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Chapter 1: Problem 49

Carry out the following operations and express the answers with the appropriate number of significant figures. $$ \begin{array}{ll}{\text { (a) } 14.3505+2.65} & {\text { (b) } 952.7-140.7389} \\\ {\text { (c) }\left(3.29 \times 10^{4}\right)(0.2501)} & {\text { (d) } 0.0588 / 0.677}\end{array} $$

Short Answer

Expert verified
(a) \(16.00\) (b) \(812.0\) (c) \(8.22 \times 10^{4}\) (d) \(0.0869\)

Step by step solution

01

(a) Addition

First, add the given numbers: \(14.3505 + 2.65 = 16.0005\) The least number of decimal places in the problem is 2 (2.65). Therefore, the answer must have two decimal places as well. The final answer should be rounded to two decimal places: \(16.0005 \approx 16.00\)
02

(b) Subtraction

Subtract the given numbers: \(952.7 - 140.7389 = 811.9611\) The least number of decimal places in the problem is 1 (952.7). Therefore, the answer must have one decimal place as well. The final answer should be rounded to one decimal place: \(811.9611 \approx 812.0\)
03

(c) Multiplication

Multiply the given numbers: \((3.29 \times 10^{4})(0.2501) = 82229.0\) The least number of significant figures in the problem is 3 (3.29). Therefore, the answer must have three significant figures as well. The final answer should be rounded to three significant figures: \(82229.0 \approx 8.22 \times 10^{4}\)
04

(d) Division

Divide the given numbers: \(0.0588 / 0.677 = 0.086838399\) The least number of significant figures in the problem is 3 (0.0588). Therefore, the answer must have three significant figures as well. The final answer should be rounded to three significant figures: \(0.086838399 \approx 0.0869\)

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Most popular questions from this chapter

A 40 -lb container of peat moss measures \(14 \times 20 \times 30\) in. A 40 -lb container of topsoil has a volume of 1.9 gal. (a) Calculate the average densities of peat moss and topsoil in units of\(\mathrm{g} / \mathrm{cm}^{3} .\) Would it be correct to say that peat moss is "lighter" than topsoil? (b) How many bags of peat moss are needed to cover an area measuring 15.0 \(\mathrm{ft} \times 20.0 \mathrm{ft}\) to a depth of 3.0 in.?

A package of aluminum foil contains 50 \(\mathrm{ft}^{2}\) of foil, which weighs approximately 8.0 oz. Aluminum has a density of 2.70 \(\mathrm{g} / \mathrm{cm}^{3} .\) What is the approximate thickness of the foil in millimeters?

Indicate the number of significant figures in each of the following measured quantities: (a) \(3.774 \mathrm{km},\) (b) 205 \(\mathrm{m}^{2}\) , (c) \(1.700 \mathrm{cm},(\mathbf{d}) 350.00 \mathrm{K},(\mathbf{e}) 307.080 \mathrm{g},(\mathbf{f}) 1.3 \times 10^{3} \mathrm{m} / \mathrm{s}\) .

In \(2009,\) a team from Northwestern University and Western Washington University reported the preparation of a new "spongy" material composed of nickel, molybdenum, and sulfur that excels at removing mercury from water. The density of this new material is 0.20 \(\mathrm{g} / \mathrm{cm}^{3}\) , and its surface area is 1242 \(\mathrm{m}^{2}\) per gram of material. (a) Calculate the volume of a 10.0 -mg sample of this material. (b) Calculate the surface area for a 10.0 -mg sample of this material. (c) A 10.0 -mL sample of contaminated water had 7.748 mg of mercury in it. After treatment with 10.0 \(\mathrm{mg}\) of the new spongy material, 0.001 \(\mathrm{mg}\) of mercury remained in the contaminated water. What percentage of the mercury was removed from the water? (d) What is the final mass of the spongy material after the exposure to mercury?

What type of quantity (for example, length, volume, density) do the following units indicate? (a) \(\mathrm{mL},(\mathbf{b}) \mathrm{cm}^{2}\) ,\((\mathbf{c}) \mathrm{mm}^{3},(\mathbf{d}) \mathrm{mg} / \mathrm{L},(\mathbf{e}) \mathrm{ps},(\mathbf{f}) \mathrm{nm},(\mathrm{g}) \mathrm{K}\)
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