Answer:
2 CH2 + 3 O2 = 2 CO2 + 2 H2O
Explanation:
This is what I think that you meant by the question listed. When balancing a chemical equation, you want to make sure that there are equal amounts of each element on each side.
Originally, the equation's elements looked like this: 1 C on left & 1 C on right; 2 H on left & 2 H on right; 2 O on left and 3 O on right. Because these are not balanced, you need to add coefficients.
When adding coefficients, you need to make sure that all of the elements stay balanced, not just one that you are trying to fix. I know that some equations are really difficult to balance, and when that is the case, there are equation balancing websites that can help out.
However, what always helps me is making a chart and continuing to keep up with the changes I am making. It is a trial and error process.
Answer:
Part A: 47.8 mi/h
Part B: 0.072 M/s
Part C: 0.144 M/s
Explanation:
Part A
The average speed or velocity (V) is the variation of the space divided by the variation of the time:
V = (241 - 2)/(8 -3)
V = 47.8 mi/h
Part B
As Part A, the average rate (r) of formation of I2 is the variation of the concentration divided by the variation of time:
r = (1.83 - 1.11)/(15 - 5)
r = 0.072 M/s
Part C
The rates of the substances are proportional of their number of moles (n) which are their coefficient, so:
rI2/nI2 = rHCl/nHCl
0.072/1 = rHCl/2
rHCl = 2*0.072
rHCl = 0.144 M/s
