Question #973fb

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Question #973fb

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Answer 1 · 2016-02-13T16:05:48

$8.98$ $8.98$ $\times$ $xx$ $N_{A}$ $N_A$ , where $N_{A}$ $N_A$ is Avogadro's number, $6.022 \times 10^{23}$ $6.022xx10^23$ $m o l^{- 1}$ $mol^-1$ .

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So the answer is $≅$ $~=$ $54 \times 10^{23}$ $54xx10^23$ individual formula units of $A l C r_{2} O_{7} C l$ $AlCr_2O_7Cl$ . In this quantity, how many individual ATOMS of aluminum, of chromium, of oxygen?

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Question #973fb

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