C111 FA 2010 Homework 1 Solution Guide and Answer Key
1.
The calculation to perform:
(
29
x
3
1905
.
0
3
.
1
+
=
(
29
3
178
.
0
09
.
7
simplify numerator on the left side,
and cube the 0.178 term on the right side:
1.4905
=
7.09
Do not round off until the end of the calculation.
3x
0.005639752
But remember the 1.4905 only has one decimal pl.
which means a datum with
only 2 sig figs
multiply both sides by x and by 0.005639752; divide both sides by 7.09 to isolate the “x”:
(0.005639752)(1.491)
=
x
(3)(7.09)
x = 3.95339 x 10

4
(report 2
sig figs because limit is set by the original 1.3 term with only 2 sig figs.)
=
0.00040
(standard notation)
or
4.0 x 10

4
(scientific notation )
2.
atom
g
10
x
271
.
3
atoms
10
x
022
.
6
mole
1
x
mole
g
967
.
196
22
23

=
moles in the denominator of first term cancel moles in the numerator of second term.
Multiply by 5 atoms, to get 1.635 x 10

21
g.
(4 sig figs)
(Notice 5 is an integer number of objects and does not limit sig. figs.).
3. With only 4 sig figs in the density data, the answer may have a maximum of 4 sig figs.
Assume only
platinum (Pt) present in the sample, and
the purity of the sample is 100%, so this density can be used.
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 Chemistry, Chemical substance, Chemical compound, Sig Figs, $0.82, $ 540

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