Silver Atoms on a Surface

Given:
Atomic weight of Silver (Ag) = 108 g/mol
Density of Ag = 10.5 g/cm³
Surface area = 10¹² m²

To find:
Number of Ag atoms on the given surface area (in scientific notation)

Solution:
1. Find the volume of Ag required to cover the given surface area:
Density of Ag = Mass / Volume
Volume = Mass / Density
Volume of Ag required to cover the surface = Mass of Ag / Density of Ag
Mass of Ag = Atomic weight of Ag * Number of Ag atoms
Number of Ag atoms = Mass of Ag / Atomic weight of Ag
Volume of Ag required to cover the surface = (Atomic weight of Ag * Number of Ag atoms) / Density of Ag
Volume of Ag required to cover the surface = (108 g/mol * Number of Ag atoms) / 10.5 g/cm³
Volume of Ag required to cover the surface = (108 g/mol * Number of Ag atoms) / (10.5 * 10⁶ g/m³)
Volume of Ag required to cover the surface = (1.02857 * 10⁻⁷ mol * Number of Ag atoms) / m³

2. Convert the given surface area from m² to cm²:
1 m² = 10⁴ cm²
Surface area = 10¹² m² = 10¹⁶ cm²

3. Find the thickness of the Ag layer required to cover the surface:
Volume of Ag required to cover the surface = Surface area * Thickness of Ag layer
Thickness of Ag layer = Volume of Ag required to cover the surface / Surface area
Thickness of Ag layer = (1.02857 * 10⁻⁷ mol * Number of Ag atoms) / (10¹⁶ cm²)
Thickness of Ag layer = (1.02857 * 10⁻²³ mol * Number of Ag atoms) / cm²

4. Assume that the Ag layer is one atom thick:
Thickness of Ag layer = Diameter of Ag atom
Diameter of Ag atom = 2 * Radius of Ag atom
Radius of Ag atom = Atomic radius / 2
Atomic radius of Ag = 144 pm (given)
Radius of Ag atom = 72 pm = 72 * 10⁻¹² cm

5. Equate the thickness of the Ag layer to the radius of an Ag atom:
(1.02857 * 10⁻²³ mol * Number of Ag atoms) / cm² = 72 * 10⁻¹² cm
Number of Ag atoms = (72 * 10⁻¹² cm * cm²) / (1.02857 * 10⁻²³ mol)
Number of Ag atoms = 7.0 * 10²²

Answer:
The number of Ag atoms on a surface of area 10¹² m² is 7.0 x 10⁻²².