Suppose you had a 2.4 g cracker burn down to 1.3 g, which raised the temperature of 50.1 g of water by 12.0 degrees Celsius. How many kilocalories of heat energy was released by the cracker (or absorbed by the water) per gram of cracker

We know that the specific heat of water is:

c = 1cal/g*°C

This means, that we need 1 cal to increase the temperature of 1 gram of water by 1°C

Here, we increased the temperature of 50.1g of water by 12°C

Then the number of calories needed to do this is given by:

x = (mass of water in grams)*(how much increased the temp in °C)*1cal/g*°C

x = (50.1g*12°C)*1cal/g*°C = 601.2 cal

But we want this in Kcal, remember that:

1Kcal = 1000cal

Then:

601.2 cal = (601.2/1000) Kcal = 0.6012 Kcal

Now, for the cracker part, the energy was released by the amount of cracker that was burned.

The original mass was 2.4g

the final mass ios 1.3g

the difference is:

2.4g - 1.3g = 1.1g

This means that 1.1g was the burned mass.

The number of kilocalories of heat per gram released by the cracker is just:

n = (0.6012 Kcal)/(1.1 g) = 0.547 Kcal/g

0.547 kilocalories per gram.

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explanation:

"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"