Jawaban:
Tentukan bilangan oksidasi dari unsur dibawah ini :
A. HNO₃
biloks HNO₃ = Biloks H + biloks N + (3 × biloks O)
0 = 1 + biloks N + (3 × (-2))
0 = 1 + biloks N - 6
0 = biloks N - 5
biloks N = +5
B. H₂SO₄
Biloks H₂SO₄ = (2 × biloks H) + biloks S + (4 × biloks O)
0 = (2 × 1) + biloks S + (4 × (-2))
0 = 2 + biloks S - 8
0 = biloks S - 6
biloks S = +6
C. H₃PO₄
Biloks H₃PO₄ = (3 × biloks H) + biloks P + (4 ×biloks O)
0 = (3 × 1) + biloks P + (4 × (-2))
0 = 3 + biloks P - 8
0 = biloks P - 5
biloks P = +5
D. K₂Cr₂O₇
Biloks K₂Cr₂O₇ = (2 × biloks K) + (2 × biloks Cr) + (7 × biloks O)
0 = (2 × 1) + (2 × biloks Cr) + (7 × (-2))
0 = 2 + (2 × biloks Cr) - 14
0 = (2 × biloks Cr) - 12
2 × biloks Cr = +12
biloks Cr = +12/2
biloks Cr = +6
E. (NH₄)₂CO₃
Biloks (NH₄)₂CO₃ = (2 × biloks N) + (8 × biloks H) + biloks C + (3 × biloks O)
0 = (2 × (-3)) + (8 × 1) + biloks C + (3 × (-2))
0 = -6 + 8 + biloks C - 6
0 = biloks C - 4
biloks C = +4
F. Cr₂O₇²⁻
Biloks Cr₂O₇²⁻ = (2 × biloks Cr) + (7 × biloks O)
-2 = (2 × biloks Cr) + (7 × (-2))
-2 = (2 × biloks Cr) - 14
2 × biloks Cr = -2 + 14
2 × biloks Cr = +12
biloks Cr = +12/2
biloks Cr = +6
G. Na₂C₂O₄
Biloks Na₂C₂O₄ = (2 × biloks Na) + (2 × biloks C) + (4 × biloks O)
0 = (2 × 1) + (2 × biloks C) + (4 × (-2))
0 = 2 + (2 × biloks C) - 8
0 = (2 × biloks C) - 6
2 × biloks C = +6
biloks C = +6/2
biloks C = +3
h. Na₂S₄O₆
Biloks Na₂S₄O₆ = (2 × biloks Na) + (4 × biloks S) + (6 × biloks O)
0 = (2 × 1) + (4 × biloks S) + (6 × (-2))
0 = 2 + (4 × biloks S) - 12
0 = (4 × biloks S) - 10
4 × biloks S = +10
biloks S = +10/2
biloks S = +5
bantu likee!!
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