In the beaker, add a 25.0-g sample of aluminum at 180°c, and in the calorimeter, add 110-g of water at 20.0°c. select “show graph view” and “show microscopic view,” and then click “start” to run the reaction. (use the simulation to obtain the final temperature of the solution.) calculate the change in heat of the aluminum; show all calculations. calculate the change in heat of the water; show all calculations. are the two values the same? why or why not?

Mass of methane takne = 1.5g

moles of methane used = masss / molar mass = 1.5 / 16 = 0.094 moles

mass of water = 1000 g

Initial temperature of water = 25 C

final temperature = 37 C

specific heat of water = 4.184 J /g C

1) Heat absorbed by water = q =m• C• ΔT = 1000 X 4.184 x (37-25) =  50208 Joules

2) Heat absorbed by calorimeter = Heat capacity X ΔT = 695 X (37-25) = 8340 J

3) Total heat of combustion = heat absorbed by water + calorimeter = 50208 + 8340 = 58548 Joules

This heat is released by 0.094 moles of methane

So heat released by one mole of methane =

                           - 622851.06 Joules = 622.85 kJ / mole

4) standard enthalpy of combustion = -882 kJ / mole

Error = (882-622.85) X 100 / 882 = 24.84 %

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

he made 11 calls in one day