After rounding, the answer to equation B should have

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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equation B:
(6.022 x 1023 atoms/mol) (0.541 g)
20.18 g/mol
After rounding, the answer to equation B should have
O3 significant figures.
O4 significant figures.
2 significant figures.
5 significant figures.
1 significant figure.
Incorrect
= 1.614 x 1022 atoms (unrounded)
Transcribed Image Text:equation B: (6.022 x 1023 atoms/mol) (0.541 g) 20.18 g/mol After rounding, the answer to equation B should have O3 significant figures. O4 significant figures. 2 significant figures. 5 significant figures. 1 significant figure. Incorrect = 1.614 x 1022 atoms (unrounded)
To how many significant figures should each answer be rounded?
(6.626 × 10-34 J·s) (2.9979 x 108 m/s)
4.830 x 10-7 m
equation A:
After rounding, the answer to equation A should have
O4 significant figures.
O 5 significant figures.
O 3 significant figures.
O2 significant figures.
O 1 significant figure.
equation B:
(6.022 x 1023 atoms/mol) (0.541 g)
20.18 g/mol
= 4.112647080745 × 10-19 J (unrounded)
= 1.614 x 1022 atoms (unrounded)
Transcribed Image Text:To how many significant figures should each answer be rounded? (6.626 × 10-34 J·s) (2.9979 x 108 m/s) 4.830 x 10-7 m equation A: After rounding, the answer to equation A should have O4 significant figures. O 5 significant figures. O 3 significant figures. O2 significant figures. O 1 significant figure. equation B: (6.022 x 1023 atoms/mol) (0.541 g) 20.18 g/mol = 4.112647080745 × 10-19 J (unrounded) = 1.614 x 1022 atoms (unrounded)
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