Question #35b10

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Question #35b10

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Answer 1 · 2017-04-25T21:33:34

Approx. $5.5 \times N_{A}$ $5.5xxN_A$ ............

Explanation:

In $2.74 \cdot m o l$ $2.74*mol$ fluorine gas, i.e. $F_{2}$ $F_2$ , there are $5.5 \times N_{A}$ $5.5xxN_A$ $fluorine atoms$ $"fluorine atoms"$ , i.e. $5.5 \times 6.022 \times 10^{23}$ $5.5xx6.022xx10^23$ fluorine atoms.

In fact ALL the elemental gases, save the Noble Gases, are bimolecular, i.e. $F_{2}, C l_{2}, N_{2}, O_{2}, H_{2}$ $F_2, Cl_2, N_2, O_2, H_2$ .

As always, $N_{A} \equiv Avogadro's number$ $N_A-="Avogadro's number"$ , $6.022 \times 10^{23} \cdot m o l^{- 1}$ $6.022xx10^23*mol^-1$

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Question #35b10

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