9.86 x 10²⁸ O-atoms require what volume (L) N₂O₂ at STP?

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Chemistry

Cheyanne Grimes

Today, 05:26

9.86 x 10²⁸ O-atoms require what volume (L) N₂O₂ at STP?

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Beaux · Answer 1

First, we use avogadro's number to convert atoms into moles. Then, relate the number of moles from elemental to the compound. Lastly, we use conditions at STP to calculate the volume. We do as follows:

9.86 x 10²⁸ O-atoms (1 mol / 6.022x10^23 atoms O) (1 mol N2O2 / 2 mol O) (22.4 L / 1 mol) = 1833809.37 L needed