In problems I, II, and III: Calorimeter constant is 0.00 j/degree C Specific Hea
ID: 1007196 • Letter: I
Question
In problems I, II, and III: Calorimeter constant is 0.00 j/degree C Specific Heat of water or aqueous solutions is 4.18 j/g degree C I. 76.9619 g of metal were heated to 100.7 degree C and poured into a calorimeter containing 46.30 g of water at 25.00 degree C. After stirring, the temperature of the water rose rapidly to 32.40 degree C before slowly starting to fall. Calculate: a) The Specific Heat of the metal b) The Atomic Weight Calculated using the rule of Dulong and Petit II. To a calorimeter containing 47.00 g of water at 25.05 degree C, 3.0888 g of a salt was added with stirring. As the salt dissolved, the temperature rapidly changed to 31.60 degree C before slowly returning to room temperature. Calculate: c) The Heat of Solution Given that the Heat of Formation of the solid salt is.... -210.01 Kj/mol and the Molecular Mass of the solid salt is.......... 146.30 g/mol Calculate: d) The Heat of Formation of the Solution Kj/mol III. Given the reaction... AAA + BBB rightarrow CCC You add to a calorimeter 12.0 mL of 4.30 M AAA, 9.0 mL of water, and 11.0 mL of 5.79 M BBB. All of the above solutions were initially at 26.85 degree C. After mixing, the temperature changed to 17.35 degree C. Assume the density of all solutions is 1.000 g/mL and all solutions have the same specific heat as pure water. Calculate: e) The Heat of Reaction in Kj/molExplanation / Answer
1. from the data
msDT = msDT
(76.9619*x*(100.7-32.4) = (46.3*4.18(32.4 - 25)
x = Specific Heat of the metal = 0.272 j/g.c
The Atomic Weight = 170 g/mol
atomicweight = 91.2 g/mol
76.9619*0.272 = 20.934 j/c
20.934/N = 3*8.314
N = 0.84 mol
molarmass = 76.9619/0.84 = 91.2 g/mol
11.
q = msDT
= (47+3.088)*4.18*(31.6-25.05)
=1371.36 joule = 1.371 kj
No of mol of salt = 3.0888/146.3 = 0.0211 mol
c) heat of solution = 1.371/0.0211 = 64.97 kj/mol
d) heat of formation of solution = 210+64.97 = 274.97 kj/mol
iii.
No of mol of AAA = 12/1000*4.3 = 0.0516 mol
No of mol of bbb = 11/1000*5.79 = 0.06369 mol
mass of solution = (11+12+9) = 32 grams
heat released = 32*4.18*(26.85-17.35) = 1270.72 joule.
= 0.127 kj
heat of reaction = 0.127/0.0516 = 2.46 kj/mol