llowing calculations. Round each result so that it contains only significant digits. The mbers in parentheses are absolute standard deviations. 2 (1.203 (+0.004) x 103 + 18.25 (±0.005) x 102 – 34.012(±0.001)\ y = 3.9201 (+0.0006) x 10-3

llowing calculations. Round each result so that it contains only significant digits. The mbers in parentheses are absolute standard deviations. 2 (1.203 (+0.004) x 103 + 18.25 (±0.005) x 102 – 34.012(±0.001)\ y = 3.9201 (+0.0006) x 10-3

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

2. Estimate the absolute deviation and the coefficient of variation for the results of the
following calculations. Round each result so that it contains only significant digits. The
numbers in parentheses are absolute standard deviations.
2
(1.203 (+0.004) x 103 + 18.25 (10.005) x 102 – 34.012(±0.001))
3.9201 (+0.0006) x 10-3
y =

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