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Chemistry, 10.09.2019 00:20 disneyshree9427
The reaction described by the equation ch 3 cl + naoh β ch 3 oh + nacl follows the second-order rate law, rate = k [ ch 3 cl ] [ naoh ] . when this reaction is carried out with starting concentrations [ ch 3 cl ] = 0.2 m and [ naoh ] = 1.0 m , the measured rate is 1 Γ 10 β 4 mol l β 1 s β 1 . what is the rate after one-half of the ch 3 cl has been consumed? (caution: the initial concentrations of the starting materials are not identical in this experiment. hint: determine how much of the naoh has been consumed at this point and what its new concentration is, compared with its initial concentration.)
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The reaction described by the equation ch 3 cl + naoh β ch 3 oh + nacl follows the second-order rate...
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