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 ozone layer acts as a filter for the shorter wavelength and highly hazardous ultraviolet radiation (uvr) from the sun, protecting life on earth from its potentially harmful effects. when the sky is clear, there is an inverse relationship between stratospheric ozone and solar uvr measured at the earth's surface.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :