2.4 Matter,Measurement, and Dimensional Analysis - Pratice Problems
- Page ID
- 218039
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Exercise \(\PageIndex{1}\)
A1
- Answer
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a) and c) (Sorry, I can't take out the repeat!) 9.00 x 1015 only 3 sig dig because of denominator
b) 9600 or 9.6 x 103 only 2 sig dig because of 7.1 x 102
Exercise \(\PageIndex{2}\)
A2
- Answer
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a) 6.501 ft; 1.981 m; 198.1 cm; 1981 mm; 1.981 x 10-3 km
b) 2.7483 miles; 4.4230 km; 4423.0 m
c) 0.01542 km; 1542 cm; 15,420 mm; 1.542 x 108 \(\mu\)m; 1.542 x 1011 pm
d) 37.0oC; 310.2 K
e) 1.2 x 102 km/h; 34 m/s
f) 2.492 x 102 ft2; 2.315 x 104 cm2; 3.588 x 103 in2
Exercise \(\PageIndex{3}\)
A3
- Answer
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2.222 x 103kg; To get the volume you need the density of gold, which is 19.32 g/ml. The volume is 1.150 x 102 L; the value today is $2.015 x 107
Exercise \(\PageIndex{4}\)
A4
- Answer
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The density is found by first finding the mass of the liquid ( 74.6215-54.6789 = 19.9426 g), then dividing the mass by the volume (19.9426/20.00 = 0.9771 g/mL). That is the density of water at 25 oC. There are not any other common liquids with this density. You could create a mixture that has this density.
Exercise \(\PageIndex{5}\)
A5
- Answer
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The density is found by first finding the mass of the liquid ( 72.7364-54.6789 = 18.0575 g), then dividing the mass by the volume (18.0575/21.3 = 0.848 g/mL). That is the not density of ethanol at 20 oC. The liquid could be mineral oil, which a mixture of many CxHy compounds.