Skill Builder ES1

Convert each number to scientific notation.

1. 637.8
2. 0.0479
3. 7.86
4. 12,378
5. 0.00032
6. 61.06700
7. 2002.080
8. 0.01020

Solution

1. To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left:

   $$6\underset{\nwarrow }{3}\underset{\nwarrow }{7}.8$$

   Because the decimal point was moved two places to the left, n = 2. In scientific notation, 637.8 = 6.378 × 10².

2. To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right:

   $$0.\underset{\nearrow }{0}\underset{\nearrow }{4}79$$

   Because the decimal point was moved two places to the right, n = −2. In scientific notation, 0.0479 = 4.79 × 10⁻².

3. 7.86 × 10⁰: this is usually expressed simply as 7.86. (Recall that 10⁰ = 1.)
4. 1.2378 × 10⁴; because the decimal point was moved four places to the left, n = 4.
5. 3.2 × 10⁻⁴; because the decimal point was moved four places to the right, n = −4.
6. 6.106700 × 10¹: this is usually expressed as 6.1067 × 10.
7. 2.002080 × 10³
8. 1.020 × 10⁻²

Skill Builder ES2

Carry out the appropriate operation on each number and then express the answer in scientific notation.

1. (1.36 × 10²) + (4.73 × 10³)
2. (6.923 × 10⁻³) − (8.756 × 10⁻⁴)

Solution

1. Both exponents must have the same value, so these numbers are converted to either (1.36 × 10²) + (47.3 × 10²) or (0.136 × 10³) + (4.73 × 10³). Choosing either alternative gives the same answer, reported to two decimal places:

   (1.36 × 10²) + (47.3 × 10²) = (1.36 + 47.3) × 10² = 48.66 × 10² = 4.87 × 10³ (0.136 × 10³) + (4.73 × 10³) = (0.136 + 4.73) × 10³ = 4.87 × 10³

   In converting 48.66 × 10² to scientific notation, n has become more positive by 1 because the value of N has decreased.

2. Converting the exponents to the same value gives either (6.923 × 10⁻³) − (0.8756 × 10⁻³) or (69.23 × 10⁻⁴) − (8.756 × 10⁻⁴). Completing the calculations gives the same answer, expressed to three decimal places:

   (6.923 × 10⁻³) − (0.8756 × 10⁻³) = (6.923 − 0.8756) × 10⁻³ = 6.047 × 10⁻³ (69.23 × 10⁻⁴) − (8.756 × 10⁻⁴) = (69.23 − 8.756) × 10⁻⁴ = 60.474 × 10⁻⁴ = 6.047 × 10⁻³

Skill Builder ES3

Perform the appropriate operation on each expression and express your answer in scientific notation.

1. (6.022 × 10²³)(6.42 × 10⁻²)
2. $\frac{1.67\times {10}^{-24}}{9.12\times {10}^{-28}}$
3. $\frac{(6.63\times {10}^{-34})(6.0\times 10)}{8.52\times {10}^{-2}}$

Solution

1. In multiplication, we add the exponents:

   $$(\text{6}.0\text{22 }\times \text{ 1}{0}^{\text{23}})(\text{6}.\text{42 }\times \text{ 1}{0}^{-\text{2}})=\text{ (6}\text{.022)(6}\text{.42) }\times \text{ 1}{0}^{[23+(-2)]}=\text{ 38}.\text{7 }\times \text{ 1}{0}^{\text{21}}=\text{ 3}.\text{87 }\times \text{ 1}{0}^{\text{22}}$$

2. In division, we subtract the exponents:

   $$\frac{1.67\times {10}^{-24}}{9.12\times {10}^{-28}}=\frac{1.67}{9.12}\times {10}^{[-24-(-28)]}=0.183\times {10}^4=1.83\times {10}^3$$

3. This problem has both multiplication and division:

   $$\frac{(6.63\times {10}^{-34})(6.0\times 10)}{(8.52\times {10}^{-2})}=\frac{39.78}{8.52}\times {10}^{[-34+1-(-2)]}=4.7\times {10}^{-31}$$

Skill Builder ES6

Complete the calculations and report your answers using the correct number of significant figures.

1. 87.25 mL + 3.0201 mL
2. 26.843 g + 12.23 g
3. 6 × 12.011
4. 2(1.008) g + 15.99 g
5. 137.3 + 2(35.45)
6. $\frac{\text{118}.\text{7}}{\text{2}}\text{ g}-\text{35}.\text{5 g}$
7. $\text{47}.\text{23 g}-\frac{\text{2}0\text{7}.\text{2}}{\text{5}.\text{92}}\text{ g}$
8. $\frac{\text{77}.\text{6}0\text{4}}{\text{6}.\text{467}}-\text{4}.\text{8}$
9. $\frac{\text{24}.\text{86}}{\text{2}.0}-\text{3}.\text{26(}0.\text{98})$
10. (15.9994 × 9) + 2.0158

Solution

1. 90.27 mL
2. 39.07 g
3. 72.066 (See rule 5 under “Significant Figures.”)
4. 2(1.008) g + 15.99 g = 2.016 g + 15.99 g = 18.01 g
5. 137.3 + 2(35.45) = 137.3 + 70.90 = 208.2
6. 59.35 g − 35.5 g = 23.9 g
7. 47.23 g − 35.0 g = 12.2 g
8. 12.00 − 4.8 = 7.2
9. 12 − 3.2 = 9
10. 143.9946 + 2.0158 = 146.0104

Skill Builder ES8

1. A 1-carat diamond has a mass of 200.0 mg. When a jeweler repeatedly weighed a 2-carat diamond, he obtained measurements of 450.0 mg, 459.0 mg, and 463.0 mg. Were the jeweler’s measurements accurate? Were they precise?
2. A single copper penny was tested three times to determine its composition. The first analysis gave a composition of 93.2% zinc and 2.8% copper, the second gave 92.9% zinc and 3.1% copper, and the third gave 93.5% zinc and 2.5% copper. The actual composition of the penny was 97.6% zinc and 2.4% copper. Were the results accurate? Were they precise?

Solution

1. The expected mass of a 2-carat diamond is 2 × 200.0 mg = 400.0 mg. The average of the three measurements is 457.3 mg, about 13% greater than the true mass. These measurements are not particularly accurate.

   The deviations of the measurements are 7.3 mg, 1.7 mg, and 5.7 mg, respectively, which give an average deviation of 4.9 mg and a precision of

   $$\frac{\text{4}\text{.9}\cancel{\text{mg}}}{\text{457}\text{.3}\cancel{\text{mg}}}\times \text{100 }=\text{ 1}\text{.1\%}$$

   These measurements are rather precise.

2. The average values of the measurements are 93.2% zinc and 2.8% copper versus the true values of 97.6% zinc and 2.4% copper. Thus these measurements are not very accurate, with errors of −4.5% and + 17% for zinc and copper, respectively. (The sum of the measured zinc and copper contents is only 96.0% rather than 100%, which tells us that either there is a significant error in one or both measurements or some other element is present.)

   The deviations of the measurements are 0.0%, 0.3%, and 0.3% for both zinc and copper, which give an average deviation of 0.2% for both metals. We might therefore conclude that the measurements are equally precise, but that is not the case. Recall that precision is the average deviation divided by the average value times 100. Because the average value of the zinc measurements is much greater than the average value of the copper measurements (93.2% versus 2.8%), the copper measurements are much less precise.

   $$\begin{array}{l}\text{precision (Zn) }=\frac{\text{0}\text{.2\%}}{\text{93}\text{.2\%}}\times \text{100 }=\text{ 0}\text{.2\%}\\ \text{precision (Cu) }=\frac{\text{0}\text{.2\%}}{\text{2}\text{.8\%}}\times \text{100 }=\text{ 7\%}\end{array}$$