Oblicz jaki procent masy stanowi azot w związkach: N2O5, NO2

N2O5 M=2*14+5*16=28+80=108 108  -  100% 28  -  x% x=(28*100)/108=2800/108=25,93%   NO2 M=14+2*16=14+32=46 46  -  100% 14  -  x% x=(14*100)/46=1400/46=30,43%  

A) N₂O₅ Masa N₂O₅ = 14 * 2 + 16 * 5 = 108 [u] Masa N₂ = 14 * 2 = 28[u] 108u = 100% 28u = x% x * 108 = 28 *100 x = 2800 / 108 x ≈ 25,93 [%] B) No₂ Masa NO₂ = 14u + 16u * 2 = 46u Masa N = 14u 46u = 100% 14u = x% x = 1400 / 46 x ≈ 30,43[%]

N2O5 mN205=28u+80u=108u mN=2*14u=28u 108u-100% 28u-x 108x=2800|:108 x=25,9 około azotu NO2 mNO2=14u+32u=46u 46u-100% 14u-x 46x=1400|:46 x=30,4 około azotu