3 вариант 1 CaO 2 Ca(OH)2 3 CaS 4 Ca Cl2 5 CaSO3 6 CaSO4 7 CaCO3 8 Ca(NO3)2 9 Ca3(PO4)2

Определим массовую долю веществ в соединениях:

1. CaO
   • Mr(CaO) = 40 + 16 = 56
   • w(Ca) = (40/56) * 100% = 71.43%
   • w(O) = (16/56) * 100% = 28.57%
2. Ca(OH)2
   • Mr(Ca(OH)2) = 40 + 2(16+1) = 40 + 34 = 74
   • w(Ca) = (40/74) * 100% = 54.05%
   • w(O) = (32/74) * 100% = 43.24%
   • w(H) = (2/74) * 100% = 2.7%
3. CaS
   • Mr(CaS) = 40 + 32 = 72
   • w(Ca) = (40/72) * 100% = 55.56%
   • w(S) = (32/72) * 100% = 44.44%
4. CaCl2
   • Mr(CaCl2) = 40 + 2*35.5 = 40 + 71 = 111
   • w(Ca) = (40/111) * 100% = 36.04%
   • w(Cl) = (71/111) * 100% = 63.96%
5. CaSO3
   • Mr(CaSO3) = 40 + 32 + 3*16 = 40 + 32 + 48 = 120
   • w(Ca) = (40/120) * 100% = 33.33%
   • w(S) = (32/120) * 100% = 26.67%
   • w(O) = (48/120) * 100% = 40%
6. CaSO4
   • Mr(CaSO4) = 40 + 32 + 4*16 = 40 + 32 + 64 = 136
   • w(Ca) = (40/136) * 100% = 29.41%
   • w(S) = (32/136) * 100% = 23.53%
   • w(O) = (64/136) * 100% = 47.06%
7. CaCO3
   • Mr(CaCO3) = 40 + 12 + 3*16 = 40 + 12 + 48 = 100
   • w(Ca) = (40/100) * 100% = 40%
   • w(C) = (12/100) * 100% = 12%
   • w(O) = (48/100) * 100% = 48%
8. Ca(NO3)2
   • Mr(Ca(NO3)2) = 40 + 2*(14 + 3*16) = 40 + 2*(14 + 48) = 40 + 2*62 = 40 + 124 = 164
   • w(Ca) = (40/164) * 100% = 24.39%
   • w(N) = (28/164) * 100% = 17.07%
   • w(O) = (96/164) * 100% = 58.54%
9. Ca3(PO4)2
   • Mr(Ca3(PO4)2) = 3*40 + 2*(31 + 4*16) = 120 + 2*(31 + 64) = 120 + 2*95 = 120 + 190 = 310
   • w(Ca) = (120/310) * 100% = 38.71%
   • w(P) = (62/310) * 100% = 20%
   • w(O) = (128/310) * 100% = 41.29%

Ответ: Массовые доли элементов определены.