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CHAPTER 6 ANSWERS TO SELECTED PROBLEMS SAMPLE PROBLEMS Try it yourself 61 a This is a chemical change When you digest food you break down large molecules such as starches fats and proteins into simpler molecules b This is a physical change The chemical formula of alcohol does not change when you dissolve it in water 62 The first chemical equation is correct Sodium is a metal and metal atoms do not form chemical bonds to one another Writing Nag implies that three sodium atoms link together to form a molecule which is not the case 63 This expression contains a total of six copper atoms and three oxygen atoms 64 This chemical equation is not balanced There are four oxygen atoms on the left side of the arrow note that we have two formula units of FeO each of which contains an oxygen atom but only three oxygen atoms on the right side The iron atoms are balanced with two on each side but every element must be balanced in a chemical equation 65 2A1 6HCl gt 2A1Cl3 3H2 66 Mgs 2HClaq gt MgCl2aq H2g 67 First we must work out the masses of aluminum and chlorine in the balanced equation 2 Al represent two aluminum atoms which weigh 5396 amu 2698 amu x 2 3 C12 represents three molecules of C12 which contain a total of six chlorine atoms The total weight of six chlorine atoms is 2127 amu 3545 amu x 6 Using these numbers we can express the mass relationships In amu 5396 amu of Al reacts with 2127 amu of C12 In grams 5396 g of Al reacts with 2127 g of C12 68 To solve this problem we must first work out the mass relationship between sucrose and oxygen in the chemical equation We already know that one molecule of sucrose C12H22011 weighs 342296 amu The chemical equation calls for twelve molecules of 02 each of which contains two oxygen atoms and weighs 3200 amu 1600 amu x 2 Therefore twelve 02 molecules weigh atotal of3840 amu 3200 amu x 12 We now know that the mass relationship between sucrose and oxygen in this reaction is 342296 g of sucrose to 3840 g of oxygen Finally we use this relationship to translate 100 g of sucrose into the corresponding mass of oxygen 3840 g 02 342296 g sucrose Rounding this to three significant figures we see that your body uses 112 g of 02 when it burns 100 g of sucrose 100 g sucrose X 112183607 g 02 calculator answer Chapter 6 7 Selected Answers 2 69 The formula weight of NiOH3 is 109714 amu The balanced equation contains two formula units of NiOH3 which weigh a total of 219428 amu Therefore we can state the relationship between the mass of NiOH3 and the heat as follows The reaction absorbs 252 kcal of heat when it forms 219428 g of NiOH3 Since this reaction absorbs heat AH is a positive number 610 First we need the relative mass of Hz in this reaction The balanced equation calls for two molecules of Hz One molecule weighs 2016 amu 1008 amu X 2 so two molecules weigh 4032 amu 2016 amu X 2 Using this mass we can now state the relationship between mass and energy if we use 4032 grams of Hz we get 137 kcal of energy We can use this as a conversion factor to translate 151 kcal of energy into the corresponding mass of hydrogen 151 kcal X 44440292 g of H2 calculator answer 137 kcal Rounding this to three significant figures tells us that we must burn 444 g of H2 611 We must calculate the Calorie value of each nutrient then add them up carbohydrate 7 g X 4 Cal g 28 Cal fat 16 g X 9 Calg 144 Cal protein 8 g X 4 Calg 32 Cal Total 28 Cal 144 Cal 32 Cal 204 Cal Rounding this to the nearest ten Calories gives us 200 Calories A nutritional Calorie spelled with a capital C is the same as a kilocalorie so the energy content is 200 kilocalories 612 Start with an unbalanced reaction that contains the correct reactants and products C4H100 02 gt C02 H2O Balance the carbon atoms by writing a 4 in front of the C02 C4H100 02 gt 4 C02 H2O Balance the hydrogen atoms by writing a 5 in front of the H20 C4H100 02 gt 4 C02 5 H2O Finally balance the 0Xygen atoms by writing a 6 in front of the 02 Note that C4H100 contains an 0Xygen atom The balanced equation is C4H100 6 02 gt 4 C02 5 H2O 613 The second reaction is a precipitation reaction because the reactants are all in solution but one of the products is a solid In the first reaction there is a solid reactant so the first reaction is not a precipitation 614 The product is FeOH2 which is made from Fe2 and OH ions This gives us an unbalanced form of the equation Fe2aq OHaq gt FeOHzs To balance the equation we must write a 2 in front of the hydr0Xide ion The balanced equation is Fe2aq 2 OH aq gt FeOH2s 615 The first reaction is faster because it has the lower activation energy Chapter 6 7 Selected Answers 3 Energy Product energy 8 kcal 7 km Reacmm activation T heat of energy energy reaction Start of reaction gt End of reaction 617 The reaction will go forward so it can get rid of the excess HCOg ions It will make C3H503 and H2C03 and it will consume HC3H503 and HCOgT 618 The reaction will go forward so it can replace the H2C03 that you removed It will make C3H503 and H2C03 and it will consume HC3H503 and HCOg END OF SECTION PROBLEMS Section 61 61 a This is a physical change The water changes from a liquid to a solid but it remains H20 b This is a chemical change The electrical current breaks the chemical bonds between the hydrogen atoms and the oxygen atoms in H20 c Dissolving a solid in water is generally described as a physical change However when the solid dissociates it is reasonable to call the process a chemical change 62 The steam weighs 500 grams This is an example of the law of mass conservation 63 When gasoline burns the products are invisible gases These gases do not remain in the container so we cannot see them but they still obey the law of mass conservation The mass of the gases equals the mass of the original gasoline plus the oxygen that reacted with it 64 Lead and gold are different elements No chemical reaction can change an atom of one element into an atom of another element Chapter 6 7 Selected Answers 4 Section 62 65 a This equation is not balanced because it has two oxygen atoms on the left side of the arrow and only one oxygen atom on the right side b This equation is not balanced because it has four oxygen atoms on the right side of the arrow and only two oxygen atoms on the left side c This equation is not balanced because it has four silver atoms on the right side of the arrow and only two silver atoms on the left side d This equation is balanced Each side contains two iron atoms and six chlorine atoms 66 a Mg Cl2 gt MgCl2 The original equation was balanced b 2A1 3S gt A1283 c P4 6C12 gt 4PC13 d 2MgI2 02 gt 2MgO 212 e Na2S 2AgF gt 2NaF Ag2S 67 When you write 2 N you mean that the two nitrogen atoms are not bonded to one another Writing N2 means that the atoms are linked by a chemical bond 68 The answer is incorrect because the student has changed a chemical formula When you balance a chemical equation you cannot change any of the subscript numbers in the formulas because these tell you how the atoms are bonded to one another Aluminum and chlorine can only combine to form AlC13 never AlCl2 69 2Nas 2H2Ol gt 2NaOHaq H2g 610 One molecule of gaseous sulfur trioxide reacts with two formula units of aqueous potassium hydroxide forming one formula unit of aqueous potassium sulfate and one molecule of liquid water Section 63 611 The reaction calls for one molecule of CH4 which weighs 16042 amu adding up the atomic weights of one carbon and four hydrogen atoms The reaction produces two molecules of H20 Each molecule of H20 weighs 18016 amu so two molecules weigh 36032 amu We now know that when we burn 16042 amu of CH4 we make 36032 amu of H20 This mass ratio can also be expressed in grams so 16042 g of CH4 reacts with oxygen to form 36032 g of H20 612 a We start with the relative masses of HCl and AlOH3 using the formula weights of these compounds and the balanced equation 3 HCl the formula weight of HCl is 36458 amu so three molecules of HCl weigh 109374 amu AlOH3 the formula weight of AlOH3 is 78004 amu so one formula unit of AlOH3 weighs 78004 amu Chapter 6 7 Selected Answers 5 Now we express this mass relationship in grams 109374 g of HCl reacts with 78004 g of Al0H3 This gives us the conversion factor we need 250 g Al0H3 X 350539716 g HCl calculator answer 78 004 g AlOH3 Rounding this to three signi cant gures gives us 351 g of HCl b In this problem we need the relative masses of Al0H3 and H20 We already have the mass of a formula unit of Al0H3 from part a 78004 amu 3 H20 the formula weight of H20 is 18016 amu so three molecules of H20 weigh 54048 amu We now know that we make 54048 g of H20 when we use 78004 g of Al0H3 We can use this as a conversion factor 250 g AlOH3 X 173221886 g H20 78 004 g AlOH3 Rounding this to three significant figures gives us 173 g of H20 613 a The formula weight of alcohol C2H50 is 46048 amu so two molecules of alcohol weigh 92136 amu The formula weight of fruit sugar C5H1205 is 180156 amu and the reaction calls for only one molecule of this compound Therefore we make 92136 g of alcohol when we break down 180156 g of sugar 750 g alcohol X W 146 6495181 g of sugar calculator answer 92136 g alcohol Rounding this answer to three significant figures gives us 147 g of fruit sugar b The formula weight of carbon dioxide is 4401 amu so two molecules of C02 weigh 8802 amu Therefore we make 8802 g of C02 when we make 92136 g of alcohol We worked out the relative mass of alcohol in part a 750 g alcohol X 92136 g alcohol Rounding this answer to three significant figures gives us 716 g of carbon dioxide 2 716495181 g C02 calculator answer Section 64 614 a In the balanced equation the energy is listed as a reactant Therefore the reaction consumes energy in the form of heat so the surroundings become cooler b This reaction is endothermic c AH is a positive number whenever heat is a reactant 615 The formula weight of Ba0H2 is 171316 amu Therefore when 171316 grams of Ba0H2 reacts with NH4Cl 55 kcal of heat is absorbed 616 The formula weight of C5H5 is 78108 amu so two molecules of C5H5 weigh 156216 amu Therefore the relationship between the mass of benzene and the energy is when 156216 Chapter 6 4 Selected Answers 6 g of benzene burns the reaction produces 1562 kcal of heat Using this relationship as a conversion factor we get 432 g C6H6 X 4319557536 kcal calculator answer 156216 g C6H6 Rounding this answer to three signi cant gures gives us 432 kcal of heat 617 We can use the same relationship between benzene and heat that we used in Problem 616 125 kcal X 156216 g C6H6 1562 kcal Rounding this answer to three signi cant gures gives us 125 g of C H u 12 50128041 g C6H6 calculator answer 618 a The formula weight of glucose C5H1205 is 180156 amu so our relationship is we get 163 kcal of heat when we break down 180156 g of glucose Using this relationship as a conversion factor 100 g glucose X Lkwl 009047714 kcal calculator answer 180156 g glucose Rounding this answer to three signi cant gures gives us 00905 kcal of heat when we break down one gram of glucose into lactic acid b 4 kcal 00905 kcal 442 so your body gets around 44 times as much energy from burning glucose as it does from breaking down glucose into lactic acid 619 We must calculate the Calorie value of each nutrient then add them up carbohydrate 35 g X 4 Cal g 140 Cal fat 9 g X 9 Calg 81 Cal protein 12 g X 4 Calg 48 Cal Total 140 Cal 81 Cal 48 Cal 269 Cal Rounding this to the nearest ten Calories gives us 270 Calories 620 Removing one gram of sugar will decrease the nutritional value by 4 Calories Therefore the manufacturer must remove 20 Cal 4 5 g of sugar Section 65 621 aC7H8 902 gt 7CO2 4H2O b2C4H10 13 02 gt 8CO2 10 H2O C C4H100 602 gt 4C02 5 H20 622 Reactions a and d are combustion reactions In reaction b one of the products does not contain 0Xygen and in reaction c the products must be CO2 not C2H4O2 and H20 623 Reaction b is the only precipitation reaction Reaction a does not make a solid product and reactions c and d use at least one reactant that is not dissolved in water Chapter 6 7 Selected Answers 7 624 The only precipitation reaction is b and the net equation for this reaction is AgaQ t BTTaQ gt AgBr S Section 66 625 a The rate increases because the concentration of 02 has increased b The rate decreases All reactions slow down when they are cooled c The rate increases The bubbles provide a larger surface area where the gaseous oxygen and the wine are in contact 626 The powdered limestone dissolves faster because it has a larger surface area 627 Stirring breaks the fat into small droplets increasing the surface area where the fat is in contact with the NaOH solution 628 The second reaction using FeClz has the larger activation energy Slow reactions have larger activation energies than fast reactions 629 no catalyst red curve gt I I a I I with catalyst green curve L 21 I In La kcall Ikcal I I energy of energy of products reactants Section 67 630 An equilibrium mixture is a stable mixture of products and reactants that is formed by a reversible reaction 631 Only statement c is correct 632 Statement a is correct 633 a The reaction goes backward to remove the excess glucose b The reaction goes forward to remove the excess sucrose c The reaction goes forward to make more fructose Chapter 6 7 Selected Answers 8 634 The concentration of fructose will decrease When we add glucose the reaction goes backward to remove some of the glucose we added The backward direction consumes both glucose and fructose so the number of molecules of fructose in our mixture decreases CUMULATIVE PROBLEMS Odd numbered problems only 635 In a chemical change the chemical formulas of the starting materials and products are different In a physical change the starting materials and the products have the same chemical formulas 637 a physical b chemical c chemical d physical 639 a physical b chemical c physical 64l Rusting is a chemical reaction in which iron combines with oxygen from the air to form a compound The compound weighs more than the original iron because it also contains some oxygen 643 When your body burns food to obtain energy one of the products is water The difference between the mass of water you take in and the mass of water you put out is the water that your body makes as it burns food 645 a This reaction is not balanced There are two Cl atoms on the left side but four on the right side b This is a balanced equation There are one Mg two 0 four H and two F atoms on each side c This is a balanced equation There are four Na four Cr fortynine O thirtyfour H eight S and six C atoms on each side d This reaction is not balanced There are seventeen O atoms on the left side but only sixteen on the right side 647 This is not a reasonable answer because the student has changed the chemical formula of the product 649 aS 2012 gt 014 b2Ag s gt Ang c2K CI2 gt 2KC1 651 a Ca0 2HCl gt CaClz H20 b4Fe 302 gt 2Fe203 cCH4 202 gt C0 2HzO 653 a2C4H10 13 02 gt 8002 10 H20 b2A1C13 3 H20 gt A1203 6HC1 c2AgNO3 MgIz gt 2AgI MgN03z Chapter 6 7 Selected Answers 9 655 a2AlOH3 3 H2304 gt AlzSO43 6H20 b4C5H11N02 27 02 gt 20 C02 1 22 H20 2N2 657 PC13s 3 H2O1 gt H3P03aq 3 HClaq 659 a The molecule contains eight sulfur atoms so its formula is Ss bSg 802 gt 8SO2 661 The reaction calls for four Fe atoms Each Fe weighs 5585 amu so four Fe atoms weigh 2234 amu The reaction makes two formula units of Fe203 Each formula unit weighs 1597 amu so two formula units weigh 3194 amu Expressing these relative weights in grams we can say that 2234 g of Fe reacts with oxygen to form 3194 g of Fe203 663 a For glycine C2H5NO2 one molecule weighs 7507 amu so two molecules weigh 15014 amu For ammonia NH3 one molecule weighs 17034 amu so two molecules weigh 34068 amu These masses tell us that when a fish burns 7507 grams of glycine it makes 3406 grams of ammonia We can use this relationship as a conversion factor 34 068 g ammonia 100 g glycine X 022690822 g ammonia calculator answer 15014 g glycine Rounding this answer to three significant figures gives us 0227 g of ammonia b For water one molecule weighs 18016 amu so two molecules weigh 36032 amu We already worked out the mass of two molecules of ammonia in part a 34068 amu These masses tell us that when a fish makes 34068 grams of ammonia it also makes 36032 grams of water Once again we use this relationship as a conversion factor 36032 g H20 250 g NH3 X 2644123518 g of H20 calculator answer 34 068 g NH3 Rounding this answer to three significant figures gives us 264 g of H20 665 a For trilaurin C39H7405 one molecule weighs 638982 amu so two molecules weigh 1277964 amu For oxygen 02 one molecule weighs 3200 amu so 109 molecules weigh 3488 amu These masses tell us that we use 3488 g of oxygen when we burn 1277964 g oftrilaurin Using this relationship as a conversion factor 3488 g 02 1277964 g trilaurin Rounding this answer to three significant figures gives us 341 g of 02 125 g trilaurin X 3411676699 g of 02 calculator answer b For CO2 one molecule weighs 4401 amu so 78 molecules weigh 343278 amu For H2O one molecule weighs 18016 amu so 74 molecules weigh 1333184 amu These masses tell us that when we burn 1277964 g of trilaurin from part a we make 343278 g of CO2 and 1333184 g of H20 Now we can use the relationship between trilaurin and CO2 to calculate the mass of CO2 we make when we burn 125 g of trilaurin Chapter 6 4 Selected Answers 10 343278 g C02 1277964 g trilaurin We can use the relationship between trilaurin and H20 to calculate the mass of H20 1304011694 g H20 calculator answer 1277964 g trllaurm Rounding these values to three signi cant figures we nd that we will make 336 g of CO2 and 130 g of H20 c We used a total of 466 g of reactants 125 g oftrilaurin and 341 g of oxygen and we made a total of 466 g of products 336 g of carbon dioxide and 130 g of water The reactants and products have equal masses obeying the law of mass conservation 125 g trilaurin X 3357665005 g of C02 calculator answer 125 g trilaurin X 667 We need to work out each reaction separately For the reaction of alanine C3H7N02 one molecule of alanine weighs 89096 amu so two molecules weigh 178192 amu For urea N2H4C0 one molecule weighs 60062 amu Therefore we make 60062 g of urea when we burn 178192 g of alanine Using this mass relationship 60 062 g urea 178192 g alanine For the reaction of asparagine C4H8N203 one molecule of asparagine weighs 132124 amu Therefore we make 60062 g of urea when we burn 132124 g of asparagine Using this mass relationship 100 g alanine X 337063392 g of urea calculator answer 60062 g urea 132124 g asparagine From these answers we can see that mammals make more urea when they burn 100 g of asparagine 100 g asparagine X 454588114 g of urea calculator answer 669 a One molecule of vitamin C C5H805 weighs 176124 amu so two molecules weigh 352248 amu 0ne molecule of 02 weighs 3200 amu Therefore 352238 g ofvitamin C reacts with 3200 g of oxygen Using this mass relationship as a conversion factor and being sure to convert 500 mg of vitamin C into grams first 32 00 g 02 352238 g Vitamin C Rounding this answer to three significant figures gives us 00454 g of 02 b The density tells us that one liter 1000 mL of oxygen weighs 13 grams We can use this relationship as a conversion factor to translate 00454 g of oxygen into the corresponding number of milliliters To avoid a rounding error we use the calculator value from part a 004542254 g x M 05 g Vitamin C X 004542254 g of 02 calculator value 3494041785 mL calculator answer Rounding this answer to three significant figures gives us 349 mL of 02 671 a One formula unit of CaC03 weighs 10009 amu 0ne molecule of HCl weighs 36458 amu so two molecules weigh 72916 amu Therefore we know that 10009 g of CaC03 reacts with 72916 g of HCl Using this relationship as a conversion factor 100 g CaC03 X 072850435 g of HCl calculator answer 10009 g CaC03 Chapter 6 7 Selected Answers 11 Rounding this answer to three signi cant gures gives us 0729 g of HCl b We already know that 0729 g of HCl reacts with one gram of CaC03 All we need to do is convert this mass into moles using the fact that one mole of HCl weighs 36458 g To avoid a rounding error we use the calculator value from part a 072850435 g X L01 001998202 moles of HCl calculator answer 36458 g Rounding this answer to three signi cant gures gives us 00200 moles of HCl c The molarity tells us that one liter of the HCl solution in your stomach contain 001 moles of HCl We can use this relationship to translate 00200 moles of HCl into the corresponding number of liters of solution Since the molarity is only given to one signi cant gure there is no reason to use an extremely precise value for the moles of HCl 00200 mol HCl X L 2 L of solution 001 mol HCl Therefore 100 g of CaC03 can neutralize roughly 2000 mL of 001 M HCl 673 The products are C02 and H20 carbon dioxide and water 675 a C7H15 1102 gt 7C02 8H20 b2C5H120 1502 gt 10 C02 12 H20 677 We must start by writing the balanced equation for the combustion reaction The correct equation is 2 C2H2 5 02 gt 4 C02 2 H20 Using this equation we next work out the mass relationship between acetylene and C02 0ne molecule of C2H2 weighs 26036 amu so two molecules weighs 52072 amu 0ne molecule of C02 weighs 4401 amu so four molecules weigh 17604 amu Therefore we know that we make 17604 g of C02 when we burn 52072 g of acetylene Using this relationship as a conversion factor 500 g C2H2 x 1763904 g C02 52072 g C2H2 Rounding this answer to three signi cant gures we get 169 g of C02 1690351821 g C02 calculator answer 679 Reactions b c and e are precipitation reactions In reaction a one of the reactants is a solid and none of the products is a solid In reaction d the reactants are not dissolved in water 681 a Ca2aq 2 F aq gt CaF2s b2Agaq s2 aq a Agzss c Fe3aq 3 0Haq gt Fe0H3s 683 a This is an endothermic reaction The heat appears on the reactant side of the balanced equation so this reaction consumes heat b The surroundings become cooler c AH is a positive number Chapter 6 7 Selected Answers 12 685 The formula weight ofNaHS03 is 104058 amu Therefore 104 kcal of heat is absorbed when 104058 g ofNaHS03 reacts with HCl 687 a The surroundings become hotter AH is a negative number which tells us that the reaction produces heat b Heat is a product in this reaction c 2 Mgs Ozg gt 2 MgO 287 kcal 689 One molecule of sugar C5H1205 weighs 180156 amu Therefore this reaction produces 166 kcal of heat when 180156 g of sugar breaks down into alcohol and carbon dioxide We can use this relationship as a conversion factor 28 g sugar X Lkwl 25799862 kcal of heat calculator answer 180156 g sugar Rounding this answer to two significant figures gives us 26 kcal of heat 691 One molecule of ethanol CszO weighs 46068 amu so two molecules of ethanol weigh 92136 amu Therefore this reaction produces 166 kcal of heat when it makes 92136 g of ethanol We can use this relationship as a conversion factor 275 kcal X W 1526349398 g of ethanol calculator answer 166 kcal Rounding this answer to three significant figures gives us 153 g of ethanol 693 The balanced equation calls for one molecule of propane C3H8 which weighs 44094 amu Therefore to complete the chemical equation we need to know how much heat the reaction produces when we burn 44094 grams of propane According to the information in the problem we get 120 kcal of heat when we burn 100 g of propane We can use this relationship as a conversion factor to calculate the amount of heat produced by 44094 g of propane 44094 g propane X 529128 kcal calculator answer 100 g propane Rounding this answer to three significant figures gives us 529 kcal Since the reaction produces heat the actual heat of reaction AH is a negative number 7529 kcal We can write the complete chemical equation as C3H8 5 02 gt 3 C02 4 H20 529 kcal 695 a One molecule of alanine C3H7NOz weighs 89096 amu so two molecules of alanine weigh 178192 amu Therefore we know that a mammal produces 624 kcal of heat when it burns 178192 grams of alanine Using this relationship as a conversion factor 100 g alanine X w1 350184071 kcal of heat calculator answer 178192 g alanlne Rounding this answer to three significant figures gives us 350 kcal of heat b In part a we found that one gram of alanine produces 350 kcal of energy when it is burned by a mammal including a human being Since a nutritional Calorie is the same as a kilocalorie the nutritional value of alanine is 350 Calories per gram This is somewhat lower than the Calorie value for proteins The nutritional value of protein is an average value for all of the amino acids in typical proteins Chapter 6 7 Selected Answers 13 697 a In Problem 350 we found that our bodies produce 350 kcal of heat when we burn one gram of alanine Now let us calculate the amount of heat that we get from the normal combustion reaction One molecule of alanine C3H7NOz weighs 89096 amu so four molecules of alanine weigh 356384 amu Therefore we know that we get 1552 kcal ofheat when we burn 356384 grams of alanine Using this relationship as a conversion factor 100 g alanine X 435485319 kcal of heat calculator answer 356384 g alanme Rounding this answer to three significant figures gives us 435 kcal of heat b Our bodies obtain 350 g of heat when we burn a gram of alanine Normal combustion would give us 435 kcal so we lose 085 kcal ofheat 435 kcal 7 350 kcal 085 kcal To calculate the percentage divide the amount of heat lost by the total amount of heat in normal combustion and then multiply the answer by 100 7085 kcal x 100 1954022989 calculator value 435 kcal We lose roughly 20 of the heat of combustion 699 We must calculate the Calorie value of each nutrient then add them up carbohydrate 6 g X 4 Cal g 24 Cal fat 10 g X 9 Calg 90 Cal protein 28 g X 4 Calg 112 Cal Total 24 Cal 90 Cal 112 Cal 226 Cal Rounding this to the nearest ten Calories gives us 230 Calories 6101 First we can work out the number of Calories from the protein and the carbohydrate carbohydrate 27 g X 4 Calg 108 Cal protein 1 g X 4 Calg 4 Cal Adding these up gives us 112 Calories The Twinkie contains 160 Calories so the remaining Calories must be from fat 160 Cal 7 112 Cal 48 Calories from fat One gram of fat contains 9 Calories 48 Cal 7 9 Calg 533333333 g offat calculator answer Masses of nutrients are normally only measured to the nearest gram so the Twinkie contains 5 grams of fat 6103 One molecule of ethanol CszO weighs 46068 amu so our bodies get 327 kcal of energy when we burn 46068 grams of ethanol We can use this relationship as a conversion factor to translate 100 g of ethanol into the corresponding amount of energy 100 g ethanol X ch 709820266 kcal of energy calculator answer 46 068 g ethanol Rounding this answer to three significant figures gives us 710 kcal which is the same as 710 nutritional Calories Therefore the Calorie value of ethanol is 710 Calories per gram This is substantially more energy than you obtain from a gram of carbohydrate Chapter 6 7 Selected Answers 14 6105 a 500g X y 400 calories g b To relate our answer in part a to nutritional Calories we must convert 400 calories to kilocalories 400 cal is the same as 04 kcal Akilocalorie is the same as a nutritional Calorie so your body needs just 04 Calories to melt 500 g of ice We get 4 Calories of energy from one gram of carbohydrate 04 Cal X W 01 g carbohydrate Cal Your body must burn 01 g of carbohydrate 6107 When we say 30 of your Calories come from fat we mean that when you consume 100 Calories 30 Calories come from fat We can use this relationship as a conversion factor 2000 Calories X W 600 fat Calories 100 Calorles One gram of fat gives you 9 Calories 600 Cal X 19g 2 6666666667 g of fat calculator answer a You should eat no more than 67 grams of fat according to the US government guidelines 6109 a The rate will increase b The rate will decrease c The rate will increase d The rate will increase 6111 Catalysts lower the activation energy of the reaction and they position the reactants in the correct orientation Lowering the activation energy has a greater impact on the reaction rate 6113 Bacteria cannot regulate their temperature so they cool down when you put the food into the refrigerator All reactions slow down at colder temperatures including reactions in living organisms so the reactions that produce the unpleasant products occur more slowly in refrigerated food 6115 a The HClcatalyzed reaction is slower so it must have the larger activation energy red curve catalyzed by HCl high activation energy green curve catalyzed by trypsin low activation energy ENERGY energy of reactants energy of products Chapter 6 7 Selected Answers 15 6117 Na0H is a more effective catalyst because the reaction has a lower activation energy when you use NaOH The lower the activation energy the faster the reaction 6119 The energy of the reactants is 20 kcal and the top of curve is 60 kcal so the activation energy is 40 kcal 60 7 20 40 The reactant energy is 20 kcal and the product energy is 0 kcal so the heat of reaction is 720 kcal The heat of reaction is negative because the energy of the products is lower than the energy of the reactants This is an exothermic reaction 6121 a This reaction is reversible The ethanol is continually breaking down into C2H4 and H20 so we also have some C2H4 and H20 in our mixture b Some of the ethanol will break down into C2H4 and H20 6123 a The reaction goes forward consuming some of the C0 that we added b The reaction goes backward consuming some of the C02 that we added c The reaction goes forward making H2 to replace some of the H2 we removed 6125 a Adding C0 will drive the reaction forward making some H2 and C02 so adding C0 will increase the amount of H2 in the equilibrium mixture b Removing C0 will drive the reaction backward consuming some of the H2 in our original mixture so removing C0 will not increase the amount of H2 in the equilibrium mixture c Adding C02 will drive the reaction backward consuming some of the H2 in our original mixture so adding C02 will not increase the amount of H2 in the equilibrium mixture d Removing C02 will drive the reaction forward making some H2 and C02 so adding C02 will increase the amount of H2 in the equilibrium mixture