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Problem

A 25.5-g aluminum block is warmed to 65.4 C and …

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Eugene S.
University of Minnesota - Twin Cities

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 52 Problem 53 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100 Problem 101 Problem 102 Problem 103 Problem 104 Problem 105 Problem 106 Problem 107 Problem 108 Problem 109 Problem 110 Problem 111 Problem 112 Problem 113 Problem 114 Problem 115 Problem 116 Problem 117 Problem 118 Problem 119 Problem 120 Problem 121 Problem 122 Problem 123 Problem 124 Problem 125 Problem 126 Problem 127 Problem 128 Problem 129 Problem 130 Problem 131 Problem 132 Problem 133 Problem 134 Problem 135 Problem 136 Problem 137 Problem 138 Problem 139 Problem 140 Problem 141 Problem 142 Problem 143 Problem 144 Problem 145 Problem 146 Problem 147 Problem 148

Problem 102 Hard Difficulty

Dry ice is solid carbon dioxide. Instead of melting, solid carbon
dioxide sublimes according to the equation:

When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry
ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. A small Styrofoam cooler holds 15.0 L of water heated to 85 C. Use standard enthalpies of formation to calculate the change in enthalpy for dry ice sublimation, and calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 25 C. Assume no heat loss to the surroundings. (The Hf for CO2(s) is -427.4 kJ>mol.)

Answer

4888.6

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Video Transcript

So in this problem, we're given the Formula four sublime ing Dry Ice Co. Two solid to co two gas, and we want to find the mass of CO two solid. That would be dry ice needed in order to put in our fifteen liters ofwater such that it brings the temperature down from eighty five degrees Celsius to twenty five degrees Celsius and completely uses up our mass of dry ice. And so the first thing we want to do is find the Delta h of this reaction and so we can take the Delta H of formation of CO two gas and subtract the Delta H of formation of CO two solid. And when we do that, we take negative for twenty seven point four, killed Jules per mall and subtract negative three ninety three point five killer Jules Permal. And that gives us thirty three point nine killer Jules Permal. And so now that we have that, we can find the heat that the heat of the water a zits could in orderto cool it down by that temperature. And so how we do that is the equation Q equal, C m delta t the heat capacity sea of water is four point one eight four. We have fifteen thousand grams because we have fifteen leaders. And so we convert leaders to mill leaders and then two grams. And then our temperature goes from eighty five to twenty five degrees Celsius. And so doing this gives us three, seven, six, five, six zero zero Jules or thirty seven sixty five point six. Kill a jewel old. So now we can take our thirty seven sixty five point six killer Jules, and divide that by our thirty three point nine killer Jules per mole that we found in the previous step from our reaction. And that gives us that we have won eleven point zero eight moles. Now this is the mole's of dry ice that we have. And so, in order to get the mass, we just multiplied by the molar mass of co two. Forty four point zero one grams parole. And we get that We have four eight, eight, eight point six grams. And that is our final answer for the mass of dry ice

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