Math Review Study GuideKEY – General Chemistry
1. Identify the standard SI unit for time, length, mass, and temperature.
Second, meter, kilogram, Kelvin
2. Convert the following to scientific notation:
a. 10 000 = 1 X 104
i. 26 000 000 000 000 = 2.6 X 1013
-4
b. 0.000 1 = 1 X 10
j. 0.000 000 045 = 4.5 X 10-8
c. 10 000 000 000 = 1 X 1010
k. 48 000 000 = 4.8 X 107
d. 50 000 = 5 X 104
l. 0.000 000 000 001 3=1.3 X 10-12
9
e. 2 000 000 000 = 2 X 10
m. 0.017 5 = 1.75 X 10-2
-7
f. 0.000 000 4 = 4 X 10
n. 0.000 046 2 = 4.62 X 10-5
g. 0.000 3 = 3 X 10-4
o. 3 000 000 = 3 X 106
h. 790 000 = 7.9 X 105
3. Express the following numbers in “long form”
a. 1.00 X 105 =
f. 7.95 X 10-1 =
100000
0.795
-12
b. 1 X 10 =
g. 6.854 X 1014 =
0.000000000001
c. 4 X 102 = 400
d. 5 X 10-6
=0.000005
e. 3.2 X 10-2 =
0.032
4. Convert the following measurements:
a. 73 g to mg = 73000 mg
b. 938 mg to cg = 93.8 cg
c. 73.456 cg to dg = 7.3456 dg
d. 41.5 cg to g = 0.415 g
e. 2450 g to kg = 2.45 kg
f. 14 cg to kg = 0.00014 kg
g. 23.7 dag to g = 237 g
h. 182 mL to cL = 18.2cL
i. 14.2 L to mL = 14200mL
j. 16 um to mm = 0.016 mm
k. 3.67 dm to cm = 36.7 cm
l. 14.5 m to dm = 145 dm
m. 1.0 km to m = 1000m
685400000000000
h. 14.3 X 102
=1430
i. 9.065 X 10-4=
0.0009065
j. 4.3 X 103 =
4300
n.
o.
p.
q.
r.
s.
t.
u.
v.
w.
x.
k. 44.2 X 107 =
442000000
l. 3.08 X 10-3 =
.00308
m. 16.9 X 100 =
16.9
n. 66.3 X 104
663000
o. 4.00 X 10-6=
.00000400
23.896 cm to km = 0.00023896 km
245 mm to cm = 24.5 cm
5.67 m to dm = 56.7 dm
13 m to km = 0.013 km
2.9 X 106 m to mm = 2.9 X 109 mm
5.79 X 10-12 cg to kg =5.79 X 10-17kg
6.9845 X 103 kg to mg =6.9845 X 109 mg
6.4 X 10-6 cm to dm = 6.4 X 10-7dm
2.156 X 108 L to mL = 2.156 X 1011mL
1.0 X 101 dg to cg = 1.0 X 102 cg
1.02 X 10-2 cg to g 1.02 X 10-4 g
5. Using dimensional analysis, solve the following problems:
a. 22.5 km/hr to m/sec
f. 8.05 X 105 g/mL to kg/L
(22.5km/hr) X (1000m/1km) X
(1hr/3600sec) = 6.25 m/sec
b. 94.862 km to cm
94.862km X (105 cm/1km)= 9486200 cm
c. 8.35 X
10-5
km/sec to mm/min
(8.35 X 10-5km/sec) X (106mm/km) X
(60sec/1 min)= 5010 mm/min
d. 6.957 dg to cg
9.657dg X (10cg/1dg)= 96.57cg
e. 3.42 X 103 kg/day to g/year
(3.42 X 103 kg/day) X (1000g/1kg) X
(365days/ 1 year) = 1.2483 X 109g/year
(8.05 X 105g/mL) X (1kg/1000g) X (1000
mL/1 L) = 8.05 X 105kg/L
g. 987.62 kL to mL
987.62 kL X (106mL/1 kL) = 9.8762 X 108 mL
h. 7.3 X 10-4 cm/s to cm/week
i.
j.
(7.3 X 10-4cm/s) X (60s/min) X (60min/h) X
(24h/d) X (7d/wk)=441.5cm/wk=440cm/wk
14.7 g/cm3 to kg/cm3
(14.7g/cm3) X (1kg/1000g) = 0.0147kg/cm3
How many hours are in 3 weeks?
3wks X (7d/wk) X (24h/d)= 504h=500h
6. Calculate the percent error in the following problems:
a. What is the % error for a mass measurement of 15.6 g, given that the correct
value is 26.9g?
/15.6-26.9/ ÷ 26.9 X 100 = 42.0%
b. A volume is measured experimentally as 4.26 mL. What is the % error, given
that the correct value is 4.15mL?
/4.26-4.15/ ÷ 4.15 X 100 = 2.65%
c. A student measures the mass and volume of a substance and calculates its
density as 1.40 g/mL. The correct value of the density is 1.36 g/mL. What is
the percent error of the student’s measurement?
/1.40-1.36/ ÷ 1.36 X 100 = 2.94%
d. A student measures the mass of a sample as 9.67 g. Calculate the percent
error, given that the correct mass is 9.82 g.
/9.67-9.82/ ÷ 9.82 X 100 = 1.53%
e. What is the % error of a length measurement of 0.229 cm if the correct value is
0.225 cm? /0.229-0.225/ ÷ 0.225 X 100 = 1.78%
7. Determine the number of significant figures in each of the following numbers:
a. 5.432 g
4
i. 144 kg
b. 40.319 g
5
j. 2500 cm
3
c. 146 cm
3
k. 2500.0 cm
d. 0.189 kg
3
l. 1.04 X 10 14 g
e. 429.3 g
4
m. 3.58 X 10-9 nm
f. 2873.0 cm3 5
n. 48.571 93 kg
3
g. 99.9 cm
3
o. 8365.6 g
h. 0.000 235 g 3
p. 0.002 300 mg
3
2
5
3
3
7
5
4
8. Round each figure to three significant figures:
a. 0.003 210 g = 0.00321 g
b. 3.875 4 kg = 3.88 kg
c. 219 034 m =219000 m
d. 25.38 L = 25.4 L
e. 0.087 63 cm = 0.0876 cm
f. 0.003 109 mg = 0.00311 mg
9. Round each number to four significant figures:
a. 431 801 kg = 431800 kg
b. 10 235.0 mg = 10240 mg
c. 1.0348 m = 1.025 m
d. 0.004 387 010 cm = 0.004387cm
e. 0.000 781 00 mL=0.0007810mL
f. 0.009 864 1 cg = 0.009864 cg
10. Round each of the answers in the following problems to the correct number of
significant figures:
a. 7.31 X 104 + 3.23 X 103
89.5 cm3
= 7.63 X 104
e. 3.40 mg + 7.34 mg + 6.45 mg
b. 8.54 X 10-3 – 3.41 X 10-4
17.19 mg
-3
= 8.20 X 10
f. 45 m X 72 m X 132 m
c. 4.35 dm X 2.34 dm X 7.35 dm
430000 m3
3
= 74.8 dm
g. 38736 km / 4784 km
d. 4.78 cm X 3.218 cm X 5.82 cm
8.097
11. Write
a.
b.
c.
the conversion factor that converts:
Grams to kilograms 1kg/1000g
Kilograms to grams 1000g/1kg
Millimeters to meters 1m/1000mm
d. Meters to millimeters 1000mm/1m
e. Centiliters to liters 1L/100cL
f. Liters to centiliters 100cL/1L
12. Calculate the density of the following:
a. A sample of aluminum metal has a mass of 8.4 g. The volume of the sample is 3.1
cm3. Calculate the density of aluminum. D = m/V = 8.4g/3.1cm3 = 2.7g/cm3
b. 8.0 cm3 of platinum are found to have a mass of 171.2 g. Determine the density.
D = 171.2 g/8.0 cm3 = 21 g/cm3
c. 50.0 cm3 of lead have a mass of 565 g. What is the density of the lead?
D = 565 g/50.0 cm3 = 11.3 g/cm3
d. Diamond has a density of 3.26 g/cm3. What is the mass of a diamond that has a
volume of 0.350 cm3? m = DV = 3.26 g/cm3 X 0.350 cm3 = 1.14 g
e. What is the volume of a sample of liquid mercury that has a mass of 76.2 g, given that
the density of mercury is 13.6 g/mL? V = m/D = 76.2 g/13.6 g/mL = 5.60 mL