# The diameter of a chlorine atom is 200. pm. How many chlorine atoms are required to line up end to end to stretch a distance of 1.0 mm?

Good question. $\frac{1 \times {10}^{-} 3 \cdot m}{200 \times {10}^{-} 12 \cdot m}$
$\frac{1 \times {10}^{-} 3 \cdot m}{200 \times {10}^{-} 12 \cdot m}$ $=$ ??
A more realistic measurement might consider the chlorine molecule. $C {l}_{2}$ has a bond length of $2.00 \times {10}^{-} 10$ $m$. How many of these ened to end will fit in a $m m$. Pretty vast but not infinite.