How many atoms are there in a 58.7*g58.7⋅g mass of "ammonium nitrite"ammonium nitrite?

1 Answer
anor277
Sep 10, 2017

Well, there are (58.7*g)/(64.06*g*mol^-1)=0.916*mol58.7⋅g64.06⋅g⋅mol−1=0.916⋅mol with respect to "ammonium nitrite"ammonium nitrite.....we get approx. 42xx10^2342×1023 individual atoms.

Explanation:

And so..................

"no. of atoms"="moles of ammonium nitrite"xx"7 atoms mol"^-1xxN_Ano. of atoms=moles of ammonium nitrite×7 atoms mol−1×NA

where N_A=6.022xx10^23........

And so we gots 7xx6.022xx10^23*mol^-1xx0.916*mol=??

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