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Chapter 1: Problem 63

The thin outer layer of Earth, called the crust, contains only 0.50 percent of Earth's total mass and yet is the source of almost all the elements (the atmosphere provides elements such as oxygen, nitrogen, and a few other gases). Silicon (Si) is the second most abundant element in Earth's crust (27.2 percent by mass). Calculate the mass of silicon in kilograms in Earth's crust. (The mass of Earth is \(5.9 \times 10^{21}\) tons. 1 ton \(=2000 \mathrm{lb}\); \(1 \mathrm{lb}=453.6 \mathrm{~g} .)\)

Short Answer

Expert verified
The mass of silicon in Earth's crust is approximately \(3.65 \times 10^{21}\) kg.

Step by step solution

01

Convert the mass of the Earth to kilograms

To begin, convert the mass of Earth from tons to kilograms. There are 2,000 lbs in a ton, and 453.6 g in a lb. Therefore, 1 ton = \(2000 \times 453.6\ g\), and, as 1 kg = 1000 g, 1 ton = \( \frac{2000 \times 453.6}{1000}\ kg\). Then, the mass of Earth is \(5.9 \times 10^{21}\ \times \frac{2000 \times 453.6}{1000} = 2.68 \times 10^{24}\) kg.
02

Calculate the mass of the crust

Next, determine the mass of the crust, which is said to be 0.50 percent of Earth's total mass. Therefore, the mass of the crust is \(2.68 \times 10^{24} \times 0.005 = 1.34 \times 10^{22}\) kg.
03

Calculate the mass of silicon

Finally, the mass of silicon in the crust can be calculated. The problem states that silicon comprises 27.2 percent of the crust's mass by mass. As a percentage, this is equivalent to 0.272. Therefore, the mass of silicon is \(1.34 \times 10^{22} \times 0.272 = 3.65 \times 10^{21}\) kg.

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Most popular questions from this chapter

A slow jogger runs a mile in \(13 \mathrm{~min} .\) Calculate the speed in (a) in/s, (b) \(\mathrm{m} / \mathrm{min}\), (c) \(\mathrm{km} / \mathrm{h} .(1 \mathrm{mi}=1609 \mathrm{~m}\) \(1 \mathrm{in}=2.54 \mathrm{~cm} .)\)

Express the answers to these in scientific notation: (a) \(145.75+\left(2.3 \times 10^{-1}\right)\) (b) \(79,500 \div\left(2.5 \times 10^{2}\right)\) (c) \(\left(7.0 \times 10^{-3}\right)-\left(8.0 \times 10^{-4}\right)\) (d) \(\left(1.0 \times 10^{4}\right) \times\left(9.9 \times 10^{6}\right)\)

Which of these properties are intensive and which are extensive? (a) length, (b) volume, (c) temperature, (d) mass.

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) \((7.8 \mathrm{~m}-0.34 \mathrm{~m}) /(1.15 \mathrm{~s}+0.82 \mathrm{~s})\)

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