What is the number of representative particles in $4.80 \cdot 10^{- 3}$ $4.80 * 10^-3$ mol $N a I$ $NaI$ ?

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What is the number of representative particles in $4.80 \cdot 10^{- 3}$ $4.80 * 10^-3$ mol $N a I$ $NaI$ ?

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Answer 1 · 2016-04-06T20:03:31

$9.6 \times 10^{- 3}$ $9.6xx10^-3$ total moles of ions.

Explanation:

A sodium iodide unit is conceived to derive from one $N a^{+}$ $Na^+$ and one $I^{-}$ $I^-$ . Because matter is electrically neutral these combine in a 1:1 fashion.

You specified $4.80 \times 10^{- 3}$ $4.80xx10^-3$ $m o l$ $mol$ of $N a I$ $NaI$ . In this molar quantity there are $4.80 \times 10^{- 3}$ $4.80xx10^-3$ $m o l$ $mol$ of $N a^{+}$ $Na^+$ ions, and $4.80 \times 10^{- 3}$ $4.80xx10^-3$ $m o l$ $mol$ of $I^{-}$ $I^-$ ; i.e. $9.6 \times 10^{- 3}$ $9.6xx10^-3$ of individual ions.

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What is the number of representative particles in $4.80 \cdot 10^{- 3}$ $4.80 * 10^-3$ mol $N a I$ $NaI$ ?

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What is the number of representative particles in $4.80 \cdot 10^{- 3}$ $4.80 * 10^-3$ mol $N a I$ $NaI$ ?