# 2.4 Matter,Measurement, and Dimensional Analysis - Pratice Problems

- Page ID
- 218039

## Answers

Exercise \(\PageIndex{1}\)

A1

**Answer**-
a) and c) (Sorry, I can't take out the repeat!) 9.00 x 10

^{15}only 3 sig dig because of denominatorb) 9600 or 9.6 x 10

^{3}only 2 sig dig because of 7.1 x 10^{2}

Exercise \(\PageIndex{2}\)

A2

**Answer**-
a) 6.501 ft; 1.981 m; 198.1 cm; 1981 mm; 1.981 x 10

^{-3}kmb) 2.7483 miles; 4.4230 km; 4423.0 m

c) 0.01542 km; 1542 cm; 15,420 mm; 1.542 x 10

^{8}\(\mu\)m; 1.542 x 10^{11}pmd) 37.0

^{o}C; 310.2 Ke) 1.2 x 10

^{2}km/h; 34 m/sf) 2.492 x 10

^{2}ft^{2}; 2.315 x 10^{4}cm^{2}; 3.588 x 10^{3}in^{2}

Exercise \(\PageIndex{3}\)

A3

**Answer**-
2.222 x 10

^{3}kg; To get the volume you need the density of gold, which is 19.32 g/ml. The volume is 1.150 x 10^{2}L; the value today is $2.015 x 10^{7}

Exercise \(\PageIndex{4}\)

A4

**Answer**-
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

^{o}C. There are not any other common liquids with this density. You could create a mixture that has this density.

Exercise \(\PageIndex{5}\)

A5

**Answer**-
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

^{o}C. The liquid could be mineral oil, which a mixture of many C_{x}H_{y}compounds.