Questions
Q1.1.1
Express the following values in scientific notation.
- 150,000,000
- 0.000043
- 332000
- 0.0293
- 932
- 0.1873
- 78,000
- 0.0001
- 4500
- 0.00290
- 6281
- 0.00700
Q1.1.2
Express the following values in decimal notation.
- 3.8 x 10-3
- 9.21 x 105
- 7.91 x 10-2
- 2.5 x 106
- 3.42 x 10-8
- 5.4 x 105
- 3 x 10-3
- 7.34 x 102
- 9.8 x 10-4
- 6 x 107
- 4.20 x 10-6
- 4.20 x 106
Q1.1.3
What SI base unit would be appropriate for each measurement?
- the length of a room
- the amount of carbon in a diamond
- the mass of NaCl in a bottle
Q1.1.4
List the meaning of each abbreviation of the base units.
- m
- K
- kg
- s
- mol
Q1.1.5
What is the the derived unit from the SI base units for the relationship of each pair of quantities?
- mass and volume
- distance and time
- amount of substance and volume
- area
Q1.1.6
Give the meaning and name of each metrix prefix abbreviation.
- M
- m
- n
- d
Q1.1.7
Give the abbreviation and meaning of each metrix prefix.
- kilo
- centi
- micro
- giga
Q1.1.8
Name the prefix with the following numerical meaning.
- 1/10
- 1,000,000
- 1/1,000,000
- 1/100
- 1
Q1.1.9
Convert each temperature to the missing one between Celsius and Fahrenheit.
- 77°F
- 212°F
- 37°C
- 22°C
- 95°F
- 15°C
- 0°F
- 0°C
- –10°C
- –10°F
Q1.2.1
Explain the similarities and differences between accuracy and precision.
Q1.2.2
The density of a copper sample was determined by three different students (shown below). Each performed the measurement three times and is reported below (all values in units of g/cm3). The accepted value for the density of copper is 8.92 g/cm3.
- Determine if each student's data is accurate, precise, neither or both.
- What is the average density based on Justin's data?
- What is the average density based on Jane's data?
- Jane: 8.94, 8.89, 8.91
- Justin: 8.32, 8.31, 8.34
- Julia: 8.64, 9.71, and 9.10
Q1.2.3
Determine the number of significant figures in each of the following values.
- 406
- 3.00
- 3.20
- 0.25
- 0.0689
- 0.002910
- 3941
- 46.250
- 30.21
- 0.10300
Q1.2.4
Write each value with three significant figures, use scientific notation if necessary.
- 34500
- 24
- 0.0345
- 0.012
- 612.8
- 98.22
- 0.14928
- 300
Q1.2.5
Give three examples of exact numbers.
Q1.2.6
Find the result of each of the following calculations and report the value with the correct number of significant figures.
- 0.23 + 12.2 =
- 13 - 1.03 =
- 0.839 + 0.28925 =
- 28 + 34.4 =
- 0.8 + 2.3 =
- 34.9 - 0.583 =
- 21 - 0.132 =
- 0.840 + 0.9334
Q1.2.7
Find the result of each of the following calculations and report the value with the correct number of significant figures.
- 34 x 0.12 =
- 68.2 / 0.78 =
- 3.29 x 104 x 16.2 =
- 0.8449 x 29.7 =
- 5.92 x 103 / 0.628 =
- 3.00 x 2.6 =
- 2.50 x 9.331 =
- 3.20 / 12.75 =
Q1.3.1
What is a conversion factor?
Q1.3.2
What is the conversion factor between each pair of units?
- feet and inches
- mL and cm3
- kg and g
- cm and m
- mm and cm
- inches and centimeters
- grams and pounds
- g and µg (mcg)
Q1.3.3
Complete each of the following conversions.
- 34 cm to m
- 3.7 ft to in
- 345 mg to Mg
- 5.3 km to mm
- 4.0 L to mL
- 3.45 x 103 mm to km
- 78 cm3 to mL
- 0.85 kg to dg
- 13 pints to gallon
- 0.35 L to cm3
Q1.3.4
Complete each of the following conversions.
- 342 cm3 to dm3
- 2.70 g/cm3 to kg/L
- 34 mi/hr to km/min
- 0.00722 km2 to m2
- 4.9 x 105 mcm3 to mm3
- 80. km/hr to mi/hr
Q1.4.1
Solve each of the following.
- What percent of 35 is 8.2?
- What percent of 56 is 12?
- What percent of 923 is 38?
- What percent of 342 is 118?
Q1.4.2
Solve each of the following?
- What is 42% of 94?
- What is 83% of 239?
- What is 16% of 45?
- What is 38% of 872?
Q1.4.3
Solve each of the following?
- 42 is 34% of what number?
- 73 is 82% of what number?
- 13 is 57% of what number?
- 75 is 25% of what number?
- 25 is 15% of what number?
- 98 is 76% of what number?
Q1.4.4
A patient originally weighs 182 pounds and loses 15.0% of their body weight. What is their final weight?
Q1.4.5
A patient's original weight was 135 pounds and they lose 12 pounds. What percent of their body weight did they lose?
Q1.4.6
A patient needs to increase their calcium supplement by 25% a week. If they are currently taking a 300. mg supplement, how much more will they need to take?
Q1.4.7
An infant's birth weight is 7 pounds, 1 ounce. Her discharge weight is 6 pounds, 13 ounces. What percent of her birth weight did she lose?
Q1.4.8
A patient needs a 20.% decrease in their medication dosage from 125 mg. What will his dosage be after the decrease?
Answers
Q1.1.1
- 1.5 × 108
- 4.3 × 10–5
- 3.32 × 105
- 2.93 × 10–2
- 9.32 × 102
- 1.873 × 10–1
- 7.8 × 104
- 1 × 10–4
- 4.5 × 103
- 2.9 × 10–3
- 6.281 × 103
- 7 × 10–3
Q1.1.2
- 0.0038
- 921000
- 0.0791
- 2500000
- 0.0000000342
- 540000
- 0.003
- 734
- 0.00098
- 60000000
- 0.00000420
- 4200000
Q1.1.3
- meter
- mole
- kilogram
Q1.1.4
- meter
- Kelvin
- kilogram
- second
- mole
Q1.1.5
- kg/m3
- m/s
- mol/m3 is based on SI base units, but mol/L is also acceptable
- m2
Q1.1.6
- Mega, 106
- milli, 10–3
- nano, 10–9
- deci, 10–1
Q1.1.7
- k, 103
- c, 10–2
- µ (or mc), 10–6
- G, 109
Q1.1.8
- deci
- mega
- micro
- centi
- none (base unit)
Q1.1.9
- 77°F = 25°C
- 212°F = 100°C
- 37°C = 98.6°F
- 22°C = 72°F
- 95°F = 35°C
- 15°C = 59°F
- 0°F = –18°C
- 0°C = 32°F
- –10°C = 14°F
- –10°F = –23°C
Q1.2.1
Accuracy is a measure of how close the values are close to the correct value while precision is a measure of how close values are to each other.
Q1.2.2
-
- Jane: 8.94, 8.89, 8.91 - accurate and precise
- Justin: 8.32, 8.31, 8.34 - precise
- Julia: 8.64, 9.71, and 9.10 - neither accurate nor precise
- 8.32 g/cm3
- 8.91 g/cm3
Q1.2.3
- 3
- 3
- 3
- 2
- 3
- 4
- 4
- 5
- 4
- 5
Q1.2.4
- 3.45 × 104
- 2.40 × 101
- 3.45 × 10–2
- 1.20 × 10–2
- 613 or 6.13 × 102
- 9.82 × 101
- 0.149 or 1.49 × 10–1
- 300. or 3.00 × 102
Q1.2.5
Answers will vary. 12 eggs, 100 cm = 1 m, 1 inch = 2.54 cm, 4 people
Q1.2.6
- 0.23 + 12.2 = 12.43 = 12.4
- 13 - 1.03 = 11.97 = 12
- 0.839 + 0.28925 = 1.12825 = 1.128
- 28 + 34.4 = 62.4 = 62
- 0.8 + 2.3 = 3.1
- 34.9 - 0.583 = 34.317 = 34.3
- 21 - 0.132 = 20.868 = 21
- 0.840 + 0.9334 = 1.7734 = 1.773
Q1.2.7
- 34 x 0.12 = 4.08 = 4.1
- 68.2 / 0.78 = 87.4358974 = 87
- 3.29 x 104 x 16.2 = 5.32980 × 105 = 5.33 × 105
- 0.8449 x 29.7 = 25.09353 = 25.1
- 5.92 x 103 / 0.628 = 9.4267515 × 103 = 9.43 × 103
- 3.00 x 2.6 = 7.8
- 2.50 x 9.331 = 23.3275 = 23.3
- 3.20 / 12.75 = 0.25098 = 0.251
Q1.3.1
A conversion factor is a relationship between two units. The value in the numerator has some equivalence to the value in the denominator.
Q1.3.2
- 1 foot = 12 inches
- 1 mL = 1 cm3
- 1 kg = 1000 g or 1 × 10–3 kg = 1 g
- 100 cm = 1 m or 1 cm = 1 × 10–2 m
- 10 mm = 1 cm
- 1 inch = 2.54 cm
- 454 grams = 1 pound
- 1 g = 1 × 106 µg (mcg) or 1 × 10–6 g = 1 µg (mcg)
Q1.3.3
- \(34 \; cm \times \frac{1 \; m}{100\;cm} = 0.34\;m\)
- \(3.7 \; ft \times \frac{12 \; in}{1\;ft}=44.4\;in=44\;in\)
- \(345\;mg \times \frac{1\;g}{1000\;mg} \times \frac{1\;Mg}{1\times {10}^6\;g}=3.45 \times {10}^{-7}\;Mg\)
- \(5.3\;km\times\frac{1000\;m}{1\;km}\times\frac{1000\;mm}{1\;m}=5.3\times{10}^6\;mm\)
- \(4.0\;L\times\frac{1000\;mL}{1\;L}=4.0\times{10}^3\;mL\)
- \(3.45\times{10}^3\;mm\times\frac{1\;m}{1000\;mm}\times\frac{1\;km}{1000\;m}=3.45\times{10}^{-3}\;km\)
- \(78\;{cm}^3\times\frac{1\;mL}{{cm}^3}=78\;mL\)
- \(0.85\;kg\times\frac{1000\;g}{1\;kg}\times\frac{10\;dg}{1\;g}=8.5\times{10}^3\;dg\)
- \(13\;pints\times\frac{1\;quart}{2\;pints}\times\frac{1\;gallon}{4\;quarts}=1.6\;gallons\)
- \(0.35\;L\times\frac{1000\;mL}{1\;L}\times\frac{1\;mL}{1\;{cm}^3}=3.5\times{10}^2\;{cm}^3\)
Q1.3.4
- \(342\;{cm}^3\times\frac{1\;dm}{10\;cm}\times\frac{1\;dm}{10\;cm}\times\frac{1\;dm}{10\;cm}=0.342\;{dm}^3\) or \(342\;{cm}^3\times {\left( \frac{1\;dm}{10\;cm} \right)}^3=342\;{cm}^3\times \frac{{1}^3\;{dm}^3}{{10}^3\;{cm}^3}=0.342\;{dm}^3\)
- \(\frac{2.70\;g}{{cm}^3}\times\frac{1\;kg}{1000\;g}\times\frac{1\;{cm}^3}{1\;mL}\times\frac{1000\;mL}{1\;L}=\frac{2.70\;kg}{L}\)
- \(\frac{34\;mi}{hr} \times \frac{5280\;ft}{1\;mi} \times \frac{12\;in}{1\;ft}\times \frac{2.54\;cm}{1\;in} \times \frac{1\;m}{100\;cm} \times \frac{1\;km}{1000\;m} \times \frac{1\;hr}{60\;min} = \frac{0.91\;km}{min}\)
- \(0.00722\;k{m^2} \times \frac{1000\;m}{1\;km} \times \frac{1000\;m}{1\;km} = 7.22 \times {10^3}\;{m^2}\)
- \(4.95 \times {10^5}\;mc{m^3} \times \frac{1\;mm}{1000\;mcm} \times \frac{1\;mm}{1000\;mcm} \times \frac{1\;mm}{1000\;mcm} = 4.95 \times {10^ - }^4\;m{m^3}\)
- \(\frac{80.\;km}{hr}\times\frac{1000\;m}{1\;km}\times\frac{100\;cm}{1\;m}\times\frac{1\;in}{2.54\;cm}\times\frac{1\;ft}{12\;in}\times\frac{1\;mi}{5280\;ft}=\frac{50.\;mi}{hr}\)
Q1.4.1
- \(\%= \frac{part}{whole} \times 100= \frac{8.2}{35} \times 100= 23\% \)
- \(\%= \frac{part}{whole} \times 100= \frac{12}{56} \times 100= 21\% \)
- \(\%= \frac{part}{whole} \times 100= \frac{38}{923} \times 100= 4.1\% \)
- \(\%= \frac{part}{whole} \times 100= \frac{118}{342} \times 100= 34.5\% \)
Q1.4.2
- \(\begin{array}{c}
\% = \frac{part}{whole} \times 100\\
42\% = \frac{part}{94} \times 100\\
part = 39
\end{array}\)
- \(\begin{array}{c}
\quad \\ \% = \frac{part}{whole} \times 100\\
83\% = \frac{part}{239} \times 100\\
part = 198=2.0\times{10}^2
\end{array}\)
- \(\begin{array}{c}
\quad \\ \% = \frac{part}{whole} \times 100\\
16\% = \frac{part}{45} \times 100\\
part = 7.2\\
\end{array}\)
- \(\begin{array}{c}
\quad \\ \% = \frac{part}{whole} \times 100\\
38\% = \frac{part}{872} \times 100\\
part = 3.3\times{10}^2
\end{array}\)
Q1.4.3
- \(\begin{array}{c}
\quad \\
\% = \frac{part}{whole} \times 100\\
34\% = \frac{42}{whole} \times 100\\
whole = 1.2\times{10}^2
\end{array}\)
- \(\begin{array}{c}
\quad \\
\% = \frac{part}{whole} \times 100\\
82\% = \frac{73}{whole} \times 100\\
whole = 89
\end{array}\)
- \(\begin{array}{c}
\quad \\
\% = \frac{part}{whole} \times 100\\
57\% = \frac{13}{whole} \times 100\\
whole = 23
\end{array}\)
- \(\begin{array}{c}
\quad \\
\% = \frac{part}{whole} \times 100\\
25\% = \frac{75}{whole} \times 100\\
whole = 3.0\times{10}^2
\end{array}\)
- \(\begin{array}{c}
\quad \\
\% = \frac{part}{whole} \times 100\\
15\% = \frac{25}{whole} \times 100\\
whole = 1.7\times{10}^2
\end{array}\)
- \(\begin{array}{c}
\quad \\
\% = \frac{part}{whole} \times 100\\
76\% = \frac{98}{whole} \times 100\\
whole = 129
\end{array}\)
Q1.4.4
\(\begin{array}{l}
\% = \frac{part}{whole} \times 100\\
15.0\% = \frac{part}{182\;pounds} \times 100\\
part = 27.3\;pounds\;lost\\
\\
182\;pounds - 27.3\;pounds = 154.7\;pounds = 155\;pounds
\end{array}\)
Q1.4.5
\(\begin{array}{l}
\% = \frac{part}{whole} \times 100\\
\% = \frac{12\;pounds}{135\;pounds} \times 100\\
\% = 8.9\%\;lost\\
\end{array}\)
Q1.4.6
\(\begin{array}{l}
\% = \frac{part}{whole} \times 100\\
25\% = \frac{part}{300.\;mg} \times 100\\
part = 75\;mg\;more\\
\end{array}\)
Q1.4.7
Convert both weights to ounces, find the ounces lost, and then find the percent lost.
Birth weight: \(\left( 7\;pounds\times 16 \right) + 1\;ounce=113\;ounces\)
Discharge weight: \(\left( 6\;pounds\times 16 \right) + 13\;ounces=109\;ounces\)
Weight lost: \(113\;ounces-109\;ounces=4\;ounces\)
Percent lost from original brith weight.
\(\begin{array}{l}
\% = \frac{part}{whole} \times 100\\
\% = \frac{4\;ounces}{113\;ounces} \times 100\\
\% = 3.5\% =4\%
\end{array}\)
Q1.4.8
\(\begin{array}{l}
\% = \frac{part}{whole} \times 100\\
20\% = \frac{part}{125\;mg} \times 100\\
part = 25\;mg\;lost\\
\\
125\;mg - 25\;mg = 100.\;mg
\end{array}\)