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How are xenon fluorides XeF_2, XeF_(4) and XeF_(6)obtained ? |
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Answer» Solution :`underset("excess")(XE(g) + F_(2)(g) overset(673 K)underset(1 BAR) to XeF_(2)(s)` `Xe(g) + underset(1:5 "RATIO") (2F_(2)) (g) overset(873 K)underset(7 bar) to XeF_(4)(s)` `Xe(g) + underset(1:20 "ratio")(3F_(2)) overset(60-70 "bar")underset(573 k) to XeF_(6)(s)` |
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