The document provides instructions on how to properly complete different sections of a data collection and processing (DCP) practical experiment for chemistry. It emphasizes including raw data, units, uncertainties, calculations, assumptions, and qualitative observations. Completing the sections with these details can result in a higher mark than sections missing any of this essential information.

1. How to get 6 in DCP
Mr.T’s powerpoint for cheaters in chemistry

2. DCP1 presenting data
Trial Diameter of circle (cm)
1 17.10
2 23.60
3 16.20
4 21.30
NOT AT ALL
Volume of oil drop = 0.01cm 3
•What is missing?
•Are there any mistakes?
•What mark would this section be awarded?

3. DCP1 presenting data
Trial number Diameter of circle (cm) ±0.05cm
1 17.1
2 23.6
3 16.2
4 21.3
109 drops in 1cm3 ± 0.05cm3 COMPLETE!
Qualitative data:
When the oil drop made contact with the surface of the water it
immediately spread out into a distorted circle. The lycopodium
powder was pushed back by the spreading oil.

4. Summary
• Remember units
• Include uncertainties
• Include ALL raw data
• Include qualitative data

5. DCP2 processing data
Trial 1 diameter = 17.1
Area = 918
Diameter = 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. 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 10.65
Average radius
(8.55 + 11.53 + 8.1 + 10.65) / 4 = 9.71cm ± 0.05cm
109 drops in 1cm3 ± 0.05cm3
Therefore 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. DCP2 Processing data
Average radius = 9.71cm
Average volume of 1 drop = 1/109 = 0.00917cm3
Area of oil layer = πr2 = π x 9.712 = 264.17cm2
Height of cylinder = Volume of cylinder / Area
Volume of oil layer = volume of oil drop
Height of oil layer = 0.00917cm3 / 264.17cm2 = 3.47 x 10-5 cm
Assuming the oil layer is 1 molecule in height, and each molecule contains 12 carbon
Atoms, diameter of a carbon atom = 3.47 x 10-5 cm / 12 = 2.89 x 10-6 cm
COMPLETE!

8. Summary
• Show all steps in calculations
• Convert units where necessary
• Show units throughout
• State equations and define terms
• Explain assumptions

9. DCP3 Presenting data
Average 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 / Area
Volume of oil layer = volume of oil drop
Height 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 carbon
Atoms, diameter of a carbon atom = 3.47 x 10-5 cm / 12 = 2.89 x 10-6cm ± 6.10%
COMPLETE!

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