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(0.33-2x) = .33176.33-2x) X.1094-.6634* - .134x3.1094 6634x 66347 - 6634x = 1099 x=1649 x-.1649 3. At 1200°C, the value of Ke
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equilibrium constant Keg = 2.57X109 and Ch 0.250 mol are present in IL . So the concentration of Hz and Ch hf 0.250M (as M=mo; equilibrium concentration FHF 0.25-X = 0-25-0·2469 M = 0.003) M [C127= 0-25-x= 0.25- 0.2469 M - 0.0031M [Hell= 2X0:2469 M =

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Im struggling, please help. (0.33-2x) = .33176.33-2x) X.1094-.6634* - .134x3.1094 6634x 66347 - 6634x = 1099...
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