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Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO4 , Cr2O72- and NO3-. |
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Answer» H2SO4 (2 × +1) + x+ (4 × -2) = 0 +2 + x – 8 = 0 x – 6 = 0 x = +6 Cr2O72- 2x + 7 ×-2 = -2 2x = -2 + 14 2x = 12 ∴ x = +6 NO3- x + 3 × -2 = -1 x = -1 + 5 x = +4 |
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