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Chapter 1: Problem 30

Carry out these operations as if they were calculations of experimental results, and express each answer in the correct units and with the correct number of significant figures: (a) \(7.310 \mathrm{~km} \div 5.70 \mathrm{~km}\) (b) \(\left(3.26 \times 10^{-3} \mathrm{mg}\right)-\left(7.88 \times 10^{-5} \mathrm{mg}\right)\) (c) \(\left(4.02 \times 10^{6} \mathrm{dm}\right)+\left(7.74 \times 10^{7} \mathrm{dm}\right)\) (d) \((7.8 \mathrm{~m}-0.34 \mathrm{~m}) /(1.15 \mathrm{~s}+0.82 \mathrm{~s})\)

Short Answer

Expert verified
The solutions for the parts are as follows: (a) 1.28 (dimensionless), (b) 0.00318 mg, (c) \(8.14 \times 10^{7} \mathrm{dm}\), and (d) 4.74 m/s.

Step by step solution

01

Problem (a) Solving

Firstly, solve the division for (a): \(7.310 \mathrm{~km} \div 5.70 \mathrm{~km} = 1.283333333... \mathrm{~km}\). The answer should have three significant figures; thus, rounding will be necessary. This leads to the final answer of 1.28 (dimensionless as the units cancel out) which is to 3 significant figures.
02

Problem (b) Solving

Next, resolve the subtraction for (b): \( \left(3.26 \times 10^{-3} \mathrm{mg}\right)-\left(7.88 \times 10^{-5} \mathrm{mg}\right) = 0.00318212 \mathrm{mg} \). The answer should have three significant figures, thus rounded to 0.00318 mg.
03

Problem (c) Solving

Resolve the addition for (c): \(\left(4.02 \times 10^{6} \mathrm{dm}\right)+\left(7.74 \times 10^{7} \mathrm{dm}\right) = 8.142 \times 10^{7} \mathrm{dm}\) . The answer should have three significant figures, thus the final answer is \(8.14 \times 10^{7} \mathrm{dm}\).
04

Problem (d) Solving

Resolve the calculation for (d): \((7.8 \mathrm{~m} - 0.34 \mathrm{~m}) /(1.15 \mathrm{~s} + 0.82 \mathrm{~s}) = 4.73809524... \mathrm{m/s}\). The answer should have three significant figures, thus the final answer is 4.74 m/s.

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Most popular questions from this chapter

Define these terms: (a) matter, (b) mass, (c) weight, (d) substance, (e) mixture.

Give the SI units for expressing these: (a) length, (b) area, (c) volume, (d) mass, (e) time, (f) force, (g) energy, (h) temperature.

(a) Convert the following temperatures to kelvin: (i) \(113^{\circ} \mathrm{C},\) the melting point of sulfur, (ii) \(37^{\circ} \mathrm{C}\), the normal body temperature, (iii) \(357^{\circ} \mathrm{C},\) the boiling point of mercury. (b) Convert the following temperatures to degrees Celsius: (i) \(77 \mathrm{~K}\), the boiling point of liquid nitrogen, (ii) \(4.2 \mathrm{~K},\) the boiling point of liquid helium, (iii) \(601 \mathrm{~K}\), the melting point of lead.

Magnesium (Mg) is a valuable metal used in alloys, in batteries, and in chemical synthesis. It is obtained mostly from seawater, which contains about \(1.3 \mathrm{~g}\) of Mg for every kilogram of seawater. Calculate the volume of seawater (in liters) needed to extract \(8.0 \times 10^{4}\) tons of \(\mathrm{Mg}\), which is roughly the annual production in the United States. (Density of seawater = \(1.03 \mathrm{~g} / \mathrm{mL} .)\)

This procedure was carried out to determine the volume of a flask. The flask was weighed dry and then filled with water. If the masses of the empty flask and the filled flask were \(56.12 \mathrm{~g}\) and \(87.39 \mathrm{~g},\) respectively, and the density of water is \(0.9976 \mathrm{~g} / \mathrm{cm}^{3},\) calculate the volume of the flask in cubic centimeters.
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