I need help with C,D and E. The answers I got for A and B are a=0.147j/g-C and b

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I need help with C,D and E. The answers I got for A and B are a=0.147j/g-C and b=170g/mol

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. 136.8587 g of metal were heated to 101.2 °C and poured into a calorimeter containing 43.20 g of water at 29.35 °C.

After stirring, the temperature of the water rose rapidly to 36.55 °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 41.40 g of water at 29.15 °C, 5.4876 g of a salt was added with stirring. As the salt dissolved, the temperature rapidly changed to 33.60 °C before slowly returning to room temperature. Calculate:

c) The Heat of Solution ______________

Given that the Heat of Formation of the solid salt is.... -64.89 Kj/mol and the Molecular Mass of the solid salt is.......... 147.87 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 14.0 mL of 2.49 M AAA, 9.0 mL of water, and 7.0 mL of 3.25 M BBB. All of the above solutions were initially at 25.55 °C. After mixing, the temperature changed to 28.85 °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

c) For calorimeter containg 41.60 water

dH=MCdT

dH=41.40(4,184J/0C)(29.15-33.60)

dH=41.40(4,184J/0C)(-4.45)

dH= -770J gined by water lost by dissolution

Heat of solution =- 770J/0C

Heat of solution per mol

147.87@ -770J/0C

=-113859J/0C

= -113.85KJ/0C

D)

as given heat of formation of solid salt is -64.89kJ/mol,calulate heat of formation of solution

dH reaction =dH(product- reactant)

dH reaction =heat of formation of solution -heat of formation of solid

-113.859kJ/0C=HEat of formation of solution -(-64.89)

-113.859kJ/0C=HEat of formation of solution+-64.89kJ/0C

HEat of formation of solution=-178kJ/Mol

Foe E

Calculate the heat released, q, in Joules (J), by the precipitation reaction:

q = mass x specific heat capacity x change in temperature

Q=(14+7)*4.184*(28.85-25.55)

q=289.90J

Calculate the moles of species specified,

from eqation 1M AAA gives 1 M CCC

n(AAA)=n(CCC)

nAAA=molarity x volume

=2.49*14*10-3

=0.0348mol

Calculate Heat of Reaction in Kj/mol H, in kJ mol-1

H = -q/1000 ÷ n(solute)

=-289.89/1000÷ 0.0348

- 0.289/0.0348

=-8.3kJ/Mol

H is negative because the reaction is exothermic.

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