initial commit / add all books

9 files changed, 214 insertions, 0 deletions

diff --git a/2465/CH17/EX17.1/Example_1.sce b/2465/CH17/EX17.1/Example_1.sce
new file mode 100644
index 000000000..ae0a60afd
--- /dev/null
+++ b/2465/CH17/EX17.1/Example_1.sce
@@ -0,0 +1,29 @@
+//Chapter-17,Example 1,Page 369

+clc();

+close();

+

+m1 = 146    //mass of Mg(HCO3)2

+

+m2 = 162    //mass of Ca(HCO3)2

+

+m3 = 95    //mass of MgCl2

+

+m4 = 136    //mass of CaSO4

+

+amnt_1 = 7.5     //amount of Mg(HCO3)2 in mg/l

+

+amnt_2 = 16     //amount of Ca(HCO3)2 in mg/l

+

+amnt_3 = 9     //amount of MgCl2 in mg/l

+

+amnt_4 = 13.6     //amount of CaSO4 in mg/l

+

+temp_hard= (amnt_1*100/m1)+(amnt_2*100/m2)

+

+perm_hard= (amnt_3*100/m3)+(amnt_4*100/m4)

+

+total= temp_hard +perm_hard

+

+printf("the temporary hardness is = %.2f mg/l",temp_hard)

+

+printf("\n the total  hardness is = %.2f mg/l",total)

diff --git a/2465/CH17/EX17.2/Example_2.sce b/2465/CH17/EX17.2/Example_2.sce
new file mode 100644
index 000000000..60ecef2e8
--- /dev/null
+++ b/2465/CH17/EX17.2/Example_2.sce
@@ -0,0 +1,14 @@
+//Chapter-17,Example 2,Page 369

+clc();

+close();

+

+m1= 136   // mass of FeSO4

+

+m2 = 100  //mass of CaCO3

+

+//for 100 ppm hardness FeSO4 required per 10^6 parts of water is 136 parts

+//for 200 ppm hardness

+

+amt= m1*200/m2

+

+printf("the amount of FeSO4 required is = %.f mg/l",amt)

diff --git a/2465/CH17/EX17.3/Example_3.sce b/2465/CH17/EX17.3/Example_3.sce
new file mode 100644
index 000000000..11296ac69
--- /dev/null
+++ b/2465/CH17/EX17.3/Example_3.sce
@@ -0,0 +1,11 @@
+//Chapter-17,Example 3,Page 369

+clc();

+close();

+

+conc = 15.6 *10^-6    //concentration of (CO3)-2

+

+m = 60   //mass of CO3

+

+Molarity= conc*100/m

+

+printf("the molarity of (CO3)-2  is = %.6f M",Molarity)

diff --git a/2465/CH17/EX17.4/Example_4.sce b/2465/CH17/EX17.4/Example_4.sce
new file mode 100644
index 000000000..a0af0e9db
--- /dev/null
+++ b/2465/CH17/EX17.4/Example_4.sce
@@ -0,0 +1,41 @@
+//Chapter-17,Example 4,Page 370

+clc();

+close();

+

+m1 = 146    //mass of Mg(HCO3)2

+

+m2 = 162    //mass of Ca(HCO3)2

+

+m3 = 111    //mass of CaCl2

+

+m4 = 120    //mass of MgSO4

+

+m5 = 136    //mass of CaSO4

+

+amnt_1 = 12.5     //amount of Mg(HCO3)2 in ppm

+

+amnt_2 = 10.5     //amount of Ca(HCO3)2 in ppm

+

+amnt_3 = 8.2     //amount of CaCl2 in ppm 

+

+amnt_4 = 2.6     //amount of MgSO4 in ppm

+

+amnt_5 = 7.5     //amount of CaSO4 in ppm

+

+temp_hard= (amnt_1*100/m1)+(amnt_2*100/m2)

+

+perm_hard= (amnt_3*100/m3)+(amnt_4*100/m4)+(amnt_5*100/m5)

+

+total= temp_hard +perm_hard

+

+printf("the temporary hardness is = %.3f mg/l",temp_hard)

+

+printf("\n the permanent hardness is = %.3f mg/l",perm_hard)

+

+printf("\n the total  hardness is = %.3f mg/l",total)

+

+v= 100    //volume of sample

+

+v_EDTA = total*v/1000   //volume of EDTA 

+

+printf("\n the volume of M/100 EDTA required is = %.3f ml",v_EDTA)

diff --git a/2465/CH17/EX17.5/Example_5.sce b/2465/CH17/EX17.5/Example_5.sce
new file mode 100644
index 000000000..81bbb40e5
--- /dev/null
+++ b/2465/CH17/EX17.5/Example_5.sce
@@ -0,0 +1,25 @@
+//Chapter-17,Example 5,Page 370

+clc();

+close();

+

+v= 50000  //volume of water

+

+m1 = 84    //mass of MgCO3

+

+m2 = 100    //mass of CaCO3

+

+m3 = 95    //mass of MgCl2

+

+m4 = 111    //mass of CaCl2

+

+amnt_1 = 144     //amount of MgCO3 in ppm

+

+amnt_2 = 25     //amount of CaCO3 in ppm

+

+amnt_3 = 95     //amount of MgCl2 in ppm 

+

+amnt_4 = 111     //amount of CaCl2 in ppm

+

+lime = (74/100)*[2*(amnt_1*100/m1)+(amnt_2*100/m2)+(amnt_3*100/m3)]*v

+

+printf("the lime required is = %.3f mg",lime)

diff --git a/2465/CH17/EX17.6/Example_6.sce b/2465/CH17/EX17.6/Example_6.sce
new file mode 100644
index 000000000..f138207aa
--- /dev/null
+++ b/2465/CH17/EX17.6/Example_6.sce
@@ -0,0 +1,29 @@
+//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)

diff --git a/2465/CH17/EX17.7/Example_7.sce b/2465/CH17/EX17.7/Example_7.sce
new file mode 100644
index 000000000..870f634e3
--- /dev/null
+++ b/2465/CH17/EX17.7/Example_7.sce
@@ -0,0 +1,17 @@
+//Chapter-17,Example 7,Page 371

+clc();

+close();

+

+v= 150    //volume of NaCl 

+

+conc = 150   //concentration of NaCl

+

+amnt =v*conc *100/117    //amnt of NaCl

+

+hard = 600   //hardness of water

+

+vol= amnt*1000/hard

+

+printf("the volume of water is = %.2f litres",vol)

+

+//calculation mistake in textbook

diff --git a/2465/CH17/EX17.8/Example_8.sce b/2465/CH17/EX17.8/Example_8.sce
new file mode 100644
index 000000000..3642e0698
--- /dev/null
+++ b/2465/CH17/EX17.8/Example_8.sce
@@ -0,0 +1,17 @@
+//Chapter-17,Example 8,Page 371

+clc();

+close();

+

+strength = 10*0.85/9    //strength of EDTA

+

+//1000 ml EDTA solution == 1 g CaCO3

+

+//for 20 ml EDTA solution

+

+amnt= 20*strength/1000

+

+//50 ml smple of water contains amnt CaCO3

+

+hard= amnt*10^6/50    //hardness of water 

+

+printf("the hardness of water is = %.2f ppm", hard)

diff --git a/2465/CH17/EX17.9/Example_9.sce b/2465/CH17/EX17.9/Example_9.sce
new file mode 100644
index 000000000..0436f50e0
--- /dev/null
+++ b/2465/CH17/EX17.9/Example_9.sce
@@ -0,0 +1,31 @@
+//Chapter-17,Example 9,Page 372

+clc();

+close();

+

+m1 = 146    //mass of Mg(HCO3)2

+

+m2 = 162    //mass of Ca(HCO3)2

+

+m3 = 111    //mass of CaCl2

+

+m4 = 120    //mass of MgSO4

+

+m5 = 136    //mass of CaSO4

+

+amnt_1 = 12.5     //amount of Mg(HCO3)2 in ppm

+

+amnt_2 = 10.5     //amount of Ca(HCO3)2 in ppm

+

+amnt_3 = 8.2     //amount of CaCl2 in ppm 

+

+amnt_4 = 2.6     //amount of MgSO4 in ppm

+

+amnt_5 = 7.5     //amount of CaSO4 in ppm

+

+temp_hard= [(amnt_1*100/m1)+(amnt_2*100/m2)]*0.1

+

+perm_hard= [(amnt_3*100/m3)+(amnt_4*100/m4)+(amnt_5*100/m5)]*0.1

+

+printf("the temporary hardness is = %.4f degree Fr",temp_hard)

+

+printf("\n the permanent hardness is = %.4f degree Fr",perm_hard)