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To find the number of atoms in 179.0 g of iridium (Ir), first determine its molar mass, which is approximately 192.22 g/mol. Next, calculate the number of moles in 179.0 g by dividing the mass by the molar mass: 179.0 g / 192.22 g/mol ≈ 0.933 moles. Finally, multiply the number of moles by Avogadro's number (approximately (6.022 \times 10^{23}) atoms/mol) to find the total number of atoms: (0.933 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 5.61 \times 10^{23}) atoms.

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How many atoms are present in 179.0 g of iridium?

To find the number of atoms in 179.0 g of iridium, first determine the molar mass of iridium, which is approximately 192.22 g/mol. Then, calculate the number of moles in 179.0 g by dividing the mass by the molar mass: ( \text{moles} = \frac{179.0 , \text{g}}{192.22 , \text{g/mol}} \approx 0.93 , \text{mol} ). Finally, multiply the number of moles by Avogadro's number (( 6.022 \times 10^{23} , \text{atoms/mol} )) to find the total number of atoms: ( 0.93 , \text{mol} \times 6.022 \times 10^{23} \approx 5.59 \times 10^{23} , \text{atoms} ).


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