Question

Find the number of atoms in a copper rod with a length l of 9.85 cm and a radius r of 1.05 cm . the density of copper is 8.96 g/cm3 . (the volume of a cylinder is v

Answer 1

Answer is: number of atoms in a copper rod is 2.89·10²⁴.

r(Cu) = 1.05 cm; radius of copper rod.

l(Cu) = 9.85 cm; lenght of copper rod.

V(Cu) = r²(Cu) · π· l(Cu).

V(Cu) = (1.05 cm)² · 3.14 · 9.85 cm.

V(Cu) = 34.1 cm³, volume of copper rod.

m(Cu) = d(Cu) · V(Cu).

m(Cu) = 8.96 g/cm³ · 34.1 cm³.

m(Cu) = 305.53 g; mass of copper.

n(Cu) = m(Cu) ÷ M(Cu).

n(Cu) = 305.53 g ÷ 63.55 g/mol.

n(Cu) = 4.81 mol; amount of substance.

N(Cu) = n(Cu) · Na (Avogadro constant).

N(Cu) = 4.81 mol · 6.022·10²³ 1/mol.

N(Cu) = 2.89·10²⁴.

Answer 2

Hope this helps you.