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How are XeO_(3) and XeOF_(4) prepared ? |
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Answer» SOLUTION :Hydrolysis of `XeF_(4) and XeF_(6)` with water gives `XeO_(3)` `6 XeF_(4) + 12 H_(2) O RARR 4 Xe + 2 XeO_(3) + 24 HF + 3 O_(2),""XeF_(6) + 3H_(2)O rarr XeO_(3) + 6HF` On the other hand, partial hydrolysis of `XeF_(6)` gives `XeOF_(4)`. `XeF_(6) + H_(2)O rarr XeOF_(4) + 2HF` |
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