Calculate the amount of energy that is absorbed by 75 g.. water
from 22.5 °c to 37.8 °c. the specific heat of water is 1 cal/ g x °c

Q = 1147.5 cal

Explanation:

Given data:

Energy absorbed = ?

Mass of water = 75 g

Initial temperature = 22.5°C

Final temperature = 37.8°C

Specific heat of water = 1 cal/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT  = T2- T1

ΔT  = 37.8°C - 22.5°C

ΔT  = 15.3 °C

Q = m.c. ΔT

Q = 75 g . 1 cal/g.°C . 15.3 °C

Q = 1147.5 cal

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"