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5 April, 15:37

A certain analytical method for the determination of lead yields masses for lead that are low by 0.5 g. Calculate the percent relative error caused by this deviation for each measured mass of lead. Report the percent relative error with the correct number of significant figures.

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  1. 5 April, 18:12
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    Question Continuation

    if the measured weight of lead in the sample is

    a.) 764.9g lead

    b.) 226.3g lead

    c.) 53.5g lead

    Answer:

    a.

    Relative Error = 0.065

    b.

    Relative Error = 0.221

    c.

    Relative Error = 0.935

    Step-by-step explanation:

    Given

    Absolute Error = 0.5g

    Relative error = absolute error/magnitude of measurement.

    Relative error % = Relative error * 100

    a.

    Relative Error = 0.5/764.9 * 100

    Relative Error = 50/764.9

    Relative Error = 0.065

    b.

    Relative Error = 0.5/226.3 * 100

    Relative Error = 50/226.3

    Relative Error = 0.221

    c.

    Relative Error = 0.5/53.5 * 100

    Relative Error = 50/53.5

    Relative Error = 0.935
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