Chemistry: The Molecular Nature of Matter and Change Canadian Edition (2013) Solution Manual

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Chemistry: The Molecular Nature of Matter and Change Canadian Edition (2013) Solution Manual - Page 1 preview image

Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-1Solution ManualCHAPTER 1KEYS TO THE STUDY OFCHEMISTRYEND–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.Chemical 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.

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Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-2Solution Manual1.5Plan: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.1.12The 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).

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Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-3Solution Manual1.14Observationsare 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.1.18The unit you begin with (kilometres) must be in the denominator to cancel. The unit desired (centimetres) mustbe in the numerator. The kilometres will cancel leaving centimetres. If the conversion is inverted the answerwould be in units of kilometres squared per centimetre.1.19Plan: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 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 (km to m) and one for time (h to s). It does notmatter which conversion is done first and alternate methods may be used.To convert distance, km to m, use:1000 m1 km= 103m/kmTo convert time, h to s, use:1 h1 min1 h60 min60 s3600 s =Therefore, the complete conversion factor is:1000 m1 h1 m hkm3600 s3 6 km s.⋅ =⋅Do the units cancel when you start with units of km/h?d) To convert from kg/m3to g/cm3requires two conversion factors:To convert mass, kg to g:

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Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-4Solution Manual31000 gg101 kgkg =To convert volume from cm3to m3use,()()336331 mm=10cm100 cm−.The complete conversion is:3336333gmg m101010kgcmkg cm−−⋅=⋅Do the units cancel when you start with units of kg/m3?1.20Plan: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 mm, use:10 mm1 cmTo convert time, s to min, use:1 min60 sThe complete conversion is:10 mm1 min1 mm min1 cm60 s6 cm s⋅ =⋅b) To convert from m3to cm3, use()()33100 cm1 mc) This problem requires two conversion factors: one for distance and one for time squared. It does not matterwhichconversion is done first. Alternate methods may be used.To convert distance, m to km, use:1 km1000 m=10-3km/mTo convert time, s2to h2, use:2260 min60 s1 h1 min   =1.296 x 107s2/1 h2Therefore, the complete conversion factor is310km1.29610s1 m1 h××⋅=⋅72102221.29610km sm h.Do the units cancel when you start with a measurement of m/s2?d) 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, mL to L, use:1 L1000 mLTo convert time, s to min, use:60 s1 minThe complete conversion factor is:1 L60 s6 L s1000 mL1 min100 mL min⋅ =⋅

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Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-5Solution Manual1.21Plan: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.1.22Plan: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.1.23Plan:massDensity = volume. An increase in mass or a decrease in volume will increase the density. Adecreasein 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.e) Densityremains the same. The water does not alter either the mass or the volume of the diamond.1.24Plan: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 K:T(in K) =T(in °C) + 273.15 = (65°C) + 273.15 = 338 KA temperature of 65°C is 338 K. Heat will flow from the hot water (65°C or 338 K) to the cooler water (65 K).The 65°C water contains more heat than the cooler water.1.25When we have a set of ratios, the units will cancel out as they are multiplicative. For example, m=d/V; if weexpress mass in kg and density in kg/m3, it would give us the identical volume (but in different units) than if weused mass in g and density in g/cm3. In the first case, the volume will have units of m3and in the second it willhave units of cm3; however, the actual volume will be the same (if you convert one to the other). Whentemperature is one of the variables, however, the conversion between Celsius and Kelvin is anadditiveconversion, not multiplicative. Hence, in the equation PV=nRT, we must have the units for T match the units forR (which contains the temperature unit to be cancelled). This temperature, with only one or two exceptions, willbe a temperature in kelvin.1.26Plan: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 nm

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Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-6Solution Manual1.27Plan:Use conversion factors from the inside back cover: 10–12m = 1 pm; 10-9m = 1 nm.Solution:Radius (Å) =()()102121091 pm2.22x10m2.2210pm10m1 nm2.22x10m0.222 nm10m−−−− =× =1.28Plan:Use conversion factors: 1 m = 10-9nmSolution:Length (nm) =()91110nm100. m10nm1 m =1.29Plan:Use the conversion factor 1 km = 106mm to convert km to height in mm.Solution:Height (mm) =()610mm0.00196 km=1960 mm1 km1.30Plan: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 =()()()222210 mm$3.2520 7 cm1 mm1cm×3.= $6.73101.31Plan:Use conversion factors (1 mm)2= (10–3m)2; (0.01 m)2= (1 cm)2;Solution:a) Area (m2) =()()()232210m7903 mm.1 mm327 90310m−− =×b) Time (s) =()2245 s7903 mm135 mm= 2.634333x103=2.6x103s1.32Plan:Use conversion factor 1 g = 1 paper clip.Solution:total mass (g) =()()mass per clipnumber of clips()333.56×10g3.56×10clipsmass per clip1 g/cliTotal massgNumber of Clipsp===1.33Plan:Use conversion factor 1000 kg = 1 metric ton.Solution:Mass (T) =()2131 kg1 T2.36 x10g1000 g10 kg=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 tomg

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Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-7Solution Manualin part b) with the conversion factors 1000 mg = 1 g. Volume in cm3is converted to m3with theconversion factor (1 cm)3= (0.01 m)3and to mm3with the conversion factors (10 mm)3= (1 cm)3. Theconversions may be performed in any order.Solution:a) Density (kg/m3) =()()3331 cm5.52 g1 kg1000 gcm0.01 m=5.52x103kg/m3b) Density (mg/mm3) =()()33331 cm5.52 g1000 mg5.52 mg / mm1 gcm10 mm=1.35Plan: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 (cm/min) =8122.998 x10 m60 s100 cm1.799x10cm / min1 s1 min1m =1.36Plan: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 mL to L (1 mL = 1x10–3L) and then to m3(1 L = 10–3m3).For part b), convert from mm3to L using the following conversion factors: (10 mm)3= (1 cm)3, 1 mL = 1 cm3and1 mL = 10–3L.Solution:a) Volume (m3) =33310L10m946.4 mL1 mL1 L−−=9.464x10–4m3b) Volume (L) =()33333(1 cm)1mL10L835 mm1 mL(10 mm)1 cm−.xL48 35 10−=1.38Plan: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 g

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Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-8Solution ManualVolume (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 usedto 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/cm3

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Chemistry: The Molecular Nature of Matter and Change Canadian Edition (2013) Solution Manual - Page 10 preview image

Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-9Solution Manual1.42Plan:Use the equations given in the text for converting between the three temperature scales.Solution:a)T(in K) =T(in °C) + 273.15 = 18°C + 273.15 = 291.15 =291 Kb)T(in K) =T(in °C) + 273.15= –164°C + 273.15 = 109.15 =109 Kc)T(in °C) =T(in K) – 273.15 = 0 K – 273.15 = –273.15 =–273°C1.43Plan:Use the equations given in the text for converting between the three temperature scales.Solution:a)T(in K) =T(in °C) + 273.15 = 37°C + 273.15 = 310.15 =310 Kb)T(in K) =T(in °C) + 273.15 = 3410°C + 273 =3683 Kc)T(in°C) =T(in K) –273.15 = 6.1x103K – 273 = 5.827x103=5.8 x 103°C1.44Plan:Find the volume occupied by each metal by taking the difference between the volume of water and metaland the initial volume of the water (25.0 mL).Divide the mass of the metal by the volume of the metal tocalculate density. Use the density value of each metal to identify the metal.Solution:Cylinder A: volume of metal = [volume of water + metal] – [volume of water]volume of metal = 28.2 mL – 25.0 mL = 3.2 mLDensity =mass25.0 g=volume3.2 mL= 7.81254 =7.8 g/mLCylinder A containsiron.Cylinder B: volume of metal = [volume of water + metal] – [volume of water]volume of metal = 27.8 mL – 25.0 mL = 2.8 mLDensity =mass25.0 g=volume2.8 mL= 8.92857 =8.9 g/mLCylinder B containsnickel.Cylinder C: volume of metal = [volume of water + metal] – [volume of water]volume of metal = 28.5 mL – 25.0 mL = 3.5 mLDensity =mass25.0 g=volume3.5 mL= 7.14286 =7.1 g/mLCylinder C containszinc.1.45Plan:Use 1 nm = 10–9m to convert wavelength in nm to m. To convert wavelength in pm to nm, use1000 pm = 1 nm.Solution:a) Wavelength (m) =()910m247 nm1 nm−=2.47x10–7mb) Wavelength (Å ) =()1 nm6760 pm1000 pm=6.76 nm1.46Plan:The liquid with the larger density will occupy the bottom of the beaker, while the liquid with thesmaller density volume will be on top of the more dense liquid.Solution:a) Liquid A is more dense than water; liquids B and C are less dense than water.b) Density of liquid B could be0.94 g/mL. Liquid B is more dense than C so its density must be greater than0.88 g/mL. Liquid B is less dense than water so its density must be less than 1.0 g/mL.

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Chemistry: The Molecular Nature of Matter and Change Canadian Edition (2013) Solution Manual - Page 11 preview image

Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-10Solution Manual1.47Plan:Calculate the volume of the cylinder in cm3by using the equation for the volume of a cylinder. Thediameter of the cylinder must be halved to find the radius. Convert the volume in cm3to dm3by using theconversion factors (1 cm)3= (10–2m)3and (10–1m)3= (1 dm)3.Solution:Radius = diameter/2 = 0.85 cm/2 = 0.425 cmVolume (cm3) =πr2h=π(0.425 cm)2(9.5 cm) = 5.3907766 cm3Volume (dm3) =()3323110m1 dm5.3907766 cm1 cm10m−−   = 5.39078x10–3=5.4x10–3dm31.48Plan:Use the percent of copper in the ore to find the mass of copper in 5.01 kg of ore. Convert the mass in kg tomass in g. The density of copper is used to find the volume of that mass of copper. Use the volume equation for acylinder to calculate the height of the cylinder (the length of wire); the diameter of the wire is used to find theradius which must be expressed in units of cm. Length of wire in cm must be converted to m.Solution:Mass (kg) of copper =()66%5.01 kg Covellite100%= 3.3066 kg copperMass (g) of copper =()1000 g3.3066 kg1 kg= 3.3066 x103gVolume (cm3) of copper =()33cmCu3.3066x10g Cu8.95 g Cu= 369.453 cm3CuV=πr2hRadius (cm) =0.1601 mm1 cm210 mm= 8.005x10–3cmHeight (length) in cm =2πVr=()()323369.453 cm8.005 x10cm−π= 1.835 x 106cmLength (m) =()2610m1.835x10cm1 cm−= 1.835x104=1.84x104m1.49An exact number is defined to have a certain value (exactly). There is no uncertainty in an exact number. Anexact number is considered to have an infinite number of significant figures and, therefore, does not limit thedigits in the calculation.1.50Random error of a measurement is decreased by(1)taking the average of more measurements. Moremeasurements allow a more precise estimate of the true value of the measurement. Calibrating the instrument willallow greater accuracy but not necessarily greater precision.1.51a) If the number is an exact count then there are an infinite number of significant figures. If it is not an exactcount, there are only 5 significant figures.b) Other things, such as number of tickets sold, could have been counted instead.c) A value of 15,000 to two significant figures is 1.5x104.Values would range from 14,501 to 15,499. Both of these values round to 1.5x104.1.52Plan:Review the rules for significant zeros.Solution:a) No significant zeros (leading zeros are not significant)b) No significant zeros (leading zeros are not significant)c) 0.0410 (terminal zeros to the right of the decimal point are significant)d) 4.0100x104(zeros between nonzero digits are significant; terminal zeros to the right of the decimal point aresignificant)

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Chemistry: The Molecular Nature of Matter and Change Canadian Edition (2013) Solution Manual - Page 12 preview image

Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-11Solution Manual1.53Plan:Review the rules for significant zeros.Solution:a) 5.08 (zeros between nonzero digits are significant)b) 508 (zeros between nonzero digits are significant)c) 5.080x103(zeros between nonzero digits are significant; terminal zeros to the right of the decimal point aresignificant)d) 0.05080 (leading zeros are not significant; zeros between nonzero digits are significant; terminal zeros to theright of the decimal point are significant)1.54Plan:Review the rules for rounding.Solution:(significant figures are underlined)a) 0.0003554: the extra digits are 54 at the end of the number. When the digit to be removed is 5 and that 5 isfollowed by nonzero numbers, the last digit kept is increased by 1:0.00036b)35.8348: the extra digits are 48. Since the digit to be removed (4) is less than 5, the last digit kept isunchanged:35.83c)22.4555: the extra digits are 555. When the digit to be removed is 5 and that 5 is followed by nonzeronumbers, the last digit kept is increased by 1:22.51.55Plan:Review the rules for rounding.Solution:(significant figures are underlined)a)231.554: the extra digits are 54 at the end of the number. When the digit to be removed is 5 and that 5 isfollowed by nonzero numbers, the last digit kept is increased by 1:231.6b) 0.00845: the extra digit is 5 at the end of the number. When the digit to be removed is 5 and that 5 is notfollowed by nonzero numbers, the last digit kept remains unchanged if it is even and increased by 1 if it is odd:0.0084c)144,000: the extra digits are 4000 at the end of the number. When the digit to be removed (4) is less than 5,the last digit kept remains unchanged:140,000(or 1.4x105)1.56Plan:Review the rules for rounding.Solution:19 rounds to 20: the digit to be removed (9) is greater than 5 so the digit kept is increased by 1.155 rounds to 160: the digit to be removed is 5 and the digit to be kept is an odd number, so that digit kept isincreased by 1.8.3 rounds to 8: the digit to be removed (3) is less than 5 so the digit kept remains unchanged.3.2 rounds to 3: the digit to be removed (2) is less than 5 so the digit kept remains unchanged.2.9 rounds to 3: the digit to be removed (9) is greater than 5 so the digit kept is increased by 1.4.7 rounds to 5: the digit to be removed (7) is greater than 5 so the digit kept is increased by 1.20 x 160 x 83 x 3 x 5= 568.89 =6x102Since there are numbers in the calculation with only one significant figure, the answer can be reported only toone significant figure. (Note that the answer is 560 using the original numbers.)1.57Plan:Review the rules for rounding.Solution:10.8 rounds to 11: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1.6.18 rounds to 6.2: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1.2.381 rounds to 2.38: the digit to be removed (1) is less than 5 so the digit kept remains unchanged.24.3 rounds to 24: the digit to be removed (3) is less than 5 so the digit kept remains unchanged.1.8 rounds to 2: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1.19.5 rounds to 20: the digit to be removed is 5 and the digit to be kept is an odd number, so that digit kept isincreased by 1.11 x 6.2 x 2.3824 x 2 x 20= 0.1691 =0.2Since there is a number in the calculation with only one significant figure, the answer can be reported only to

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Chemistry: The Molecular Nature of Matter and Change Canadian Edition (2013) Solution Manual - Page 13 preview image

Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-12Solution Manualone significant figure. (Note that the answer is 0.19 with original number of significant figures.)1.58Plan:Use a calculator to obtain an initial value. Use the rules for significant figures and rounding to get the finalanswer.Solution:a)()()2.795 m310 m6.48 m= 133.71 =134 m(maximum of 3 significant figures allowed since two of the originalnumbers in the calculation have only 3 significant figures)b) V =()3417.282 mm3 π= 21,620.74 =21,621 mm3(maximum of 5 significant figures allowed)c) 1.110 cm + 17.3 cm + 108.2 cm + 316 cm = 442.61 =443 cm(no digits allowed to the right of the decimalsince 316 has no digits to the right of the decimal point)1.59Plan:Use a calculator to obtain an initial value. Use the rules for significant figures and rounding to get the finalanswer.Solution:a)2.420 g15.6 g4.8 g+= 3.7542 =3.8(maximum of 2 significant figures allowed since one of the originalnumbers in the calculation has only 2 significant figures)b)7.87 mL16.1 mL8.44 mL−= 1.0274 =1.0(After the subtraction, the denominator has 2 significant figures; only onedigit is allowed to the right of the decimal in the value in the denominator since 16.1 has only one digit to the rightof the decimal.)c) V =π(6.23 cm)2(4.630 cm) = 564.556 =565 cm3(maximum of 3 significant figures allowed since one of theoriginal numbers in the calculation has only 3 significant figures)1.60Plan:Review the procedure for changing a number to scientific notation. There can be only 1 nonzero digit to theleft of the decimal point in correct scientific notation. Moving the decimal point to the left results in a positiveexponent while moving the decimal point to the right results in a negative exponent.Solution:a)1.310000x105(Note that all zeros are significant.)b)4.7x10–4(No zeros are significant.)c)2.10006x105d)2.1605x1031.61Plan:Review the procedure for changing a number to scientific notation. There can be only 1 nonzero digit to theleft of the decimal point in correct scientific notation. Moving the decimal point to the left results in a positiveexponent while moving the decimal point to the right results in a negative exponent.Solution:a)2.820x102(Note that the zero is significant.)b)3.80x10–2(Note the one significant zero.)c)4.2708x103d)5.82009x1041.62Plan:Review the examples for changing a number from scientific notation to standard notation. If the exponent ispositive, move the decimal back to the right; if the exponent is negative, move the decimal point back to the left.Solution:a)5550(Do not use terminal decimal point since the zero is not significant.)b)10070.(Use terminal decimal point since final zero is significant.)c)0.000000885d)0.003004

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Chemistry: The Molecular Nature of Matter and Change Canadian Edition (2013) Solution Manual - Page 14 preview image

Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-13Solution Manual1.63Plan:Review the examples for changing a number from scientific notation to standard notation. If the exponent ispositive, move the decimal back to the right; if the exponent is negative, move the decimal point back to the left.Solution:a)6500.(Use terminal decimal point since the final zero is significant.)b)0.0000346c)750(Do not use terminal decimal point since the zero is not significant.)d)188.561.64Plan:In most cases, this involves a simple addition or subtraction of values from the exponents. There can beonly 1 nonzero digit to the left of the decimal point in correct scientific notation.Solution:a)8.025x104(The decimal point must be moved an additional 2 places to the left: 102+ 102= 104)b)1.0098x10–3(The decimal point must be moved an additional 3 places to the left: 103+ 10–6= 10–3)c)7.7x10–11(The decimal point must be moved an additional 2 places to the right: 10–2+ 10–9= 10–11)1.65Plan:In most cases, this involves a simple addition or subtraction of values from the exponents. There can beonly 1 nonzero digit to the left of the decimal point in correct scientific notation.Solution:a)1.43x102(The decimal point must be moved an additional 1 place to the left: 101+ 101= 102)b)8.51(The decimal point must be moved an additional 2 places to the left: 102+ 10–2= 100)c)7.5(The decimal point must be moved an additional 3 places to the left: 103+ 10–3= 100)1.66Plan:Calculate a temporary answer by simply entering the numbers into a calculator. Then you will need toround the value to the appropriate number of significant figures. Cancel units as you would cancel numbers, andplace the remaining units after your numerical answer.Solution:a)() ()34896.626 x10J s2.9979 x10 m/s489 x10m−−= 4.062185x10–19J4.06x10–19J(489x10–9m limits the answer to 3 significant figures; units of m and s cancel)b)()()2326.022 x 10molecules/mol1.23 x 10g46.07 g/mol= 1.6078x1024molecules1.61x1024molecules(1.23x102g limits answer to 3 significant figures; units of mol and g cancel)c)()()231822116.022 x 10atoms/mol2.18 x 10J/atom23−−= 1.82333x105J/mol1.82x105J/mol(2.18x10–18J/atom limits answer to 3 significant figures; unit of atoms cancels)1.67Plan:Calculate a temporary answer by simply entering the numbers into a calculator. Then you will need toround the value to the appropriate number of significant figures. Cancel units as you would cancel numbers, andplace the remaining units after your numerical answer.Solution:a)()()7324.32 x10 g43.14161.95x10 cm3= 1.3909 =1.39 g/cm3(4.32x107g limits the answer to 3 significant figures)b)()()221.84x10g44.7 m/s2= 1.8382x105=1.84x105g·m2/s2(1.84x102g limits the answer to 3 significant figures)

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Chemistry: The Molecular Nature of Matter and Change Canadian Edition (2013) Solution Manual - Page 15 preview image

Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-14Solution Manualc)()()()()24335221.07 x10mol / L3.8 x 10mol / L8.35 x10mol / L1.48 x 10mol / L−−−−= 1.9248 x 103=1.9 x 103L2/mol2(3.8x10–3mol/L limits the answer to 2 significant figures; mol3/L3in the numerator cancels mol5/L5in thedenominator to leave mol2/L2in the denominator or units of L2/mol2)1.68Plan:Exact numbers are those which have no uncertainty. Unit definitions and number counts of items in a groupare examples of exact numbers.Solution:a) The height of Horseshoe Falls is a measured quantity. This isnotan exact number.b) The number of planets in the solar system is a number count. Thisisan exact number.c) The number of grams in a pound is not a unit definition. This isnotan exact number.d) The number of millimeters in a meter is a definition of the prefix “milli–.” Thisisan exact number.1.69Plan:Exact numbers are those which have no uncertainty. Unit definitions and number counts of items in a groupare examples of exact numbers.Solution:a) The speed of light is a measured quantity. It isnotan exact number.b) The density of mercury is a measured quantity. It isnotan exact number.c) The number of seconds in an hour is based on the definitions of minutes and hours. Thisisan exact number.d) The number of provinces and territories is a counted value. These are exact numbers.1.70Plan:Observe the figure, and estimate a reading the best you can.Solution:The scale markings are 0.2 cm apart. The end of the metal strip falls between the mark for 7.4 cm and 7.6 cm. Ifwe assume that one can divide the space between markings into fourths, the uncertainty is one-fourth theseparation between the marks. Thus, since the end of the metal strip falls between 7.45 and 7.55 we can report itslength as7.50 ± 0.05 cm. (Note: If the assumption is that one can divide the space between markings into halvesonly, then the result is 7.5 ± 0.1 cm.)1.71Plan:You are given the density values for five solvents. Use the mass and volume given to calculatethe density of the solvent in the cleaner and compare that value to the density values given to identify the solvent.Use the uncertainties in the mass and volume to recalculate the density.Solution:a)mass11.775 gDensity (g/mL)volume15.00 mL=== 0.7850 g/mL. The closest value isisopropanol.b) Ethanol is denser than isopropanol. Recalculating the density using the maximum mass = (11.775 + 0.003) gwith the minimum volume = (15.00 – 0.02) mL, givesmass11.778 gDensity (g/mL)volume14.98 mL=== 0.7862 g/mL. This result is still clearly not ethanol.Yes, the equipment is precise enough.1.72Plan:Calculate the average of each data set. Remember that accuracy refers to how close a measurement is tothe actual or true value while precision refers to how close multiple measurements are to each other.Solution:a) Iavg=8.72 g8.74 g8.70 g3++= 8.7200 =8.72 gIIavg=8.56 g8.77 g8.83 g3++= 8.7200 =8.72 gIIIavg=8.50 g8.48 g8.51 g3++= 8.4967 =8.50 g

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Chemistry: The Molecular Nature of Matter and Change Canadian Edition (2013) Solution Manual - Page 16 preview image

Silberberg, Lavieri, & Venkateswaran,Chemistry: The Molecular Nature of Matter and Change, Canadian EditionPage1-15Solution ManualIVavg=8.41 g8.72 g8.55 g3++= 8.5600 =8.56 gSetsIandIIare most accurate since their average value, 8.72 g, is closest to the true value, 8.72 g.b) To get an idea of precision, calculate the range of each set of values: largest value – smallest value. A smallrange is an indication of good precision since the values are close to each other.Irange= 8.74 g – 8.70 g = 0.04 gIIrange= 8.83 g – 8.56 g = 0.27 gIIIrange= 8.51 g – 8.48 g = 0.03 gIVrange= 8.72 g – 8.41 g = 0.31 gSet IIIis the most precise (smallest range), but is the least accurate (the average is the farthest from the actualvalue).c)Set Ihas the best combination of high accuracy (average value = actual value) and high precision (relativelysmall range).d)Set IVhas both low accuracy (average value differs from actual value) and low precision (has the largestrange).1.73Plan:Remember that accuracy refers to how close a measurement is to the actual or true value; since the bull’s-eye represents the actual value, the darts that are closest to the bull’s-eye are the most accurate. Precision refers tohow close multiple measurements are to each other; darts that are positioned close to each other on the target havehigh precision.Solution:a)Experiments II and IV— the averages appear to be near each other.b)Experiments III and IV— the darts are closely grouped.c)Experiment IV and perhaps Experiment II— the average is in or near the bull’s-eye.d)Experiment III— the darts are close together, but not near the bull’s-eye.1.74Plan:If it is necessary to force something to happen, the potential energy will be higher.Solution:a)b)a) The balls on the relaxed spring have a lower potential energy and are more stable. The balls on the compressedspring have a higher potential energy, because the balls will move once the spring is released. This configurationis less stable.b) The two + charges apart from each other have a lower potential energy and are more stable. The two + chargesnear each other have a higher potential energy, because they repel one another. This arrangement is less stable.Potential EnergyPotential Energy+++ +

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