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Get Full Access to Chemistry: The Central Science - 12 Edition - Chapter 10 - Problem 8e
Get Full Access to Chemistry: The Central Science - 12 Edition - Chapter 10 - Problem 8e

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# Visualizing ConceptsOn a single plot, qualitatively sketch

ISBN: 9780321696724 27

## Solution for problem 8E Chapter 10

Chemistry: The Central Science | 12th Edition

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Problem 8E

Problem 8E

Visualizing Concepts

On a single plot, qualitatively sketch the distribution of molecular speeds for (a) Kr(g) at -50 °C, (b) Kr(g) at 0 °C, (c) Ar(g) at 0 °C. [Section]

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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)) = 12,010 amu = 12.01 x 10 amu Conversion factor: 1 Carbon atom 12.01 amu 20 19 3.00 x 10 amu x 1 Carbon atom = 2.50 x 10 Carbon atoms 12.01 amu Ex #2: Mass of 75 Al atoms Average Mass for 1 Al atom: 26.98 amu Equivalence Statement: 1 Al atom = 26.98 amu 75 Al atoms x 26.98 amu = 2024 amu 1 Al atom Pg 172. THE MOLE *”samples in which the ratio of the masses is the same as the ratio of the masses of the individual atoms always contain the same number of ratios” Mole: the quantity of anything that has the same number of particles Avogadro’s Number: the number of “elementary” particles (molecules, atoms, compounds, etc.) per mole of a substance 

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