Liquid c liquid t liquid n solid 4423 mol2849 j mol k

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liquid C liquid T liquid n solid = (4.423 mol)(28.49 J mol K )(126 K) (2.266 mol) = 7.01 × 10 3 J mol = 7.01 kJ mol If you solve for the q liquid part first, that should be 1.588 × 10 4 J.
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NAME:_________________________________ © 2015 L.S. Brown A9 (10 pts) 8. Some “hot packs” are based on the exothermic dissolution of calcium chloride in water. CaCl 2 (s) Ca 2+ (aq) + 2 Cl (aq) Δ H ° = –81.3 kJ A particular hot pack contains 50.0 g of CaCl 2 and 200.0 g of water, originally separated by a partition. When the pack is squeezed, the partition breaks and the CaCl 2 dissolves in the water. If the pack and its contents are initially at 22.0 ° C, what is the maximum temperature that this bag could reach? (Assume that the calcium chloride solution has a specific heat of 4.25 J g –1 K –1 , and that the heat capacity of the bag itself is small enough to be neglected.) I’ll start by finding q (or Δ H) for the amount of CaCl 2 dissolving. Note that the given value for Δ H is for dissolving 1 mole, so for our 50.0 g sample we will get: Δ H = 50.0 g × 1 mol 110.98 g × –81.3 kJ 1 mol = –36.628 kJ That will be the amount of energy we have available to heat up the solution. And there will be 250 g of solution, since it includes both the 50 g of CaCl 2 and the 200 g of water. We can find Δ T: q = mc Δ T, so Δ T = q mc = 36,628 J (250 g)(4.25 J g ° C ) = 34.5 ° C We started at 22 ° C, so the final T will be 56.5 ° C.
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