Student has calibrated his/her calorimeter and finds the heat capacity to be 14.2 j/°c. s/he then determines the molar heat capacity of aluminum. the data are: 25.5 g al at 100.0°c are put into the calorimeter, which contains 99.0 g h2o at 18.6°c. the final temperature comes to 22.7°c. calculate the heat capacity of al in j/mol·°c.

24.03 J/mol.ºC

Explanation:

For a calorimeter, the heat lost must be equal to the heat gained from water plus the heat gained from calorimeter, which has the same initial temperature as the water.

-Qal = Qw + Qc (minus signal represents that the heat is lost)

-mal*Cal*ΔTal = mw*Cw*ΔTw + Cc*ΔTc

Where m is the mass, C is the specific heat, ΔT is the temperature variation, al is from aluminum. w from water and c from the calorimeter. Cw = 4.186 J/gºC

-25.5*Cal*(22.7 - 100) = 99.0*4.186*(22.7 - 18.6) + 14.2*(22.7 - 18.6)

1971.15Cal = 1699.10 + 58.22

1971.15Cal = 1757.32

Cal = 0.89 J/g.ºC

The molar mass of Al is 27 g/mol

Cal = 0.89 J/g.ºC * 27 g/mol

Cal = 24.03 J/mol.ºC

the number of protons in the nucleus of a given element. has slight luster and conducts electricity at high temperatures like metal.but like a nonmetal,boron is brittle and does not conduct electricity well at low temperatures. in graphite, carbon is a soft,brittle solid.it is reflective,ductile,or malleable.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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