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Chapter 3: Problem 20

A modern penny weighs \(2.5 \mathrm{~g}\) but contains only \(0.063 \mathrm{~g}\) of copper (Cu). How many copper atoms are present in a modern penny?

Short Answer

Expert verified
A modern penny contains approximately \(5.98 \times 10^{20}\) copper atoms.

Step by step solution

01

Calculate the number of moles

First, you need to calculate the number of moles of Copper in the 0.063 grams of copper. This can be done by using the atomic mass of copper which is \(63.5 \mathrm{~g/mol}\). The number of moles, \(n\), is calculated using the formula \(n = \frac{m}{M}\), where \(m\) is the mass of the substance and \(M\) is the molar mass. Therefore, \(n = \frac{0.063 \mathrm{~g}}{63.5 \mathrm{~g/mol}} = 0.000992 \mathrm{~mol}\).
02

Apply Avogadro's number

Next, you must convert moles of copper to atoms of copper. Avogadro's number, \(6.022 \times 10^{23} \mathrm{~atoms/mol}\), is used to do this. The number of atoms, \(N\), is calculated by multiplying Avogadro's number by the number of moles. So \(N = n \times \text{Avogadro's number} = 0.000992 \mathrm{~mol} \times 6.022 \times 10^{23} \mathrm{~atoms/mol} = 5.98 \times 10^{20} \mathrm{~atoms}\).
03

Summarize the solution

Therefore, in a modern penny which contains 0.063 g of copper, there are approximately \(5.98 \times 10^{20}\) copper atoms. This answer provides an approximation due to the rounding of Avogadro's number and the atomic mass of copper.

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Most popular questions from this chapter

An impure sample of zinc (Zn) is treated with an excess of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) to form zinc sulfate \(\left(\mathrm{ZnSO}_{4}\right)\) and molecular hydrogen \(\left(\mathrm{H}_{2}\right) .\) (a) Write a balanced equation for the reaction. (b) If \(0.0764 \mathrm{~g}\) of \(\mathrm{H}_{2}\) is obtained from \(3.86 \mathrm{~g}\) of the sample, calculate the percent purity of the sample. (c) What assumptions must you make in (b)?

What mole ratio of molecular chlorine \(\left(\mathrm{Cl}_{2}\right)\) to molecular oxygen \(\left(\mathrm{O}_{2}\right)\) would result from the breakup of the compound \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) into its constituent elements?

Why is the theoretical yield of a reaction determined only by the amount of the limiting reactant?

Which of the following has the greater mass: \(0.72 \mathrm{~g}\) of \(\mathrm{O}_{2}\) or 0.0011 mole of chlorophyll \(\left(\mathrm{C}_{55} \mathrm{H}_{72} \mathrm{MgN}_{4} \mathrm{O}_{5}\right) ?\)

Industrially, nitric acid is produced by the Ostwald process represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\\2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{HNO}_{3}(a q)+\mathrm{HNO}_{2}(a q) \end{aligned}$$What mass of \(\mathrm{NH}_{3}\) (in grams) must be used to produce 1.00 ton of \(\mathrm{HNO}_{3}\) by the above procedure, assuming an 80 percent yield in each step? ( 1 ton = \(2000 \mathrm{lb} ; 1 \mathrm{lb}=453.6 \mathrm{~g} .)\)
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