일반 화학 자유 5판 대학교재솔루션 등록

일반 화학 자유 5판 대학교재솔루션 등록

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일반 화학 자유 5판 대학교재솔루션

일반화학 자유 5판 챕터3

chapter 3
stoichiometry

3.5 (34.968 amu)(0.7553) ? (36.956 amu)(0.2447) ? 35.45 amu

3.6 Strategy: Each isotope contributes to the average atomic mass based on its relative abundance. Multiplying the mass of an isotope by its fractional abundance (not percent) will give the contribution to the average atomic mass of that particular isotope.

It would seem that there are two unknowns in this problem, the fractional abundance of 6Li and the fractional abundance of 7Li. However, these two quantities are not independent of each other; they are related by the fact that they must sum to 1. Start by letting x be the fractional abundance of 6Li. Since the sum of the two abundance’s must be 1, we can write

Abundance 7Li ? (1 ( x)

Solution:

Average atomic mass of Li ? 6.941 amu ? x(6.0151 amu) ? (1 ( x)(7.0160 amu)
6.941 ? 1.0009x ? 7.0160
1.0009x ? 0.075
x ? 0.075

x ? 0.075 corresponds to a natural abundance of 6Li of 7.5 percent. The natural a

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[문서정보]

문서분량 : 55 Page
파일종류 : DOC 파일
자료제목 : 일반 화학 자유 5판 대학교재솔루션
파일이름 : Chapter3.doc
키워드 : 일반화학,자유,chapter,일반,화학,5판,대학교재솔루션
자료No(pk) : 11005758