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!!