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.

*tilts head* sounds more like a statement

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.