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Chemistry 4th Edition Solution Manual - Document preview page 1

Chemistry 4th Edition Solution Manual - Page 1

Document preview content for Chemistry 4th Edition Solution Manual

Chemistry 4th Edition Solution Manual

Solve your textbook questions with ease using Chemistry 4th Edition Solution Manual, a comprehensive and easy-to-follow guide.

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Chemistry 4th Edition Solution Manual - Page 1 preview imageChapter 1Chemistry: The Central SciencePractice Problems C1.1(iii)1.2(i) and (i)1.3pink liquid = grey solid < blue solid < yellow liquid < blue liquid < green solid1.4physical: (iii), chemical: (i), (ii) is neither1.512 blue cubes, infinite number of significant figures;2 × 101red spheres, one significant figure1.62.4 × 102lbs.1.72.67 g/cm31.8(a) 4 red blocks/1 object(b) 1 object/1 yellow block(c) 2 white blocks/1 yellow block(d) 1 yellow block/6 grey connectors1.9375 red bars; 3500 yellow ballsApplying What You’ve Learneda)The recommended storage-temperature range for cidofovir is 20°C−25°C.b)The density of the fluid in a vial is 1.18 g/mL. (The density should be reported to three significant figures.)c)The recommended dosage of cidofovir for a 177-lb man is 4 × 102mg or 0.4 g.d)1.18 × 103g/L, 1.18 × 103kg/m3.
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Chemistry 4th Edition Solution Manual - Page 2 preview image
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Chemistry 4th Edition Solution Manual - Page 3 preview imageChapter 1 -- Chemistry: The Central Science2Questions and Problems1.1Chemistry is the study of matter and the changes that matter undergoes.Matter is anything that has mass and occupies space.1.2The scientific method is a set of guidelines used by scientists to add their experimental results to thelarger body of knowledge in a given field. The process involves observation, hypothesis,experimentation, theory development, and further experimentation.1.3A hypothesis explains observations. A theory explains data from accumulated experiments andpredicts related phenomena.1.4a.Hypothesis– This statement is an opinion.b.Law– Newton’s Law of Gravitation.c.Theory –AtomicTheory.1.5a.Law– Newton’s 2nd Law of Motion.b.Theory– Big Bang Theory.c.Hypothesis– It may be possible but we have no data to support this statement.1.6a.O and Hb.C and Hc.H and Cld.N1.7a.C and Ob.F and Hc.N and Hd.O1.8a. Matter is anything that has mass and occupies space. Examples includeair, seawater, concrete, anautomobile, or a dog.b. A substance is a form of matter that has definite (constant) composition and distinct properties.Examples includeiron, silver, water, or sugar.
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Chemistry 4th Edition Solution Manual - Page 4 preview imageChapter 1 -- Chemistry: The Central Science3c. A mixture is a combination of two or more substances in which the substances retain their distinctidentities. Examples includemilk, salt water, air, or steel.1.9Examples of homogeneous mixtures:apple juice or root beer.Examples of heterogeneous mixtures:chocolate chip cookie or vinaigrette salad dressing.1.10Examples of elements (see the front cover for a complete list):oxygen, platinum, sodium, cobalt.Examples of compounds:sugar, salt, hemoglobin, citric acid.An element cannot be separated into simpler substances by chemical means. A compound can be separatedinto its constituent elements by a chemical reaction.1.11There are118known elements.1.12Li:LithiumF:FluorineP:PhosphorusCu:CopperAs:ArsenicZn:ZincCl:ChlorinePt:PlatinumMg:MagnesiumU:UraniumAl:AluminumSi:SiliconNe:Neon1.13a.K(potassium)b.Sn(tin)d.B(boron)e.Ba(barium)g.S(sulfur)h.Ar (argon)
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Chemistry 4th Edition Solution Manual - Page 5 preview imageChapter 1 -- Chemistry: The Central Science4c.Cr(chromium)f.Pu(plutonium)i.Hg(mercury)1.14a. hydrogen:elementb. water:compoundc. gold:elementd. sugar:compound1.15a.The sea is a heterogeneous mixture of seawater and biological matter, but seawater, with the biomassfiltered out, is ahomogeneous mixture.b.elementc.compoundd.homogeneous mixturee.heterogeneous mixturef.homogeneous mixtureg.heterogeneous mixture1.16a.liquidb.gasc.mixtured.solid1.17a.elementb.compoundc.compoundd.element1.18a. Chemistry Units:meter (m), centimeter (cm), millimeter (mm)SI Base Unit:meter (m)b. Chemistry Units:cubic decimeter (dm3) or liter (L), milliliter (mL), cubic centimeter (cm3)SI Base Unit:cubic meter (m3)
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Chemistry 4th Edition Solution Manual - Page 6 preview imageChapter 1 -- Chemistry: The Central Science5c. Chemistry Units:gram (g)SI Base Unit:kilogram (kg)d. Chemistry Units:second (s)SI Base Unit:second (s)e. Chemistry Units:kelvin (K) or degrees Celsius (°C)SI Base Unit:kelvin (K)1.19a.1 × 106b.1 × 103c.1 × 10−1d.1 × 10−2e.1 × 10−3f.1 × 10−6g.1 × 10−9h.1 × 10−121.20For liquids and solids, chemists normally useg/mL or g/cm3as units for density.For gases, chemists normally useg/Las units for density. Gas densities are generally very low, so thesmaller unit of g/L is typically used. 1 g/L = 0.001 g/mL.1.21Weight is the force exerted by an object or sample due to gravity. It depends on the gravitational forcewhere the weight is measured.Mass is a measure of the amount of matter in an object or sample. It remains constant regardless of where itis measured.Since gravity on the moon is about one sixth that on Earth,1weight on the moon168 lbs on Earth628 lbs1.22Kelvin is known as the absolute temperature scale, meaning the lowest possible temperature is 0 K.The units of the Celsius and Kelvin scales are the same, so conversion between units is a matter of addition:KC273.15The freezing point of water is defined as 0°C. The boiling point of water is defined as 100°C.In the Fahrenheit scale, the freezing point of water is 32°F and the boiling point of water is 212°F. Since thedifference is 180°F, compared to 100°C between the freezing and boiling points of water, one degreeFahrenheit represents a smaller change in temperature than one degree Celsius. To convert between these
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Chemistry 4th Edition Solution Manual - Page 7 preview imageChapter 1 -- Chemistry: The Central Science6two temperature scales, use:5 Ctemperature in Celsiustemperature in F32 F9 F1.23Strategy:Use the density equation:mdVSolution:586 g188 mLmV3.12g/ mLd1.24mass of ethanolൌ଴.଻ଽ଼ ௚ଵ ௠௅ൈ 205 ܮ ݉ൌ ૚૟૝ ࢍ1.25Strategy:Find the appropriate equations for converting between Fahrenheit and Celsius and betweenCelsius and Fahrenheit given in Section 1.3 of the text. Substitute the temperature values givenin the problem into the appropriate equation.Setup:Conversion from Fahrenheit to Celsius:5 CC( F32 F)9 F Conversion from Celsius to Fahrenheit:9 FFC32 F5 CSolution:a.5 CC(95 F32 F)9 F35 Cb.5 CC(12 F32 F)9 F 11 Cc.5 CC(102 F32 F)9 F39 C
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Chemistry 4th Edition Solution Manual - Page 8 preview imageChapter 1 -- Chemistry: The Central Science7d.5 CC(1852 F32 F)9 F1011 Ce.9 FF273.15 C32 F5 C459.67 F1.26a.5 CC105 F32 F9 F41°Cb.9 FF11.5 C32 F5 C11.3°Fc.39 FF6.310C32 F5 C41.1×10 °F1.27Strategy:Use the density equation.Solution:volume of water =ܸ ൌ௠ௗൌଶ଻.଴ ௚଴.ଽଽଶ ௚ ௠௅⁄ൌ 27.2ܮ݉1.28387.6 gvolume of platinum21.5 g/cm34.07 cm1.29Strategy:Use the equation for converting °C to K.Setup:Conversion from Celsius to Kelvin:K = °C + 273.15Solution:a.K= 115.21C + 273.15 =388.36 Kb.K= 37°C + 273 =3.10 × 102KAMPS Solution Co
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Chemistry 4th Edition Solution Manual - Page 9 preview imageChapter 1 -- Chemistry: The Central Science8c.K= 357°C + 273 =6.30 × 102KNote that when there are no digits to the right of the decimal point in the original temperature,we use 273 instead of 273.15.1.30a.°C= K – 273 = 77 K – 273 =–196°Cb.C= 4.22 K273.15 =268.93Cc.C= 600.61 K273.15 =327.46CAMPS Solution Co1.31The picture on the right best illustrates the measurement of the boiling point of water using theCelsius and Kelvin scales. A temperature on the Kelvin scale is numerically equal to the temperaturein Celsius plus 273.15.1.32The relative densities of the aluminum differ based upon the volume the shapes occupy. The largerand flatter ball spreads its mass over a larger area, creating a larger volume and lowering the overalldensity relative to the same volume of water causing the material to float. The smaller rolled up ballminimizes its mass over a small volume creating a larger overall density relative to the same volume ofwater, causing it to sink. The density of the aluminum remains constant.1.33Qualitative data does not require explicit measurement. Quantitative data requires measurement andis expressed with a number.1.34Physical properties can be observed and measured without changing the identity of a substance.Forexample, the boiling point of water can be determined by heating a container of water and measuring thetemperature at which the liquid water turns to steam. The water vapor (steam) is still H2O, so the identity ofthe substance has not changed. Liquid water can be recovered by allowing the water vapor to contact a coolsurface, on which it condenses to liquid water.Chemical properties can only be observed by carrying out a chemical change.During themeasurement, the identity of the substance changes. The original substance cannot be recovered by anyphysical means. For example, when iron is exposed to water and oxygen, it undergoes a chemical change toproduce rust. The iron cannot be recovered by any physical means.1.35An extensive property depends on the amount of substance present. An intensive property isindependent of the amount of substance present.
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Chemistry 4th Edition Solution Manual - Page 10 preview imageChapter 1 -- Chemistry: The Central Science91.36a.extensiveb.extensivec.intensived.extensive1.37a.Quantitative. This statement 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. The statement is a value judgment.e.Qualitative. Even though numbers are involved, they are not the result of measurement.1.38a.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.1.39a.Physical Change.The material is helium regardless of whether it is located inside or outside the balloon.b.Chemical changein the battery.
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Chemistry 4th Edition Solution Manual - Page 11 preview imageChapter 1 -- Chemistry: The Central Science10c.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.1.40Mass is extensive and additive: 44.3 + 115.2 =159.5 gTemperature is intensive:10°CDensity is intensive:1.00 g/mL1.41Mass is extensive and additive: 37.2 + 62.7 =99.9 gTemperature is intensive:20°CDensity is intensive:11.35 g/cm31.42a.Exact.The number of tickets is determined by counting.b.Inexact.The volume must be measured.c.Exact.The number of eggs is determined by counting.d.Inexact.The mass of oxygen must be measured.e.Exact.The number of days is a defined value.1.43Using scientific notation avoids the ambiguity associated with trailing zeros.1.44Significant figures are the meaningful digits in a reported number. They indicate the level ofuncertainty in a measurement. Using too many significant figures implies a greater certainty in ameasured or calculated number than is realistic.1.45Accuracy tells us how close a measurement is to the true value. Precision tells us how close multiplemeasurements are to one another. Having precise measurements does not always guarantee anaccurate result, because there may be an error made that is common g to all the measurements.
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Chemistry 4th Edition Solution Manual - Page 12 preview imageChapter 1 -- Chemistry: The Central Science111.46a.The decimal point must be moved eight places to the right, making the exponent –8.0.000000027 =2.7 × 10–8b.The decimal point must be moved two places to the left, making the exponent 2.356 =3.56 × 102c.The decimal point must be moved four places to the left, making the exponent 4.47,764 =4.7764× 104d.The decimal point must be moved two places to the right, making the exponent –2.0.096 =9.6 × 10–21.47Strategy:To convert an exponential numberN× 10nto a decimal number, move the decimalnplaces tothe left ifn< 0, or move itnplaces to the right ifn> 0. While shifting the decimal, add place-holding zeros as needed.Solution:a.1.5210–2=0.0152b.7.7810–8=0.0000000778c.1 × 10–6=0.000001d.1.6001 × 103=1600.11.48a.1145.752.310145.750.2321.4598×10b.422795007.95×102.5×102.5×1023.2×10c.34337.0108.0107.0100.801036.2×10
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Chemistry 4th Edition Solution Manual - Page 13 preview imageChapter 1 -- Chemistry: The Central Science12d.461.0×109.9×10109.9×101.49a.Addition using scientific notation.Strategy:A measurement is inscientific notationwhen it is written in the formN× 10n, where 0≤N<10 andnis an integer. When adding measurements that are written in scientific notation,rewrite the quantities so that they share a common exponent. To get the “Npart” of theresult, we simply add the “Nparts” of the rewritten numbers. To get the exponent of theresult, we simply set it equal to the common exponent. Finally, if need be, we rewrite theresult so that its value ofNsatisfies 0≤N< 10.Solution:Rewrite the quantities so that they have a common exponent. In this case, choose thecommon exponentn= –3.0.0095 = 9.5×10–3Add the “Nparts” of the rewritten numbers and set the exponent of the result equal to thecommon exponent.3339.5108.51018.010Rewrite the number so that it is in scientific notation (so that010N).18.010–3=1.810–2b.Division using scientific notation.Strategy:When dividing two numbers using scientific notation, divide the “Nparts” of the numbers inthe usual way. To find the exponent of the result,subtractthe exponent of the devisor fromthat of the dividend.Solution:Make sure that all numbers are expressed in scientific notation.653 = 6.53102Divide the “Nparts” of the numbers in the usual way.6.535.75 = 1.14Subtractthe exponents.
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Chemistry 4th Edition Solution Manual - Page 14 preview imageChapter 1 -- Chemistry: The Central Science131.1410+2 – (–8)= 1.1410+2 + 8=1.141010c.Subtraction using scientific notation.Strategy:When subtracting two measurements that are written in scientific notation, rewrite thequantities so that they share a common exponent. To get the “N part” of the result, wesimply subtract the “N parts” of the rewritten numbers. To get the exponent of the result, wesimply set it equal to the common exponent. Finally, if need be, we rewrite the result so thatits value of N satisfies010N.Solution:Rewrite the quantities sot that they have a common exponent. Rewrite 850,000 in such a waythatn= 5.850,000 = 8.5105Subtract the “Nparts” of the numbers and set the exponent of the result equal to the commonexponent.5558.5109.0100.510Rewrite the number so that010N(ignore the sign ofNwhen it is negative).0.5105=5104d.Multiplication using scientific notation.Strategy:When multiplying two numbers using scientific notation, multiply the “Nparts” of thenumbers in the usual way. To find the exponent of the result,addthe exponents of the twomeasurements.Solution:Multiply the “Nparts” of the numbers in the usual way.3.63.6 = 13Addthe exponents.1310–4 + (+6)= 13102Rewrite the number so that it is in scientific notation (so that010N).
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Chemistry 4th Edition Solution Manual - Page 15 preview imageChapter 1 -- Chemistry: The Central Science1413102= 1.31031.50a.fourb.twoc.fived.two, three, or foure.threef.oneg.oneh.two1.51a.oneb.threec.threed.foure.threef.oneg.one or two1.52a.10.6 mb.0.79 gc.16.5 cm1.53a.DivisionStrategy:The number of significant figures in the answer is determined by the original number havingthe smallest number of significant figures.Solution:7.310 km5.70 km1.283The 3 (bolded) is a nonsignificant digit because the original number 5.70 only has threesignificant digits. Therefore, the answer has only three significant digits.The correct answer rounded off to the correct number of significant figures is:1.28ThinkAbout It:Why are there no units?
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Chemistry 4th Edition Solution Manual - Page 16 preview imageChapter 1 -- Chemistry: The Central Science15b.SubtractionStrategy:The number of significant figures to the right of the decimal point in the answer isdetermined by the lowest number of digits to the right of the decimal point in any of theoriginal numbers.Solution:Writing both numbers in the decimal notation, we have0.00326 mg0.0000788 mg0.0031812 mgThe bold numbers are nonsignificant digits because the number 0.00326 has five digits to theright of the decimal point. Therefore, we carry five digits to the right of the decimal point inour answer.The correct answer rounded off to the correct number of significant figures is:0.00318 mg = 3.18×10–3mgc.AdditionStrategy:The number of significant figures to the right of the decimal point in the answer isdetermined by the lowest number of digits to the right of the decimal point in any of theoriginal numbers.Solution:Writing both numbers with exponents = +7, we have(0.402107dm) + (7.74107dm) =8.14107dmSince 7.74107has only two digits to the right of the decimal point, two digits are carried tothe right of the decimal point in the final answer.1.54Student A’s results are neither precise nor accurate. Student B’s results are both precise andaccurate. Student C’s results are precise but not accurate.1.55Tailor Z’s measurements are the most accurate. Tailor Y’s measurements are the least accurate.Tailor X’s measurements are the most precise. Tailor Y’s measurements are the least precise.
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Subject
Chemistry
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Chemistry by Julia R. Burdge

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