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Chapter 2: Problem 81

How many atoms are present in a \(1.00 \mathrm{m}\) length of 20-gauge copper wire? A 20-gauge wire has a diameter of 0.03196 in., and the density of copper is \(8.92 \mathrm{g} / \mathrm{cm}^{3}\)

Short Answer

Expert verified
Finally, calculate the number of atoms by substituting the values in the formula obtained in the previous step.

Step by step solution

01

Conversion of Diameter to Radius and Length to Volume

Start by converting the diameter of wire into radius as the formula to find volume of cylinder (which the wire essentially is) takes radius as a parameter. Given diameter is 0.03196 in., converting to cm by multiplying it with 2.54 gives approximately 0.0812 cm. Hence, the radius becomes approximately \(0.0812 / 2 = 0.0406\) cm. Also, convert 1.00 m of length to cm by multiplying it by 100 to get 100.0 cm. Now, the volume \(V\) of wire can be calculated using the formula for volume of a cylinder \(V= \pi r^{2} h\), where \(r\) is radius and \(h\) is height. Substituting the values, we get \(V = \pi * (0.0406)^{2} * 100.0\) cm³.
02

Calculate Mass Using Density

Knowing that the density \(d\) of an object equals its mass \(m\) divided by its volume \(V\), rearrange the formula to solve for the mass: \(m = d * V\). Substitute the given density of copper (8.92 g/cm³) and the calculated volume from the previous step to find the mass of copper in the wire.
03

Determine the Number of Atoms

First, find the number of moles (\(n\)) in the copper wire by using the formula \(n = m / M\), where \(M\) is the molar mass of copper, which is approximately 63.546 g/mol. After finding the number of moles, use Avogadro's number \(6.022 * 10^{23}\) to determine the number of atoms present in the wire. The formula is \(N = n * N_{A}\), where \(N_{A}\) is Avogadro's number and \(n\) is the number of moles calculated earlier.

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Most popular questions from this chapter

A 5.585-kg sample of iron (Fe) contains (a) \(10.0 \mathrm{mol} \mathrm{Fe}\) (b) twice as many atoms as does \(600.6 \mathrm{g} \mathrm{C}\) (c) 10 times as many atoms as does \(52.00 \mathrm{g} \mathrm{Cr}\) (d) \(6.022 \times 10^{24}\) atoms

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