Solution Manual For Chemistry: The Molecular Nature Of Matter And Change, 7th Edition

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1-1CHAPTER 1KEYS TO THE STUDY OFCHEMISTRYFOLLOW–UP PROBLEMS1.1APlan:The real question is “Does the substance change composition or just change form?” A change incomposition is a chemical change while a change in form is a physical change.Solution:The figure on the left shows red atoms and molecules composed of one red atom and one blue atom. The figureon the right shows a change to blue atoms and molecules containing two red atoms. The change ischemicalsincethe substances themselves have changed in composition.1.1BPlan:The real question is “Does the substance change composition or just change form?” A change incomposition is a chemical change while a change in form is a physical change.Solution:The figure on the left shows red atoms that are close together, in the solid state. The figure on the right shows redatoms that are far apart from each other, in the gaseous state. The change isphysicalsince the substancesthemselves have not changed in composition.1.2APlan:The real question is “Does the substance change composition or just change form?” A change incomposition is a chemical change while a change in form is a physical change.Solution:a) Both the solid and the vapor are iodine, so this must be aphysicalchange.b) The burning of the gasoline fumes produces energy and products that are different gases. This is achemicalchange.c) The scab forms due to achemicalchange.1.2BPlan:The real question is “Does the substance change composition or just change form?” A change incomposition is a chemical change while a change in form is a physical change.Solution:a) Clouds form when gaseous water (water vapor) changes to droplets of liquid water. This is aphysicalchange.b) When old milk sours, the compounds in milk undergo a reaction to become different compounds (as indicatedby a change in the smell, the taste, the texture, and the consistency of the milk). This is achemicalchange.c) Both the solid and the liquid are butter, so this must be aphysicalchange.1.3APlan:We need to find the amount of time it takes for the professor to walk 10,500 m. We know how many milesshe can walk in 15 min (her speed), so we can convert the distance the professor walks to miles and use her speedto calculate the amount of time it will take to walk 10,500 m.Solution:Time (min) = 10,500 m�1 km1000 m� �1 mi1.609 km� �15 min1 mi�= 97.8869 =98 minRoad map:1000 m = 1 km1.609 km = 1 miDistance (m)Distance (km)Distance (mi)

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1-21 mi = 15 min1.3BPlan: We need to find the number of virus particles that can line up side by side in a 1 inch distance. We knowthe diameter of a virus in nm units. If we convert the 1 inch distance to nm, we can use the diameter of the virus tocalculate the number of virus particles we can line up over a 1 inch distance.Solution:No. of virus particles =1.0 in�2.54 cm1 in� �1 x107nm1 cm� �1 virus particle30 nm�=8.4667 x 105=8.5 x 105virus particlesRoad map:1 in = 2.54 cm1 cm = 1 x 107nm30 nm = 1 particle1.4APlan:The diameter in nm is used to obtain the radius in nm, which is converted to the radius in dm. The volume ofthe ribosome in dm3is then determined using the equation for the volume of a sphere given in the problem. Thisvolume may then be converted to volume in μL.Solution:Radius (dm) =diameter2=921.4 nm1 m1 dm20.1 m1 x 10nm= 1.07 x 10–7dmVolume (dm3) =()()33744 3.141591.07 x 10dm33−π=r= 5.13145 x 10–21=5.13 x 10–21dm3Volume (μL) =()213361 L1μL5.13145 x 10dm(1 dm)10L−−= 5.13145 x 10–15=5.13 x 10–15μLRoad map:diameter = 2rV= 4/3πr3Diameter (dm)Radius (dm)Volume (dm3)Time (min)Length (in)Length (cm)Length (nm)No. of particles

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1-31 dm3= 1 L1 L = 106μL1.4BPlan:We need to convert gallon units to liter units. If we first convert gallons to dm3, we can then convert to L.Solution:Volume (L) = 8400 gal�3.785dm31 gal� �1 L1dm3�= 31,794 =32,000 LRoad map:1 gal = 3.785 dm31 dm3= 1 L1.5APlan:The time is given in hours and the rate of delivery is in drops per second. Conversions relating hours toseconds are needed. This will give the total number of drops, which may be combined with their mass to get thetotal mass. The mg of drops will then be changed to kilograms.Solution:Mass (kg) =3360 min60s1.5 drops65 mg10g1 kg8.0 h1 h1 min1 s1 drop1 mg10 g−= 2.808 =2.8 kgRoad map:1 hr = 60 min1 min = 60 s1 s = 1.5 drops1 drop = 65 mg1 mg = 103gVolume (μL)Time (hr)Time (min)Time (s)No. of dropsMass (mg) of solutionMass (g) of solutionVolume (gal)Volume (dm3)Volume (L)

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1-4103g = 1 kg1.5BPlan:We have the mass of apples in kg and need to find the mass of potassium in those apples in g. The numberof apples per pound and the mass of potassium per apple are given. Convert the mass of apples in kg to pounds.Then use the number of apples per pound to calculate the number of apples. Use the mass of potassium in oneapple to calculate the mass (mg) of potassium in the group of apples. Finally, convert the mass in mg to g.Solution:Mass (g) = 3.25 kg�1 lb0.4536 kg� �3 apples1 lb� �159 mg potassium1 apple� �1 g103mg�= 3.4177 =3.42 gRoad map:0.4536 kg = 1 lb1 lb = 3 apples1 apple = 159 mg potassium103mg = 1 g1.6APlan: We know the area of a field in m2. We need to know how many bottles of herbicide will be needed to treatthat field. The volume of each bottle (in fl oz) and the volume of herbicide needed to treat 300 ft2of field aregiven. Convert the area of the field from m2to ft2(don’t forget to square the conversion factor when convertingfrom squared units to squared units!). Then use the given conversion factors to calculate the number of bottles ofherbicide needed. Convert first from ft2of field to fl oz of herbicide (because this conversion is from a squaredunit to a non-squared unit, we do not need to square the conversion factor). Then use the number of fl oz perbottle to calculate the number of bottles needed.Solution:No. of bottles = 2050 m2�1 ft2(0.3048)2m2� �1.5 fl oz300 ft2� �1 bottle16 fl oz�= 6.8956 =7 bottlesRoad map:(0.3048)2m2= 1 ft2Mass (kg) of solutionMass (kg) of applesMass (lb) of applesNo. of applesMass (mg) potassiumMass (g) potassiumArea (m2)Area (ft2)

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1-5300 ft2= 1.5 fl oz16 oz = 1 bottle1.6BPlan:Calculate the mass of mercury in g. Convert the surface area of the lake form mi2to ft2. Find the volume ofthe lake in ft3by multiplying the surface area (in ft2) by the depth (in ft). Then convert the volume of the lake tomL by converting first from ft3to m3, then from m3to cm3, and from cm3to mL. Finally, divide the mass in g bythe volume in mL to find the mass of mercury in each mL of the lake.Solution:Mass (g) = 75,000 kg�1000 g1 kg�= 7.5 x 107gVolume (mL) = 4.5 mi2�(5280)2ft21 mi2� �35 ft� �0.02832 m31 ft3� �1 x 106cm31 m3� �1 mL1 cm3�= 1.2 x 1014mLMass (g) of mercury per mL =7.5 x 107g1.2 x 1014mL=6.2 x 10–7g/mLRoad map:1 kg = 103g1 mi2= (5280)2ft2V = area (ft2) x depth (ft)1 ft3= 0.02832 m31 m3= 106cm31 cm3= 1 mLVolume (fl oz)No. of bottlesMass (kg)Mass (g)Area (mi2)Area (ft2)Volume (ft3)Volume (m3)Volume (cm3)Volume (mL)Mass (g) of mercury in 1 mL of waterdivide mass by volume

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1-21.7APlan:Find the mass of Venus in g. Calculate the radius of Venus by dividing its diameter by 2. Convert the radiusfrom km to cm. Use the radius to calculate the volume of Venus. Finally, find the density of Venus by dividingthe mass of Venus (in g) by the volume of Venus (in cm3).Solution:Mass (g) = 4.9 x 1024kg�103g1 kg�= 4.9 x 1027gRadius (cm) =�12,100 km2� �103m1 km� �102cm1 m�= 6.05 x 108cmVolume (cm3) =43𝜋𝑟3=43(3.14)(6.05 x 108cm)3= 9.27 x 1026cm3Density (g/cm3) =4.9x 1027g9.27 x 1026cm3=5.3 g/cm3Road map:1 kg = 103gd = 2r1 km = 103m1 m = 102cmV =43𝜋𝑟31.7BPlan:The volume unit may be factored away by multiplying by the density. Then it is simply a matter ofchanging grams to kilograms.Solution:Mass (kg) =()337.5 g1 kg4.6 cm1000 gcm= 0.0345 =0.034 kgRoad map:multiply by density (1 cm3= 7.5 g)Volume (cm3)Mass (g)Mass (kg)Mass (g)Diameter (km)Radius (km)Radius (m)Radius (cm)Volume (cm3)divide mass by volumeDensity (g/cm3)divide mass by volume

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1-2103g = 1 kg1.8APlan:Using the relationship between the Kelvin and Celsius scales, change the Kelvin temperature to the Celsiustemperature. Then convert the Celsius temperature to the Fahrenheit value using the relationship between thesetwo scales.Solution:T(in °C) =T(in K) – 273.15 = 234 K – 273.15 = –39.15 =–39°CT(in °F) =59T (in °C) + 32 =59(–39.15°C) + 32 = –38.47 =–38°FCheck:Since the Kelvin temperature is below 273, the Celsius temperature must be negative. The low Celsiusvalue gives a negative Fahrenheit value.1.8BPlan:Convert the Fahrenheit temperature to the Celsius value using the relationship between these two scales.Then use the relationship between the Kelvin and Celsius scales to change the Celsius temperature to the Kelvintemperature.Solution:T(in °C) =59(T (in °F) – 32) =59(2325°F – 32) = 1273.8889 = 1274°CT(in K) =T(in°C) + 273.15 = 1274°C+ 273.15 = 1547.15 = 1547 KCheck:Since the Fahrenheit temperature is large and positive, both the Celsius and Kelvin temperatures shouldalso be positive. Because the Celsius temperature is greater than 273, the Kelvin temperature should be greaterthan 273, which it is.1.9APlan:Determine the significant figures by counting the digits present and accounting for the zeros. Zerosbetween non-zero digits are significant, as are trailing zeros to the right of a decimal point. Trailing zeros to theleft of a decimal point are only significant if the decimal point is present.Solution:a) 31.070 mg;fivesignificant figuresb) 0.06060 g;foursignificant figuresc) 850.°C;threesignificant figures — note the decimal point that makes the zero significant.Check:All significant zeros must come after a significant digit.1.9BPlan:Determine the significant figures by counting the digits present and accounting for the zeros. Zerosbetween non-zero digits are significant, as are trailing zeros to the right of a decimal point. Trailing zeros to theleft of a decimal point are only significant if the decimal point is present.Solution:a) 2.000x 102mL;foursignificant figuresb) 3.9 x 10–6m;twosignificant figures — note that none of the zeros are significant.c) 4.01 x 10–4L;threesignificant figuresCheck:All significant zeros must come after a significant digit.1.10APlan:Use the rules presented in the text. Add the two values in the numerator before dividing. The timeconversion is an exact conversion and, therefore, does not affect the significant figures in the answer.Solution:The addition of 25.65 mL and 37.4 mL gives an answer where the last significant figure is the one after thedecimal point (giving three significant figures total):25.65 mL + 37.4 mL = 63.05 = 63.0 mLWhen a four significant figure number divides a three significant figure number, the answer must round to threesignificant figures. An exact number (1 min / 60 s) will have no bearing on the number of significant figures.63.0 mL1 min73.55 s60 s= 51.394 =51.4 mL/minMass (kg)

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1-31.10BPlan:Use the rules presented in the text. Subtract the two values in the numerator and multiply the numbers in thedenominator before dividing.Solution:The subtraction of 35.26 from 154.64 gives an answer in which the last significant figure is two places after thedecimal point (giving five significant figures total):154.64 g – 35.26 g = 119.38 gThe multiplication of 4.20 cm (three significant figures) by 5.12 cm (three significant figures) by 6.752 cm (foursignificant figures) gives a number with three significant figures.4.20 cm x 5.12 cm x 6.752 cm = 145.1950 = 145 cm3When a three significant figure number divides a five significant figure number, the answer must round to threesignificant figures.119.38 g145 cm3= 0.8233 =0.823 g/cm3END–OF–CHAPTER PROBLEMS1.1Plan:If only the form of the particles has changed and not the composition of the particles, a physical change hastaken place; if particles of a different composition result, a chemical change has taken place.Solution:a) The result in C represents achemical changeas the substances in A (red spheres) and B (blue spheres) havereacted to become a different substance (particles consisting of one red and one blue sphere) represented in C.There are molecules in C composed of the atoms from A and B.b) The result in D represents achemical changeas again the atoms in A and B have reacted to form molecules ofa new substance.c) The change from C to D is aphysical change. The substance is the same in both C and D (moleculesconsisting of one red sphere and one blue sphere) but is in the gas phase in C and in the liquid phase in D.d) The sample has thesame chemical propertiesin both C and D since it is the same substance but hasdifferentphysical properties.1.2Plan:Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples.Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids havea definite volume. The volume of the container does not affect the volume of a solid or liquid.Solution:a) The helium fills the volume of the entire balloon. The addition or removal of helium will change the volume ofa balloon. Helium is agas.b) At room temperature, the mercury does not completely fill the thermometer. The surface of theliquidmercuryindicates the temperature.c) The soup completely fills the bottom of the bowl, and it has a definite surface. The soup is aliquid, though it ispossible that solid particles of food will be present.1.3Plan:Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples.Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids havea definite volume. The volume of the container does not affect the volume of a solid or liquid.Solution:a) The air fills the volume of the room. Air is agas.b) The vitamin tablets do not necessarily fill the entire bottle. The volume of the tablets is determined by thenumber of tablets in the bottle, not by the volume of the bottle.The tablets aresolid.c) The sugar has a definite volume determined by the amount of sugar, not by the volume of the container. Thesugar is asolid.1.4Plan:Define the terms and apply these definitions to the examples.Solution:Physical property– A characteristic shown by a substance itself, without interacting with or changing into othersubstances.

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1-4Chemical property– A characteristic of a substance that appears as it interacts with, or transforms into, othersubstances.a) The change in color (yellow–green and silvery to white), and the change in physical state (gas and metal tocrystals) are examples ofphysical properties. The change in the physical properties indicates that a chemicalchange occurred. Thus, the interaction between chlorine gas and sodium metal producing sodium chloride is anexample of achemical property.b) The sand and the iron are still present. Neither sand nor iron became something else. Colors along withmagnetism arephysical properties. No chemical changes took place, so there are no chemical properties toobserve.1.5Plan:Define the terms and apply these definitions to the examples.Solution:Physical change– A change in which the physical form (or state) of a substance, but not its composition, isaltered.Chemical change– A change in which a substance is converted into a different substance with differentcomposition and properties.a) The changes in the physical form arephysical changes. The physical changes indicate that there is also achemical change. Magnesium chloride has been converted to magnesium and chlorine.b) The changes in color and form arephysical changes. The physical changes indicate that there is also achemical change. Iron has been converted to a different substance, rust.1.6Plan:Apply the definitions of chemical and physical changes to the examples.Solution:a) Not a chemical change, but aphysical change— simply cooling returns the soup to its original form.b) There is achemical change— cooling the toast will not “un–toast” the bread.c) Even though the wood is now in smaller pieces, it is still wood. There has been no change in composition, thusthis is aphysical change, and not a chemical change.d) This is achemical changeconverting the wood (and air) into different substances with different compositions.The wood cannot be “unburned.”1.7Plan:If there is a physical change, in which the composition of the substance has not been altered, the processcan be reversed by a change in temperature. If there is a chemical change, in which the composition of thesubstance has been altered, the process cannot be reversed by changing the temperature.Solution:a) and c)can be reversedwith temperature; the dew can evaporate and the ice cream can be refrozen.b) and d) involve chemical changes andcannot be reversedby changing the temperature since a chemical changehas taken place.1.8Plan:A system has a higher potential energy before the energy is released (used).Solution:a) The exhaust is lower in energy than the fuel by an amount of energy equal to that released as the fuel burns.Thefuelhas a higher potential energy.b)Wood, like the fuel, is higher in energy by the amount released as the wood burns.1.9Plan:Kinetic energy is energy due to the motion of an object.Solution:a)The sled sliding down the hillhas higher kinetic energy than the unmoving sled.b)Thewater falling over the dam(moving) has more kinetic energy than the water held by the dam.1.10Alchemical: chemical methods – distillation, extraction; chemical apparatusMedical: mineral drugsTechnological: metallurgy, pottery, glass1.11Combustion released the otherwise undetectable phlogiston. The more phlogiston a substance contained; themore easily it burned. Once all the phlogiston was gone, the substance was no longer combustible.

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1-51.12The mass of the reactants and products are easily observable quantities. The explanation of combustion mustinclude an explanation of all observable quantities. Their explanation of the mass gain required phlogiston tohave a negative mass.1.13Lavoisier measured the total mass of the reactants and products, not just the mass of the solids. The total mass ofthe reactants and products remained constant. His measurements showed that a gas was involved in the reaction.He called this gas oxygen (one of his key discoveries).1.14Observationsare the first step in the scientific approach. The first observation is that the toast has not popped outof the toaster. The next step is ahypothesis(tentative explanation) to explain the observation. The hypothesis isthat the spring mechanism is stuck. Next, there will be atestof the hypothesis. In this case, the test is anadditional observation — the bread is unchanged. This observation leads to a new hypothesis — the toaster isunplugged. This hypothesis leads to additional tests — seeing if the toaster is plugged in, and if it works whenplugged into a different outlet. The final test on the toaster leads to a new hypothesis — there is a problem with thepower in the kitchen. This hypothesis leads to the final test concerning the light in the kitchen.1.15A quantitative observation is easier to characterize and reproduce. A qualitative observation may be subjectiveand open to interpretation.a) This isqualitative. When has the sun completely risen?b) The astronaut’s mass may be measured; thus, this isquantitative.c) This isqualitative. Measuring the fraction of the ice above or below the surface would make this aquantitative measurement.d) The depth is known (measured) so this isquantitative.1.16A well-designed experiment must have the following essential features:1) There must be two variables that are expected to be related.2) There must be a way to control all the variables, so that only one at a time may be changed.3) The results must be reproducible.1.17A model begins as a simplified version of the observed phenomena, designed to account for the observed effects,explain how they take place, and to make predictions of experiments yet to be done. The model is improved byfurther experiments. It should be flexible enough to allow for modifications as additional experimental results aregathered.1.18Plan:Review the definitions of mass and weight.Solution:Massis the quantity of material present, whileweightis the interaction of gravity on mass. An object has adefinite mass regardless of its location; its weight will vary with location. The lower gravitational attraction onthe Moon will make an object appear to have approximately one-sixth its Earth weight. The object has the samemass on the Moon and on Earth.1.19The unit you begin with (feet) must be in the denominator to cancel. The unit desired (inches) must be in thenumerator. The feet will cancel leaving inches. If the conversion is inverted the answer would be in units of feetsquared per inch.1.20Plan:massDensity = volume. An increase in mass or a decrease in volume will increase the density. A decreasein density will result if the mass is decreased or the volume increased.Solution:a) Densityincreases. The mass of the chlorine gas is not changed, but its volume is smaller.b) Densityremains the same. Neither the mass nor the volume of the solid has changed.c) Densitydecreases. Water is one of the few substances that expands on freezing. The mass is constant, but thevolume increases.d) Densityincreases. Iron, like most materials, contracts on cooling; thus the volume decreases while the massdoes not change.

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1-6e) Densityremains the same. The water does not alter either the mass or the volume of the diamond.1.21Plan:Review the definitions of heat and temperature. The two temperature values must be compared using onetemperature scale, either Celsius or Fahrenheit.Solution:Heat is the energy that flows between objects at different temperatures while temperature is the measure ofhow hot or cold a substance is relative to another substance. Heat is anextensive propertywhile temperature isanintensive property. It takes more heat to boil a gallon of water than to boil a teaspoon of water. However,both water samples boil at the same temperature.Convert 65°C to °F:T(in °F) =95T(in °C) + 32 =95(65°C) + 32 = 149°FA temperature of 65°C is 149°F. Heat will flow from the hot water (65°C or 149°F) to the cooler water (65°F).The 65°C water contains more heat than the cooler water.1.22There are two differences in the Celsius and Fahrenheit scales (size of a degree and the zero point), so a simpleone-step conversion will not work. The size of a degree is the same for the Celsius and Kelvin scales; only thezero point is different so a one-step conversion is sufficient.1.23Plan:Review the definitions of extensive and intensive properties.Solution:An extensive property depends on the amount of material present. An intensive property is the same regardless ofhow much material is present.a) Mass is anextensive property. Changing the amount of material will change the mass.b) Density is anintensive property. Changing the amount of material changes both the mass and the volume, butthe ratio (density) remains fixed.c) Volume is anextensive property. Changing the amount of material will change the size (volume).d) The melting point is anintensive property. The melting point depends on the substance, not on the amount ofsubstance.1.24Plan:Review the table of conversions in the chapter or inside the back cover of the book. Write the conversionfactor so that the unit initially given will cancel, leaving the desired unit.Solution:a) To convert from in2to cm2, use()()222.54 cm1 in;to convert from cm2to m2, use()()221 m100 cmb) To convert from km2to m2, use()()221000 m1 km;to convert from m2to cm2, use()()22100 cm1 mc) This problem requires two conversion factors: one for distance and one for time. It does not matter whichconversion is done first. Alternate methods may be used.To convert distance, mi to m, use:km1m1000mi1km609.1=1.609x103m/miTo convert time, h to s, use:s60min1min60h1=1 h/3600 sTherefore, the complete conversion factor is31.609 x 10m1 h1 mi3600 s =  0.4469 m•hmi•s.Do the units cancel when you start with a measurement of mi/h?d) To convert from pounds (lb) to grams (g), uselb2.205g1000.

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1-7To convert volume from ft3to cm3use,()()()()33331 ft1 in12 in2.54 cm=3.531x10–5ft3/cm3.1.25Plan:Review the table of conversions in the chapter or inside the back cover of the book. Write the conversionfactor so that the unit initially given will cancel, leaving the desired unit.Solution:a) This problem requires two conversion factors: one for distance and one for time. It does not matter whichconversion is done first. Alternate methods may be used.To convert distance, cm to in, use:1 in2.54 cmTo convert time, min to s, use:1 min60 sb) To convert from m3to cm3, use()()33100 cm1 m;to convert from cm3to in3, use()()331 in2.54 cmc) This problem requires two conversion factors: one for distance and one for time. It does not matter whichconversion is done first. Alternate methods may be used.To convert distance, m to km, use:1 km1000 mTo convert time, s2to h2, use:()()()()222260 s60 min1 min1 h=223600 shd) This problem requires two conversion factors: one for volume and one for time. It does not matter whichconversion is done first. Alternate methods may be used.To convert volume, gal to qt, use:4 qt1 gal; to convert qt to L, use:1 L1.057 qtTo convert time, h to min, use:1 h60 min1.26Plan:Use conversion factors from the inside back cover: 1 pm = 10–12m; 10–9m = 1 nm.Solution:Radius (nm) =()12910m1 nm1430 pm1 pm10m−−=1.43 nm1.27Plan:Use conversion factors from the inside back cover: 10–12m = 1 pm; 1 pm = 0.01 Å.Solution:Radius (Å) =()10121 pm0.01 Å2.22x10m1 pm10m−−  =2.22 Å1.28Plan:Use conversion factors: 0.01 m = 1 cm; 2.54 cm = 1 in.Solution:Length (in) =()1 cm1 in100. m0.01 m2.54 cm   = 3.9370x103=3.94x103in1.29Plan:Use the conversion factor 12 in = 1 ft to convert 6 ft 10 in to height in inches. Then use theconversion factors 1 in = 2.54 cm; 1 cm = 10 mm.Solution:

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Solution Manual For Chemistry: The Molecular Nature Of Matter And Change, 7th Edition - Page 14 preview image

1-8Height (in) =()12 in6 ft10 in1 ft+= 82 inHeight (mm) =()2.54 cm10 mm82 in1 in1 cm= 2.0828x103=2.1x103mm1.30Plan:Use conversion factors (1 cm)2= (0.01 m)2; (1000 m)2= (1 km)2to express the area in km2. To calculatethe cost of the patch, use the conversion factor: (2.54 cm)2= (1 in)2.Solution:a) Area (km2) =()()()()()222220.01 m1 km20.7 cm1 cm1000 m=2.07x10–9km2b) Cost =()()()22221in$3.2520.7 cm1in2.54 cm=10.4276 =$10.431.31Plan:Use conversion factors (1 mm)2= (10–3m)2; (0.01 m)2= (1 cm)2; (2.54 cm)2= (1 in)2; (12 in)2= (1 ft)2toexpress the area in ft2.Solution:a) Area (ft2) =()()()()()()()()()232222222210m1 cm1in1ft7903 mm1 mm0.01 m2.54 cm12 in−= 8.5067x10–2=8.507x10–2ft2b) Time (s) =()2245 s7903 mm135 mm= 2.634333x103=2.6x103s1.32Plan:Use conversion factor 1 kg = 2.205 lb. The following assumes a body weight of 155 lbs. Use your ownbody weight in place of the 155 lbs.Solution:Body weight (kg) =()1 kg155 lb2.205 lb=70.3 kgAnswers will vary, depending on the person’s mass.1.33Plan:Use conversion factor 1 short ton = 2000 lb; 2.205 lb = 1 kg; 1000 kg = 1 metric ton.Solution:Mass (T) =()1532000 lb1 kg1 T2.60 x10ton1 ton2.205 lb10 kg= 2.35828x1015=2.36x1015T1.34Plan:Mass in g is converted to kg in part a) with the conversion factor 1000 g = 1 kg; mass in g is converted to lbin part b) with the conversion factors 1000 g = 1 kg; 1 kg = 2.205 lb. Volume in cm3is converted to m3with theconversion factor (1 cm)3= (0.01 m)3and to ft3with the conversion factors (2.54 cm)3= (1 in)3; (12 in)3= (1 ft)3.The conversions may be performed in any order.Solution:a) Density (kg/m3) =()()3331 cm5.52 g1 kg1000 gcm0.01 m=5.52x103kg/m3b) Density (lb/ft3) =()()()()333332.54 cm12 in5.52 g1 kg2.205 lb1000 g1 kgcm1 in1 ft= 344.661 =345 lb/ft3

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Solution Manual For Chemistry: The Molecular Nature Of Matter And Change, 7th Edition - Page 15 preview image

1-91.35Plan:Length in m is converted to km in part a) with the conversion factor 1000 m = 1 km; length in m isconverted to mi in part b) with the conversion factors 1000 m = 1 km; 1 km = 0.62 mi. Time is converted usingthe conversion factors 60 s = 1 min; 60 min = 1 h. The conversions may be performed in any order.Solution:a) Velocity (km/h) =832.998 x10 m60 s60 min1 km1 s1 min1 h10 m   = 1.07928x109=1.079x109km/hb) Velocity (mi/min) =832.998 x10 m60 s1 km0.62 mi1 s1 min1 km10 m = 1.11526x107=1.1x107mi/min1.36Plan:Use the conversion factors (1 μm)3= (1x10–6m)3; (1x10–3m)3= (1 mm)3to convert to mm3.To convert to L, use the conversion factors (1 μm)3= (1x10–6m)3; (1x10–2m)3= (1 cm)3; 1 cm3= 1 mL;1 mL = 1x10–3L.Solution:a) Volume (mm3) =()()()()36333331x10m1 mm2.56μmcell1μm1x10m−−        =2.56x10–9mm3/cellb) Volume (L) =()()()()()36333533321x10m1 cm2.56μm1 mL1x10L10cellscell1 mL1 cm1μm1x10m−−−     = 2.56x10–10=10–10L1.37Plan:For part a), convert from qt to mL (1 qt = 946.4 mL) to L (1 mL = 1x10–3L) to m3(1 L = 10–3m3).For part b), convert from gal to qt (1 gal = 4 qt) to mL (1 qt = 946.4 mL) to L (1 mL = 10–3L).Solution:a) Volume (m3) =()333946.4 mL10L10m1 qt1 qt1 mL1 L−−=9.464x10–4m3b) Volume (L) =()34 qt946.4 mL10L835 gal1 gal1 qt1 mL−= 3.160976x103=3.16x103L1.38Plan:The mass of the mercury in the vial is the mass of the vial filled with mercury minus the mass of the emptyvial. Use the density of mercury and the mass of the mercury in the vial to find the volume of mercury and thusthe volume of the vial. Once the volume of the vial is known, that volume is used in part b. The density of wateris used to find the mass of the given volume of water. Add the mass of water to the mass of the empty vial.Solution:a) Mass (g) of mercury = mass of vial and mercury – mass of vial = 185.56 g – 55.32 g = 130.24 gVolume (cm3) of mercury = volume of vial =()31 cm130.24 g13.53 g= 9.626016 =9.626 cm3b) Volume (cm3) of water = volume of vial = 9.626016 cm3Mass (g) of water =()330.997 g9.626016 cm1 cm= 9.59714 g waterMass (g) of vial filled with water = mass of vial + mass of water = 55.32 g + 9.59714 g = 64.91714 =64.92 g1.39Plan:The mass of the water in the flask is the mass of the flask and water minus the mass of the empty flask.Use the density of water and the mass of the water in the flask to find the volume of water and thus the volume ofthe flask. Once the volume of the flask is known, that volume is used in part b. The density of chloroform is used

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Solution Manual For Chemistry: The Molecular Nature Of Matter And Change, 7th Edition - Page 16 preview image

1-10to find the mass of the given volume of chloroform. Add the mass of the chloroform to the mass of the emptyflask.Solution:a) Mass (g) of water = mass of flask and water – mass of flask = 489.1 g – 241.3 g = 247.8 gVolume (cm3) of water = volume of flask =()31 cm247.8 g1.00 g= 247.8 =248 cm3b) Volume (cm3) of chloroform = volume of flask = 247.8 cm3Mass (g) of chloroform =()331.48 g247.8 cmcm= 366.744 g chloroformMass (g) of flask and chloroform = mass of flask + mass of chloroform = 241.3 g + 366.744 g= 608.044 g =608 g1.40Plan:Calculate the volume of the cube using the relationship Volume = (length of side)3. The length of side inmm must be converted to cmso that volume will have units of cm3. Divide the mass of the cube by the volume tofind density.Solution:Side length (cm) =()3210m1 cm15.6 mm1 mm10m−−= 1.56 cm(convert to cm to match density unit)Al cube volume (cm3) = (length of side)3= (1.56 cm)3= 3.7964 cm333mass10.25 gDensity (g/cm )volume3.7964 cm=== 2.69993 =2.70 g/cm31.41Plan:Use the relationshipc= 2πrto find the radius of the sphere and the relationshipV= 4/3πr3to find thevolume of the sphere. The volume in mm3must be converted to cm3. Divide the mass of the sphere by the volumeto find density.Solution:c= 2πrRadius (mm) =32.5 mm=22ππc= 5.17254 mmVolume (mm3) =343πr=34(5.17254 mm)3π= 579.6958 mm3Volume (cm3) =()3333210m1 cm579.6958 mm1 mm10m−−   = 0.5796958 cm333mass4.20 gDensity (g/cm )volume0.5796958 cm=== 7.24518 =7.25 g/cm31.42Plan:Use the equations given in the text for converting between the three temperature scales.Solution:a)T(in °C) = [T(in °F) – 32]59= [68°F – 32]59=20.°CT(in K) =T(in °C) + 273.15 = 20.°C + 273.15 = 293.15 =293 Kb)T(in K) =T(in °C) + 273.15= –164°C + 273.15 = 109.15 =109 KT(in °F) =95T(in °C) + 32 =95(–164°C) + 32 = –263.2 =–263°Fc)T(in °C) =T(in K) – 273.15 = 0 K – 273.15 = –273.15 =–273°CT(in °F) =95T(in °C) + 32 =95(–273.15°C) + 32 = –459.67 =–460.°F

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