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

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1Chapter 1:Matter, Energy, and MeasurementQC1.1Multiplication:(a) 4.69105(b) 2.810-15Division:(a) 1.941018(b) 1.37105QC1.2(a)oF = 95oC + 32 = 95 64.0oC + 32 = 147oF(b)oC = 59 (oF - 32) = 59 (47oF - 32) = 8.3oCQC1.38.55 mi1.609 km1 mi= 13.8 kmQC1.4332 ms1 km1000 m1 mi1.609 km60 s1 min60 min1 hr= 743 mi/hrQC1.550. mLsolhr1.5 g antibiotic1000. mL sol1000 mg antibiotic1 g antibiotic1 hr60 min= 1.3 mg antibiotic/minQC1.6Mass of Ti = 17.3 mL4.54 g Ti1 mL= 78.5 g TiQC1.7d = m/V =56.8 g23.4 mL = 2.43 g/mLQC1.8The specific gravity of a substance is the density of this substance divided by thedensity of water, which is 1.000 g/mL.Specific gravity = 1.016 =d1.000 g/mLd = 1.016 g/mL1.2A major reason for the increase of the average life expectancy of humans in the last 80years has been great progresses in medical science. Diseases that were once fatal haveeither been eradicated or cures developed. The causes and cures for many major diseasesare now better understood and treatments more effective.

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Chapter 1: Matter, Energy, and Measurement21.4(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 ofmatter.(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 changein chemical composition.(g)Chemical change:the chemical components in food are converted to energy, carbondioxide, and water thus changing chemical composition.1.6(a) 403,000(b) 3,200(c) 0.0000713(d) 0.0000000005551.8(a) 2.101013(b) 1.07104(c) 2.9310-12(d) 8.4101(e) 1.3610-261.10(a) 7.74103(b) 8.80810-2(c) 1.30221021.123.2510-101.14(a) 3(b) 2(c) 1(d) 4(e) 51.16(a) 2.5104(b) 4.1(c) 15.51.18(a) 10963.1(b) 244(c) 172.341.20(a) 1 kg = 1000 g(b) 1 mg = 0.001 g1.22(a) 100 cm(b) 230 mL(c) 75 kg(d) 15 mL(e) 50 mg(f)100 mm(g)8g1.24Temperature conversions:oC = 59 (oF - 32)andK = 273 +oC(a)59 (320oF - 32) = 160oC and 273 + 160oC = 433 K(b)59 (212oF - 32) = 100oC and 273 + 100oC = 373 K(c)59 (0oF - 32) = -18oC and 273 + (-18)oC = 255 K(d)59 (-250oF - 32) = -157oC and 273 + (-157)oC = 116 K

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Chapter 1: Matter, Energy, and Measurement31.26Unit conversions:(a) 42.6 kg2.205 lb1 kg= 93.9 lb(b) 1.62 lb453.6 g1 lb= 735 g(c) 34 in2.54 cm1 in= 86 cm(d) 37.2 km1 mi1.609 km= 23.1 mi(e) 2.73 gal3.785 L1 gal= 10.3 L(f) 62 g1 oz28.35 g= 2.2 oz(g) 33.61 qt1 L1.057 qt= 31.80 L(h) 43.7 L1 gal3.785 L= 11.5 gal(i) 1.1 mi1.609 km1 mi= 1.8 km(j) 34.9 mL1 fl oz29.57 mL= 1.18 fl oz1.28First, convert one of the bottle's units to the other.9.5 fl. oz .29.57 mL1 fl. oz.1 cc1 mL= 2.8102cc, which is less than 300. cc bottle1.30The key here is to first convert speed from mph to kph.Yes, you would reach Ottawawithin an hour.Speed = 75 mi1 hr1.609 km1 mi= 120 kphTime to reach Ottawa = 80 km1 hr120 km= 0.7 hr (rounded to one significant figure)1.32Car efficiency =25.00 migal1.609 km1 mi1 gal3.785 L= 10.63 km/L1.34186 lb1 kg2.205 lb2.0 mg drug1 kg1 cc IV sol10. mg drug1 mL IV sol1 cc IV sol= 17 mL IV sol1.3636 lb child1 kg2.205 lb20. mg Velosef/day1 kg= 3.3102mg/day36 lb child1 kg2.205 lb20. mg Velosef /day1 kg1 mL sol208 mg Velosf= 1.6 mL solday

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Chapter 1: Matter, Energy, and Measurement41.38750 mg lidocaine150 mL IV sol500. mg lidocaine1 min5 mL IV sol= 5101minThe answer is rounded to one significant figure because both 1 min and 5 mL only containone significant figure.1.40Vol of pantoprazole IV = 40. mg PAN1 mL IV sol0.4 mg PAN= 1102mL PAN IV solutionVol of MgSO4IV = 5 g MgSO41 mL IV sol0.02 g MgSO4= 3102mL MgSO4IV solutionCombined volumes of IV solutions = 4102mL (rounded to one significant figure)1.42At low temperatures, most substances exist as solids.1.44drock sample= mass/volume =1.075 kg334.5 mL1000 g1 kg= 3.214 g/mL1.46Volume of Ti = 163 g Ti1 mL Ti4.54 g Ti= 35.9 mL Ti1.48Answer rounded to two significant figures:Massmethanol= 280 mL methanol0.791 g methanol1 mL methanol= 2.2102g methanol1.50(a)The densities of O2and CO2can be calculated as follows:dO210.00 g O26702 mL O21000 mL O21 L O2= 1.492 g/LdCO210.00 g CO25058 mL CO21000 mL CO21 L CO2= 1.977 g/L(b)Carbon dioxide (a gas that does not support combustion) being denser than oxygenwillsink to the level of the fire and temporarily displace the oxygen.By cutting off thesupply of oxygen (a fuel for the fire), the fire will be extinguished.

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Chapter 1: Matter, Energy, and Measurement51.52Matter-energy can neither be created nor destroyed (law of conservation of energy), solarpanels work by converting light energy into electrical energy.1.54Assuming that the 180 lb man is measured to three significant figures:135lbDrug dose= 135 lb man445 mg drug180 lb man= 334 mg drug1.56(a) Volume(b) Volume(c) Mass(d)Density(e)Temperature(f) Velocity1.58Vroom= length x height x width = (5.3 m)(2.0 m)(4.2 m) = 45 m3mair= dairVair=1.2510-3gcm3(45 m3) 100 cm1 m31 kg1000 g= 56 kg air1.60KE = 1/2(mv2)m = 127 lb453.6 g1 lb= 5.76104gv =14.7 mihr1.609 km1 mi1000 m1 km100 cm1 m1 hr60 min1 min60 sec= 657 cm/sKE = 1/2(5.76104g)(657 cm/s)2= 1.241010gcm2s-2= 1.241010ergs1.62Convert dollars/L into dollars/gal, then compare:Gasoline priceMontreal= $1.22L3.785 Lgal= $4.62/galGasoline is less expensive in Potsdam ($3.93/gal) than in Montreal ($4.62/gal).1.6475 mL1 hr10. gtts1 mL1 hr60 min= 13 gtts/min flow rate1.66dose = 14 lb cat2.5 mg1 lb cat= 35 mg per 12 hourstotal drug to be delivered = 7 days24 hrday1 dose12 hr35 mgdose= 4.9102mg

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Chapter 1: Matter, Energy, and Measurement61.6851.4 lbs1 kg2.205 lbs1.3 mg drug1 kg1 mL stock15 mg drug= 2.02 mL of stock solution1.70Convert each quantity into a common unit (grams): (a) is the largest and (d) is the smallest.(a)41 g3(b) 310mg1 g1000 mg= 3 g6(c) 8.210g61 g10g= 8.2 g-8(d) 4.131010kg1000 g1 kg-5= 4.131010g1.72Travel time = 1490. mi1.609 km1 mi1 hr220. km= 10.9 hr1.741.00 mL of butter = 0.860 g and 1.00 mL of sand = 2.28 g(a)mixture3.14 g mixtured= 1.57 g/mL2.00 mL(b)First calculate the volumes of sand and butter, which totals 1.60 mLsandV1.00 g sand1.00 mL sand2.28 g sandbutter= 0.439 mLV1.00 g butter1.00 mL butter0.860 g butter= 1.16 mLmixture2.00 gd== 1.25 g/mL1.60 mL1.7660. mg meperidine1 mL sol75 mg meperidine= 0.8 mL solution injected1.78Aspirin tablets that contain 325 mg of aspirin contain more aspirin than 81 mg tablets. The325 mg dose of aspirin is usually taken for pain, while the 81 mg daily dose of aspirin isusually taken to help prevent heart attacks.1.80First, it is important to know exactly what the treatment involves and the mechanism ofaction. Also, did the study compare the treatment group with a control group consisting ofear infection patients that didn’t receive the treatment?Finally, did the scientist test thetreatment for safety?

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Chapter 1: Matter, Energy, and Measurement71.82The folk medicine will be extracted and separated into pure materials. Each pure materialwill be tested for biological activity. The pure materials that are identified as biologicallyactive will have their chemical structures determined, and often, a chemical synthesisdeveloped for mass production.1.84flow rate =1 L sol12 hr1000 mL1 L sol1 hr60 min15 gttsmL= 21 gtts/min1.86(a) Dose (in milligrams) of Tylenol calculated to two significant figures:Dose = 42 lb child1 kg child2.205 lb child15 mg Tylenol1 kg child= 2.9102mg Tylenol(b) Dose in milliliters of Tylenol solution calculated to two significant figures:Dose = 42 lb child1 kg child2.205 lb child15 mg Tyl1 kg child5.0 mL sol160 mg Tyl= 8.9 mL sol

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8Chapter 2: AtomsQC2.1(a)NaClO3(b) AlF3QC2.2(a)The mass number is 15 + 16 = 31.(b)The mass number is 86 + 136 = 222.QC2.3(a)The element has 15 protons, making it phosphorus (P); its symbol is1531P.(b)The element has 86 protons, making it radon (Rn); its symbol is86222Rn.QC2.4(a)The atomic number of mercury (Hg) is 80; that oflead (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 thisisotope of Pb is 82 + 120 = 202.(d)The symbols of these isotopes are80200Hgand82202Pb.QC2.5The 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 twoisotopes are53125Iand53131I.QC2.6The atomic weight is 6.941 amu, which is nearer to 7 amu than 6 amu. Therefore,lithium-7 is the more abundant isotope. The relative abundances for these two isotopesare 92.50 percent for lithium-7 and 7.50 percent for lithium-6.QC2.7This element has 13 electrons and, therefore, 13 protons. The element with atomicnumber 13 is Aluminum (Al).2.2(b), (c), (d), (f), (g), (h), and (k): True(a)False: matter is divided into pure substances and mixtures.(e)False: mixtures can be separated into their component pure substances.(i)False: technetium, promethium, and all of the elements beyond uranium areartificial.(j)False: H, O, C, N, Ca, and P are the six most important elements in the human body.(l)False: the combining ratio is based on a ratio of atoms, not a ratio of masses.Al

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Chapter 2: Atoms92.4(a) Oxygen(b) Lead(c) Calcium(d) Sodium(e) Carbon(f) Titanium(g) Sulfur(h) Iron(i) Hydrogen(j) Potassium(k) Silver(l) Gold2.6Given 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)(l)Lutetium (Lu, 71)(m)Magnesium (Mg, 12)(n)Polonium (Po, 84)(o)Rhenium Re, 75)(p)Ruthenium (Ru, 44)(q)Scandium (Sc, 21)(r)Strontium (Sr, 38)(s)Ytterbium (Yb, 70), Terbium (Tb, 65), and Yttrium (Y, 39)(t)Thulium (Tm, 69)2.8(a) K2O(b) Na3PO4(c) LiNO32.10(a)Thelaw of conservation of mass states that matter can be neither created nor destroyed.Dalton's theory explains this because if all matter is made up of indestructible atoms,then any chemical reaction just changes the attachments between atoms and does notdestroy the atoms themselves.(b)The law of constant composition states that any compound is always made up ofelements in the same proportion by mass. Dalton's theory explains this becausemolecules consist of tightly bound groups of atoms, each of which has a particularmass. Therefore, each element in a compound always constitutes a fixed proportion ofthe total mass.2.12No.CO and CO2are different compounds, and each obeys the law of constant compositionfor that particular compound.2.14(b), (c), (e), (f), (g), (h), (k), (m), (o), (p), (q), and (s): True(a)False: electrons and protons have equal, but opposite charges. Electrons are muchlighter in mass than protons.(d)False: 1 amu = 1.660510-24grams.(i)False: opposite charges attract each other.(j)False: the size of the atom includes the space occupied by the electrons. The nucleus isa small fraction of the size of the atom.(l)False:the mass number is the number of protons and neutrons.(n)False:1H has no neutrons,2H has one neutron, and3H has two neutrons.(r)False:atomic weights areweightedaverages of the known isotopesand theirabundances.(t)False:density is mass/volume.2.16The statement is true.Theelement is determined by thenumber of protons (the atomicnumber).

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Chapter 2: Atoms102.18(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).2.20Each would still be the same element because the number of protons has not changed.(a)The element is scandium (Sc), and its symbol is2147Sc.(b)The element is titanium (Ti), and its symbol is2250Ti.(c)The element is silver (Ag), and its symbol is47109Ag.(d)The element is thorium (Th) and its symbol is90248Th.(e)The element is argon (Ar) and its symbol is1838Ar.2.22Radon (Rn) has an atomic number of 86, so each isotope has 86 protons. The number ofneutrons is mass number-atomic number.(a)Radon-210 has 210-86 = 124neutrons(b)Radon-218 has 218-86 = 132 neutrons(c)Radon-222 has 222-86 = 136 neutrons2.24Two more neutrons:tin-120Three more neutrons:tin-121Six more neutrons:tin-1242.26(a)An ion is an atom or group of bonded atoms with an unequal number of protons andelectrons.(b)Isotopes are atoms with the same number of protons in their nuclei but a differentnumber of neutrons.2.28Rounded to four significant figures, the calculated value is 12.01 amu. The value given inthe Periodic Table is 12.011 amu.98.90100´12.000 amuæèçöø÷+1.10100´13.003 amuæèçöø÷= 12.01 amu2.30Carbon-11 has 6 protons, 6 electrons, and 5 neutrons.2.32Americium-241 (Am) has atomic number 95. This isotope has 95 protons, 95 electrons and241-95 = 146 neutrons.

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Chapter 2: Atoms112.34(a), (f), (g),(h), and (i): True(b)False:main group elements go from groups 1A through 8A(c)False:very roughly, the metalloids exist in a diagonal, starting in the lower right corner,moving up to upper middle of the Periodic Table. Nonmetals exist above thediagonal, metals below it.(d)False:there are more metals than nonmetals(e)False:horizontal rows are called periods.2.36(a)Groups 2A(2), 3B(3), 4B(4), 5B(5), 6B(6), 7B(7), 8B(8,9, and 10), 1B(11), and 2B(12)contain only metals.Note that Group 1A contains one nonmetal, hydrogen.(b)No group contains only metalloids.(c)Only Groups 7A(17)and 8A(18)contain only nonmetals.2.38In the Periodic Table, elements of the same group should have similar properties: As, N,and P; I and F; Ne and He; Mg, Ca, and Ba; K and Li.2.40(a)aluminum > silicon(b)arsenic > phosphorus(c)gallium > germanium(d)gallium > aluminum2.42(a), (b), (c), (e), (g), (h), (j), (k), (l), (m), (n), (q), (r), and (s): True(d)False:principal energy level 1 contains a maximum of two electrons, the principalenergy level 2 contains a maximum of eight electrons, the principal energylevel 3contains a maximum of 18 electrons, and the principal energy level 4contains a maximum of 32 electrons.(f)False:a 2selectron is easier to remove than a 1selectron because it is further awayfrom the influence of the positively charged nucleus.(i)False:The three 2porbitals are arranged perpendicular to each other.(o)False:paired electrons have spins in the opposite direction.(p)False:each box represents an orbital and each orbital can accommodate two electrons.When an orbital is completely filled, the electrons must be paired with spins inthe opposite direction.(t)False:group 6A elements have two unpaired electrons.2.44The group number indicates the number of electrons in the valence shell of the element.2.46(a) Li(3): 1s22s1(b) Ne(10): 1s22s22p6(c) Be(4): 1s22s2(d) C(6): 1s22s22p2(e) Mg(12): 1s22s22p63s22.48(a) He(2): 1s2(b) Na(11): 1s22s22p63s1(c) Cl(17): 1s22s22p63s23p5(d) P(15): 1s22s22p63s23p3(e) H(1): 1s12.50In (a),(b), and (c); the outer-shell electron configurations are the same. The onlydifference is the principal energy level of the valence shell being filled.

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Chapter 2: Atoms122.52The element X might be in Group 2A, all of which have two valence electrons. It mightalso 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 ofelements.(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 thePeriodic Table.2.58(a)The additional electron in the valence shell exceeds the positive charge in the nucleuscausing the electrons in the valence shell to be held less tightly, thus increasing theatomic radius. Also, the additional electron in the valence shell introduces newrepulsions causing the electron cloud to expand.(b)With the loss of a valence electron, the positive charge in the nucleus exceeds thecollective negative charge of the electrons, causing the valence electrons tobeheldmore tightly, thus decreasing the atomic radius. Also, one less electron in the valenceshell reduces the electron-electron repulsions causing the electron cloud to contract.2.60Ionization energies generally increase going from left to right in a period of the PeriodicTabledue to the increasing positive charge in the nucleus. As the nucleus becomes morepositively charge within a period, the valence electrons become more difficult to remove inan ionization.2.62Sulfur and iron are essential components of proteins, and calcium is a major component ofbones and teeth.2.64Calcium is an essential element in human bones andteeth. Because strontium behaveschemically much like calcium, strontium-90 gets into our bones and teeth and gives offradioactivity for many years directly into our bodies.2.66Copper can be made harder by hammering it.2.68(a)Metals(b)Nonmetals(c)Metals(d)Nonmetals(e)Metals(f)Metals2.70The atomic radius decreases going from left to right across a period in the PeriodicTable.Atomic radius increases going down a group (column) in the Periodic Table.(a)Radium (Ra)(d)Neon (Ne)(b)Beryllium (Be)(e)Fluorine (F)(c)Lithium (Li)(f)Astatine (At)

Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition - Page 14 preview image

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Chapter 2: Atoms132.72(a)Phosporous-32 has 15 protons, 15 electrons, and 32-15 = 17 neutrons.(b)Molybdenum-98 has 42 protons, 42 electrons, and 98-42 = 56 neutrons.(c)Calcium-44 has 20 protons, 20 electrons, and 44-20 = 24 neutrons.(d)Hydrogen-3 has 1 proton, 1 electron, and 3-1 = 2 neutrons.(e)Gadolinium-158 has 64 protons, 64 electrons, and 158-64 = 94 neutrons.(f)Bismuth-212 has 83 protons, 83 electrons, and 212-83 = 129 neutrons.2.74For elements with atomic numbers less than that of iron (Fe), the number of neutrons isclose to the number of protons. Elements with atomic numbers greater than that of ironhave more neutrons than protons. Therefore, heavy elements typically have more neutronsthan protons.2.76Rounded to four significant figures, the atomic weight of naturally occurring boron is 10.81amu. The value given in the Periodic Table is 10.811amu.19.9100´10.013 amuæèçöø÷+80.1100´11.009 amuæèçöø÷= 10.8 amu2.78It would take 6.0 x 1021protons to equal the mass of a grain of salt.2.80The atomic number of this element is 54, which means that it is xenon (Xe). This isotopeof xenon has 54 protons, 54 electrons, and 131-54 = 77 neutrons.2.82(a)Ionization energy generally decreases down a column in the Periodic Table, so theionization energy ofTennessineshould be less than that of At(85).(b)Ionization energy generally increases from left to right across a row in the PeriodicTable, so ionization energy ofTennessineshould be greater than Ra(88).2.84(a)The first electron is removed from the 2sorbital. The removal of each subsequentelectron requires more energy because, after the first electron is removed, eachsubsequent electron is removed from a positive ion, which strongly attracts theremaining electrons. The second ionization energy is especially large because theelectron is removed from the filled first principal energy level, meaning that it isremoved from an ion that has the same electron configuration as helium.(b)Li+ion has a smaller radius than Li because with the loss of a valence electron fromLi, the positive charge in the nucleus exceeds the collective negative charge of theelectrons, causing the valence electrons to be held more tightly, thus decreasing theatomic radius. Also, one less electron in the valence shell of Li+reduces the electron-electron repulsions causing the electron cloud to contract.1.0 x 10-2g NaCl1.67 x 10-2 4g/ proton = 6.0 x 1021protons

Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition - Page 15 preview image

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Chapter 2: Atoms142.86Increasing size: C < B < Al < NaThere are relatively small decreases in atomic radii going from left to right in the PeriodicTable due to the increasingly stronger pull on the electrons because of the additionalprotons, therefore, the size of the atomic radii can be compared as C < B and Al < Na.Going to a higher numbered period in the PeriodicTable results inamuch larger increasein atomic radii, therefore, both Na and Al are larger than C and B.2.88Going down Group 3A, we have boron, aluminum, gallium, indium, and thallium. Theelectron configuration of each 3A element tells us that the valence shellsorbitals arecompletely filled and that the valence shellporbital has one electron.2.90The Ca3+ion is not found in chemical compounds because calcium only needs to lose twoelectrons to achieve a stablenoble gas electronic configuration. Losing three electronsbrings the electronic configuration away from a stable noble gas electronic configuration.2.92If the bromine is all used up, it is assumed that it is the limiting reagent. The reaction ofmagnesium and bromine is as follows:M g + Br2®MgBr2Mg reacted = 1.80 g Br21 mol Br2159.8 g Br2æèççöø÷÷1 mol Mg1 mol Br2æèççöø÷÷24.305 g Mg1 mol Mgæèçöø÷= 0.274 gMg remaining = 7.12 g (Mg initial mass) - 0.274 g (Mg reacted) = 6.85 g2.94Answers in bold. The mass numbers derivedreflect the most abundant isotope.SymbolAtomicnumberAtomicweightMassnumber# ofprotons# ofneutrons# ofelectronsH11.00791101Li36.9417343Al1326.981527131413Fe2655.84558263226Pt78195.0841957811778Ca2040.07837201720S1632.066321616162.960.4818109Ag atomic mass()+ 0.5182(106.905 amu) = 107.868 amu109Ag atomic mass = 52.47 amu0.4818109Ag atomic mass = 108.9 amu

Solution Manual For Introduction to General, Organic, and Biochemistry, 12th Edition - Page 16 preview image

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Chapter 2: Atoms152.98If30Si has an isotopic abundance of3.09%, then28Si +29Si = 96.91% of the mix.Therefore,if28Si = xand29Si = y,then(x + y) = 0.9691and y = (0.9691–x ).(1) 27.977(x) amu + 28.086 (y) amu + 29.974´(0.0309) amu = 28.086 amu(2) replace y with (0.9691–x) and then solve for x(3) 27.977(x) amu + 28.086 (0.9691–x) amu + 29.974´(0.0309) amu = 28.086 amux = 0.9217The isotopic abundance is:28Si= 92.17%,29Si = 4.74%, and30Si = 3.09%(given)2.100(a) Ar(b) Al(c) Si(d) Cl(e) Mg(f) P(g) Ar(h) S

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