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Question #4ef43

1 Answer

Mar 9, 2017

$10.2*molxxN_A$ , where $N_A="Avogadro's Number.."$

Explanation:

$10.2*molxxN_A$ , where $N_A="Avogadro's Number.."$

And $"Avogadro's Number"$ $=$ $6.022xx10^23*mol^-1$

And thus in $10.2*mol$ zinc, there are.........

$~=$ $6.022xx10^22$ $"zinc atoms"$ ,

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