The total number of electrons in 4.2g of N-3 ion
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N= 1s² 2s² 2p³
N-³ = 1s² 2s² 2p6
14 g of N contain 10NA electrons.
Therefore 4.2 g of N contains (10NA/14)*4.2= 3NA electrons.
N-³ = 1s² 2s² 2p6
14 g of N contain 10NA electrons.
Therefore 4.2 g of N contains (10NA/14)*4.2= 3NA electrons.
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The ans is 3*6.022*10^23
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