A sample of aluminum weighs 27.0 g and has a volume of 10.0 cm^3. The atomic radius of metallic aluminum has been determined by x-ray diffraction to be 0.143 nm. There are 6.02 x 10^23 total atoms in the piece of metal. Using the total number of atoms in the sample and the volume of an individual atom, calculate the actual volume occupied by the atoms in cm^3 and the percent empty space in the aluminum sample. Compare this with the theoretical value for the different cubic structures and identify what kind of unit cell aluminum will exhibit.
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The atomic radius of aluminum is given as 0.143 nm. Convert this radius to cm: \[ 0.143 \, \text{nm} = 0.143 \times 10^{-7} \, \text{cm} = 1.43 \times 10^{-8} \, \text{cm} \] Show more…
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Determine the volume of one atom of aluminum. Determine the number of atoms of aluminum in the foil sample. Calculate Avogadro’s number. Trial: Length of Aluminum Foil: 28 cm Width of Aluminum Foil: 10 cm Area of Aluminum Foil: 28 cm x 10 cm = 280 cm^2 Mass of Aluminum Foil: 1.06 g Density of Aluminum: 2.70 g/cm^3 Atomic Mass of Aluminum: 26.98 g/mol Atomic Radius of Aluminum (picometer = 10^-12 meters): 10,143 pm Number of Layers of Aluminum Atoms: 690 Number of Aluminum Atoms: 2.36 x 10^22 atoms Avogadro’s Number: Number of Aluminum Atoms / Moles of Aluminum = 6.022 x 10^23
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