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?

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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?

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Answer 1 · 2016-08-28T20:54:48

Good question. $\frac{1 \times 10^{- 3} \cdot m}{200 \times 10^{- 12} \cdot m}$ $(1xx10^-3*m)/(200xx10^-12*m)$

Explanation:

$\frac{1 \times 10^{- 3} \cdot m}{200 \times 10^{- 12} \cdot m}$ $(1xx10^-3*m)/(200xx10^-12*m)$ $=$ $=$ $? ?$ $??$

A more realistic measurement might consider the chlorine molecule. $C l_{2}$ $Cl_2$ has a bond length of $2.00 \times 10^{- 10}$ $2.00xx10^-10$ $m$ $m$ . How many of these ened to end will fit in a $m m$ $mm$ . Pretty vast but not infinite.

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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?

1 Answer

Explanation:

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