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What affect does the surface area of a beet have on the rate of osmosis taking place through an intact cell membrane?

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Introduction

Question: What affect does the surface area of a beet have on the rate of osmosis taking place through an intact cell membrane?

Materials:

Red Beet (half a beet)

Scalpel

Beaker (4)

Thermometer

Water

Electronic Balance

Paper Towel

Stop watch

Stirring Rod

Procedure:

  1. Obtain 4 X 500 ml beakers from the front desk and rinse them thoroughly in high pressured water to ensure that there are no contaminants in the beaker
  2. Fill the beakers with 450 ml of water (at room temperature)
  3. Obtain a piece of freshly cut red beet and rinse it thoroughly to ensure that all the  Anthocyanin that has leaked out during cutting is washed away
  4. Measure the temperature of the water by inserting a thermometer into the beakers filled with water. Make sure that the temperatures are the same otherwise it will affect the rate of osmosis providing you with incorrect results
...read more.

Middle

Diphenylamine test

A drop or two of diphenylamine reagent on the cut surface of the root will result in a blue color if nitrate is present. This easy test can be made in the field and is useful in diagnosing possible causes for the yellowing of sugar beet leaves. If the reagent remains colorless, this indicates the yellowing is due to nitrogen deficiency. If nitrate is present, then the yellowing is due to other causes, sulfur deficiency, for instance.

A source of error would be the judgment of color intensities. A solution to this problem would be to use a colorimeter. Another source of error would be due to the physical damage done to the beetroot cells when handling them. The solution to this would be to handle the beetroot with care.

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Conclusion

d_measures/fencing_problem/32459/html/images/image05.png" style="width:75px;height:50px;margin-left:0px;margin-top:0px;" alt="image05.png" />

parallelogram = bh image06.png

trapezoid = h/2 (b1 + b2) image07.png

circle = pi r 2image03.png

ellipse = pi r1 r2image08.png

triangle = (1/2) b h image09.png

equilateral triangle = (1/4)image02.png(3) a 2

triangle given SAS = (1/2) a b sin C

triangle given a,b,c = image02.png[s(s-a)(s-b)(s-c)] when s = (a+b+c)/2 (Heron's formula)

regular polygon = (1/2) n sin(360°/n) S2
   when n = # of sides and S = length from center to a corner

Source: math.com

...read more.

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