Thursday, February 10, 2011

Thursday February 10th

We continued discussion of the topic from the end of class yesterday. About how the number of atoms of each type must equal out on each side of the equation.
For example: H2+O2 yields H2O had to be changed to 2H2+O2 yields 2H2O
The three methods we learned to solve these types of problems are using pictures, writing the numbers of each element, and using ratios. The first two are the easiest methods to use.
I cannot create the picture method here but ill show the other ones.

H2+O2 yields H2o start write the number of each type of atom for each side
H=2 H=2 2. you can see that there is a different number of oxygens on
O=2 O=1 each side. To correct this you need to add more molecules.
H2+O2 yields 2H2O 3. Now the oxygens are equal but the hydrogens are different. To
H=2 H=4 correct this you need another Hydrogen molecule.
O=2 O=2
2H2+O2 yields 2H2O 4. This is you final answer.

A major part of this is that you cannot have fractions such as H1/2 because half of H would become a different element.

All of this has to do with conservation of mass. There must be and equal number of each type on each side otherwise something has gone wrong. Imagine you have just finished chocolate chip cookies. You put them in the oven and when you come back later the chocolate chips are gone. You absolutely know that you put the chocolate chips in the cookies so you start to freak out because this is impossible.
That is basically what we are dealing with when we do these problems. The mass must stay the same. Nothing can just vanish.

Tomorrow we are starting to work with Bunsen burners. We will deal with the type of chemical reaction called decomposition.


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