Before you can start this experiment, you must: Solve problems I, II, and III Pa

Question

Before you can start this experiment, you must: Solve problems I, II, and III Pass an 8 question on-screen computer quiz In problems I, II, and III:

Calorimeter constant is 0.00 j/°C Specific Heat of water or aqueous solutions is 4.18 j/g °C I. 89.6542 g of metal were heated to 101.1 °C and poured into a calorimeter containing 49.60 g of water at 27.35 °C. After stirring, the temperature of the water rose rapidly to 34.75 °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 41.90 g of water at 25.00 °C, 7.6039 g of a salt was added with stirring. As the salt dissolved, the temperature rapidly changed to 27.95 °C before slowly returning to room temperature.

Calculate:c) The Heat of Solution ______________ Given that the Heat of Formation of the solid salt is.... -228.24 Kj/mol and the Molecular Mass of the solid salt is.......... 67.22 g/mol

Calculate: d) The Heat of Formation of the Solution ______________ Kj/mol III. Given the reaction... AAA + BBB ---> CCC You add to a calorimeter 10.0 mL of 4.19 M AAA, 9.0 mL of water, and 12.0 mL of 5.14 M BBB. All of the above solutions were initially at 29.95 °C. After mixing, the temperature changed to 32.80 °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/mol ______________

Explanation / Answer

a)   Q = m Cp dT

here heat loss by the metal is equal to heat gain by the water

89.6542 x Cp x (101.1-34.75) = 49.60 x 4.184 x (34.75 -27.35)

Cp = 0.258 J / g oC

specific heat of metal = 0.258 J / g oC

this value is nearly corresponds to the metal Zirconium

atomic weight of Zr = 91.22 g / mol

b)

Q = m Cp dT

      = 41.90 x 4.184 x (27.95 -25)

      = 517.2 J

heat of solution = Q = 517.2 J

moles = 7.6039 / 67.22 = 0.113

heat formation of solution = 517.2 x 10^-3 / 0.113

heat formation of solution = - 4.60 kJ /mol

d)

solution mass = 10 + 9 + 12 = 31 g (because density =1 g/mL)

dT = 32.80 - 29.95 = 2.85

Q = m Cp dT = 31 x 4.184 x 2.85 = 369.7 J = 0.3697 kJ

limiting reagent moles = 0.0419

heat of reaction =- 0.3697 / 0.0419 = - 8.81

heat of reaction = - 8.81 kJ / mol

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