## Alberta Free Tutoring And Homework Help For Science 10

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### how do you balance equations?

5 months ago Assuming this is for chemistry

Say you have an equation like this:

(W)C6H12O6 + (X)O2--> (Y)CO+ (Z)H2O

You need to figure out how many of each molecule are needed to make this equal. W is your number of glucose molecules (that's the C6H12O6), X is your number of oxygen molecules, Y is the number of carbon dioxide molecules, and Z is the number of water molecules. You can see that carbon, hydrogen, and oxygen are all on both sides of the equation- so now you have to change W,X  Y, and Z so that there are the same number of each type of molecule on both sides.

Now because you automatically start with 6 carbons in a glucose molecules, you know you'll need at least 6 molecules of carbon dioxide- same logic for the hydrogens in glucose and water. Assuming that W is 1, then Y and Z must be 6. Next you count up your oxygens, I like to start on the right for this. If Y is 6, and there are two oxygens per CO2, then we have 12 oxygens from that plus an extra 6 from the water. That is a total of 18 oxygen molecules on the right side. On the left side there are currently 6 from the glucose, meaning that there must be 6 molecules of O2  and X = 6. Your coefficients in order are 1, 6, 6, 6. Don't forget to always double check your math with these balancing questions!

5 months ago My favourite way of doing it is drawing a table underneath the equation. This table should be elements or compounds that don't change (so SO42- would be considered its own element if it occurred on both sides of the equation).

C3H8 + O2 -------> CO2 + H2O

C = 3                  |    C = 1

H = 8                  |    H = 2

O = 2                  |    O = 3

Pick one of the elements to start balancing from (so let's do carbon first), and adjust any elements also attached.

C3H8 + O2 -------> 3CO2 + H2O

C = 3                  |    C = 1 x3 => 3

H = 8                  |    H = 2

O = 2                  |    O = 3 +2x2 => 7

Choose another element to balance (I chose hydrogen)

C3H8 + O2 --------> 3CO2 + 4H2O

C = 3                  |    C = 3

H = 8                  |    H = 2 x 4 => 8

O = 2                  |    O = 7 + 3 => 10

Balance out the leftover element

C3H8 + 5O2 ------> 3CO2 + 4H2O

C = 3                  |    C = 3

H = 8                  |    H = 8

O = 2 x5 =>10    |    O = 10

5 months ago I forgot to mention: there are a BUNCH of websites online for practice balancing equations. My personal favourite is Khan Academy

5 months ago Final Energy = Initial Energy + Work - Energy Loss

eg: if I have a 30kg sled at the top of a 10m tall hill, add 10kJ of energy by pushing it, and it loses about 5kJ of energy through friction, how fast does it go at the bottom of the hill?

Final Energy = kinetic energy = 0.5mv2

Work = 10,000 J, Energy Loss = 5000 J

Initial Energy = potential energy = mgh

m = 30kg

h = 10m

0.5(30kg)v2 = 30kg(9.81m/s2)(10m) + 10,000 J - 5,000 J

0.5(30)v2 = 7,943 J

Divide both sides by (0.5*30)

v2 = 529.53 J

Square Root both sides

v = 23 m/s (really fast for a sled)!