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CHAPTER 3
MASS RELATIONSHIPS IN
CHEMICAL REACTIONS
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 itsrelative 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 thatthere 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
factthat 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 massof 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 naturalabundance of 7Li is
(1 − x) = 0.925 or 92.5 percent.
3.7
6.022 1023 amu The unit factor required is
1 g
⎛ × ⎞
⎜⎜ ⎟⎟
⎝ ⎠
23
13.2 amu 1 g
6.022 10 amu
= × =
×
? g 2.19 × 10−23 g
3.8
6.022 1023amu The unit factor required is
1 g
⎛ × ⎞
⎜⎜ ⎟⎟
⎝ ⎠
6.022 1023 amu 8.4 g =
1 g
×
? amu = × 5.1 × 1024 amu
3.11 In one year:
(6.5 109 people) 365 days 24 hr 3600 s 2 particles 4.1 1017particles/yr
1 yr 1 day 1 hr 1 person
× × × × × = ×
23
17
6.022 10 particles
4.1 10 particles/yr
×
= =
×
Total time 1.5 × 106 yr
CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
36
3.12 Thethickness of the book in miles would be:
0.0036 in 1 ft 1 mi (6.022 1023 pages) = 3.42 1016 mi
1 page 12 in 5280 ft
× × × × ×
The distance, in miles, traveled by light in one year is:
8
1.00 yr365 day 24 h 3600 s 3.00 10 m 1 mi 5.88 1012 mi
1 yr 1 day 1 h 1 s 1609 m
×
× × × × × = ×
The thickness of the book in light-years is:
16
12
(3.42 10 mi) 1 light-yr
5.88 10 mi
× × =
×
5.8 ×...
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