What is the number of valence electrons in 0.53g of Na2CO3?
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configuration of Na = 1s², 2s², 2p^6, 3s¹
so, valence electron in Na = 1
configuration of C = 1s² , 2s² , 2p²
so, valence electrons = 4
configuration of O = 1s², 2s² , 2p⁴
so, valance electrons = 6
hence,
in 1 molecule of Na2CO3 total number of valance electrons = 1 × 2 + 4 + 6 × 3
= 24
hence, in 1 mole of Na2CO3 total number of valance electrons = 24× 6.023 × 10²³
no of mole of Na2CO3 = 0.53/106
= 5 × 10-³
so, total number of valance electrons = 5 × 10-³ × 24 × 6.023 × 10²³
=722.76 × 10^20
= 7.2276 × 10²²
so, valence electron in Na = 1
configuration of C = 1s² , 2s² , 2p²
so, valence electrons = 4
configuration of O = 1s², 2s² , 2p⁴
so, valance electrons = 6
hence,
in 1 molecule of Na2CO3 total number of valance electrons = 1 × 2 + 4 + 6 × 3
= 24
hence, in 1 mole of Na2CO3 total number of valance electrons = 24× 6.023 × 10²³
no of mole of Na2CO3 = 0.53/106
= 5 × 10-³
so, total number of valance electrons = 5 × 10-³ × 24 × 6.023 × 10²³
=722.76 × 10^20
= 7.2276 × 10²²
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