Контрольные задания
Вопрос:Фото задания: Найти Mr, W каждого элемента: 1) NaClO; 2) K₂SO₄; 3) HNO₂; 4) Cu(OH)₂; 5) Ba₃(PO₄)₂; 6) FeSO₄; 7) CaCl₂; 8) KMnO₄.Фото задания

Ответ:

Используем относительные атомные массы: H = 1, C = 12, N = 14, O = 16, Na = 23, Cl = 35,5, K = 39, S = 32, Cu = 64, Ba = 137, P = 31, Fe = 56, Ca = 40, Mn = 55. Массовая доля элемента: W = (число его атомов × Ar элемента) / Mr вещества.

  1. NaClO: Mr = 23 + 35,5 + 16 = 74,5.

    W(Na) = 23/74,5 ≈ 30,87%; W(Cl) = 35,5/74,5 ≈ 47,65%; W(O) = 16/74,5 ≈ 21,48%.

  2. K₂SO₄: Mr = 2 × 39 + 32 + 4 × 16 = 174.

    W(K) = 78/174 ≈ 44,83%; W(S) = 32/174 ≈ 18,39%; W(O) = 64/174 ≈ 36,78%.

  3. HNO₂: Mr = 1 + 14 + 2 × 16 = 47.

    W(H) = 1/47 ≈ 2,13%; W(N) = 14/47 ≈ 29,79%; W(O) = 32/47 ≈ 68,09%.

  4. Cu(OH)₂: Mr = 64 + 2 × (16 + 1) = 98.

    W(Cu) = 64/98 ≈ 65,31%; W(O) = 32/98 ≈ 32,65%; W(H) = 2/98 ≈ 2,04%.

  5. Ba₃(PO₄)₂: Mr = 3 × 137 + 2 × (31 + 4 × 16) = 601.

    W(Ba) = 411/601 ≈ 68,39%; W(P) = 62/601 ≈ 10,32%; W(O) = 128/601 ≈ 21,30%.

  6. FeSO₄: Mr = 56 + 32 + 4 × 16 = 152.

    W(Fe) = 56/152 ≈ 36,84%; W(S) = 32/152 ≈ 21,05%; W(O) = 64/152 ≈ 42,11%.

  7. CaCl₂: Mr = 40 + 2 × 35,5 = 111.

    W(Ca) = 40/111 ≈ 36,04%; W(Cl) = 71/111 ≈ 63,96%.

  8. KMnO₄: Mr = 39 + 55 + 4 × 16 = 158.

    W(K) = 39/158 ≈ 24,68%; W(Mn) = 55/158 ≈ 34,81%; W(O) = 64/158 ≈ 40,51%.