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# Radius of a carbon atom dcp practice

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### Radius of a carbon atom dcp practice

1. 1. How to get 6 in DCPMr.T’s powerpoint for cheaters in chemistry
2. 2. DCP1 presenting data Trial Diameter of circle (cm) 1 17.10 2 23.60 3 16.20 4 21.30 NOT AT ALLVolume of oil drop = 0.01cm 3 •What is missing? •Are there any mistakes? •What mark would this section be awarded?
3. 3. DCP1 presenting data Trial number Diameter of circle (cm) ±0.05cm 1 17.1 2 23.6 3 16.2 4 21.3109 drops in 1cm3 ± 0.05cm3 COMPLETE!Qualitative data:When the oil drop made contact with the surface of the water itimmediately spread out into a distorted circle. The lycopodiumpowder was pushed back by the spreading oil.
4. 4. Summary• Remember units• Include uncertainties• Include ALL raw data• Include qualitative data
5. 5. DCP2 processing dataTrial 1 diameter = 17.1Area = 918Diameter = V/A = 918 / 0.00917 = 0.0000404cm •What is missing? •Are there any mistakes? •What mark would this section be awarded? Not at all
6. 6. DCP2 processing data Diameter of circle (cm) ±0.1cm Radius of circle (cm) ±0.05cm 17.1 8.55 23.6 11.53 16.2 8.10 21.3 10.65Average radius(8.55 + 11.53 + 8.1 + 10.65) / 4 = 9.71cm ± 0.05cm109 drops in 1cm3 ± 0.05cm3Therefore average volume of 1 drop = 1/109 = 0.00917cm 3% uncertainty volume = (absolute uncertainty / volume measured) x 100= (+/-0.05cm3 / 1cm3) x 100 = 5%
7. 7. DCP2 Processing dataAverage radius = 9.71cmAverage volume of 1 drop = 1/109 = 0.00917cm3Area of oil layer = πr2 = π x 9.712 = 264.17cm2Height of cylinder = Volume of cylinder / AreaVolume of oil layer = volume of oil dropHeight of oil layer = 0.00917cm3 / 264.17cm2 = 3.47 x 10-5 cmAssuming the oil layer is 1 molecule in height, and each molecule contains 12 carbonAtoms, diameter of a carbon atom = 3.47 x 10-5 cm / 12 = 2.89 x 10-6 cm COMPLETE!
8. 8. Summary• Show all steps in calculations• Convert units where necessary• Show units throughout• State equations and define terms• Explain assumptions
9. 9. DCP3 Presenting dataAverage radius = 9.71cm ± 0.05cm%uncertainty = (0.05cm/9.17cm) x 100 = 0.55%Average volume of 1 drop = 1/109 = 0.00917cm3 ± 4.59 x 10-4cm3%uncertainty = 5%Area of oil layer = πr2 = π x 9.712 = 264.17cm2 0.55% + 0.55% = 1.10%Height of cylinder = Volume of cylinder / AreaVolume of oil layer = volume of oil dropHeight of oil layer = 0.00917cm3 / 264.17cm2 = 3.47 x 10-5 cm 1.10% + 5.00% = 6.10%Assuming the oil layer is 1 molecule in height, and each molecule contains 12 carbonAtoms, diameter of a carbon atom = 3.47 x 10-5 cm / 12 = 2.89 x 10-6cm ± 6.10% COMPLETE!
10. 10. Summary EVERYONE misses uncertainties for a COMPLETE• Absolute uncertainties do not change in the average• Convert to % uncertainty after calculating average• If data is multiplied or divided % uncertainties add• Quote percentage uncertainty with final answer