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IS <br /> 30.04+2913X =20.74 +5019X H <br /> 30.04-20.74+2913X-2913X=20 74-20.74+5019X-2913X <br /> 93 =2106X <br /> 9.3!2106 =2106 X/2106 <br /> X= 0 0044 <br /> X=The product of Pb/n*OC so it is then possible to solve for the groundwater velocity from one <br /> of the compounds retarded velocities by substitution of X for(Pb/n*Oc) <br /> Using the velocity of benzene <br /> V =30.04+2913 *0 0044 <br /> V =42 85 feet/year <br /> From the groundwater velocity and the Koc of the other compounds of interest it is possible to <br /> calculate their respective retarded groundwater velocities <br /> Vmtbe =V/(1+X*Kmtbe) <br /> Vmtbe=42 851(1+0 0044*12) <br /> Vmtbe= 40 7 feetlyear <br /> Veb=V/(1+X*Keb) <br /> Veb=42 85/(1+0.0044* 622) <br /> Veb= 1146 <br /> Vxylene =V/(1+X*Kxlyene) <br /> Vxylene=42 851(1+0 0044* 552) <br /> Vxylene= 12. 50 feet/year <br /> By assuming the groundwater velocity calculated of the sample is as a result of the average <br /> groundwater flow for the period the two peak events it is possible to estimate the hydraulic <br /> conductivity of the transport sand from the average gradient <br /> Average gradient (1) from Table # =0.0034 <br /> . Assuming n=0.3 <br /> V =KI /n <br />