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Solution Manual For Chemistry, 13th Edition - Document preview page 1

Solution Manual For Chemistry, 13th Edition - Page 1

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Solution Manual For Chemistry, 13th Edition

Gain deeper insight into your textbook problems with Solution Manual For Chemistry, 13th Edition, featuring well-explained solutions.

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Solution Manual For Chemistry, 13th Edition - Page 1 preview imageCHAPTER 1CHEMISTRY: THE STUDY OF CHANGEProblem CategoriesBiological: 1.26, 1.50, 1.71, 1.72, 1.80, 1.86, 1.96, 1.97, 1.105, 1.114.Conceptual: 1.3, 1.4, 1.17, 1.18, 1.11, 1.12, 1.56, 1.64, 1.91, 1.94, 1.101, 1.103, 1.117.Environmental: 1.72, 1.89, 1.91, 1.98, 1.109, 1.112.Industrial: 1.53, 1.57, 1.83.Difficulty LevelEasy: 1.3, 1.9, 1.10, 1.11, 1.12, 1.17, 1.23, 1.24, 1.25, 1.26, 1.27, 1.28, 1.31, 1.32, 1.33, 1.34, 1.35, 1.36, 1.56, 1.57,1.66, 1.79, 1.82, 1.86, 1.91.Medium: 1.4, 1.18, 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, 1.53,1.54, 1.55, 1.58, 1.59, 1.60, 1.61, 1.62, 1.63, 1.64, 1.65, 1.72, 1.73, 1.74, 1.75, 1.76, 1.77, 1.78, 1.80, 1.81, 1.83, 1.84,1.85, 1.87, 1.93, 1.96, 1.97, 1.98.Difficult: 1.67, 1.68, 1.69, 1.70, 1.71, 1.88, 1.89, 1.90, 1.92, 1.94, 1.95, 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 artisticexcellence.(c)Qualitative. If the numerical values for the densities of ice and water were given, it would be aquantitative statement.(d)Qualitative. Another value judgment.(e)Qualitative. Even though numbers are involved, they are not the result of measurement.1.4(a)hypothesis(b)law(c)theory1.9Li, 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.1.10(a)Cs(b)Ge(c)Ga(d)Sr(e)U(f)Se(g)Ne(h)Cd1.11(a)element(b)compound(c)element(d)compound
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Solution Manual For Chemistry, 13th Edition - Page 2 preview image
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Solution Manual For Chemistry, 13th Edition - Page 3 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE21.12(a)homogeneous mixture(b)element(c)compound(d)homogeneous mixture(e)heterogeneous mixture(f)heterogeneous mixture(g)element1.17(a)Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity arechanged.(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 orcomposition.(d)Physical property. The measurement of the densities of lead and aluminum does not change theircomposition.(e)Chemical property. When uranium undergoes nuclear decay, the products are chemically differentsubstances.1.18(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.1.23586 gmassvolume188 mL===density3.12 g/mL1.24Strategy: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.=massdensityvolumeSolution:mass=density´volume=´=0.7918 g89.9 mL1 mLmass of methanol71.2 g
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Solution Manual For Chemistry, 13th Edition - Page 4 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE31.25=-´5 C?C( F32 F)9 F(a)=-´=5 C?C(9532) F9 F35 C(b)=-´= -5 C?C(1232) F9 F11 C(c)=-´=5 C?C(10232) F9 F39 C(d)=-´=5 C?C(185232) F9 F1011 C(e)9 F? FC32 F5 Cæö÷ç÷=´+ç÷ç÷çèøæö÷ç÷= -´+= -ç÷ç÷çèø9 F?F273.15 C32 F5 C459.67 F1.26Strategy:Find the appropriate equations for converting between Fahrenheit and Celsius and betweenCelsius and Fahrenheit given in Section 1.7 of the text. Substitute the temperature values given in theproblem into the appropriate equation.(a)Conversion from Fahrenheit to Celsius.=-´5 C?C( F32 F)9 F=-´=5 C(10532) F9 F?C41 C(b)Conversion from Celsius to Fahrenheit.æö÷ç÷=´+ç÷ç÷çèø9 F? FC32 F5 Cæö÷ç÷= -´+=ç÷ç÷çèø9 F11.5 C32 F5 C?F11.3 F(c)Conversion from Celsius to Fahrenheit.æö÷ç÷=´+ç÷ç÷çèø9 F? FC32 F5 Cæö÷ç÷=´´+=ç÷ç÷çèø39 F6.310C32 F5 C4?F1.110F´(d)Conversion from Fahrenheit to Celsius.=-´5 C?C( F32 F)9 F=-´=5 C(45132) F9 F?C233 C
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Solution Manual For Chemistry, 13th Edition - Page 5 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE41.27=+1 KK( C273 C) 1 C(a)K=113C+273C=386 K(b)K=37C+273C=3.10´102K(c)K=357C+273C=6.30´102K1.28(a)=+1 KK( C273 C) 1 CC=K-273=77 K-273=-196C(b)C=4.2 K-273=-269C(c)C=601 K-273=328C1.31(a)2.7´10-8(b)3.56´102(c)4.7764´104(d)9.6´10-21.32(a)10-2indicates that the decimal point must be moved two places to the left.1.52´10-2=0.0152(b)10-8indicates that the decimal point must be moved 8 places to the left.7.78´10-8=0.00000007781.33(a)145.75+(2.3´10-1)=145.75+0.23=1.4598´102(b)4227.9510795002.5102.510´==´´23.2 × 10(c)(7.0´10-3)-(8.0´10-4)=(7.0´10-3)-(0.80´10-3)=6.2´10-3(d)(1.0´104)´(9.9´106)=9.9´10101.34(a)Addition using scientific notation.Strategy:Let’s express scientific notation asN´10n. When adding numbers using scientific notation, wemust write each quantity with the same exponent,n. We can then add theNparts of the numbers, keeping theexponent,n, the same.Solution:Write each quantity with the same exponent,n.
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Solution Manual For Chemistry, 13th Edition - Page 6 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE5Let’s write 0.0095 in such a way thatn=-3. We have decreased 10nby 103, so we must increaseNby 103.Move the decimal point 3 places to the right.0.0095=9.5´10-3Add theNparts of the numbers, keeping the exponent,n, the same.9.5´10-3+8.5´10-318.0´10-3The usual practice is to expressNas a number between 1 and 10. Since we mustdecreaseNby a factor of 10to expressNbetween 1 and 10 (1.8), we mustincrease10nby a factor of 10. The exponent,n, is increased by1 from-3 to-2.18.0´10-3=1.8´10-2(b)Division using scientific notation.Strategy:Let’s express scientific notation asN´10n. When dividing numbers using scientific notation,divide theNparts of the numbers in the usual way. To come up with the correct exponent,n, wesubtracttheexponents.Solution:Make sure that all numbers are expressed in scientific notation.653=6.53´102Divide theNparts of the numbers in the usual way.6.53¸5.75=1.14Subtractthe exponents,n.1.14´10+2-(-8)=1.14´10+2+8=1.14´1010(c)Subtraction using scientific notation.Strategy:Let’s express scientific notation asN´10n. When subtracting numbers using scientific notation,we must write each quantity with the same exponent,n. We can then subtract theNparts of the numbers,keeping the exponent,n, the same.Solution:Write each quantity with the same exponent,n.Let’s write 850,000 in such a way thatn=5. This means to move the decimal point five places to the left.850,000=8.5´105
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Solution Manual For Chemistry, 13th Edition - Page 7 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE6Subtract theNparts of the numbers, keeping the exponent,n, the same.8.5´105-9.0´105-0.5´105The usual practice is to expressNas a number between 1 and 10. Since we mustincreaseNby a factor of 10to expressNbetween 1 and 10 (5), we mustdecrease10nby a factor of 10. The exponent,n, is decreased by1 from 5 to 4.-0.5´105=-5´104(d)Multiplication using scientific notation.Strategy:Let’s express scientific notation asN´10n. When multiplying numbers using scientificnotation, multiply theNparts of the numbers in the usual way. To come up with the correct exponent,n, weaddthe exponents.Solution:Multiply theNparts of the numbers in the usual way.3.6´3.6=13Addthe exponents,n.13´10-4+(+6)=13´102The usual practice is to expressNas a number between 1 and 10. Since we mustdecreaseNby a factor of 10to expressNbetween 1 and 10 (1.3), we mustincrease10nby a factor of 10. The exponent,n, is increased by1 from 2 to 3.13´102=1.3´1031.35(a)four(b)two(c)five(d)two, three, or four(e)three(f)one(g)one(h)two1.36(a)one(b)three(c)three(d)four(e)two or three(f)one(g)one or two1.37(a)10.6 m(b)0.79 g(c)16.5 cm2(d)1 × 106g/cm31.38(a)DivisionStrategy:The number of significant figures in the answer is determined by the original number having thesmallest number of significant figures.
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Solution Manual For Chemistry, 13th Edition - Page 8 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE7Solution:7.310 km5.70 km=1.283The 3 (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?)(b)SubtractionStrategy:The number of significant figures to the right of the decimal point in the answer is determined bythe 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 have0.00326 mg-0.0000788 mg0.0031812mgThe bolded numbers are nonsignificant digits because the number 0.00326 has five digits to the right of thedecimal 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:0.00318 mg=3.18´10-3mg(c)AdditionStrategy:The number of significant figures to the right of the decimal point in the answer is determined bythe 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(0.402´107dm)+(7.74´107dm)=8.14´107dmSince 7.74´107has only two digits to the right of the decimal point, two digits are carried to the right of thedecimal point in the final answer.(d)Subtraction, addition, and divisionStrategy:For subtraction and addition, the number of significant figures to the right of the decimal point inthat part of the calculation is determined by the lowest number of digits to the right of the decimal point in
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Solution Manual For Chemistry, 13th Edition - Page 9 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE8any of the original numbers. For the division part of the calculation, the number of significant figures in theanswer is determined by the number having the smallest number of significant figures. First, perform thesubtraction and addition parts to the correct number of significant figures, and then perform the division.Solution:(7.8 m0.34 m)7.5 m(1.15 s0.82 s)1.97 s-==+3.8 m/s1.39Calculating the mean for each set of data, we find:Student A: 87.6 mLStudent B: 87.1 mLStudent C: 87.8 mLFrom these calculations, we can conclude that the volume measurements made by Student B were the mostaccurate of the three students. The precision in the measurements made by both students B and C are fairlyhigh, while the measurements made by student A are less precise. In summary:Student A: neither accurate nor preciseStudent B: both accurate and preciseStudent C: precise, but not accurate1.40Calculating the mean for each set of data, we find:Tailor X: 31.5 inTailor Y: 32.6 inTailor Z: 32.1 inFrom these calculations, we can conclude that the seam measurements made by Tailor Z were the mostaccurate 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 preciseTailor Y: least accurate and least preciseTailor Z: most accurate1.41(a)=´=1 dm22.6 m0.1 m? dm226 dm(b)0.001g1kg25.4 mg1mg1000 g=´´=5? kg2.5410kg-´(c)3110L556 mL1 mL-´=´=? L0.556 L
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Solution Manual For Chemistry, 13th Edition - Page 10 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE9(d)32310.6 kg1000 g110m1 kg1 cm1 m-æö÷´ç÷ç÷ç=´´=÷ç÷ç÷÷çèø33g?0.0106 g cmcm/1.42(a)Strategy:The problem may be stated as? mg=242 lbA relationship between pounds and grams is given on the end sheet of your text (1 lb=453.6 g). Thisrelationship will allow conversion from pounds to grams. A metric conversion is then needed to convertgrams to milligrams (1 mg=1´10-3g). Arrange the appropriate conversion factors so that pounds andgrams cancel, and the unit milligrams is obtained in your answer.Solution:The sequence of conversions islbgramsmgUsing the following conversion factors,453.6 g1 lb31 mg110g-´we obtain the answer in one step:-=´´=´31 mg453.6 g242 lb1 lb110g8? mg1.1010mg´Check:Does your answer seem reasonable? Should 242 lb be equivalent to 110 million mg? How manymg are in 1 lb? There are 453,600 mg in 1 lb.(b)Strategy:The problem may be stated as? m3=68.3 cm3Recall that 1 cm=1´10-2m. We need to set up a conversion factor to convert from cm3to m3.Solution:We need the following conversion factor so that centimeters cancel and we end up with meters.-´2110m1 cm
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Solution Manual For Chemistry, 13th Edition - Page 11 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE10Since this conversion factor deals with length and we want volume, it must therefore be cubed to give----æö÷´´´´ç÷ç÷ç´´=÷ç÷ç÷÷çèø32222110m110m110m110m1 cm1 cm1 cm1 cmWe can write-æö÷´ç÷ç÷ç=´=÷ç÷ç÷÷çèø323110m68.3 cm1 cm353? m6.8310m-´Check:We know that 1 cm3=1´10-6m3. We started with 6.83´101cm3. Multiplying this quantity by1´10-6gives 6.83´10-5.(c)Strategy:The problem may be stated as? L=7.2 m3In Chapter 1 of the text, a conversion is given between liters and cm3(1 L=1000 cm3). If we can convert m3to cm3, we can then convert to liters. Recall that 1 cm=1´10-2m. We need to set up two conversionfactors to convert from m3to L. Arrange the appropriate conversion factors so that m3and cm3cancel, andthe unit liters is obtained in your answer.Solution:The sequence of conversions ism3cm3LUsing the following conversion factors,-æö÷ç÷ç÷ç÷ç÷ç÷´÷çèø321 cm110m31 L1000 cmthe answer is obtained in one step:-æö÷ç÷ç÷ç=´´=÷ç÷ç÷´÷çèø33231 cm1 L7.2 m110m1000 cm3? L7.210L´Check:From the above conversion factors you can show that 1 m3=1´103L. Therefore, 7 m3wouldequal 7´103L, which is close to the answer.
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Solution Manual For Chemistry, 13th Edition - Page 12 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE11(d)Strategy:The problem may be stated as? lb=28.3mgA relationship between pounds and grams is given on the end sheet of your text (1 lb=453.6 g). Thisrelationship will allow conversion from grams to pounds. If we can convert frommg to grams, we can thenconvert from grams to pounds. Recall that 1mg=1´10-6g. Arrange the appropriate conversion factors sothatmg and grams cancel, and the unit pounds is obtained in your answer.Solution:The sequence of conversions ismgglbUsing the following conversion factors,-´m6110g1g1 lb453.6 gwe can write-´=m´´=m6110g1 lb28.3g1g453.6 g8? lb6.2410lb-´Check:Does the answer seem reasonable? What number does the prefixmrepresent? Should 28.3mg be avery small mass?1.431255 m1 mi3600 s1 s1609 m1 h´´=2808 mi h/1.44Strategy:The problem may be stated as? s=365.24 daysYou should know conversion factors that will allow you to convert between days and hours, between hoursand 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 isdayshoursminutessecondsUsing the following conversion factors,24 h1 day60 min1 h60 s1 min
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Solution Manual For Chemistry, 13th Edition - Page 13 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE12we can write24 h60 min60 s365.24 days1 day1 h1 min=´´´=´7? s3.155710sCheck:Does your answer seem reasonable? Should there be a very large number of seconds in 1 year?1.45´´´´´=´681.609 km1000 m1 s1 min(9310mi)1 mi1 km60 s3.0010m8.3 min1.46(a)? in/s=´´´1 mi5280 ft12 in1 min8.92 min1 mi1 ft60 s=118 in/s(b)? m/min=´1 mi1609 m8.92 min1 mi=1.80´210m/min(c)? km/h=´´´1 mi1609 m1 km60 min8.92 min1 mi1000 m1 h=10.8 km/h1.47´=1 m6.0 ft3.28 ft1.8 m´´=453.6 g1 kg168 lb1 lb1000 g76.2 kg1.48? mph=´286 km1 mi1 h1.609 km=178 mph1.49´´=62 m1 mi3600 s1 s1609 m1 h21.410mph´1.50=´60.62 g Pb0.62 ppm Pb110g blood´´=´´360.62 g Pb6.010g of blood110g blood33.710g Pb-1.51(a)8365 days24 h3600 s3.0010m1 mi1.42 yr1 yr1 day1 h1 s1609 m´´´´´´=128.3510mi´(b)´´36 in2.54 cm32.4 yd1 yd1 in32.9610cm=´
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Solution Manual For Chemistry, 13th Edition - Page 14 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE13(c)103.010cm1 in1 ft1 s2.54 cm12 in´´´=89.810ft /s´1.52(a)=´=1 lb70 kg0.4536 kg2? lbs1.510lbs´(b)=´´´´=9365 d24 h3600 s1410yr1 yr1 d1 h17? s4.410s´(c)-´=´´22.54 cm110m90 in1 in1 cm? m2.3 m=(d)-æö÷ç÷=´´ç÷ç÷çèø´33231 cm1 L88.6 m110m1000 cm4? L8.8610L=´1.53332.70 g1 kg1 cm1000 g0.01 m1 cmæö÷ç÷ç=´´=÷ç÷ç÷÷çèø33density2.7010kg m´/1.5430.625 g1 L1 mL1 L1000 mL1 cm-=´´=43density6.2510g cm´/1.55SubstanceQualitative StatementQuantitative Statement(a)watercolorless liquidfreezes at 0C(b)carbonblack solid (graphite)density=2.26 g/cm3(c)ironrusts easilydensity=7.86 g/cm3(d)hydrogen gascolorless gasmelts at-255.3C(e)sucrosetastes sweetat 0C, 179 g of sucrose dissolves in 100 g of H2O(f)table salttastes saltymelts at 801C(g)mercuryliquid at room temperatureboils at 357C(h)golda precious metaldensity=19.3 g/cm3(i)aira mixture of gasescontains 20% oxygen by volume1.56See Section 1.6 of your text for a discussion of these terms.(a)Chemical property. Iron has changed its composition and identity by chemically combining withoxygen and water.(b)Chemical property. The water reacts with chemicals in the air (such as sulfur dioxide) to produce acids,thus changing the composition and identity of the water.
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Solution Manual For Chemistry, 13th Edition - Page 15 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE14(c)Physical property. The color of the hemoglobin can be observed and measured without changing itscomposition or identity.(d)Physical property. The evaporation of water does not change its chemical properties. Evaporation is achange in matter from the liquid state to the gaseous state.(e)Chemical property. The carbon dioxide is chemically converted into other molecules.1.57´´=´931 ton(95.010lb of sulfuric acid)2.010lb74.7510tons of sulfuric acid´1.58Volume of rectangular bar=length´width´height52.7064 g= (8.53 cm)(2.4 cm)(1.0 cm)==3density2.6 g cmmV/1.59mass=density´volume(a)mass=(19.3 g/cm3)´[43p(10.0 cm)3]=8.08´104g(b)-æö÷ç÷ç=´´=÷ç÷ç÷÷çèø331 cm(21.4 g/cm )0.040 mm10 mm6mass1.410g´(c)mass=(0.798 g/mL) (50.0 mL)=39.9 g1.60You are asked to solve for the inner diameter of the bottle. If we can calculate the volume that the cookingoil occupies, we can calculate the radius of the cylinder. The volume of the cylinder is,Vcylinder=pr2h(risthe inner radius of the cylinder, andhis the height of the cylinder). The cylinder diameter is 2r.=mass of oilvolume of oil filling bottledensity of oil==´=´3331360 gvolume of oil filling bottle1.4310mL1.4310cm0.953 g/mLNext, solve for the radius of the cylinder.Volume of cylinder=pr2hvolume=p ´rh´==p ´331.4310cm4.60 cm21.5 cmrThe inner diameter of the bottle equals 2r.Bottle diameter=2r=2(4.60 cm)=9.20 cm
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Solution Manual For Chemistry, 13th Edition - Page 16 preview imageCHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE151.61From the mass of the water and its density, we can calculate the volume that the water occupies. The volumethat the water occupies is equal to the volume of the flask.=massvolumedensityMass of water=87.39 g-56.12 g=31.27 g==331.27 gmass=density0.9976 g/cm3Volume of the flask31.35 cm1.62´´=343 m1 mi3600 s1 s1609 m1 h767 mph1.63The volume of silver is equal to the volume of water it displaces.Volume of silver=260.5 mL-242.0 mL=18.5 mL=18.5 cm33194.3 g18.5 cm==3density10.5 g cm/1.64In order to work this problem, you need to understand the physical principles involved in the experiment inProblem 1.61. The volume of the water displaced must equal the volume of the piece of silver. If the silverdid not sink, would you have been able to determine the volume of the piece of silver?The liquid must beless densethan the ice in order for the ice to sink. The temperature of the experiment mustbe maintained at or below0°Cto prevent the ice from melting.1.65The volume of a sphere is:æö÷ç÷ç=p=p=´÷ç÷ç÷çèø334348.6 cm446.0110cm332Vr´===´5436.85210gmassvolume6.0110cmdensity11.4g/cm31.66=massVolumedensity´==331.2010g0.53 g / cm33Volume occupied by Li2.310 cm´
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