Medium2 marksStructured
AQA GCSE · Question 07.3 · The rate and extent of chemical change
Table 5 shows the student's results for the experiment with hydrochloric acid of a lower concentration. Calculate mean value X in Table 5. Do not include the anomalous result in your calculation. Give your answer to 2 significant figures.
Table 5 shows the student's results for the experiment with hydrochloric acid of a lower concentration. Calculate mean value X in Table 5. Do not include the anomalous result in your calculation. Give your answer to 2 significant figures.
How to approach this question
1. Look at the results for time = 40 seconds: 46, 30, 47, 49.
2. Identify the anomalous result. An anomaly is a result that does not fit the pattern. 46, 47, and 49 are all close together, while 30 is much lower. So, 30 is the anomaly.
3. Calculate the mean of the other three results (46, 47, 49). Add them together and divide by the number of results (3).
4. (46 + 47 + 49) / 3 = 142 / 3 = 47.333...
5. Round your answer to 2 significant figures.
Full Answer
Anomalous result is 30.
Mean X = (46 + 47 + 49) / 3
Mean X = 142 / 3
Mean X = 47.333...
To 2 significant figures, X = 47 cm³
To calculate the mean value X, we first need to identify the anomalous result in the row for 40 seconds. The values are 46, 30, 47, and 49. The value 30 is significantly different from the other three, so it is the anomaly and should be excluded from the calculation.
We then calculate the mean of the remaining (concordant) results:
Mean = (46 + 47 + 49) / 3
Mean = 142 / 3
Mean = 47.333... cm³
The question asks for the answer to 2 significant figures. The first two significant figures are 4 and 7. The next digit is 3, which is less than 5, so we round down.
Mean value X = 47 cm³.
Common mistakes
✗ Including the anomalous result in the calculation.
✗ Dividing by 4 instead of 3 after excluding the anomaly.
✗ Incorrectly rounding to 2 significant figures.
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