Question

срочно дам 50 баллов​

Answer 1

1.P2O5(V II)
KCL (I. I)
Mn2O7(VII II)
K2O(I II)
Ag2O(I II)
P3N5 (V III)
3. KCL=39+35=74
R2O5= 31*2+16*5=142
Fe(OH)2=56+(1+16)*2=90
Al(NO3)2=27+(14+16*3)*2=166
Na3PO4=23*3+31+16*4=164
4. Mr(N2O5) = 108
W(N) = (Ar(N) * 2) / Mr(N2O5) = (14 * 2) / 108 = 0.25926 = 25.926%
W(O) = (Ar(O) * 5) / Mr(N2O5) = (16 * 5) / 108 = 0.74074 = 74.074%
Ответ: W(N) = 25.926%; W(O) = 74.074%

Mr (H3PO4) = 3 Ar (H) + Ar (P) + 4 Ar (O) = 3 * 1 + 31 + 4 * 16 = 98 г/моль

Находим массовые доли элементов :

W (H) = ( 3 Ar (H) : Mr (H3PO4)) * 100 = ((3 * 1) : 98) * 100 = 3,06 %

W (P) = (Ar (P) : Mr (H3PO4)) * 100 = (31 : 98) * 100 = 31,63 %

W (O) = (4 Ar (O) : Mr (H3PO4)) * 100 = ((4 * 16) : 98) * 100 = 65,31 %