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The Periodic Table, Molecules and Molecular | Ch 2 - 42E

Chemistry: The Central Science | 12th Edition | ISBN: 9780321696724 | Authors: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward ISBN: 9780321696724 27

Solution for problem 42E Chapter 2

Chemistry: The Central Science | 12th Edition

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Chemistry: The Central Science | 12th Edition | ISBN: 9780321696724 | Authors: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward

Chemistry: The Central Science | 12th Edition

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Problem 42E

The Periodic Table, Molecules and Molecular Compounds, and Ions and Ionic Compounds (Sections)Two compounds have the same empirical formula. One substance is a gas, whereas the other is a viscous liquid. How is it possible for two substances with the same empirical formula to have markedly different properties?

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th th Chapter 8 from “Introductory Chemistry” by Zumdahl and Decoste, 7 /8 edition Pg 167. COUNTING BY WEIGHING Average mass = total mass Number of item/variables Pg 170. ATOMIC MASSES: COUNTING ATOMS BY WEIGHING Want to know how many molecules are needed to make carbon dioxide C (s) + O2(g)  CO 2g) 1 atom + 1 molecule  1 molecule *because atoms weigh so little and kilograms would be too large of a measurement, chemists use Atomic Mass Unit (amu) In terms of grams: 1 amu = 1.66 x 10 -24g Average Atomic Mass: (means what its named…duh) Ex: Mass of 1000 natural C atoms = (1000 atoms) (12.01 (amu/atom))

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Chapter 2, Problem 42E is Solved
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Textbook: Chemistry: The Central Science
Edition: 12
Author: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward
ISBN: 9780321696724

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The Periodic Table, Molecules and Molecular | Ch 2 - 42E