Using the formula (Q =MCvT), perform the calculations below. 3.100g of water at 4*C is heated to 29*C. If the specific heat of water is 4.18j/g*C.
Calculate the amount of energy needed. Hint find the change in temperature before solve!
Q= M c v T. m= mass (g), c = heat capacity, v T = change in temp (T f-Ti) .
Q= (100).(4.18).(29*C- 4.0*C)

"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"

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