Calorimetry: Calculate Specific Heat, Atomic Weight, Heat of Solution & Reaction

[purpletouch]

to the admin. pls. don't delete this i figure it be easier to understand this if it's set up this way because its a prelab.thanks

I. 89.7716 g of metal were heated to 99.8 °C and poured into a calorimeter containing 46.20 g of water at 28.45 °C.
After stirring, the temperature of the water rose rapidly to 36.25 °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 48.80 g of water at 28.85 °C, 7.0545 g of a salt was added with stirring. As the salt dissolved, the temperature rapidly changed to 31.45 °C before slowly returning to room temperature.
Calculate:
c) The Heat of Solution ______________

Given that the Heat of Formation of the solid salt is... -180.59 Kj/mol
and the Molecular Mass of the solid salt is... 143.69 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 8.0 mL of 5.65 M AAA, 5.0 mL of water, and 12.0 mL of 5.51 M BBB. All of the above solutions were initially at 29.10 °C. After mixing, the temperature changed to 30.60 °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 ______________

i don't know how to get B D and E

for A i got .264
for B i divided .264 from 6 but the answer is wrong. i tried it with 6.2 but it I'm still not getting the right answer
C=-75.180
D. i really don't have a clue
E. I'm given this formula but I'm stuck on limiting reagant
(Volume in mL)(Molarity) = millimoles
VolumeHCl + VolumeNaOH+ Volumewater = Volumeliquids
Temperaturefinal-Temperatureinitial = Temp. Changeliquids
(Massliquids)(-Temp. Changeliquids)(4.18 j/gºC) = Energy Change
Energy Change/[(1000 j/Kj)(Moles of Limiting Reagent)] = DH Reaction

thanks!

 

[Redbelly98]

for A i got .264

Can you show how you got that? And please include units. I think you did a unit conversion incorrectly, but without seeing your work and what units you mean to use, there is no way to tell.

 

[Blueshift5]

Dear admin, thank you for providing clear instructions for the prelab. I will do my best to assist you with your calculations for calorimetry.

a) The Specific Heat of the metal can be calculated using the formula q = mCΔT, where q is the heat absorbed by the water, m is the mass of the water, C is the specific heat of water (4.18 J/g°C), and ΔT is the change in temperature of the water. Rearranging this formula, we get C = q/(mΔT). Plugging in the values given, we get C = (46.20 g)(4.18 J/g°C)(36.25 °C - 28.45 °C)/(89.7716 g)(99.8 °C - 36.25 °C) = 0.264 J/g°C. Therefore, the specific heat of the metal is 0.264 J/g°C.

b) The Atomic Weight of the metal can be calculated using the rule of Dulong and Petit, which states that the atomic weight of an element is equal to the product of its specific heat and its atomic mass. In this case, the specific heat (0.264 J/g°C) and atomic mass are given, so we can calculate the atomic weight by dividing the specific heat by the atomic mass. Therefore, the atomic weight of the metal is 0.264 J/g°C divided by 6 g/mol = 0.044 g/J.

c) The Heat of Solution can be calculated using the formula ΔH = q/m, where q is the heat absorbed by the water and m is the mass of the salt added. In this case, the heat absorbed by the water can be calculated using the formula q = mcΔT, where m is the mass of the water, c is the specific heat of water (4.18 J/g°C), and ΔT is the change in temperature of the water. Plugging in the values given, we get q = (48.80 g)(4.18 J/g°C)(31.45°C - 28.85°C) = 439.16 J. Therefore, the Heat of Solution is ΔH = 439.16 J/7.0545 g = -62.18 J/g.

d) The Heat of Formation of the Solution can be calculated using the formula ΔH = ΔHf° + n