Analytical Chem istry - DePauw University

38 Analytical Chemistry 2.02KSolutions to Practice ExercisesPractice Exercise 2.1The correct answer to this exercise is 1.9×10 –2 . To see why this is correct,let’s work through the problem in a series of steps. Here is the originalproblem−3 −20. 250× ( 993 . × 10 ) − 0. 100× ( 1. 927×10 )=−3 −2993 . × 10 + 1.927×10Following the correct order of operations we first complete the two multiplicationsin the numerator. In each case the answer has three significantfigures, although we retain an extra digit, highlight in red, to avoid roundofferrors.−248 . 2× 10 − 1.927×103 −3−993 . × 10 + 1.927×103 −2=Completing the subtraction in the numerator leaves us with two significantfigures since the last significant digit for each value is in the hundredthsplace.055 . 5×10−3−993 . × 10 + 1.927×103 −2=The two values in the denominator have different exponents. Because weare adding together these values, we first rewrite them using a commonexponent.055 . 5×10−3−0. 993× 10 + 1.927×102 −2=The sum in the denominator has four significant figures since each valuehas three decimal places.055 . 5×102.920×10−3−2=Finally, we complete the division, which leaves us with a result having twosignificant figures.055 . 5×102.920×10−3−2= 19 . × 10−2Click here to return to the chapter.Practice Exercise 2.2The concentrations of the two solutions are