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21 July, 23:06

32. A solution suspected of containing copper is analyzed. To determine the standard curve, the absorbance of five standards ranging from 0.125 mg/L Cu to 1.25 mg/L Cu was measured and graphed against the concentration. The slope of the linear best-fit line was determined to be 0.6136 and the y-intercept was 0.0142. If three samples of the unknown solution have absorbance values of 0.438, 0.434, and 0.439 what is the concentration of the unknown

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  1. 21 July, 23:33
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    0.689 mg/L Cu

    Explanation:

    The equation of the best-fit line is:

    y = 0.6136x + 0.0142

    Where y is the absorbance and x is the concentration in mg/L.

    Now using that equation we calculate x when y is equal to 0.438, 0.434 & 0.439.

    1) 0.438 = 0.6136x + 0.0142

    x = 0.691 mg/L

    2) 0.434 = 0.6136x + 0.0142

    x = 0.684 mg/L

    3) 0.439 = 0.6136x + 0.0142

    x = 0.692 mg/L

    Finally we calculate the mean of the three concentrations:

    (0.691 + 0.684 + 0.692) / 3 = 0.689 mg/L
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