Monday, March 26, 2012


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Obtaining evidence

During the investigation I did the experiment three times with each set of pieces, my results are shown below -

Pieces of chips cut from 5cm No. of oxygen produced in mins (cm) Average (cm)

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Whole 4. .6 .6 .7

Half 5 5.5 5.4 5.

Quarter 11. 4.7 5.1 7

Eighth 11.5 11.7 10.6 11.

Sixteenth .7 16. 16.4 14.1

My results show me that the bigger the surface area then the more oxygen produced. The result which I have highlighted is different to the other two results in the column because that experiment was done on a completely different day, because the temperature changed so radically the result has been affected by a large margin.

Analysing and considering evidence

To analyse my evidence I will need to construct a bar chart to show how the number of oxygen produced varies as I cut my 5cm chip into smaller pieces, this chart is shown on the following page.

Using my scientific knowledge I can construct a line graph to show how the change in surface area causes more oxygen to be produced. To work out the surface area I will need to do the following sums-

For the surface area of a whole chip = 5x4+=

For the surface area of half the chip = +=4

For the surface area of a quarter chip = 4+4=8

For the surface area of an eighth of a chip = 8+8=6

For the surface area of a sixteenth of a chip = 6+16=5

I worked out the whole surface area by time sing the length of my potato chip by the amount of rectangular sides it had, then I added the other two sides to the answer. For the surface area for the other sizes of chips I just added the previous answer by the number of pieces that the side had.

The line graph will follow the bar chart.


When I have drawn the line of best fit in my graph, I found out that as the surface areas of my chips increase, the volume of oxygen collected in minutes has also increased. This is because when the chips have more surface areas then there will be more areas for the hydrogen peroxide to spread into the potato, so more catalse enzymes are working as the same time, therefore the faster the rate of reaction.

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