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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)
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