 IV solution
Combined volumes of IV solutions = 4  102 mL (rounded to one significant figure)
1.42 At low temperatures, most substances exist as solids.
1.44 d rock sample = mass/volume = 1.075 kg
334.5 mL





 1000 g
1 kg





 = 3.214 g/mL
1.46 Volume of Ti = 163 g Ti 1 mL Ti
4.54 g Ti





 = 35.9 mL Ti
1.48 Answer rounded to two significant figures:
Mass methanol = 280 mL methanol 0.791 g methanol
1 mL methanol



 = 2.2  10 2 g methanol
1.50 (a) The densities of O2 and CO2 can be calculated as follows:
d O2  10.00 g O 2
6702 mL O 2








1000 mL O 2
1 L O 2







 = 1.492 g/L
d CO2  10.00 g CO 2
5058 mL CO 2








1000 mL CO 2
1 L CO 2







 = 1.977 g/L
(b) Carbon dioxide (a gas that does not support combustion) being denser than oxygen will
sink to the level of the fire and temporarily displace the oxygen. By cutting off the
supply of oxygen (a fuel for the fire), the fire will be extinguished.

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Chapter 1: Matter, Energy, and Measurement
6
1.68 51.4 lbs 1 kg
2.205 lbs





 1.3 mg drug
1 kg





 1 mL stock
15 mg drug





 = 2.02 mL of stock solution
1.70 Convert each quantity into a common unit (grams): (a) is the largest and (d) is the smallest.
(a) 41 g
3
(b) 3 10 mg 1 g
1000 mg = 3 g
 
 
 
 
6
(c) 8.2 10 g
 6
1 g
10 g
 = 8.2 g
 
 
 


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Chapter 1: Matter, Energy, and Measurement
7
1.82 The folk medicine will be extracted and separated into pure materials. Each pure material
will be tested for biological activity. The pure materials that are identified as biologically
active will have their chemical structures determined, and often, a chemical synthesis
developed for mass production.
1.84 flow rate = 1 L sol
12 hr





 1000 mL
1 L sol





 1 hr
60 min





 15 gtts
mL



 = 21 gtts/min
1.86 (a) Dose (in milligrams) of Tylenol calculated to two significant figures:
Dose = 42 lb child 1 kg child
2.205 lb child





 15 mg Tylenol
1 kg child





 = 2.9  10 2 mg Tylenol
(b) Dose in milliliters of Tylenol solution calculated to two significant figures:
Dose = 42 lb child 1 kg child
2.205 lb child





 15 mg Tyl
1 kg child





 5.0 mL sol
160 mg Tyl





 = 8.9 mL sol

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8
Chapter 2: Atoms
QC 2.1 (a) NaClO3 (b) AlF3
QC 2.2 (a) The mass number is 15 + 16 = 31.
(b) The mass number is 86 + 136 = 222.
QC 2.3 (a) The element has 15 protons, making it phosphorus (P); its symbol is15
31 P .
(b) The element has 86 protons, making it radon (Rn); its symbol is86
222 Rn .
QC 2.4 (a) The atomic number of mercury (Hg) is 80; that of lead (Pb) is 82.
(b) An atom of Hg has 80 protons; an atom of Pb has 82 protons.
(c) The mass number of this isotope of Hg is 80 + 120 = 200; the mass number for this
isotope of Pb is 82 + 120 = 202.
(d) The symbols of these isotopes are80
200 Hg and82
202 Pb .
QC 2.5 The atomic number of iodine (I) is 53. The number of neutrons in each isotope is 125 -
53 = 72 for iodine-125 and 131 - 53 = 78 for iodine-131. The symbols for these two
isotopes are53
125 I and53
131 I .
QC 2.6 The atomic weight is 6.941 amu, which is nearer to 7 amu than 6 amu. Therefore,
lithium

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Chapter 2: Atoms
9
2.4 (a) Oxygen (b) Lead (c) Calcium (d) Sodium (e) Carbon (f) Titanium
(g) Sulfur (h) Iron (i) Hydrogen (j) Potassium (k) Silver (l) Gold
2.6 Given here is the element, its symbol, and its atomic number:
(a) Americium (Am, 95) (b) Berkelium (Bk, 97) (c) Californium (Cf, 98)
(d) Dubnium (Db, 105) (e) Europium (Eu, 63) (f) Francium (Fr, 87)
(g) Gallium (Ga, 31) (h) Germanium (Ge, 32) (i) Hafnium (Hf, 72)
(j) Hassium (Hs, 108) (k) Holmium (Ho, 67)

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Chapter 2: Atoms
10
2.18 (a) The element with 22 protons is titanium (Ti).
(b) The element with 76 protons is osmium (Os).
(c) The element with 34 protons is selenium (Se).
(d) The element with 94 protons is plutonium (Pu).
2.20 Each would still be the same element because the number of protons has not changed.
(a) The element is scandium (Sc), and its symbol is21
47 Sc .
(b) The element is titanium (Ti), and its symbol is22
50 Ti .
(c) The element is silver (Ag), and its symbol is47
109 Ag .
(d) The element is thorium (Th) and its symbol is

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Chapter 2: Atoms
12
2.52 The element X might be in Group 2A, all of which have two valence electrons. It might
also be helium.
2.54 (a), (b), (c), (f), and (h): True
(d) False: helium is a group 8A element and has two valence electrons.
(e) False: the period number has nothing to do with the number of elements.
(g) False: period 3 has eight elements.
2.56 (a), (b), (c), (d), and (f): True
(e) False: ionization energy decreases going from top to bottom within a column of the
Periodic Table.
2.58 (a) The additional electron in the valence shell exceeds the positive charge in the nucleus
causing the electrons in the valence shell to be held less tightly, thus increasing the
atomic radius. Also, the additional electron in the valence shell introduces new
repulsions causing the electron cloud to expand.
(b) With the loss of a valence electron, the positive charge in the nucleus exceeds the
collective negative charge of the electrons, causing the valence electrons to be held
more tightly, thus decreasing the atomic radius. Also, one less electron in the valence
shell reduces the electron-electron repulsions causing the electron cloud to contract.
2.60 Ionization energies generally increase going from left to right in a period of the Periodic
Table due to the increasing positive charge in the nucleus. As the nucleus becomes more
positively charge within a period, the valence electrons become more difficult to remove in
an ionization.
2.62 Sulfur and iron are essential components of proteins, and calcium is a major component of
bones and teeth.
2.64 Calcium is an essential element in human bones and teeth. Because strontium behaves
chemically much like calcium, strontium-90 gets into our bones and teeth and gives off
radioactivity for many years directly into our bodies.
2.66 Copper can be made harder by hammering it.
2.68 (a) Metals (b) Nonmetals (c) Metals
(d) Nonmetals (e) Metals (f) Metals

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Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition

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