From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

---
 2465/CH10/EX10.1/Example_1.sce   | 13 ++++++
 2465/CH10/EX10.10/Example_10.sce | 12 ++++++
 2465/CH10/EX10.11/Example_11.sce | 16 ++++++++
 2465/CH10/EX10.12/Example_12.sce | 30 ++++++++++++++
 2465/CH10/EX10.2/Example_2.sce   | 21 ++++++++++
 2465/CH10/EX10.3/Example_3.sce   | 25 ++++++++++++
 2465/CH10/EX10.4/Example_4.sce   | 23 +++++++++++
 2465/CH10/EX10.5/Example_5.sce   | 13 ++++++
 2465/CH10/EX10.6/Example_6.sce   | 14 +++++++
 2465/CH10/EX10.7/Example_7.sce   | 19 +++++++++
 2465/CH10/EX10.8/Example_8.sce   | 30 ++++++++++++++
 2465/CH10/EX10.9/Example_9.sce   | 25 ++++++++++++
 2465/CH11/EX11.1/Example_1.sce   | 17 ++++++++
 2465/CH11/EX11.2/Example_2.sce   | 19 +++++++++
 2465/CH17/EX17.1/Example_1.sce   | 29 ++++++++++++++
 2465/CH17/EX17.2/Example_2.sce   | 14 +++++++
 2465/CH17/EX17.3/Example_3.sce   | 11 +++++
 2465/CH17/EX17.4/Example_4.sce   | 41 +++++++++++++++++++
 2465/CH17/EX17.5/Example_5.sce   | 25 ++++++++++++
 2465/CH17/EX17.6/Example_6.sce   | 29 ++++++++++++++
 2465/CH17/EX17.7/Example_7.sce   | 17 ++++++++
 2465/CH17/EX17.8/Example_8.sce   | 17 ++++++++
 2465/CH17/EX17.9/Example_9.sce   | 31 +++++++++++++++
 2465/CH18/EX18.1/Example_1.sce   | 27 +++++++++++++
 2465/CH18/EX18.2/Example_2.sce   | 35 ++++++++++++++++
 2465/CH18/EX18.3/Example_3.sce   | 27 +++++++++++++
 2465/CH18/EX18.4/Example_4.sce   | 17 ++++++++
 2465/CH18/EX18.5/Example_5.sce   | 86 ++++++++++++++++++++++++++++++++++++++++
 2465/CH18/EX18.6/Example_6.sce   | 37 +++++++++++++++++
 2465/CH22/EX22.2/Example_2.sce   | 15 +++++++
 2465/CH22/EX22.4/Example_4.sce   | 26 ++++++++++++
 2465/CH22/EX22.5/Example_5.sce   | 22 ++++++++++
 2465/CH3/EX3.1/Example_1.sce     | 16 ++++++++
 2465/CH3/EX3.10/Example_10.sce   | 17 ++++++++
 2465/CH3/EX3.11/Example_11.sce   | 11 +++++
 2465/CH3/EX3.12/Example_12.sce   | 11 +++++
 2465/CH3/EX3.13/Example_13.sce   | 19 +++++++++
 2465/CH3/EX3.14/Example_14.sce   | 19 +++++++++
 2465/CH3/EX3.15/Example_15.sce   | 20 ++++++++++
 2465/CH3/EX3.16/Example_16.sce   | 22 ++++++++++
 2465/CH3/EX3.17/Example_17.sce   | 18 +++++++++
 2465/CH3/EX3.18/Example_18.sce   | 25 ++++++++++++
 2465/CH3/EX3.19/Example_19.sce   | 20 ++++++++++
 2465/CH3/EX3.2/Example_2.sce     | 20 ++++++++++
 2465/CH3/EX3.4/Examlpe_4.sce     | 23 +++++++++++
 2465/CH3/EX3.5/Example_5.sce     | 27 +++++++++++++
 2465/CH3/EX3.6/Example_6.sce     | 18 +++++++++
 2465/CH3/EX3.7/Example_7.sce     | 19 +++++++++
 2465/CH3/EX3.8/Example_8.sce     | 23 +++++++++++
 2465/CH3/EX3.9/Example_9.sce     | 25 ++++++++++++
 2465/CH4/EX4.1/Example_1.sce     | 17 ++++++++
 2465/CH4/EX4.10/Example_10.sce   | 19 +++++++++
 2465/CH4/EX4.11/Example_11.sce   | 17 ++++++++
 2465/CH4/EX4.12/Example_12.sce   | 13 ++++++
 2465/CH4/EX4.13/Example_13.sce   | 14 +++++++
 2465/CH4/EX4.15/Example_15.sce   | 19 +++++++++
 2465/CH4/EX4.16/Example_16.sce   | 27 +++++++++++++
 2465/CH4/EX4.17/Example_17.sce   | 19 +++++++++
 2465/CH4/EX4.18/Example_18.sce   | 21 ++++++++++
 2465/CH4/EX4.19/Example_19.sce   | 19 +++++++++
 2465/CH4/EX4.2/Example_2.sce     | 11 +++++
 2465/CH4/EX4.3/Example_3.sce     | 13 ++++++
 2465/CH4/EX4.4/Example_4.sce     | 21 ++++++++++
 2465/CH4/EX4.5/Example_5.sce     | 19 +++++++++
 2465/CH4/EX4.6/Example_6.sce     | 18 +++++++++
 2465/CH4/EX4.7/Example_7.sce     | 16 ++++++++
 2465/CH4/EX4.8/Example_8.sce     | 23 +++++++++++
 2465/CH4/EX4.9/Example_9.sce     | 15 +++++++
 2465/CH5/EX5.1/Example_1.sce     | 23 +++++++++++
 2465/CH5/EX5.10/Example_10.sce   | 29 ++++++++++++++
 2465/CH5/EX5.11/Example_11.sce   | 23 +++++++++++
 2465/CH5/EX5.12/Example_12.sce   | 23 +++++++++++
 2465/CH5/EX5.15/Example_15.sce   | 17 ++++++++
 2465/CH5/EX5.16/Example_16.sce   | 19 +++++++++
 2465/CH5/EX5.17/Example_17.sce   | 17 ++++++++
 2465/CH5/EX5.2/Example_2.sce     | 19 +++++++++
 2465/CH5/EX5.3/Example_3.sce     | 19 +++++++++
 2465/CH5/EX5.4/Example_4.sce     | 15 +++++++
 2465/CH5/EX5.5/Example_5.sce     | 17 ++++++++
 2465/CH5/EX5.6/Example_6.sce     | 17 ++++++++
 2465/CH5/EX5.7/Example_7.sce     | 27 +++++++++++++
 2465/CH5/EX5.9/Example_9.sce     | 15 +++++++
 2465/CH8/EX8.1/Example_1.sce     | 11 +++++
 2465/CH8/EX8.2/Example_2.sce     | 21 ++++++++++
 2465/CH8/EX8.3/Example_3.sce     | 32 +++++++++++++++
 2465/CH8/EX8.4/Example_4.sce     | 24 +++++++++++
 2465/CH8/EX8.5/Example_5.sce     | 15 +++++++
 2465/CH8/EX8.6/Example_6.sce     | 17 ++++++++
 2465/CH8/EX8.7/Example_7.sce     | 14 +++++++
 2465/CH8/EX8.8/Example_8.sce     | 19 +++++++++
 2465/CH9/EX9.1/Example_1.sce     | 19 +++++++++
 2465/CH9/EX9.10/Example_10.sce   | 22 ++++++++++
 2465/CH9/EX9.11/Example_11.sce   | 17 ++++++++
 2465/CH9/EX9.12/Example_12.sce   | 21 ++++++++++
 2465/CH9/EX9.2/Example_2.sce     | 16 ++++++++
 2465/CH9/EX9.3/Example_3.sce     | 17 ++++++++
 2465/CH9/EX9.4/Example_4.sce     | 25 ++++++++++++
 2465/CH9/EX9.5/Example_5.sce     | 17 ++++++++
 2465/CH9/EX9.6/Example_6.sce     | 11 +++++
 2465/CH9/EX9.7/Example_7.sce     | 16 ++++++++
 2465/CH9/EX9.8/Example_8.sce     | 15 +++++++
 2465/CH9/EX9.9/Example_9.sce     | 18 +++++++++
 102 files changed, 2104 insertions(+)
 create mode 100644 2465/CH10/EX10.1/Example_1.sce
 create mode 100644 2465/CH10/EX10.10/Example_10.sce
 create mode 100644 2465/CH10/EX10.11/Example_11.sce
 create mode 100644 2465/CH10/EX10.12/Example_12.sce
 create mode 100644 2465/CH10/EX10.2/Example_2.sce
 create mode 100644 2465/CH10/EX10.3/Example_3.sce
 create mode 100644 2465/CH10/EX10.4/Example_4.sce
 create mode 100644 2465/CH10/EX10.5/Example_5.sce
 create mode 100644 2465/CH10/EX10.6/Example_6.sce
 create mode 100644 2465/CH10/EX10.7/Example_7.sce
 create mode 100644 2465/CH10/EX10.8/Example_8.sce
 create mode 100644 2465/CH10/EX10.9/Example_9.sce
 create mode 100644 2465/CH11/EX11.1/Example_1.sce
 create mode 100644 2465/CH11/EX11.2/Example_2.sce
 create mode 100644 2465/CH17/EX17.1/Example_1.sce
 create mode 100644 2465/CH17/EX17.2/Example_2.sce
 create mode 100644 2465/CH17/EX17.3/Example_3.sce
 create mode 100644 2465/CH17/EX17.4/Example_4.sce
 create mode 100644 2465/CH17/EX17.5/Example_5.sce
 create mode 100644 2465/CH17/EX17.6/Example_6.sce
 create mode 100644 2465/CH17/EX17.7/Example_7.sce
 create mode 100644 2465/CH17/EX17.8/Example_8.sce
 create mode 100644 2465/CH17/EX17.9/Example_9.sce
 create mode 100644 2465/CH18/EX18.1/Example_1.sce
 create mode 100644 2465/CH18/EX18.2/Example_2.sce
 create mode 100644 2465/CH18/EX18.3/Example_3.sce
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 create mode 100644 2465/CH18/EX18.6/Example_6.sce
 create mode 100644 2465/CH22/EX22.2/Example_2.sce
 create mode 100644 2465/CH22/EX22.4/Example_4.sce
 create mode 100644 2465/CH22/EX22.5/Example_5.sce
 create mode 100644 2465/CH3/EX3.1/Example_1.sce
 create mode 100644 2465/CH3/EX3.10/Example_10.sce
 create mode 100644 2465/CH3/EX3.11/Example_11.sce
 create mode 100644 2465/CH3/EX3.12/Example_12.sce
 create mode 100644 2465/CH3/EX3.13/Example_13.sce
 create mode 100644 2465/CH3/EX3.14/Example_14.sce
 create mode 100644 2465/CH3/EX3.15/Example_15.sce
 create mode 100644 2465/CH3/EX3.16/Example_16.sce
 create mode 100644 2465/CH3/EX3.17/Example_17.sce
 create mode 100644 2465/CH3/EX3.18/Example_18.sce
 create mode 100644 2465/CH3/EX3.19/Example_19.sce
 create mode 100644 2465/CH3/EX3.2/Example_2.sce
 create mode 100644 2465/CH3/EX3.4/Examlpe_4.sce
 create mode 100644 2465/CH3/EX3.5/Example_5.sce
 create mode 100644 2465/CH3/EX3.6/Example_6.sce
 create mode 100644 2465/CH3/EX3.7/Example_7.sce
 create mode 100644 2465/CH3/EX3.8/Example_8.sce
 create mode 100644 2465/CH3/EX3.9/Example_9.sce
 create mode 100644 2465/CH4/EX4.1/Example_1.sce
 create mode 100644 2465/CH4/EX4.10/Example_10.sce
 create mode 100644 2465/CH4/EX4.11/Example_11.sce
 create mode 100644 2465/CH4/EX4.12/Example_12.sce
 create mode 100644 2465/CH4/EX4.13/Example_13.sce
 create mode 100644 2465/CH4/EX4.15/Example_15.sce
 create mode 100644 2465/CH4/EX4.16/Example_16.sce
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 create mode 100644 2465/CH4/EX4.18/Example_18.sce
 create mode 100644 2465/CH4/EX4.19/Example_19.sce
 create mode 100644 2465/CH4/EX4.2/Example_2.sce
 create mode 100644 2465/CH4/EX4.3/Example_3.sce
 create mode 100644 2465/CH4/EX4.4/Example_4.sce
 create mode 100644 2465/CH4/EX4.5/Example_5.sce
 create mode 100644 2465/CH4/EX4.6/Example_6.sce
 create mode 100644 2465/CH4/EX4.7/Example_7.sce
 create mode 100644 2465/CH4/EX4.8/Example_8.sce
 create mode 100644 2465/CH4/EX4.9/Example_9.sce
 create mode 100644 2465/CH5/EX5.1/Example_1.sce
 create mode 100644 2465/CH5/EX5.10/Example_10.sce
 create mode 100644 2465/CH5/EX5.11/Example_11.sce
 create mode 100644 2465/CH5/EX5.12/Example_12.sce
 create mode 100644 2465/CH5/EX5.15/Example_15.sce
 create mode 100644 2465/CH5/EX5.16/Example_16.sce
 create mode 100644 2465/CH5/EX5.17/Example_17.sce
 create mode 100644 2465/CH5/EX5.2/Example_2.sce
 create mode 100644 2465/CH5/EX5.3/Example_3.sce
 create mode 100644 2465/CH5/EX5.4/Example_4.sce
 create mode 100644 2465/CH5/EX5.5/Example_5.sce
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 create mode 100644 2465/CH5/EX5.7/Example_7.sce
 create mode 100644 2465/CH5/EX5.9/Example_9.sce
 create mode 100644 2465/CH8/EX8.1/Example_1.sce
 create mode 100644 2465/CH8/EX8.2/Example_2.sce
 create mode 100644 2465/CH8/EX8.3/Example_3.sce
 create mode 100644 2465/CH8/EX8.4/Example_4.sce
 create mode 100644 2465/CH8/EX8.5/Example_5.sce
 create mode 100644 2465/CH8/EX8.6/Example_6.sce
 create mode 100644 2465/CH8/EX8.7/Example_7.sce
 create mode 100644 2465/CH8/EX8.8/Example_8.sce
 create mode 100644 2465/CH9/EX9.1/Example_1.sce
 create mode 100644 2465/CH9/EX9.10/Example_10.sce
 create mode 100644 2465/CH9/EX9.11/Example_11.sce
 create mode 100644 2465/CH9/EX9.12/Example_12.sce
 create mode 100644 2465/CH9/EX9.2/Example_2.sce
 create mode 100644 2465/CH9/EX9.3/Example_3.sce
 create mode 100644 2465/CH9/EX9.4/Example_4.sce
 create mode 100644 2465/CH9/EX9.5/Example_5.sce
 create mode 100644 2465/CH9/EX9.6/Example_6.sce
 create mode 100644 2465/CH9/EX9.7/Example_7.sce
 create mode 100644 2465/CH9/EX9.8/Example_8.sce
 create mode 100644 2465/CH9/EX9.9/Example_9.sce

(limited to '2465')

diff --git a/2465/CH10/EX10.1/Example_1.sce b/2465/CH10/EX10.1/Example_1.sce
new file mode 100644
index 000000000..02cf4577c
--- /dev/null
+++ b/2465/CH10/EX10.1/Example_1.sce
@@ -0,0 +1,13 @@
+//Chapter-10,Example 1,Page 252
+clc();
+close();
+
+E = 0.296    //electrode potential at 25 degree
+
+n= 2
+
+Cu = 0.015
+
+E0=E-(0.0592/n)*log10(Cu)
+
+printf('the standard potential of Cu+2 is %.5f V ',E0)
diff --git a/2465/CH10/EX10.10/Example_10.sce b/2465/CH10/EX10.10/Example_10.sce
new file mode 100644
index 000000000..1137bac40
--- /dev/null
+++ b/2465/CH10/EX10.10/Example_10.sce
@@ -0,0 +1,12 @@
+//Chapter-10,Example 10,Page 255
+clc();
+close();
+
+//E_H = -0.0592*pH
+//E_cell = E_H = -0.0592 *pH
+
+E_cell = 0.29
+
+pH = E_cell/0.0592
+
+printf('the pH of the solution is pH = %.2f ',pH)
diff --git a/2465/CH10/EX10.11/Example_11.sce b/2465/CH10/EX10.11/Example_11.sce
new file mode 100644
index 000000000..3577f29b4
--- /dev/null
+++ b/2465/CH10/EX10.11/Example_11.sce
@@ -0,0 +1,16 @@
+//Chapter-10,Example 11,Page 255
+clc();
+close();
+
+E_cell = 0.123
+
+E_calomel = 0.2415
+
+E_Q = 0.6990
+
+//E_Q/H2Q = E_Q - 0.0592 *pH
+//E_cell= E_Q/H2Q - E_calomel
+
+pH = (E_cell + E_calomel - E_Q)/(-0.0592)
+
+printf('the pH  of solution is pH = %.2f',pH)
diff --git a/2465/CH10/EX10.12/Example_12.sce b/2465/CH10/EX10.12/Example_12.sce
new file mode 100644
index 000000000..8b2460a55
--- /dev/null
+++ b/2465/CH10/EX10.12/Example_12.sce
@@ -0,0 +1,30 @@
+//Chapter-10,Example 12,Page 255
+clc();
+close();
+
+R=8.316    //gas constant
+
+F=96500    //Farade's constant
+
+n=1
+
+T=298      //temperature in Kelvin
+
+E0_AgCl=-0.2223   
+
+E0_Ag=0.798
+
+//cell reaction...Ag + Cl- <----> AgCl
+
+E0_cell =E0_Ag + E0_AgCl
+
+//at equilibrium two electrode potential s will be equal
+//   E0_cell = (2.303*R*T/n*F)*log10(K)
+
+Ksp = 10^-(E0_cell*n*F/(2.303*R*T))
+
+printf('for AgCl solution Ksp = ')
+
+disp(Ksp)
+
+printf('    mol^2/l^2')
diff --git a/2465/CH10/EX10.2/Example_2.sce b/2465/CH10/EX10.2/Example_2.sce
new file mode 100644
index 000000000..aec51c898
--- /dev/null
+++ b/2465/CH10/EX10.2/Example_2.sce
@@ -0,0 +1,21 @@
+//Chapter-10,Example 2,Page 252
+clc();
+close();
+
+E0 = 0.34    //standard potential for copper
+
+n= 2
+
+Cu = 0.15
+
+R=8.314    //gas constant
+
+F=96500    //Farade's constant
+
+n=2
+
+T=298      //temperature in Kelvin
+
+E=E0+(2.303*R*T/(n*F))*log10(Cu)
+
+printf('the single electrode potential of copper is %.5f V ',E)
diff --git a/2465/CH10/EX10.3/Example_3.sce b/2465/CH10/EX10.3/Example_3.sce
new file mode 100644
index 000000000..e6f950fa8
--- /dev/null
+++ b/2465/CH10/EX10.3/Example_3.sce
@@ -0,0 +1,25 @@
+//Chapter-10,Example 3,Page 252
+clc();
+close();
+
+//Cell reaction is ...Zn+2  +2Ag <----> Zn + 2Ag+
+
+E0_Zn=-0.762     //standard electrode potential for Zn
+
+E0_Ag=0.798     //standard electrode potential for Ag
+
+R=8.314    //gas constant
+
+F=96500    //Farade's constant
+
+n=2
+
+T=298      //temperature in Kelvin
+
+Zn= 0.2
+
+Ag= 0.1
+
+E_cell= (E0_Zn + (R*T/(n*F))*log(Zn))-(E0_Ag + (R*T/(n*F))*log(Ag^2))
+
+printf('the cell voltage at 25 degree is %.3f V',E_cell)
diff --git a/2465/CH10/EX10.4/Example_4.sce b/2465/CH10/EX10.4/Example_4.sce
new file mode 100644
index 000000000..1b8546cea
--- /dev/null
+++ b/2465/CH10/EX10.4/Example_4.sce
@@ -0,0 +1,23 @@
+//Chapter-10,Example 4,Page 253
+clc();
+close();
+
+//Cell reaction is ...Zn+2  +2Ag <----> Zn + 2Ag+
+
+E0_cell= 1.1    //standard potential for cell 
+
+R=8.314    //gas constant
+
+F=96500    //Farade's constant
+
+n=2
+
+T=298      //temperature in Kelvin
+
+Zn= 0.001
+
+Cu= 0.1
+
+E_cell=E0_cell+(2.303*R*T/(n*F))*log10(Cu/Zn)
+
+printf('the e.m.f. of Daniel cell is %.4f V',E_cell)
diff --git a/2465/CH10/EX10.5/Example_5.sce b/2465/CH10/EX10.5/Example_5.sce
new file mode 100644
index 000000000..714124f20
--- /dev/null
+++ b/2465/CH10/EX10.5/Example_5.sce
@@ -0,0 +1,13 @@
+//Chapter-10,Example 5,Page 253
+clc();
+close();
+
+E0_Pb=-0.13   
+
+E0_Ni=-0.24   
+
+E0_cell=E0_Pb-E0_Ni
+
+printf('the e.m.f. of cell is %.4f V',E0_cell)
+printf('\n the cell reaction is')
+printf('\n  Ni + Pb+2 <----> Ni+2 + Pb')
diff --git a/2465/CH10/EX10.6/Example_6.sce b/2465/CH10/EX10.6/Example_6.sce
new file mode 100644
index 000000000..da05652c0
--- /dev/null
+++ b/2465/CH10/EX10.6/Example_6.sce
@@ -0,0 +1,14 @@
+//Chapter-10,Example 5,Page 253
+clc();
+close();
+
+E0_Zn=-0.76   
+
+E0_Ag=0.8   
+
+E0_cell=E0_Ag-E0_Zn
+
+printf('\n  the cell reaction is')
+printf('\n  2Ag+ + Zn <----> 2Ag + Zn+2')
+printf('\n  the e.m.f. of cell is %.4f V',E0_cell)
+
diff --git a/2465/CH10/EX10.7/Example_7.sce b/2465/CH10/EX10.7/Example_7.sce
new file mode 100644
index 000000000..c8f37d6ed
--- /dev/null
+++ b/2465/CH10/EX10.7/Example_7.sce
@@ -0,0 +1,19 @@
+//Chapter-10,Example 7,Page 254
+clc();
+close();
+
+R=8.314    //gas constant
+
+F=96500    //Farade's constant
+
+n=2
+
+T=298      //temperature in Kelvin
+
+C1= 0.01
+
+C2= 0.1
+
+E_cell=(2.303*R*T/(n*F))*log10(C2/C1)
+
+printf('the e.m.f. of cell is %.4f V',E_cell)
diff --git a/2465/CH10/EX10.8/Example_8.sce b/2465/CH10/EX10.8/Example_8.sce
new file mode 100644
index 000000000..ce565986f
--- /dev/null
+++ b/2465/CH10/EX10.8/Example_8.sce
@@ -0,0 +1,30 @@
+//Chapter-10,Example 8,Page 254
+clc();
+close();
+
+R=8.316    //gas constant
+
+F=96500    //Farade's constant
+
+n=2
+
+T=298      //temperature in Kelvin
+
+E0_Zn=-0.765   
+
+E0_Cu=0.337
+
+//cell reaction...Zn + Cu+2 <----> Zn+2 + Cu
+//  K = [Zn+2]*[Cu]/[Zn]*[Cu+2]...equilibrium constant
+
+E0_cell =E0_Cu - E0_Zn
+
+//at equilibrium two electrode potential s will be equal
+//   E0_cell = (2.303*R*T/n*F)*log10([Zn+2]*[Cu]/[Zn]*[Cu+2])
+//   E0_cell = (2.303*R*T/n*F)*log10(K)
+
+K = 10^(E0_cell/(2.303*R*T/(n*F)))
+
+printf('the equilibrium constant is K = ')
+
+disp(K)
diff --git a/2465/CH10/EX10.9/Example_9.sce b/2465/CH10/EX10.9/Example_9.sce
new file mode 100644
index 000000000..d9ab56f37
--- /dev/null
+++ b/2465/CH10/EX10.9/Example_9.sce
@@ -0,0 +1,25 @@
+//Chapter-10,Example 9,Page 255
+clc();
+close();
+
+E0_Ag = 0.799    //standard potential for copper
+
+Ksp=8.3*10^-17
+
+I=1
+
+Ag= Ksp/I
+
+n= 2
+
+R=8.314    //gas constant
+
+F=96500    //Farade's constant
+
+n=2
+
+T=298      //temperature in Kelvin
+
+E_Ag=E0_Ag+(2.303*R*T/(n*F))*log10(Ag)
+
+printf('the single electrode potential of Ag is %.5f V ',E_Ag)
diff --git a/2465/CH11/EX11.1/Example_1.sce b/2465/CH11/EX11.1/Example_1.sce
new file mode 100644
index 000000000..a91312a6c
--- /dev/null
+++ b/2465/CH11/EX11.1/Example_1.sce
@@ -0,0 +1,17 @@
+//Chapter-11,Example 1,Page 275
+clc();
+close();
+
+M =1000    //mass of alloy
+
+m_Cd= 0.25*M   //25% of Cd in alloy
+
+//since in the eutectic system, 40% is Cd and 60% is Bi
+
+//corresponding to m_Cd  Cd the content of Bi in eutectic is 
+
+m_Bi = m_Cd*60/40
+
+m= m_Cd+m_Bi
+
+printf('the mass of eutectic in 1 kg alloy is %.f gm ',m)
diff --git a/2465/CH11/EX11.2/Example_2.sce b/2465/CH11/EX11.2/Example_2.sce
new file mode 100644
index 000000000..a1c82f3a8
--- /dev/null
+++ b/2465/CH11/EX11.2/Example_2.sce
@@ -0,0 +1,19 @@
+//Chapter-11,Example 2,Page 275
+clc();
+close();
+
+M =1000    //mass of alloy
+
+m_A= 0.4*M   //40% of A in alloy
+
+m_B= 0.6*M   //60% of B in alloy
+
+//since in the eutectic system, 40% is B and 60% is A
+
+//corresponding to m_A  the content of m_B in eutectic is 
+
+m_Be = m_A*40/60   //in eutectic 
+
+m= m_B-m_Be  //amount of B separated out
+
+printf('the amount of B separated out is %.2f gm ',m)
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)
diff --git a/2465/CH18/EX18.1/Example_1.sce b/2465/CH18/EX18.1/Example_1.sce
new file mode 100644
index 000000000..d93962d69
--- /dev/null
+++ b/2465/CH18/EX18.1/Example_1.sce
@@ -0,0 +1,27 @@
+//Chapter-18,Example 1,Page 404
+clc();
+close();
+
+H=6
+
+W= 2200    //water equivalent of bomb calorimeter
+
+w= 550    //weight of water taken
+
+del_t = 2.42    //rise in temperature
+
+m= 0.92    //weight of coal burnt
+
+L =580    //latent heat of steam
+
+fuse = 10   //fuse correction
+
+acid =50    //acid correction
+
+HCV=((W+w)*(del_t)-(acid+fuse))/m
+
+NCV=HCV-(0.09*H*L)
+
+printf("HCV = %.2f cal/g",HCV)
+
+printf("\n NCV = %.2f cal/g",NCV)
diff --git a/2465/CH18/EX18.2/Example_2.sce b/2465/CH18/EX18.2/Example_2.sce
new file mode 100644
index 000000000..76386308f
--- /dev/null
+++ b/2465/CH18/EX18.2/Example_2.sce
@@ -0,0 +1,35 @@
+//Chapter-18,Example 2,Page 405
+clc();
+close();
+
+W1 = 2.5     //weight of coal
+
+W2 = 2.415   //weight of coal after heating at 110 C
+
+W_res= 1.528  //weight of residue
+
+W_ash= 0.245    //weight of ash
+
+Mois= W1-W2    //moisture in sample
+
+per_M=Mois*100/W1
+
+printf("the percentage of moisture is %.2f ",per_M )
+
+VCM=W2-W_res  //amount of VCM in sample
+
+per_VCM=VCM*100/W1
+
+printf("\n the percentage of VCM is %.2f ",per_VCM )
+
+per_ash=W_ash*100/W1
+
+printf("\n the percentage of ash is %.2f",per_ash )
+
+Fix_C= W_res-W_ash  //fixed carbon
+
+per_fix=Fix_C*100/W1
+
+printf("\n the percentage of fixed carbon is %.2f",per_fix )
+
+//mistake in textbook
diff --git a/2465/CH18/EX18.3/Example_3.sce b/2465/CH18/EX18.3/Example_3.sce
new file mode 100644
index 000000000..b3e14c5d3
--- /dev/null
+++ b/2465/CH18/EX18.3/Example_3.sce
@@ -0,0 +1,27 @@
+//Chapter-18,Example 3,Page 405
+clc();
+close();
+
+H=0.77
+
+W= 395    //water equivalent of bomb calorimeter
+
+w= 3500    //weight of water taken
+
+T1=26.5    //temperature
+
+T2=29.2    //temperature
+
+m= 0.83    //weight of fuel burnt
+
+L =587   //latent heat of steam
+
+HCV=((W+w)*(T2-T1))/m
+
+NCV=HCV-(0.09*H*L)
+
+printf("HCV = %.2f cal/g",HCV)
+
+printf("\n NCV = %.2f cal/g",NCV)
+
+//calculation mistake in textbook
diff --git a/2465/CH18/EX18.4/Example_4.sce b/2465/CH18/EX18.4/Example_4.sce
new file mode 100644
index 000000000..ba8dab90c
--- /dev/null
+++ b/2465/CH18/EX18.4/Example_4.sce
@@ -0,0 +1,17 @@
+//Chapter-18,Example 4,Page 406
+clc();
+close();
+
+V1= 25     //volume of H2SO4
+
+V2 =15     //volumeof NaOH
+
+v= V1*0.1-V2*0.1    //volume of H2SO4  consumed
+
+//100 cc H2SO4 ==17 g NH3 == 14 g N
+//1 cc H2SO4 = 14/1000 g N =0.014 g N
+//0.014 g N is present in 1 g coal
+
+N= 0.014*100
+
+printf("the percentage of nitrogen is %.2f ",N)
diff --git a/2465/CH18/EX18.5/Example_5.sce b/2465/CH18/EX18.5/Example_5.sce
new file mode 100644
index 000000000..811aae588
--- /dev/null
+++ b/2465/CH18/EX18.5/Example_5.sce
@@ -0,0 +1,86 @@
+//Chapter-18,Example 5,Page 406
+clc();
+close();
+
+
+H2 =0.24   //composition of H2
+
+CH4 =0.3   //composition of CH4
+
+CO =0.06   //composition of CO
+
+C2H6 =0.11   //composition of C2H6
+
+C2H4 =0.045   //composition of C2H4
+
+C4H8 =0.025   //composition of C4H8
+
+N2=0.12   //composition of N2
+
+CO2=0.08  //composition of CO2
+
+O2=0.02   //composition of O2
+
+//for reaction H2 + (1/2)O2 = H2O
+
+V1=H2*(1/2)   //volume of O2 required
+
+//for reaction CH4 + 2O2  = CO2 + 2H2O
+
+V2=CH4*2   //volume of O2 required
+vCO2_1=CH4*1   //volume of CO2
+
+//for reaction C2H6 + (7/2)O2 = 2CO2 +3H2O
+
+V3=C2H6*(7/2)   //volume of O2 required
+vCO2_2=C2H6*2   //volume of CO2
+
+//for reaction C2H4 + 3O2 = 2CO2 +2H2O
+
+V4=C2H4*3   //volume of O2 required
+vCO2_3=C2H4*2   //volume of CO2
+
+//for reaction C4H8 + 6O2 = 4CO2 +4H2O
+
+V5=C4H8*6   //volume of O2 required
+vCO2_4=C4H8*4   //volume of CO2
+
+//for reaction CO + (1/2)O2  = CO2
+
+V6=CO*(1/2)   //volume of O2 required
+vCO2_5=CO*1   //volume of CO2
+
+total_O2= V1+V2+V3+V4+V5+V6-O2  //total volume of oxygen
+
+//as air contains 21% of O2 by volume
+//when 40% excess
+
+V_air = total_O2*(100/21)*(140/100)   //volume of air 
+
+printf("the air to fuel ratio is %.3f",V_air)
+
+total_CO2 = vCO2_1+vCO2_2+vCO2_3+vCO2_4+vCO2_5+CO2  //total volume of CO2
+
+total_dry= total_CO2 +[N2+(79*V_air/100)]+[(V_air*21/100)-total_O2]
+
+printf("\n the total volume of dry products is %.4f cubicmeter ",total_dry)
+
+CO2_dry =total_CO2*100/total_dry
+
+N2_dry =[N2+(79*V_air/100)]*100/total_dry
+
+O2_dry =[(V_air*21/100)-total_O2]*100/total_dry
+
+printf("\n  Composition of products of combustion on dry basis")
+
+printf("\n  CO2 = %.3f",CO2_dry)
+
+printf("\n  N2 = %.3f",N2_dry)
+
+printf("\n  O2 = %.3f",O2_dry)
+
+//calculation mistake in textbook
+
+
+
+
diff --git a/2465/CH18/EX18.6/Example_6.sce b/2465/CH18/EX18.6/Example_6.sce
new file mode 100644
index 000000000..9441a7a58
--- /dev/null
+++ b/2465/CH18/EX18.6/Example_6.sce
@@ -0,0 +1,37 @@
+//Chapter-18,Example 6,Page 407
+clc();
+close();
+
+CO =0.46   //composition of CO
+
+CH4 =0.1   //composition of CH4
+
+H2 =0.4   //composition of H2
+
+C2H2 =0.02   //composition of C2H2
+
+N2=0.01   //composition of N2
+
+//for reaction CO + (1/2)O2  = CO2
+
+V1=CO*(1/2)   //volume of O2 required
+
+//for reaction CH4 + 2O2  = CO2 + 2H2O
+
+V2=CH4*2   //volume of O2 required
+
+//for reaction H2 + (1/2)O2 = H2O
+
+V3=H2*(1/2)   //volume of O2 required
+
+//for reaction C2H2 + (5/2)O2 = 2CO2 +H2O
+
+V4=C2H2*(5/2)   //volume of O2 required
+
+total_v= V1+V2+V3+V4
+
+//as air contains 21% of O2 by volume
+
+V_air = total_v*100/21   //volume of air 
+
+printf("the volume of air required is %.3f cubicmeter",V_air)
diff --git a/2465/CH22/EX22.2/Example_2.sce b/2465/CH22/EX22.2/Example_2.sce
new file mode 100644
index 000000000..2b2998ca9
--- /dev/null
+++ b/2465/CH22/EX22.2/Example_2.sce
@@ -0,0 +1,15 @@
+//Chapter-22,Example 2,Page 502
+clc();
+close();
+
+h=2
+
+k=2
+
+l=0
+
+a= 450   //length of cube in pm
+
+d=a/sqrt((h^2)+(k^2)+(l^2))
+
+printf("\n the spacing between planes is d = %.1f pm",d)
diff --git a/2465/CH22/EX22.4/Example_4.sce b/2465/CH22/EX22.4/Example_4.sce
new file mode 100644
index 000000000..445fdbd62
--- /dev/null
+++ b/2465/CH22/EX22.4/Example_4.sce
@@ -0,0 +1,26 @@
+//Chapter-22,Example 4,Page 502
+clc();
+close();
+
+M=58.46    //molecular weight of NaCl
+
+N= 6.023*10^23   //Avogadro number
+
+p=2.167   //density of NaCl
+
+n= 4    //number of molecules per unit cell
+
+a=nthroot((n*M/(p*N)),3)/100   //lenght of the edge
+
+h=1
+
+k=1 
+
+l=0
+
+d=a/sqrt((h^2)+(k^2)+(l^2))
+
+printf("the lattice constant is a= %.12f meter",a)
+
+printf("\n the spacing between planes is d = %.10f meter",d)
+
diff --git a/2465/CH22/EX22.5/Example_5.sce b/2465/CH22/EX22.5/Example_5.sce
new file mode 100644
index 000000000..91e96006a
--- /dev/null
+++ b/2465/CH22/EX22.5/Example_5.sce
@@ -0,0 +1,22 @@
+//Chapter-22,Example 5,Page 502
+clc();
+close();
+
+//since output current of transistor is 96% of the input current
+
+alpha = 96/100     //current gain  =  output current/input current
+
+Rout= 2000    //output resistance
+
+Rin= 20    //input resistance
+
+R_gain= Rout/Rin   //resistance gain
+
+//According to Ohm's law  V=I*R
+
+volt_gain = R_gain*alpha
+
+printf("the voltage gain = %.f",volt_gain)
+
+//voltage gain has no unit 
+//printing mistake in textbook
diff --git a/2465/CH3/EX3.1/Example_1.sce b/2465/CH3/EX3.1/Example_1.sce
new file mode 100644
index 000000000..093b57b8d
--- /dev/null
+++ b/2465/CH3/EX3.1/Example_1.sce
@@ -0,0 +1,16 @@
+
+//Chapter-3,Example 1,Page 56
+clc;
+close;
+
+M_0=200  //mass of radium
+
+total_time= 8378-1898   //in years
+
+//since t-half for radium is 1620 years
+
+t_half=6480/1620  // number of half lives
+
+m_left=M_0*(1/2)^t_half   //mass of radium left
+
+printf('mass of radium left after 6480 years is %.1f mg',m_left)
diff --git a/2465/CH3/EX3.10/Example_10.sce b/2465/CH3/EX3.10/Example_10.sce
new file mode 100644
index 000000000..334f2bd4d
--- /dev/null
+++ b/2465/CH3/EX3.10/Example_10.sce
@@ -0,0 +1,17 @@
+//Chapter-3,Example 10,Page 59
+clc;
+close;
+
+t_half = 5577    //half life of carbon(14)
+
+amnt = 1/6      // amount of carbon in fresh wood
+
+t= 2.303*t_half*log10(1/amnt)/0.693
+
+printf('the age of the wood is')
+
+disp(t)
+
+printf('years')
+
+//mistake in textbook
diff --git a/2465/CH3/EX3.11/Example_11.sce b/2465/CH3/EX3.11/Example_11.sce
new file mode 100644
index 000000000..945857c41
--- /dev/null
+++ b/2465/CH3/EX3.11/Example_11.sce
@@ -0,0 +1,11 @@
+//Chapter-3,Example 11,Page 59
+clc;
+close;
+
+t_half = 5760    //half life of carbon(14)
+
+amnt = 1/4      // amount of carbon in fresh wood
+
+t= 2.303*t_half*log10(1/amnt)/0.693
+
+printf('the age of the wood is %.f years ',t)
diff --git a/2465/CH3/EX3.12/Example_12.sce b/2465/CH3/EX3.12/Example_12.sce
new file mode 100644
index 000000000..8e01cce46
--- /dev/null
+++ b/2465/CH3/EX3.12/Example_12.sce
@@ -0,0 +1,11 @@
+//Chapter-3,Example 12,Page 60
+clc;
+close;
+
+t_half =6.13    //half life of Ac(222)
+
+t= 10   //time period
+
+amnt=1/10^(t*0.693/(2.303*t_half))
+
+printf('the amount of the substance left is %.4f ',amnt)
diff --git a/2465/CH3/EX3.13/Example_13.sce b/2465/CH3/EX3.13/Example_13.sce
new file mode 100644
index 000000000..fe6f94a34
--- /dev/null
+++ b/2465/CH3/EX3.13/Example_13.sce
@@ -0,0 +1,19 @@
+//Chapter-3,Example 13,Page 60
+clc;
+close;
+
+//Reaction.....N(14) + He(4) ---> O(17) +H(1)
+
+m_r= 18.01140     // total mass of reactants in  a.m.u.
+
+m_p= 18.01264     // total mass of product in a.m.u.
+
+m= m_p -m_r    // increase in mass
+
+Q_value= m*931   // in electron volt   since 1 a.m.u. =931 MeV
+
+//since mass is increased after reaction
+// Q value is negative
+
+printf('the Q value for the reaction is %.2f MeV',-Q_value)
+
diff --git a/2465/CH3/EX3.14/Example_14.sce b/2465/CH3/EX3.14/Example_14.sce
new file mode 100644
index 000000000..418a68b43
--- /dev/null
+++ b/2465/CH3/EX3.14/Example_14.sce
@@ -0,0 +1,19 @@
+//Chapter-3,Example 14,Page 61
+clc;
+close;
+
+//Reaction.....Li(7) + H(1) ---> He(4) +He(4)
+
+m_r= 8.02636     // total mass of reactants in  a.m.u.
+
+m_p= 8.00774     // total mass of product in a.m.u.
+
+m= m_r -m_p    // increase in mass
+
+Q_value= m*931   // in electron volt   since 1 a.m.u. =931 MeV
+
+//since mass is decreased after reaction
+// Q value is positive
+
+printf('the Q value for the reaction is %.2f MeV',Q_value)
+
diff --git a/2465/CH3/EX3.15/Example_15.sce b/2465/CH3/EX3.15/Example_15.sce
new file mode 100644
index 000000000..a9db46a01
--- /dev/null
+++ b/2465/CH3/EX3.15/Example_15.sce
@@ -0,0 +1,20 @@
+//Chapter-3,Example 15,Page 61
+clc;
+close;
+
+//Reaction.....Li(7) + D(2) ---> He(4) + He(4) + Q
+
+m_Li= 6.01702     // Isotopic  mass of Lithium in  a.m.u.
+
+m_D= 2.01474     //  Isotopic  mass of D in  a.m.u.
+
+m_He= 4.00387     //  Isotopic  mass of Helium in  a.m.u.
+
+Q_value= (m_Li + m_D - 2*m_He)*931   // in electron volt   since 1 a.m.u. =931 MeV
+
+//since mass is decreased after reaction
+// Q value is positive
+
+printf('the Q value for the reaction is %.2f MeV',Q_value)
+
+//mistake in textbook
diff --git a/2465/CH3/EX3.16/Example_16.sce b/2465/CH3/EX3.16/Example_16.sce
new file mode 100644
index 000000000..cee4cbed1
--- /dev/null
+++ b/2465/CH3/EX3.16/Example_16.sce
@@ -0,0 +1,22 @@
+//Chapter-3,Example 16,Page 61
+clc;
+close;
+
+//Reaction.....U(235) + n(1) ---> Kr(95) + Ba(139) + 2*n(1) + Q
+
+m_U= 235.124     // Isotopic  mass of Uranium in  a.m.u.
+
+m_n= 1.0099     //    mass of neutron in  a.m.u.
+
+m_Kr= 94.945     //  Isotopic  mass of Kripton in  a.m.u.
+
+m_Ba=138.954      //  Isotopic  mass of Ba in  a.m.u.
+
+Q_value= (m_U + m_n - (m_Kr + m_Ba + 2*m_n))*931   // in electron volt   since 1 a.m.u. =931 MeV
+
+//since mass is decreased after reaction
+// Q value is positive
+
+printf('the Q value for the reaction is %.3f MeV',Q_value)
+
+//mistake in textbook
diff --git a/2465/CH3/EX3.17/Example_17.sce b/2465/CH3/EX3.17/Example_17.sce
new file mode 100644
index 000000000..2138c04fb
--- /dev/null
+++ b/2465/CH3/EX3.17/Example_17.sce
@@ -0,0 +1,18 @@
+//Chapter-3,Example 17,Page 61
+clc;
+close;
+
+m_Ca = 39.975   //atomic mass of Calcium in a.m.u.
+
+a_no= 20  //atomic number of calcium
+
+m_proton = 1.0078  //mass of proton
+
+m_neutron = 1.0086    //mass of neutron\
+
+delta_m=a_no*(m_neutron + m_proton)- m_Ca  //mass defect
+
+energy= delta_m*931/40   //binding energy per nucleon
+
+printf('binding energy per nucleon is %.3f MeV',energy)
+
diff --git a/2465/CH3/EX3.18/Example_18.sce b/2465/CH3/EX3.18/Example_18.sce
new file mode 100644
index 000000000..b132f9fdc
--- /dev/null
+++ b/2465/CH3/EX3.18/Example_18.sce
@@ -0,0 +1,25 @@
+//Chapter-3,Example 18,Page 61
+clc;
+close;
+
+energy_1= 200 *1.6*10^-13    //energy released per fission of Uranium
+
+power =1   //in watt
+
+F_rate = power/energy_1   //fission rate for generation 1 watt
+
+printf('The fission rate for generation 1 watt is ')
+
+disp(F_rate)
+
+printf('    fission/sec')
+
+//1 kg atom Of U(235) =235 Kg = 6.023*10^26 atoms
+
+energy_2 = energy_1*6.023*10^26/235    //energy released per 1 kg U(235)
+
+printf('\nThe energy released per 1kg of U(235) is ')
+
+disp(energy_2)
+
+printf('    Joule')
diff --git a/2465/CH3/EX3.19/Example_19.sce b/2465/CH3/EX3.19/Example_19.sce
new file mode 100644
index 000000000..e0c98256f
--- /dev/null
+++ b/2465/CH3/EX3.19/Example_19.sce
@@ -0,0 +1,20 @@
+//Chapter-3,Example 19,Page 62
+clc;
+close;
+
+energy= (100*10^6)*24*3600    //energy comsumed in city in a day in Joule
+
+efcy=20/100   //efficiency of reactor
+
+energy_r = energy/efcy    //energy required per day
+
+energy_rl=200*1.6*10^-13   //energy released per nuclide
+
+n = energy_r/energy_rl    //number of U(235) to be fissioned 
+
+//6.023*10^26 atoms of U(235) are present in 235 kg
+//n  atoms of U(235) are present in
+
+m=235*n/(6.023*10^26)
+
+printf('the amount of fule required for one day operation is %.2f kg',m)
diff --git a/2465/CH3/EX3.2/Example_2.sce b/2465/CH3/EX3.2/Example_2.sce
new file mode 100644
index 000000000..f89b82a5e
--- /dev/null
+++ b/2465/CH3/EX3.2/Example_2.sce
@@ -0,0 +1,20 @@
+//Chapter-3,Example 2,Page 56
+clc;
+close;
+
+m_alpha=6.646*10^-24    //mass of one alpha particle
+
+n= 2300 // number of alpha particles
+
+M=1*10^-6  //mass of plutonium
+
+//as -(dM/dt)= lamda*M 
+//also (dM/dt)= n*m_alpha
+
+lamda=n*m_alpha/M
+
+t_half= 0.693/lamda   //half life of Plutonium
+
+printf('the half life of Plutonium is %.f years', t_half)
+
+//mistake in text book
diff --git a/2465/CH3/EX3.4/Examlpe_4.sce b/2465/CH3/EX3.4/Examlpe_4.sce
new file mode 100644
index 000000000..2a9fb99e1
--- /dev/null
+++ b/2465/CH3/EX3.4/Examlpe_4.sce
@@ -0,0 +1,23 @@
+//Chapter-3,Example 4,Page 57
+clc;
+close;
+
+m=234  // atomic mass of uranium
+
+M_0 = 4   // initial mass of uranium
+
+t_half= 2.48*10^5  // half life of uranium 
+
+t= 62000*365*24*3600  // time period
+
+lamda=8.88*10^-14
+
+M= M_0*exp(-lamda*t)
+
+printf('Mass of uranium left unchanged is %.3f mg', M)
+
+N= (M*6.023*10^20)/m
+
+A= lamda*N
+
+printf(' \n activity of uranium is %.3f disintigrations/sec ', A)
diff --git a/2465/CH3/EX3.5/Example_5.sce b/2465/CH3/EX3.5/Example_5.sce
new file mode 100644
index 000000000..589cf1fdd
--- /dev/null
+++ b/2465/CH3/EX3.5/Example_5.sce
@@ -0,0 +1,27 @@
+//Chapter-3,Example 5,Page 57
+clc;
+close;
+
+//Part (a)
+
+t_half= 1620  //half life of radium
+
+lamda= 0.693/t_half 
+
+//as radium lose one centigram mass
+
+N_0=100 // in centigram
+
+N_1=N_0-1
+
+t_1=log10(N_0/N_1)/(lamda*log10(%e))
+
+printf('Part (a)---total number of years required are %.2f years ',t_1)
+
+// Part (b)
+
+N_2= 1
+
+t_2=log10(N_0/N_2)/(lamda*log10(%e))
+
+printf('\n Part (b)---total number of years required are %.2f years ',t_2)
diff --git a/2465/CH3/EX3.6/Example_6.sce b/2465/CH3/EX3.6/Example_6.sce
new file mode 100644
index 000000000..92663dfd4
--- /dev/null
+++ b/2465/CH3/EX3.6/Example_6.sce
@@ -0,0 +1,18 @@
+//Chapter-3,Example 6,Page 58
+clc;
+close;
+
+M = 214  // molecular mass of RaB
+
+lamda= 4.31*10^-4
+
+//since -(dN/dt)= lamda*N =3.7 *10^10
+//N = m * 6.023*10^23/ M
+
+m=(3.7*10^10)*214/(lamda*6.023*10^23)
+
+printf('the mass of RaB is ')
+
+disp(m)
+
+printf('    gram')
diff --git a/2465/CH3/EX3.7/Example_7.sce b/2465/CH3/EX3.7/Example_7.sce
new file mode 100644
index 000000000..256697d5e
--- /dev/null
+++ b/2465/CH3/EX3.7/Example_7.sce
@@ -0,0 +1,19 @@
+//Chapter-3,Example 7,Page 58
+clc;
+close;
+
+M = 214  // molecular mass of RaB
+
+lamda= 4.31*10^-4
+
+//for 1 rd activity  (dN/dt) = 10^6 dis/sec
+// -(dN/dt)= lamda*N
+//N = m * 6.023*10^23/ M
+
+m=(10^6)*214/(lamda*6.023*10^23)
+
+printf('the mass of RaB is ')
+
+disp(m)
+
+printf('    gram')
diff --git a/2465/CH3/EX3.8/Example_8.sce b/2465/CH3/EX3.8/Example_8.sce
new file mode 100644
index 000000000..3011a501e
--- /dev/null
+++ b/2465/CH3/EX3.8/Example_8.sce
@@ -0,0 +1,23 @@
+//Chapter-3,Example 8,Page 58
+clc;
+close;
+
+//  U(238)=(U(238) + Pb(206)) * exp(-lamda*t)
+
+//  1 = (1 + Pb(206)/U(238)) * exp(-lamda*t)
+
+//since  Pb(206)/U(238) = 0.5 
+
+//  1 = (1 + 0.5) * exp(-lamda*t)
+
+t_half = 4.5 *10^9   //half life of Uranium
+
+lamda = 0.693/t_half
+
+t= log10(1.5)/(log10(%e)*lamda)
+
+printf('the age of the rock specimen is ')
+
+disp(t)
+
+printf('   years')
diff --git a/2465/CH3/EX3.9/Example_9.sce b/2465/CH3/EX3.9/Example_9.sce
new file mode 100644
index 000000000..1a02f1182
--- /dev/null
+++ b/2465/CH3/EX3.9/Example_9.sce
@@ -0,0 +1,25 @@
+//Chapter-3,Example 9,Page 59
+clc;
+close;
+
+mole_U =11.9/238 //mole of Uranium
+
+mole_Pb =10.3/206 //mole of lead
+
+t_half= 4.5*10^9   //half life of Uranium
+
+//  U(238)=(U(238) + Pb(206)) * exp(-lamda*t)
+
+//  1 = (1 + Pb(206)/U(238)) * exp(-lamda*t)
+
+//  1 = (1 + 0.5) * exp(-lamda*t)
+
+lamda = 0.693/t_half
+
+t= log10(1+ mole_Pb/mole_U)/(log10(%e)*lamda)
+
+printf('the age of the ore is ')
+
+disp(t)
+
+printf('    years')
diff --git a/2465/CH4/EX4.1/Example_1.sce b/2465/CH4/EX4.1/Example_1.sce
new file mode 100644
index 000000000..502811c64
--- /dev/null
+++ b/2465/CH4/EX4.1/Example_1.sce
@@ -0,0 +1,17 @@
+//Chapter-4,Example 1,Page 92
+clc;
+close;
+
+R= 2     // gas constant
+
+//as water temperature is 100 degree
+
+T = 273 + 100   // temperature in  Kelvin
+
+w=R*T    // work done
+
+q= 536*18   //heat in cal/mol
+
+delta_E= q-w 
+
+printf('the amount of energy increased is %.1f cal/mol',delta_E)  
diff --git a/2465/CH4/EX4.10/Example_10.sce b/2465/CH4/EX4.10/Example_10.sce
new file mode 100644
index 000000000..b22df3756
--- /dev/null
+++ b/2465/CH4/EX4.10/Example_10.sce
@@ -0,0 +1,19 @@
+//Chapter-4,Example 10,Page 95
+clc;
+close;
+
+delta_H1 = 538   //latent heat of water at 100 degree 
+
+T1= 273 + 100  //temperature in Kelvin
+
+T2= 273 +150   //temperature in Kelvin
+
+Cp_w = 1    // for water
+
+Cp_s = 8.1/18  //for steam
+
+delta_Cp = Cp_s - Cp_w 
+
+delta_H2 = delta_H1 + delta_Cp*(T2-T1)   //latent heat of water at 150 degree
+
+printf('the latent heat of  water at 150 degree is %.2f cal/g',delta_H2)
diff --git a/2465/CH4/EX4.11/Example_11.sce b/2465/CH4/EX4.11/Example_11.sce
new file mode 100644
index 000000000..da8cc6183
--- /dev/null
+++ b/2465/CH4/EX4.11/Example_11.sce
@@ -0,0 +1,17 @@
+//Chapter-4,Example 11,Page 96
+clc;
+close;
+
+R= 8.31   //gas constant
+
+T= 273+25   // temperature in Kelvin
+
+P1= 2   //pressure in atm 
+
+P2= 1   //pressure in atm
+
+w= 2.303 *R*T*log10(P1/P2)   //maximum work
+
+printf('maximum work done is %.f J', w)
+
+//mistake in textbook
diff --git a/2465/CH4/EX4.12/Example_12.sce b/2465/CH4/EX4.12/Example_12.sce
new file mode 100644
index 000000000..e245a0a62
--- /dev/null
+++ b/2465/CH4/EX4.12/Example_12.sce
@@ -0,0 +1,13 @@
+//Chapter-4,Example 12,Page 96
+clc;
+close;
+
+q_rev= 19.14    //latent heat 
+
+n= 18    //mols
+
+T= 273   //temperature in Kelvin
+
+dS= q_rev*n/T
+
+printf('the change of molar entropy is %.2f J/mol',dS)
diff --git a/2465/CH4/EX4.13/Example_13.sce b/2465/CH4/EX4.13/Example_13.sce
new file mode 100644
index 000000000..f47b7d766
--- /dev/null
+++ b/2465/CH4/EX4.13/Example_13.sce
@@ -0,0 +1,14 @@
+//Chapter-4,Example 13,Page 96
+clc;
+close;
+
+
+q_rev= 12.19    //latent heat 
+
+n= 32    //mols
+
+T= 273-182.9   //temperature in Kelvin
+
+dS= q_rev*n/T
+
+printf('the change of molar entropy is %.2f J/mol',dS)
diff --git a/2465/CH4/EX4.15/Example_15.sce b/2465/CH4/EX4.15/Example_15.sce
new file mode 100644
index 000000000..95bcb4a40
--- /dev/null
+++ b/2465/CH4/EX4.15/Example_15.sce
@@ -0,0 +1,19 @@
+//Chapter-4,Example 15,Page 96
+clc;
+close;
+
+P1= 528   // pressure in mm of Hg
+
+P2= 760   // pressure in mm of Hg
+
+T2=100+273   //teperature in Kelvin
+
+delta_Hv= 545.5 *18    // latent heat of vapourisation of water in J/mol
+
+R= 1.987  //gas constant
+
+//from the integrated form of Clausius-Clapeyron equation
+
+T1= 1/((log10(P2/P1)*2.303*R/delta_Hv)+(1/T2))
+
+printf('the temperature of water is %.f K',T1)
diff --git a/2465/CH4/EX4.16/Example_16.sce b/2465/CH4/EX4.16/Example_16.sce
new file mode 100644
index 000000000..0b2037a77
--- /dev/null
+++ b/2465/CH4/EX4.16/Example_16.sce
@@ -0,0 +1,27 @@
+//Chapter-4,Example 16,Page 97
+clc;
+close;
+
+//since the operation is isothermal & hte gas is ideal therefore..
+
+delta_E= 0  // from 1st law of thermodynamics
+
+P= 1   //pressure in atm
+
+V1= 10   // volume in cubic decimeter
+
+V2= 20   // volume in cubic decimeter
+
+W= P*(V2-V1)*(8.314/0.0821)   // work done by system
+
+q=W   //from 1st law of thermodynamics
+
+delta_H = delta_E + W
+
+printf(' q = %.2f J',q)
+
+printf('\n  W = %.2f',W)
+
+printf('\n  delta_E = %.f J',delta_E)
+
+printf('\n  delta_H = %.2f J',delta_H)
diff --git a/2465/CH4/EX4.17/Example_17.sce b/2465/CH4/EX4.17/Example_17.sce
new file mode 100644
index 000000000..483dd3afd
--- /dev/null
+++ b/2465/CH4/EX4.17/Example_17.sce
@@ -0,0 +1,19 @@
+//Chapter-4,Example 17,Page 97
+clc();
+close();
+
+q= 300   //heat energy
+
+P= 2   // pressure in atm
+
+V1= 10   // volume in litre
+
+V2= 20   //volume in litre
+
+//since 1 lit.atm = 24.25 cal
+
+W=P*(V2-V1)*24.25     //work done 
+
+delta_E= q-W   //from the 1st law of thermodynamics
+
+printf('the change in internal energy is %.f cal',delta_E)
diff --git a/2465/CH4/EX4.18/Example_18.sce b/2465/CH4/EX4.18/Example_18.sce
new file mode 100644
index 000000000..2c66dc887
--- /dev/null
+++ b/2465/CH4/EX4.18/Example_18.sce
@@ -0,0 +1,21 @@
+//Chapter-4,Example 18,Page 97
+clc();
+close();
+
+T1= 300   //temperature in Kelvin
+
+T2= 363   //temperature in Kelvin
+
+P1= 1   //pressure in atm
+
+P2=7    //pressure in atm
+
+Cv=5
+
+R=2    //gas constant
+
+Cp=Cv+R
+
+delta_S= Cp*log(T2/T1)+R*log(P1/P2)  //entropy change
+
+printf('the entropy change is %.4f cal/deg ', delta_S)
diff --git a/2465/CH4/EX4.19/Example_19.sce b/2465/CH4/EX4.19/Example_19.sce
new file mode 100644
index 000000000..ee777211a
--- /dev/null
+++ b/2465/CH4/EX4.19/Example_19.sce
@@ -0,0 +1,19 @@
+//Chapter-4,Example 19,Page 97
+clc();
+close();
+
+T1= 300   //temperature in Kelvin
+
+T2= 310   //temperature in Kelvin
+
+Kp1=3.49*10^-2    //equilibrium constant
+
+delta_H=-11200
+
+R= 1.987  //gas constant
+
+//from Van't Hoff's Equation
+
+Kp2=Kp1*10^(delta_H*((1/T1)-(1/T2))/(2.303*R))
+
+printf('the value of Kp2 = %.6f/atm ', Kp2)
diff --git a/2465/CH4/EX4.2/Example_2.sce b/2465/CH4/EX4.2/Example_2.sce
new file mode 100644
index 000000000..d2dbc03d9
--- /dev/null
+++ b/2465/CH4/EX4.2/Example_2.sce
@@ -0,0 +1,11 @@
+//Chapter-4,Example 2,Page 93
+clc;
+close;
+
+q= 990*4.2/10^3   //heat in kiloJoule
+
+w= 8.36*10^9/((10^3)*(10^7))  //work  in kiloJoule
+
+delta_E = q-w
+
+printf('the internal energy change of system is %.3f kJ',delta_E)
diff --git a/2465/CH4/EX4.3/Example_3.sce b/2465/CH4/EX4.3/Example_3.sce
new file mode 100644
index 000000000..d3037835a
--- /dev/null
+++ b/2465/CH4/EX4.3/Example_3.sce
@@ -0,0 +1,13 @@
+//Chapter-4,Example 3,Page 93
+clc;
+close;
+
+n=1     // number of mol
+
+R= 8.314     // gas constant
+
+T = 273 + 27   // temperature in  Kelvin
+
+w=n*R*T/1000    // work done in kiloJoule
+
+printf('work done by reaction ai 27 degree is %.4f kJ',w)
diff --git a/2465/CH4/EX4.4/Example_4.sce b/2465/CH4/EX4.4/Example_4.sce
new file mode 100644
index 000000000..9afb4eb21
--- /dev/null
+++ b/2465/CH4/EX4.4/Example_4.sce
@@ -0,0 +1,21 @@
+//Chapter-4,Example 4,Page 93
+clc;
+close;
+
+q_v=-97000   //in cal
+
+R= 8.314     // gas constant
+
+T = 273 + 200   // temperature in  Kelvin
+
+n_1= 1    //mols of gaseous reactant
+
+n_2= 1    // mols of gaseous product
+
+delta_n= n_2-n_1
+
+//q_p= q_v + delta_n*R*T
+
+q_p= q_v + delta_n*R*T
+
+printf('the heat combustion of carbon is %.f cals',q_p)
diff --git a/2465/CH4/EX4.5/Example_5.sce b/2465/CH4/EX4.5/Example_5.sce
new file mode 100644
index 000000000..ce36439e7
--- /dev/null
+++ b/2465/CH4/EX4.5/Example_5.sce
@@ -0,0 +1,19 @@
+//Chapter-4,Example 5,Page 93
+clc;
+close;
+
+delta_H= -109   // heat change in Kcal
+
+n_1= 2    //mols of gaseous reactant
+
+n_2= 1    // mols of gaseous product
+
+delta_n= n_2-n_1
+
+T=500
+
+R= 2*10^-3
+
+delta_E = (delta_H) - (delta_n*R*T)
+
+printf('the value of delta_E is %.f Kcal',delta_E)
diff --git a/2465/CH4/EX4.6/Example_6.sce b/2465/CH4/EX4.6/Example_6.sce
new file mode 100644
index 000000000..7509cd20a
--- /dev/null
+++ b/2465/CH4/EX4.6/Example_6.sce
@@ -0,0 +1,18 @@
+//Chapter-4,Example 6,Page 93
+clc;
+close;
+
+delta_H1= -337.2    // Heat combustion for ethylene 
+
+delta_H2=-68.3    //  Heat combustion for hudrogen
+
+delta_H3= 372.8    // Heat combustion for ethane
+
+//Given reaction is...
+//  C2H4(g) +H2(g) ---> C2H6(g)
+
+delta_H= delta_H1 + delta_H2 +delta_H3
+
+printf('the heat combustion for given reaction is %.2f Kcal',delta_H)
+
+
diff --git a/2465/CH4/EX4.7/Example_7.sce b/2465/CH4/EX4.7/Example_7.sce
new file mode 100644
index 000000000..bb3bd5922
--- /dev/null
+++ b/2465/CH4/EX4.7/Example_7.sce
@@ -0,0 +1,16 @@
+//Chapter-4,Example 7,Page 94
+clc;
+close;
+
+delta_H1= 104    //for reaction.. H2(g)---> 2H(g)
+
+delta_H2= 120/2    //for reaction.. (1/2)O2(g)---> O(g)
+
+delta_H3= -58    //for reaction.. H2(g) + (1/2)O2(g)---> H2O(g)
+
+delta_H=delta_H1 + delta_H2 - delta_H3
+
+//there are two O-H bonds 
+//therefore for one bond required heat energy is half of delta_H
+
+printf('the O-H bond energy is %.f Kcal',delta_H/2)
diff --git a/2465/CH4/EX4.8/Example_8.sce b/2465/CH4/EX4.8/Example_8.sce
new file mode 100644
index 000000000..2a642f92e
--- /dev/null
+++ b/2465/CH4/EX4.8/Example_8.sce
@@ -0,0 +1,23 @@
+//Chapter-4,Example 8,Page 94
+clc;
+close;
+
+delta_H_C= -393   // enthalpy for carbon
+
+delta_H_H2= -286   //enthalpy for hydrogen
+
+delta_H_C3H8=-2220  //enthalpy for propane
+
+// According to Hess's Law... delta_H1 = delta_H2 - delta_H3
+
+//delta_H2  for reaction... 3C +4H2 +5O2 ----> 3CO2 +4H2O
+
+delta_H2= 3*delta_H_C +4*delta_H_H2  
+
+//delta_H2  for reaction... C3H8 + 5O2 ----> 3CO2 +4H2O
+
+delta_H3= delta_H_C3H8 
+
+delta_Hf= delta_H2 - delta_H3   //enthalpy for propane at 298 K
+
+printf('the enthalpy of formation of propane at 298K is %.f Kcal', delta_Hf)
diff --git a/2465/CH4/EX4.9/Example_9.sce b/2465/CH4/EX4.9/Example_9.sce
new file mode 100644
index 000000000..39c08702c
--- /dev/null
+++ b/2465/CH4/EX4.9/Example_9.sce
@@ -0,0 +1,15 @@
+//Chapter-4,Example 9,Page 95
+clc;
+close;
+
+delta_H2= 2386   //enthalpy for.. yellow P---> H3PO4
+
+delta_H3= 2113   //enthalpy for.. red P---> H3PO4
+
+delta_HT = delta_H2- delta_H3   //enthalpy for...yellow P ---> red P
+
+// According to Hess's Law... delta_H1 = delta_H2 - delta_H3
+
+delta_HT = delta_H2 - delta_H3   // delta_H1 = delta_HT
+
+printf('the enthalpy change of transition from yellow P to red P is %.f cals',delta_HT)
diff --git a/2465/CH5/EX5.1/Example_1.sce b/2465/CH5/EX5.1/Example_1.sce
new file mode 100644
index 000000000..aa4ffc6d9
--- /dev/null
+++ b/2465/CH5/EX5.1/Example_1.sce
@@ -0,0 +1,23 @@
+//Chapter-5,Example 1,Page 121
+clc();
+close();
+
+//for 1st order reaction 
+//k = (1/t)*log(a/(a-x))
+
+a= 46.1     //time value 
+
+//time intervals
+
+t=[ 5 10 20 30 50]
+
+x=[ 37.1 29.8 19.6 12.3 5.0]
+
+k = (1 ./t).*log(a./(x))
+
+printf('value of k are ' )
+
+disp(k)
+
+printf('since k values are fairly constant by putting in 1nd order rate equation. \nHence decomposition of H2O2 is of 1st order.')
+
diff --git a/2465/CH5/EX5.10/Example_10.sce b/2465/CH5/EX5.10/Example_10.sce
new file mode 100644
index 000000000..d2c27a4d5
--- /dev/null
+++ b/2465/CH5/EX5.10/Example_10.sce
@@ -0,0 +1,29 @@
+//Chapter-5,Example 10,Page 125
+clc();
+close();
+
+T1=50   //time in sec
+
+T2 = 25   //time in sec
+
+a1=0.5   //initial concentration
+
+a2= 1 
+
+//  (T1/T2) = (a2/a1)^(n-1)
+//therefore  (50/25) =(1/0.5)^(n-1)
+//           2=2^(n-1)
+//           n=2
+//hence its 2nd order 
+
+t_half= T1
+
+k=1/(a1*t_half)
+
+//assume y= a-x
+
+y=0.2*a1     //remaining concentration
+
+t=(a1-y)/(a1*k*(y))
+
+printf('the time taken is %.f sec ',t)
diff --git a/2465/CH5/EX5.11/Example_11.sce b/2465/CH5/EX5.11/Example_11.sce
new file mode 100644
index 000000000..06b492323
--- /dev/null
+++ b/2465/CH5/EX5.11/Example_11.sce
@@ -0,0 +1,23 @@
+//Chapter-5,Example 11,Page 126
+clc();
+close();
+
+a=0.1  //initial concentration of reactants
+
+x=0.2*a
+
+t=40   //time 
+
+k=x/(a*t*(a-x))
+
+t_half=1/(a*k)
+
+x1=0.75*a
+
+t1=x1/(k*a*(a-x1))
+
+printf('the rate constant is k = %.4f l/mol.min',k)
+
+printf('\n the half life period is %.f mins',t_half)
+
+printf('\n the time required to complete 75 percent reaction is %.f mins',t1)
diff --git a/2465/CH5/EX5.12/Example_12.sce b/2465/CH5/EX5.12/Example_12.sce
new file mode 100644
index 000000000..608882bf0
--- /dev/null
+++ b/2465/CH5/EX5.12/Example_12.sce
@@ -0,0 +1,23 @@
+//Chapter-5,Example 12,Page 126
+clc();
+close();
+
+a1=100
+
+x1=1
+
+t1=1
+
+k=2.303*log10(a1/(a1-x1))/t1
+
+t2=60    //time in minutes
+
+a2=100 
+
+//assume (a2-x2)= y
+
+y= 1/(10^(k*t2/2.303)/a2)
+
+printf('the undecomposed is %.2f ',y)
+
+//mistake in textbook
diff --git a/2465/CH5/EX5.15/Example_15.sce b/2465/CH5/EX5.15/Example_15.sce
new file mode 100644
index 000000000..0b9f3a800
--- /dev/null
+++ b/2465/CH5/EX5.15/Example_15.sce
@@ -0,0 +1,17 @@
+//Chapter-5,Example 15,Page 128
+clc();
+close();
+
+K1=2.45*10^-5    //rate constant at 273 K
+
+K2=162*10^-5    //rate constant at 303 K
+
+T1=273   //temperature in Kelvin
+
+T2=303   //temperature in Kelvin
+
+R=1.987   //gas constant
+
+Ea= log10(K2/K1)*2.303*R*T1*T2/(T2-T1)
+
+printf('the activation energy is Ea = %.f cal/mole' ,Ea)
diff --git a/2465/CH5/EX5.16/Example_16.sce b/2465/CH5/EX5.16/Example_16.sce
new file mode 100644
index 000000000..90d6670a2
--- /dev/null
+++ b/2465/CH5/EX5.16/Example_16.sce
@@ -0,0 +1,19 @@
+//Chapter-5,Example 16,Page 128
+clc();
+close();
+
+t_half = 600    // half life
+
+K=0.693/t_half
+
+Ea=98600   //activation energy
+
+A= 4*10^13   //Arrhenius factor
+
+R=8.316   //gas constant
+
+T=Ea/(2.303*R*log10(A/K))
+
+printf('temperature is %.f K',T)
+
+//mistake in textbook
diff --git a/2465/CH5/EX5.17/Example_17.sce b/2465/CH5/EX5.17/Example_17.sce
new file mode 100644
index 000000000..7156929ab
--- /dev/null
+++ b/2465/CH5/EX5.17/Example_17.sce
@@ -0,0 +1,17 @@
+//Chapter-5,Example 17,Page 129
+clc();
+close();
+
+K1=5*10^-3    //rate constant at 800 degrees
+
+Ea=4.5*10^4   //activation energy
+
+T1=800+273   //temperature in Kelvin
+
+T2=875+273   //temperature in Kelvin
+
+R=8.314   //gas constant
+
+K2=K1*10^(Ea*(T2-T1)/(2.303*R*T1*T2))
+
+printf('the value of K2 = %.4f l/mol.sec',K2)
diff --git a/2465/CH5/EX5.2/Example_2.sce b/2465/CH5/EX5.2/Example_2.sce
new file mode 100644
index 000000000..e1d462398
--- /dev/null
+++ b/2465/CH5/EX5.2/Example_2.sce
@@ -0,0 +1,19 @@
+//Chapter-5,Example 2,Page 122
+clc();
+close();
+
+t=[7.18 18 27.05]      //time in minute
+
+r=[ 21.4 17.7 15]   //rotation in degrees
+
+r_0=24.09
+
+r_a=-10.74
+
+k=(1 ./t).*log10((r_0-r_a)./(r-r_a))
+
+printf('values of k')
+
+disp(k)
+
+printf('since k values are fairly constant by putting in 1nd order rate equation. \nHence hydrolysis of methyl acetate is of 1st order.')
diff --git a/2465/CH5/EX5.3/Example_3.sce b/2465/CH5/EX5.3/Example_3.sce
new file mode 100644
index 000000000..08a5f99ec
--- /dev/null
+++ b/2465/CH5/EX5.3/Example_3.sce
@@ -0,0 +1,19 @@
+//Chapter-5,Example 3,Page 122
+clc();
+close();
+
+t=[75 119 183]      //time in minute
+
+V=[24.20 26.60 29.32]   //volume of alkali used
+
+V_0=19.24
+
+V_a=42.03
+
+k=(2.303 ./t).*log10((V_a-V_0)./(V_a-V))
+
+printf('values of k')
+
+disp(k)
+
+printf('since k values are fairly constant by putting in 1nd order rate equation. \nHence hydrolysis of methyl acetate is of 1st order.')
diff --git a/2465/CH5/EX5.4/Example_4.sce b/2465/CH5/EX5.4/Example_4.sce
new file mode 100644
index 000000000..9f7f895e6
--- /dev/null
+++ b/2465/CH5/EX5.4/Example_4.sce
@@ -0,0 +1,15 @@
+//Chapter-5,Example 4,Page 123
+clc();
+close();
+
+t= 30   //time in minutes
+
+a=100
+
+x= 25
+
+k=(2.303/t)*log10(a/(a-x))
+
+t_half=0.693/k
+
+printf('the time of 50 percent completion of reaction is %.2f mins',t_half)
diff --git a/2465/CH5/EX5.5/Example_5.sce b/2465/CH5/EX5.5/Example_5.sce
new file mode 100644
index 000000000..371a55fec
--- /dev/null
+++ b/2465/CH5/EX5.5/Example_5.sce
@@ -0,0 +1,17 @@
+//Chapter-5,Example 5,Page 123
+clc();
+close();
+
+t_half=17   //half life period
+
+k=0.693/t_half   //rate constant
+
+a=100
+
+x= 75
+
+t=(2.303/k)*log10(a/(a-x))
+
+printf('the rate constant is k = %.5f /min',k)
+
+printf('\n the time taken  t = %.1f min', t)
diff --git a/2465/CH5/EX5.6/Example_6.sce b/2465/CH5/EX5.6/Example_6.sce
new file mode 100644
index 000000000..f607293bc
--- /dev/null
+++ b/2465/CH5/EX5.6/Example_6.sce
@@ -0,0 +1,17 @@
+//Chapter-5,Example 6,Page 123
+clc();
+close();
+
+t_half=1600   //half life period
+
+k=0.693/t_half   //rate constant
+
+a=100
+
+x= 80
+
+t=(2.303/k)*log10(a/(a-x))
+
+printf('\n the time taken  t = %.2f years', t)
+
+//mistake in textbook
diff --git a/2465/CH5/EX5.7/Example_7.sce b/2465/CH5/EX5.7/Example_7.sce
new file mode 100644
index 000000000..728c0cf49
--- /dev/null
+++ b/2465/CH5/EX5.7/Example_7.sce
@@ -0,0 +1,27 @@
+//Chapter-5,Example 7,Page 124
+clc();
+close();
+
+//for 2st order reaction 
+//k = (1/a*t)*(x/(a-x))
+
+a= 16    
+
+//time intervals
+
+t=[ 5 15 25 35]
+
+//assume y = a-x
+
+y=[ 10.24 6.13 4.32 3.41]    //volume of acid
+
+x=a-y
+
+k = (1 ./(a*t)).*(x./(y))
+
+printf('value of k are ' )
+
+disp(k)
+
+printf('since k values are fairly constant by putting in 2nd order rate equation. \nHence dhydrolysis of methyl acetate is of 2st order.')
+
diff --git a/2465/CH5/EX5.9/Example_9.sce b/2465/CH5/EX5.9/Example_9.sce
new file mode 100644
index 000000000..9e528c241
--- /dev/null
+++ b/2465/CH5/EX5.9/Example_9.sce
@@ -0,0 +1,15 @@
+//Chapter-5,Example 9,Page 125
+clc();
+close();
+
+ //final concentration is half of initial concentration
+//therefore t =t_half
+t= 60   //time in minutes
+
+t_half=t
+
+k=5.2*10^-3   //rste constant
+
+a=1/(k*t_half)   //for 2nd order reaction
+
+printf('the initial concentration is %.2f mol/litre',a)
diff --git a/2465/CH8/EX8.1/Example_1.sce b/2465/CH8/EX8.1/Example_1.sce
new file mode 100644
index 000000000..9917fc3f0
--- /dev/null
+++ b/2465/CH8/EX8.1/Example_1.sce
@@ -0,0 +1,11 @@
+//Chapter-8,Example 1,Page 195
+clc();
+close();
+
+OH=2*0.005   //in mol/litre
+
+pOH=-log10(OH)
+
+pH=14-pOH
+
+printf('the pH of Ca(OH)2 is pH = %.f ',pH)
diff --git a/2465/CH8/EX8.2/Example_2.sce b/2465/CH8/EX8.2/Example_2.sce
new file mode 100644
index 000000000..f6a1b0244
--- /dev/null
+++ b/2465/CH8/EX8.2/Example_2.sce
@@ -0,0 +1,21 @@
+//Chapter-8,Example 2,Page 195
+clc();
+close();
+
+H=20*0.1/1000    //as 20 ml of 0.1M HCl 
+
+pH=-log10(H)
+
+pOH=14-pH
+
+OH=10^(-pOH)
+
+printf(' the [H+] = %.4f mole/l',H)
+
+printf('\n  the [OH-] =')
+
+disp(OH)
+
+printf('  mole/l')
+
+printf('\n  the pH = %.f ',pH)
diff --git a/2465/CH8/EX8.3/Example_3.sce b/2465/CH8/EX8.3/Example_3.sce
new file mode 100644
index 000000000..6c3156976
--- /dev/null
+++ b/2465/CH8/EX8.3/Example_3.sce
@@ -0,0 +1,32 @@
+//Chapter-8,Example 3,Page 195
+clc();
+close();
+
+//solution for (a) part
+
+conc1=1*10^-8   //concentration of HCl solution
+
+//let [H+] concentration from water = x
+//so, [H+] of solution = conc1*x an [OH-] = x
+//......Kw = [H+]*[OH-] = 10^-14
+//......x^2 +(10^-8)*x -(10^-14)=0
+x = (-10^-8 + sqrt((10^-8)^2 + 4*1*10^-14))/(2*1)
+
+H=conc1 +x 
+
+pH1=-log10(H)
+
+printf('for HCl the pH = %.3f',pH1)
+
+
+//solution for (b) part
+conc2= 1*10^-8   //concentration of NaOH solution
+
+OH=x+conc2
+
+pOH2=-log10(OH)
+
+pH2=14 - pOH2
+
+printf('\n for NaOH the pH = %.3f',pH2)
+
diff --git a/2465/CH8/EX8.4/Example_4.sce b/2465/CH8/EX8.4/Example_4.sce
new file mode 100644
index 000000000..a9d10cf1d
--- /dev/null
+++ b/2465/CH8/EX8.4/Example_4.sce
@@ -0,0 +1,24 @@
+//Chapter-8,Example 4,Page 196
+clc();
+close();
+
+alpha1=0.02
+
+Ka=1.8*10^-5 
+
+//at equilibrium..
+//[CH3COOH] = C1* (1-alpha1)
+//[H+] = C1* alpha1
+//[CH3COO-] = C1* alpha1
+//  Ka =[H+] * [CH3COO-]/[CH3COOH]
+//  Ka = C1* alpha1*C1* alpha1/(C1 (1-alpha1))
+
+C1=Ka*(1-alpha1)/alpha1^2
+
+printf('the molar concentration of CH3COOH is C = %.4f molar',C1)
+
+C2=0.01
+
+alpha2= sqrt(Ka/C2)
+
+printf('\n alpha = %.4f ',alpha2)
diff --git a/2465/CH8/EX8.5/Example_5.sce b/2465/CH8/EX8.5/Example_5.sce
new file mode 100644
index 000000000..90551f959
--- /dev/null
+++ b/2465/CH8/EX8.5/Example_5.sce
@@ -0,0 +1,15 @@
+//Chapter-8,Example 5,Page 196
+clc();
+close();
+
+pKa=4.74
+
+salt=0.1
+
+acid=0.1
+
+//according to Henderson equation pH of buffer solution
+
+pH = pKa + log10(salt/acid)
+
+printf('the pH of buffer solution is pH = %.2f ',pH)
diff --git a/2465/CH8/EX8.6/Example_6.sce b/2465/CH8/EX8.6/Example_6.sce
new file mode 100644
index 000000000..984f0c31f
--- /dev/null
+++ b/2465/CH8/EX8.6/Example_6.sce
@@ -0,0 +1,17 @@
+//Chapter-8,Example 6,Page 196
+clc();
+close();
+
+pH= 7.4    //of blood
+
+H= 10^(-pH)
+
+//assume ratio of HCO3-  and  H2CO3 is r
+
+Ka= 4.5*10^-7
+
+//  Ka = [H+]*[HCO3-]/[H2CO3]
+
+r=Ka/H
+
+printf('the ratio of HCO3-  and  H2CO3 is %.f',r)
diff --git a/2465/CH8/EX8.7/Example_7.sce b/2465/CH8/EX8.7/Example_7.sce
new file mode 100644
index 000000000..1d5b25cf9
--- /dev/null
+++ b/2465/CH8/EX8.7/Example_7.sce
@@ -0,0 +1,14 @@
+//Chapter-8,Example 7,Page 196
+clc();
+close();
+
+Ksp=3.45*10^-11     //solubility product of CaF2
+
+//Ksp = [Ca+2]*[F-]^2
+//Ksp = [S]*[2*S]^2
+
+S = nthroot(Ksp,3)/4
+
+printf('the solubility of CaF2 is S = %.7f mole/litre',S)
+
+//mistake in textbook
diff --git a/2465/CH8/EX8.8/Example_8.sce b/2465/CH8/EX8.8/Example_8.sce
new file mode 100644
index 000000000..7e43b504c
--- /dev/null
+++ b/2465/CH8/EX8.8/Example_8.sce
@@ -0,0 +1,19 @@
+//Chapter-8,Example 8,Page 197
+clc();
+close();
+
+Ksp=8*10^-12  //solubility product of SrF2
+
+//Ksp= [Sr+2]*[F-]^2.....F=0.1 M
+
+F=0.1      //concentration of F in SrF2
+
+S=Ksp/F^2
+
+printf('the solubility of SrF2 is')
+
+disp(S)
+
+printf('mol/litre')
+
+//mistake in textbook
diff --git a/2465/CH9/EX9.1/Example_1.sce b/2465/CH9/EX9.1/Example_1.sce
new file mode 100644
index 000000000..8c983f2ae
--- /dev/null
+++ b/2465/CH9/EX9.1/Example_1.sce
@@ -0,0 +1,19 @@
+//Chapter-9,Example 1,Page 219
+clc();
+close();
+
+a= 1.25   //cross section area in cmsquare
+
+l= 10.5    //distance of seperation
+
+r=1996    //resistance
+
+O_cond= 1/r   //observed conductivity
+
+C_constant = l/a   //cell constant
+
+S_cond=C_constant*O_cond    //specific conductivity
+
+printf('the cell constant is %.2f /cm',C_constant)
+
+printf('\n the specific conductivity is %.5f /ohm.cm ',S_cond)
diff --git a/2465/CH9/EX9.10/Example_10.sce b/2465/CH9/EX9.10/Example_10.sce
new file mode 100644
index 000000000..dcd2b7092
--- /dev/null
+++ b/2465/CH9/EX9.10/Example_10.sce
@@ -0,0 +1,22 @@
+//Chapter-9,Example 10,Page 221
+clc();
+close();
+
+lamda_Ag = 58.3  
+
+lamda_Cl=65.3
+
+lamda_v=lamda_Ag+lamda_Cl   //Kohlrausch's law
+
+Kv=1.24*10^-6   //specific conductivity
+
+V=lamda_v/(Kv*1000)
+
+wt=143.5    //molecular weight of AgCl
+
+S=wt/V
+
+printf('the solubility off AGCl is %.5f g/l',S)
+
+
+//mistake in textbook
diff --git a/2465/CH9/EX9.11/Example_11.sce b/2465/CH9/EX9.11/Example_11.sce
new file mode 100644
index 000000000..3c674c678
--- /dev/null
+++ b/2465/CH9/EX9.11/Example_11.sce
@@ -0,0 +1,17 @@
+//Chapter-9,Example 11,Page 222
+clc();
+close();
+
+u= 0.196   //speed of Ag+
+
+v=1      //speed of NO3-
+
+t_Ag=u/(u+v)   //transport number of Ag+ ions
+
+t_NO3= 1-t_Ag   //transportnumber of NO3- ions
+
+printf('the transport number of Ag+ ions is %.3f',t_Ag)
+
+printf('\n the transport number of NO3+ ions is %.3f',t_NO3)
+
+//mistake in textbook
diff --git a/2465/CH9/EX9.12/Example_12.sce b/2465/CH9/EX9.12/Example_12.sce
new file mode 100644
index 000000000..0af82950a
--- /dev/null
+++ b/2465/CH9/EX9.12/Example_12.sce
@@ -0,0 +1,21 @@
+//Chapter-9,Example 12,Page 222
+clc();
+close();
+
+wt_Ag = 0.1351   //weight of Ag deposited in a silver coulometer
+
+Ewt_Ag = 107.88   //atomic weight of Ag
+
+Ewt_Cu = 63.6   //atomic weight of Cu
+
+wt_Cu= wt_Ag*(Ewt_Cu/2)/Ewt_Ag  //wt of Cu deposited
+
+loss=0.6350-0.6236  //loss in weight of Cu at anode
+
+t_Cu = loss/wt_Cu
+
+t_SO4= 1-t_Cu
+
+printf('the transport number of Cu+2 ion is %.3f ',t_Cu)
+
+printf('\n the transport number of SO4 ion is %.3f ',t_SO4)
diff --git a/2465/CH9/EX9.2/Example_2.sce b/2465/CH9/EX9.2/Example_2.sce
new file mode 100644
index 000000000..b2d50cde9
--- /dev/null
+++ b/2465/CH9/EX9.2/Example_2.sce
@@ -0,0 +1,16 @@
+//Chapter-9,Example 2,Page 219
+clc();
+close();
+
+R= 500   //resistance of the cell
+
+K= 0.0002765   //specific conductivity
+
+//cell constant= l/a  and R= p(l/a)
+//sice l= length  a= area  p= resistivity 
+//(1/p) = K = specific conductivity
+//(l/a) = R*K
+
+C_constant= R*K   //cell constant
+
+printf('the cell constant is %.3f /cm',C_constant)
diff --git a/2465/CH9/EX9.3/Example_3.sce b/2465/CH9/EX9.3/Example_3.sce
new file mode 100644
index 000000000..3e51aedc3
--- /dev/null
+++ b/2465/CH9/EX9.3/Example_3.sce
@@ -0,0 +1,17 @@
+//Chapter-9,Example 3,Page 220
+clc();
+close();
+
+R= 4364   //resistance of the cell
+
+K= 2.767*10^-3   //specific conductivity
+
+C_constant= R*K   //cell constant
+
+
+//cell constant= l/a = R/p
+R1 = 3050   //new resistance
+
+p= R1/C_constant
+
+printf('the specific resistance is %.3f ohm.cm ',p)
diff --git a/2465/CH9/EX9.4/Example_4.sce b/2465/CH9/EX9.4/Example_4.sce
new file mode 100644
index 000000000..fbd6e46ab
--- /dev/null
+++ b/2465/CH9/EX9.4/Example_4.sce
@@ -0,0 +1,25 @@
+//Chapter-9,Example 4,Page 220
+clc();
+close();
+
+R= 550   //resistance of the cell
+
+K=0.002768    //specific conductivity
+
+C_constant= R*K
+
+p= 72.18    //observed resistance
+
+Kv = C_constant*(1/p)
+
+C= 0.2     //concentration
+
+lamda_v= Kv*1000/C  //equivalent conductivity
+
+M= 0.1
+
+lamda_m= 1000*Kv/M   //molar conductivity
+
+printf('the equivalent conductivity of ZnSO4 is %.2f /ohm.cm^2',lamda_v)
+
+printf('\n the molar conductivity of ZnSO4 is %.2f /ohm.cm^2',lamda_m)
diff --git a/2465/CH9/EX9.5/Example_5.sce b/2465/CH9/EX9.5/Example_5.sce
new file mode 100644
index 000000000..eb9776f64
--- /dev/null
+++ b/2465/CH9/EX9.5/Example_5.sce
@@ -0,0 +1,17 @@
+//Chapter-9,Example 5,Page 220
+clc();
+close();
+
+R= 32    //resistance of solution
+
+l= 1.8    //distance between electrodes
+
+a= 5.4   //area 
+
+Kv=l/(R*a)    //specific conductivity
+
+C= 0.1     //concentration
+
+lamda_v= Kv*1000/C  //equivalent conductivity
+
+printf('the equivalent conductivity is %.3f /ohm.cm^2',lamda_v)
diff --git a/2465/CH9/EX9.6/Example_6.sce b/2465/CH9/EX9.6/Example_6.sce
new file mode 100644
index 000000000..6cec6eaf8
--- /dev/null
+++ b/2465/CH9/EX9.6/Example_6.sce
@@ -0,0 +1,11 @@
+//Chapter-9,Example 6,Page 221
+clc();
+close();
+
+lamda_v= 48.15  //equivalent conductivity
+
+lamda_v1= 390.6   //equivalent conductivity at infinity
+
+alpha= lamda_v/lamda_v1
+
+printf('the degree of dissolution of acetic acid is %.4f ',alpha)
diff --git a/2465/CH9/EX9.7/Example_7.sce b/2465/CH9/EX9.7/Example_7.sce
new file mode 100644
index 000000000..88a5053e4
--- /dev/null
+++ b/2465/CH9/EX9.7/Example_7.sce
@@ -0,0 +1,16 @@
+//Chapter-9,Example 7,Page 221
+clc();
+close();
+
+lamda_HCl=426.1    //equivalent conductance of HCl
+
+lamda_AcONa=91    //equivalent conductance of AcONa
+
+lamda_NaCl=126.5    //equivalent conductance of NaCl
+
+//  lamda_HCl + lamda_AcONa - lamda_NaCl= (lamda_H+lamda_Cl)+(lamda_Na+lamda_OAc)-(lamda_Na+lamda_Cl)
+//                                      = lamda_H +lamda_OAc  = lamda_AcOH
+
+lamda_AcOH = lamda_HCl + lamda_AcONa - lamda_NaCl
+
+printf('the equivalent conductance of AcOH = %.2f/ohm.cm^2',lamda_AcOH)
diff --git a/2465/CH9/EX9.8/Example_8.sce b/2465/CH9/EX9.8/Example_8.sce
new file mode 100644
index 000000000..522db0bde
--- /dev/null
+++ b/2465/CH9/EX9.8/Example_8.sce
@@ -0,0 +1,15 @@
+//Chapter-9,Example 8,Page 221
+clc();
+close();
+
+lamda_H=0.0348   //equivalent conductance of H+ ion
+
+lamda_CH3COO=0.004  //equivalent conductance of CH3COO- ion
+
+lamda= lamda_H+lamda_CH3COO //equivalent conductance  at infinity
+
+lamda_v= 0.018    //equvalent conductance 
+
+alpha= lamda_v/lamda   //degree of dissolution
+
+printf('the degree of dissolution is %.4f ',alpha)
diff --git a/2465/CH9/EX9.9/Example_9.sce b/2465/CH9/EX9.9/Example_9.sce
new file mode 100644
index 000000000..b0181a822
--- /dev/null
+++ b/2465/CH9/EX9.9/Example_9.sce
@@ -0,0 +1,18 @@
+//Chapter-9,Example 9,Page 221
+clc();
+close();
+
+strength =0.05   //strength of CH3COOH
+
+Ka=1.8*10^-5
+
+//              CH3COOH <---> CH3COO- +  H+
+//intially  =     0.05          0        0
+//dissolution                   a 
+//at equilibrium= 0.05(1-a)   0.05*a   0.05*a
+//Ka =(0.05*a*0.05*a)/(0.05(1-a))
+//Ka=0.05*a^2      a=negligible   1-a=1
+
+a=sqrt(Ka/strength)
+
+printf('the degree of dissolution is %.4f ',a)
-- 
cgit