Essay
Gradient, formation of stearic acid, formation of paraffin wax, evaluation and how to improve.
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- PEARSON (PEARSON)
All are distinction level.
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GRADIENTPARAFFIN WAX
Cooling at its cooling point is:
M = 2/3 (0.6666…)
Gradient at its melting point:
M=0
Gradient at the end:
M = 1/15 (0.0666…)
STEARIC ACID
Cooling it its cooling point:
M = -17/42 (-0.4047619048)
Gradient at its melting point:
M = 40/-7 (-5.714285714)
Gradient at the end:
M = -0.074
COMPARE
Comparing the graph that is determined, results were compared to other students to show if
they are accurate. Student A [3] calculated their melting point for stearic acid as 45°C,
comparing it to this, melting point for stearic acid 58°C.This could either be a cause of an
additional heat that causes the level of temperature to increase. Student A melting point is
much earlier, indicating that the substance’s (stearic) particles move closely to an arranged
pattern. Furthermore Student A’s melting point for paraffin wax is 44°C while in this
experiment it's 45°C, the closeness of melting points in both experiments portrays that the
level of accuracy is on point. Comparing this experiment with 1 other student could increase
the level of proximity of accuracy so Student B [4] melting point for stearic acid is 45°C while
melting point for paraffin is 45°C, hence melting point for paraffin wax is closely accurate
while in this experiment melting point for stearic is a human error.
Moreover the percentage difference for student A is 64.9%[S] and 4.34%[P] this is more
precise due to the % difference for the substances as a whole is 5% and for student B 35%
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