Выберите формулы сильных электролитов: а) б) в)

Mr[N2] = Ar[N] * 2 = 14.0067 * 2 = 28.0134 = 28

Mr[SO2] = Ar[S] + Ar[O] * 2 = 32.065 + 15.9994 * 2 = 64.0638 = 64

Mr[CuO] = Ar[Cu] + Ar[O] = 63.546 + 15.9994 = 79.5454 = 80

Mr[NaCl] = Ar[Na] + Ar[Cl] = 22.98976928 + 35.453 = 58.44276928 = 58

Mr[SO3] = Ar[S] + Ar[O] * 3 = 32.065 + 15.9994 * 3 = 80.0632 = 80

Mr[B2O3] = Ar[B] * 2 + Ar[O] * 3 = 10.811 * 2 + 15.9994 * 3 = 69.6202 = 70

Mr[Mn2O7] = Ar[Mn] * 2 + Ar[O] * 7 = 54.938045 * 2 + 15.9994 * 7 = 221.87189 = 222

Mr[HNO3] = Ar[H] + Ar[N] + Ar[O] * 3 = 1.00794 + 14.0067 + 15.9994 * 3 = 63.01284 = 63

Mr[H3PO4] = Ar[H] * 3 + Ar[P] + Ar[O] * 4 = 1.00794 * 3 + 30.973762 + 15.9994 * 4 = 97.995182 = 98

Mr[Na2CO3] = Ar[Na] * 2 + Ar[C] + Ar[O] * 3 = 22.98976928 * 2 + 12.0107 + 15.9994 * 3 = 105.98843856 = 106

Mr[Al(OH)3] = Ar[Al] + Ar[O] * 3 + Ar[H] * 3 = 26.9815386 + 15.9994 * 3 + 1.00794 * 3 = 78.0035586 = 78

Mr[Al2(SO4)3] = Ar[Al] * 2 + Ar[S] * 3 + Ar[O] * 12 = 26.9815386 * 2 + 32.065 * 3 + 15.9994 * 12 = 342.1508772 = 342

Объяснение:

ba(oh)2+co2=baco3+h2o     (ba)2+   +   2(oh)-     +   co2= baco3+h2o    

h2so4+2naoh= na2so4+2h2o   2(h)+   +   (so4)2-   + 2(na)+   +   2(oh)-   = 2na+   +   (so4)   2-   + 2h2o      

2al+6hbr=2albr3+3h2     2al     +     6(h)+   +   6(br)-   =   2(al)3+   +   6(br)-   +   3h2   ионы   брома сокращаем   и получаем ионные краткие    

feso4+2koh=k2so4+fe(oh)2     (fe)2+   +   (so4)2-   +   2(k)+     +     2(oh)- =2(k)+   +   (so4)2-     +   fe(oh)2   это полное .   краткое   (fe)2+   +   2(oh)-   =   fe(oh)2