1. The problem statement, all variables and given/known data A golf O-ring is used to form a gastight seal in a high-vacuum chamber. The ring is formed from a 100-mm length of 1.5-mm-diameter wire. Calculate the number of gold atoms in the O-ring. 2. Relevant equations Volume of cylinder = PI*h*r^2 3. The attempt at a solution I got the volume of the cylinder: PI*100*.75^2 = 176.7 I used the formula m = D*V: density gold = 19.28 g/cm^3 amu gold = 196.97 grams 19.28*176.7 = 3406.776 I then multiplied by avogadro's number over amu. 3406.776 *(6.02x10^23 / 196.97) = 1.04x10^25 atoms The answer should be 10.4x10^21 atoms. I'm thinking I messed up some unit conversion. I am unsure of where.