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    PracticeAQA GCSEAQA GCSE Chemistry Foundation Tier Paper 2Question 07.3
    Medium2 marksStructured
    The rate and extent of chemical changeFoundationcalculationmeananomalies

    AQA GCSE · Question 07.3 · The rate and extent of chemical change

    Table 5
    Time in seconds Volume of gas collected in cm³
    Test 1 Test 2 Test 3 Test 4 Mean
    000000
    4046304749X
    807883838282
    1209894969596
    160100100100100100

    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

    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.
    Question 07.2All questionsQuestion 07.4

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