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A Student Determined the Equilibrium Concentration in the Following Equation x2(0.05x)\frac { x ^ { 2 } } { ( 0.05 - x ) }

question 9

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A student determined the equilibrium concentration in the following equation by neglecting the x term in the denominator. Was she justified or not in this approximation and why?
K = 4.0 *10-2 = x2(0.05x) \frac { x ^ { 2 } } { ( 0.05 - x ) }


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