Question

a piece of pure aluminum foil is cut to dimensions of 5.00 cm x 8.00 cm. its mass is measured as 0.260 g. given the density of al is 2.70 g/cm³ and the diameter of a single al atom is 286 pm, how many al atoms span the thickness (height) of the foil? hint: the volume of a rectangular object can be calculated using the equation: v = l w h. 1.45×10¹⁰ 9.08×10⁴ 6.13×10⁵ 6.022×10²³ 8.42×10⁴

Explanation:

Step1: Calculate volume of foil

$V=\frac{m}{
ho}=\frac{0.260\ g}{2.70\ g/cm^{3}}\approx0.0963\ cm^{3}$

Step2: Calculate height of foil

$h = \frac{V}{l\times w}=\frac{0.0963\ cm^{3}}{5.00\ cm\times8.00\ cm}=0.00241\ cm = 2.41\times10^{7}\ pm$

Step3: Calculate number of atoms

$n=\frac{2.41\times10^{7}\ pm}{286\ pm}\approx8.42\times10^{4}$

Answer:

$8.42\times10^{4}$