Chemistry Reference
In-Depth Information
9.
[B. 1 dm
3
]—1 dm
3
represents 1000 ml, which is greater than answer choice
A; 1 dm
3
represents 1000 cm
3
, which is more than answer C; 1 dm
3
is
equivalent to 1 L, which is more than answer choice D.
10. [D. 450 g]—If we convert all of the answers into common units, we find that
1 g
100 cg
1 g
1000 mg
A.
2.6 cg ×
= 0.026 g
; B.
; and
124 mg ×
= 0.124 g
0.07 kg ×
1000 g
1 kg
C.
Answer D is more massive than the other choices.
= 70 g.
Lesson 2-3 Review
1.
[precision]
2.
[accurate]
3.
[An estimation digit is added to certain digits of each measurement in order
to report the measurement with maximum precision.]
4.
[1.13 cm]—The ribbon is clearly in between the lines indicating 1.1 and 1.2
cm. It is less than halfway beyond the 1.1 line, which would indicate 1.15 cm.
You may judge the length of the ribbon to be 1.12 cm or 1.14, and that would
be fine. What you can't do is record more or less digits, as in 1.1 cm or 1.130 cm.
5.
[1.51 cm]—The ribbon is clearly just past the 1.5 cm mark. You might
estimate its length at 1.51 cm or 1.52 cm, but you must include one, and only
one, estimation digit.
Lesson 2- 4 Review
For questions 1-6, the significant digits are shown in boldface within the pa-
rentheses.
1.
[1]—(40) In the number 40, the 4 is significant, because it is a nonzero digit.
The 0 is only a placeholder and is not significant.
2.
[3]—(602) In the number 602, the 6 and the 2 are significant, because they
are nonzero digits. The 0 is also significant, because it is set between two
significant digits.
3.
[2]—(0.000043) The number 0.000043 only shows two significant digits: the
4 and the 3. All of the zeros are placeholders. Remember: A zero must be
to the right of both a decimal place and a significant digit to be significant.
4.
[68.3]—The original answer to the calculation is 68.310 g. However, the
measurement 21.0 g from the problem is considered accurate only to the
tenth place. We needed to round to the tenth place.
5.
[16]—The original answer to the calculation is 15.92136. The measurement of
0.21 cm from the problem only shows two significant digits, so we must round
to two significant digits. The 9 in the tenth place tells us to round up to 16.
