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-//Chapter-17,Example 6,Page 371
-clc();
-close();
-
-v= 10^6 //volume of water
-
-m1 = 40 //mass of Ca+2
-
-m2 = 24 //mass of Mg+2
-
-m3 = 44 //mass of CO2
-
-m4 = 122 //mass of HCO3-
-
-amnt_1 = 20 //amount of Ca+2 in ppm
-
-amnt_2 = 25 //amount of Mg+2 in ppm
-
-amnt_3 = 30 //amount of CO2 in ppm
-
-amnt_4 = 150 //amount of HCO3- in ppm
-
-lime_1 = (74/100)*[(amnt_2*100/m2)+(amnt_3*100/m3)+(amnt_4*100/m4)]*v
-
-soda = (106/100)*[(amnt_1*100/m1)+(amnt_2*100/m2)-(amnt_4*100/m4)]*v
-
-printf("the lime required is = %.3f mg",lime_1)
-
-printf("\n the soda required is = %.3f mg",soda)