How many gold atoms are in an 0.375 ounce, 18 k gold bracelet? (18 k gold is 75% gold by

Answer is: 2,42·10²² atoms of gold.
1 ounce (oz) is equal to 28,34952 grams (g).
0,375 oz = 10,631 g.
75% gold = 10,631 g · 0,75 = 7,973 g.
n(Au) = m(Au) ÷ M(Au).
n(Au) = 7,973 g ÷ 197 g/mol.
n(Au) = 0,0404 mol.
N(Au) = n(Au) · Na.
N(Au) = 0,0404 mol · 6,023·10²³ 1/mol.
N(Au) = 2,42·10²².
n - amount of substance.
Na - Avogadro number.