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Copper Lab Conclusion

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E. Discussion: In part one of the Lab, our goal was to find the density of three pieces of copper. The density of the first piece of copper was approximately 8.2 g/mL. The density of the second piece of copper was approximately 10.6 g/mL and the density of the third piece of copper was also 8.2 g/mL. After finding the density, we used the standard deviation formula, σ = 1N i+1N(xi- )2, to tell how much the measurements, from the dataset, are spread out from the expected value. After calculating the standard deviation of our dataset, our results were =8.1. Our results were very accurate, being that our percent difference was only 0.45%. In part two of the lab, we were trying to identify and find the density of material “Unknown B.” We found …show more content…
9 g/ml-8.96 g/cm3 8.96 g/cm3x 100= 0.45%
What is the reason for this difference? The reason for this slight difference can be a systematic error from when we were reading the volume measurements off of the graduated cylinder from before we put the copper in and after we put the copper in. We each had a slight disagreement on what the measurement was.
What would you have to do in order to increase the accuracy of this experiment, i.e., decrease that standard deviation? In order to increase the accuracy of this experiment, we can try repeating the experiment a few more times.
Suppose you performed this experiment, and the percent difference from the accepted value of density of copper and your value was very large, let’s say 300%. However, the standard deviation of your dataset was tiny; let’s say 0.01 g/cm3. What conclusions can you draw about this discrepancy? We can conclude that there was an outlier that caused the difference between percent difference (300%) and the standard deviation of the dataset (0.01 g/cm3). We can conclude that errors were probably made during the process of finding the volume of the copper and finding the mass of the

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