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Chapter 3: Problem 46

Calculate the number of atoms present in each of the following: \begin{equation} \begin{array}{l}{\text { a. } 2 \text { mol Fe }} \\ {\text { b. } 40.1 \text { g Ca, which has an atomic mass }} \\ {\text { of } 40.08 \text { amu }} \\\ {\text { c. } 4.5 \text { mol of boron-11 }}\end{array} \end{equation}

Short Answer

Expert verified
The number of atoms in A. 2 mol Fe is \(1.2044 \times 10^{24}\) atoms, B. 40.1g Ca is \(5.9994 \times 10^{23}\) atoms, C. 4.5 mol of boron-11 is \(2.7099 \times 10^{24}\) atoms.

Step by step solution

01

Calculating number of atoms in 2 mol Fe

To calculate the number of atoms, we can use Avogadro’s number, given as \(6.022 \times 10^{23}\). Multiplying the number of moles by Avogadro's number gives the number of atoms: \( Number of atoms = 2 \text { mol Fe } \times 6.022 \times 10^{23} = 1.2044 \times 10^{24} \text { atoms}\)
02

Calculating number of atoms in 40.1g Ca

First, the mass of Ca in grams needs to be converted to moles using its molar mass (40.08 amu). Hence, \( Number of moles = \frac {the mass of the sample in grams}{molar mass} = \frac{40.1 g}{40.08 g/mol} = 0.999 moles Ca\). Then use Avogadro's number to convert moles to number of atoms: \( Number of atoms = 0.999 moles Ca \times 6.022 \times 10^{23} = 5.9994 \times 10^{23} \text { atoms}.\)
03

Calculating number of atoms in 4.5 moles of boron-11

Using the Avogadro's number, the number of atoms can be calculated as follows: \( Number of atoms = 4.5 \text {mol B} \times 6.022 \times 10^{23} = 2.7099 \times 10^{24} \text { atoms}. \)

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Most popular questions from this chapter

The magnetic properties of an element depend on the number of unpaired electrons it has. Explain why iron, Fe, is highly magnetic but neon, Ne, is not.

What is Avogadro's number?

How many moles are represented by each of the following. \begin{equation} \begin{array}{l}{\text { a. } 11.5 \mathrm{g} \text { Na which has an atomic mass of }} \\ {22.99 \text { amu }} \\ {\text { b. } 150 \mathrm{g} \text { S which has an atomic mass of }} \\ {32.07 \text { amu }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. } 5.87 \text { g Ni which has an atomic mass of }} \\\ {58.69 \text { amu }}\end{array} \end{equation}

If an electron is beyond the \(n=\infty\) level, is the electron a part of the hydrogen atom?

Research several elements whose symbols are inconsistent with their English names. Some examples include silver, Ag; gold, Au; and mercury, Hg. Compare the origin of these names with the origin of the symbols.
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