Question

3. Вычислить массовую долю бария (w(Ba)) в таких соединениях: a) BaCl2 б) BaCO3 B) Ba(OH)2 г) Ва3(PO4)2​

Answer 1

Ответ:

Объяснение:

Mr(BaCl₂) = Ar(Ba) + 2Ar(Cl) = 137 + 2* 35,5 = 208

w(Br) = Ar(Br) * n/Mr(BaCl₂) * 100% = 137/208 * 100% = 65,87%

Mr(BaCO3) = Ar(Ba) + 2Ar(C) + 3Ar(O) = 137 + 12 + 16*3 = 197

w(Br) = Ar(Br) * n/Mr(BaCO3) * 100% = 137/197 * 100% = 69,54%

Mr(Ba(OH)₂) = Ar(Ba) + 2Ar(O) + 2Ar (H) = 137 + 2 * 16 + 2 * 1 = 171

w(Br) = Ar(Br) * n/Mr(Ba(OH)₂) * 100% = 137/171 * 100% = 80,12%

Mr(Ba3(PO4)₂) = 3Ar(Ba) + 2Ar(P) + 8Ar(O) = 137 * 3 + 31 * 2 + 16 * 8 = 601

w(Br) = Ar(Br) * n/Mr(Ba3(PO4)₂) * 100% = 137* 3/601 * 100% = 68,39%