diff options
Diffstat (limited to '2465')
102 files changed, 0 insertions, 2104 deletions
diff --git a/2465/CH10/EX10.1/Ex10_1.sce b/2465/CH10/EX10.1/Ex10_1.sce deleted file mode 100644 index 02cf4577c..000000000 --- a/2465/CH10/EX10.1/Ex10_1.sce +++ /dev/null @@ -1,13 +0,0 @@ -//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/Ex10_10.sce b/2465/CH10/EX10.10/Ex10_10.sce deleted file mode 100644 index 1137bac40..000000000 --- a/2465/CH10/EX10.10/Ex10_10.sce +++ /dev/null @@ -1,12 +0,0 @@ -//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/Ex10_11.sce b/2465/CH10/EX10.11/Ex10_11.sce deleted file mode 100644 index 3577f29b4..000000000 --- a/2465/CH10/EX10.11/Ex10_11.sce +++ /dev/null @@ -1,16 +0,0 @@ -//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/Ex10_12.sce b/2465/CH10/EX10.12/Ex10_12.sce deleted file mode 100644 index 8b2460a55..000000000 --- a/2465/CH10/EX10.12/Ex10_12.sce +++ /dev/null @@ -1,30 +0,0 @@ -//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/Ex10_2.sce b/2465/CH10/EX10.2/Ex10_2.sce deleted file mode 100644 index aec51c898..000000000 --- a/2465/CH10/EX10.2/Ex10_2.sce +++ /dev/null @@ -1,21 +0,0 @@ -//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/Ex10_3.sce b/2465/CH10/EX10.3/Ex10_3.sce deleted file mode 100644 index e6f950fa8..000000000 --- a/2465/CH10/EX10.3/Ex10_3.sce +++ /dev/null @@ -1,25 +0,0 @@ -//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/Ex10_4.sce b/2465/CH10/EX10.4/Ex10_4.sce deleted file mode 100644 index 1b8546cea..000000000 --- a/2465/CH10/EX10.4/Ex10_4.sce +++ /dev/null @@ -1,23 +0,0 @@ -//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/Ex10_5.sce b/2465/CH10/EX10.5/Ex10_5.sce deleted file mode 100644 index 714124f20..000000000 --- a/2465/CH10/EX10.5/Ex10_5.sce +++ /dev/null @@ -1,13 +0,0 @@ -//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/Ex10_6.sce b/2465/CH10/EX10.6/Ex10_6.sce deleted file mode 100644 index da05652c0..000000000 --- a/2465/CH10/EX10.6/Ex10_6.sce +++ /dev/null @@ -1,14 +0,0 @@ -//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/Ex10_7.sce b/2465/CH10/EX10.7/Ex10_7.sce deleted file mode 100644 index c8f37d6ed..000000000 --- a/2465/CH10/EX10.7/Ex10_7.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex10_8.sce b/2465/CH10/EX10.8/Ex10_8.sce deleted file mode 100644 index ce565986f..000000000 --- a/2465/CH10/EX10.8/Ex10_8.sce +++ /dev/null @@ -1,30 +0,0 @@ -//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/Ex10_9.sce b/2465/CH10/EX10.9/Ex10_9.sce deleted file mode 100644 index d9ab56f37..000000000 --- a/2465/CH10/EX10.9/Ex10_9.sce +++ /dev/null @@ -1,25 +0,0 @@ -//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/Ex11_1.sce b/2465/CH11/EX11.1/Ex11_1.sce deleted file mode 100644 index a91312a6c..000000000 --- a/2465/CH11/EX11.1/Ex11_1.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex11_2.sce b/2465/CH11/EX11.2/Ex11_2.sce deleted file mode 100644 index a1c82f3a8..000000000 --- a/2465/CH11/EX11.2/Ex11_2.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex17_1.sce b/2465/CH17/EX17.1/Ex17_1.sce deleted file mode 100644 index ae0a60afd..000000000 --- a/2465/CH17/EX17.1/Ex17_1.sce +++ /dev/null @@ -1,29 +0,0 @@ -//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/Ex17_2.sce b/2465/CH17/EX17.2/Ex17_2.sce deleted file mode 100644 index 60ecef2e8..000000000 --- a/2465/CH17/EX17.2/Ex17_2.sce +++ /dev/null @@ -1,14 +0,0 @@ -//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/Ex17_3.sce b/2465/CH17/EX17.3/Ex17_3.sce deleted file mode 100644 index 11296ac69..000000000 --- a/2465/CH17/EX17.3/Ex17_3.sce +++ /dev/null @@ -1,11 +0,0 @@ -//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/Ex17_4.sce b/2465/CH17/EX17.4/Ex17_4.sce deleted file mode 100644 index a0af0e9db..000000000 --- a/2465/CH17/EX17.4/Ex17_4.sce +++ /dev/null @@ -1,41 +0,0 @@ -//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/Ex17_5.sce b/2465/CH17/EX17.5/Ex17_5.sce deleted file mode 100644 index 81bbb40e5..000000000 --- a/2465/CH17/EX17.5/Ex17_5.sce +++ /dev/null @@ -1,25 +0,0 @@ -//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/Ex17_6.sce b/2465/CH17/EX17.6/Ex17_6.sce deleted file mode 100644 index f138207aa..000000000 --- a/2465/CH17/EX17.6/Ex17_6.sce +++ /dev/null @@ -1,29 +0,0 @@ -//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/Ex17_7.sce b/2465/CH17/EX17.7/Ex17_7.sce deleted file mode 100644 index 870f634e3..000000000 --- a/2465/CH17/EX17.7/Ex17_7.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex17_8.sce b/2465/CH17/EX17.8/Ex17_8.sce deleted file mode 100644 index 3642e0698..000000000 --- a/2465/CH17/EX17.8/Ex17_8.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex17_9.sce b/2465/CH17/EX17.9/Ex17_9.sce deleted file mode 100644 index 0436f50e0..000000000 --- a/2465/CH17/EX17.9/Ex17_9.sce +++ /dev/null @@ -1,31 +0,0 @@ -//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/Ex18_1.sce b/2465/CH18/EX18.1/Ex18_1.sce deleted file mode 100644 index d93962d69..000000000 --- a/2465/CH18/EX18.1/Ex18_1.sce +++ /dev/null @@ -1,27 +0,0 @@ -//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/Ex18_2.sce b/2465/CH18/EX18.2/Ex18_2.sce deleted file mode 100644 index 76386308f..000000000 --- a/2465/CH18/EX18.2/Ex18_2.sce +++ /dev/null @@ -1,35 +0,0 @@ -//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/Ex18_3.sce b/2465/CH18/EX18.3/Ex18_3.sce deleted file mode 100644 index b3e14c5d3..000000000 --- a/2465/CH18/EX18.3/Ex18_3.sce +++ /dev/null @@ -1,27 +0,0 @@ -//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/Ex18_4.sce b/2465/CH18/EX18.4/Ex18_4.sce deleted file mode 100644 index ba8dab90c..000000000 --- a/2465/CH18/EX18.4/Ex18_4.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex18_5.sce b/2465/CH18/EX18.5/Ex18_5.sce deleted file mode 100644 index 811aae588..000000000 --- a/2465/CH18/EX18.5/Ex18_5.sce +++ /dev/null @@ -1,86 +0,0 @@ -//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/Ex18_6.sce b/2465/CH18/EX18.6/Ex18_6.sce deleted file mode 100644 index 9441a7a58..000000000 --- a/2465/CH18/EX18.6/Ex18_6.sce +++ /dev/null @@ -1,37 +0,0 @@ -//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/Ex22_2.sce b/2465/CH22/EX22.2/Ex22_2.sce deleted file mode 100644 index 2b2998ca9..000000000 --- a/2465/CH22/EX22.2/Ex22_2.sce +++ /dev/null @@ -1,15 +0,0 @@ -//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/Ex22_4.sce b/2465/CH22/EX22.4/Ex22_4.sce deleted file mode 100644 index 445fdbd62..000000000 --- a/2465/CH22/EX22.4/Ex22_4.sce +++ /dev/null @@ -1,26 +0,0 @@ -//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/Ex22_5.sce b/2465/CH22/EX22.5/Ex22_5.sce deleted file mode 100644 index 91e96006a..000000000 --- a/2465/CH22/EX22.5/Ex22_5.sce +++ /dev/null @@ -1,22 +0,0 @@ -//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/Ex3_1.sce b/2465/CH3/EX3.1/Ex3_1.sce deleted file mode 100644 index 093b57b8d..000000000 --- a/2465/CH3/EX3.1/Ex3_1.sce +++ /dev/null @@ -1,16 +0,0 @@ -
-//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/Ex3_10.sce b/2465/CH3/EX3.10/Ex3_10.sce deleted file mode 100644 index 334f2bd4d..000000000 --- a/2465/CH3/EX3.10/Ex3_10.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex3_11.sce b/2465/CH3/EX3.11/Ex3_11.sce deleted file mode 100644 index 945857c41..000000000 --- a/2465/CH3/EX3.11/Ex3_11.sce +++ /dev/null @@ -1,11 +0,0 @@ -//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/Ex3_12.sce b/2465/CH3/EX3.12/Ex3_12.sce deleted file mode 100644 index 8e01cce46..000000000 --- a/2465/CH3/EX3.12/Ex3_12.sce +++ /dev/null @@ -1,11 +0,0 @@ -//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/Ex3_13.sce b/2465/CH3/EX3.13/Ex3_13.sce deleted file mode 100644 index fe6f94a34..000000000 --- a/2465/CH3/EX3.13/Ex3_13.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex3_14.sce b/2465/CH3/EX3.14/Ex3_14.sce deleted file mode 100644 index 418a68b43..000000000 --- a/2465/CH3/EX3.14/Ex3_14.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex3_15.sce b/2465/CH3/EX3.15/Ex3_15.sce deleted file mode 100644 index a9db46a01..000000000 --- a/2465/CH3/EX3.15/Ex3_15.sce +++ /dev/null @@ -1,20 +0,0 @@ -//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/Ex3_16.sce b/2465/CH3/EX3.16/Ex3_16.sce deleted file mode 100644 index cee4cbed1..000000000 --- a/2465/CH3/EX3.16/Ex3_16.sce +++ /dev/null @@ -1,22 +0,0 @@ -//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/Ex3_17.sce b/2465/CH3/EX3.17/Ex3_17.sce deleted file mode 100644 index 2138c04fb..000000000 --- a/2465/CH3/EX3.17/Ex3_17.sce +++ /dev/null @@ -1,18 +0,0 @@ -//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/Ex3_18.sce b/2465/CH3/EX3.18/Ex3_18.sce deleted file mode 100644 index b132f9fdc..000000000 --- a/2465/CH3/EX3.18/Ex3_18.sce +++ /dev/null @@ -1,25 +0,0 @@ -//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/Ex3_19.sce b/2465/CH3/EX3.19/Ex3_19.sce deleted file mode 100644 index e0c98256f..000000000 --- a/2465/CH3/EX3.19/Ex3_19.sce +++ /dev/null @@ -1,20 +0,0 @@ -//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/Ex3_2.sce b/2465/CH3/EX3.2/Ex3_2.sce deleted file mode 100644 index f89b82a5e..000000000 --- a/2465/CH3/EX3.2/Ex3_2.sce +++ /dev/null @@ -1,20 +0,0 @@ -//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/Ex3_4.sce b/2465/CH3/EX3.4/Ex3_4.sce deleted file mode 100644 index 2a9fb99e1..000000000 --- a/2465/CH3/EX3.4/Ex3_4.sce +++ /dev/null @@ -1,23 +0,0 @@ -//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/Ex3_5.sce b/2465/CH3/EX3.5/Ex3_5.sce deleted file mode 100644 index 589cf1fdd..000000000 --- a/2465/CH3/EX3.5/Ex3_5.sce +++ /dev/null @@ -1,27 +0,0 @@ -//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/Ex3_6.sce b/2465/CH3/EX3.6/Ex3_6.sce deleted file mode 100644 index 92663dfd4..000000000 --- a/2465/CH3/EX3.6/Ex3_6.sce +++ /dev/null @@ -1,18 +0,0 @@ -//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/Ex3_7.sce b/2465/CH3/EX3.7/Ex3_7.sce deleted file mode 100644 index 256697d5e..000000000 --- a/2465/CH3/EX3.7/Ex3_7.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex3_8.sce b/2465/CH3/EX3.8/Ex3_8.sce deleted file mode 100644 index 3011a501e..000000000 --- a/2465/CH3/EX3.8/Ex3_8.sce +++ /dev/null @@ -1,23 +0,0 @@ -//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/Ex3_9.sce b/2465/CH3/EX3.9/Ex3_9.sce deleted file mode 100644 index 1a02f1182..000000000 --- a/2465/CH3/EX3.9/Ex3_9.sce +++ /dev/null @@ -1,25 +0,0 @@ -//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/Ex4_1.sce b/2465/CH4/EX4.1/Ex4_1.sce deleted file mode 100644 index 502811c64..000000000 --- a/2465/CH4/EX4.1/Ex4_1.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex4_10.sce b/2465/CH4/EX4.10/Ex4_10.sce deleted file mode 100644 index b22df3756..000000000 --- a/2465/CH4/EX4.10/Ex4_10.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex4_11.sce b/2465/CH4/EX4.11/Ex4_11.sce deleted file mode 100644 index da8cc6183..000000000 --- a/2465/CH4/EX4.11/Ex4_11.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex4_12.sce b/2465/CH4/EX4.12/Ex4_12.sce deleted file mode 100644 index e245a0a62..000000000 --- a/2465/CH4/EX4.12/Ex4_12.sce +++ /dev/null @@ -1,13 +0,0 @@ -//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/Ex4_13.sce b/2465/CH4/EX4.13/Ex4_13.sce deleted file mode 100644 index f47b7d766..000000000 --- a/2465/CH4/EX4.13/Ex4_13.sce +++ /dev/null @@ -1,14 +0,0 @@ -//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/Ex4_15.sce b/2465/CH4/EX4.15/Ex4_15.sce deleted file mode 100644 index 95bcb4a40..000000000 --- a/2465/CH4/EX4.15/Ex4_15.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex4_16.sce b/2465/CH4/EX4.16/Ex4_16.sce deleted file mode 100644 index 0b2037a77..000000000 --- a/2465/CH4/EX4.16/Ex4_16.sce +++ /dev/null @@ -1,27 +0,0 @@ -//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/Ex4_17.sce b/2465/CH4/EX4.17/Ex4_17.sce deleted file mode 100644 index 483dd3afd..000000000 --- a/2465/CH4/EX4.17/Ex4_17.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex4_18.sce b/2465/CH4/EX4.18/Ex4_18.sce deleted file mode 100644 index 2c66dc887..000000000 --- a/2465/CH4/EX4.18/Ex4_18.sce +++ /dev/null @@ -1,21 +0,0 @@ -//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/Ex4_19.sce b/2465/CH4/EX4.19/Ex4_19.sce deleted file mode 100644 index ee777211a..000000000 --- a/2465/CH4/EX4.19/Ex4_19.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex4_2.sce b/2465/CH4/EX4.2/Ex4_2.sce deleted file mode 100644 index d2dbc03d9..000000000 --- a/2465/CH4/EX4.2/Ex4_2.sce +++ /dev/null @@ -1,11 +0,0 @@ -//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/Ex4_3.sce b/2465/CH4/EX4.3/Ex4_3.sce deleted file mode 100644 index d3037835a..000000000 --- a/2465/CH4/EX4.3/Ex4_3.sce +++ /dev/null @@ -1,13 +0,0 @@ -//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/Ex4_4.sce b/2465/CH4/EX4.4/Ex4_4.sce deleted file mode 100644 index 9afb4eb21..000000000 --- a/2465/CH4/EX4.4/Ex4_4.sce +++ /dev/null @@ -1,21 +0,0 @@ -//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/Ex4_5.sce b/2465/CH4/EX4.5/Ex4_5.sce deleted file mode 100644 index ce36439e7..000000000 --- a/2465/CH4/EX4.5/Ex4_5.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex4_6.sce b/2465/CH4/EX4.6/Ex4_6.sce deleted file mode 100644 index 7509cd20a..000000000 --- a/2465/CH4/EX4.6/Ex4_6.sce +++ /dev/null @@ -1,18 +0,0 @@ -//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/Ex4_7.sce b/2465/CH4/EX4.7/Ex4_7.sce deleted file mode 100644 index bb3bd5922..000000000 --- a/2465/CH4/EX4.7/Ex4_7.sce +++ /dev/null @@ -1,16 +0,0 @@ -//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/Ex4_8.sce b/2465/CH4/EX4.8/Ex4_8.sce deleted file mode 100644 index 2a642f92e..000000000 --- a/2465/CH4/EX4.8/Ex4_8.sce +++ /dev/null @@ -1,23 +0,0 @@ -//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/Ex4_9.sce b/2465/CH4/EX4.9/Ex4_9.sce deleted file mode 100644 index 39c08702c..000000000 --- a/2465/CH4/EX4.9/Ex4_9.sce +++ /dev/null @@ -1,15 +0,0 @@ -//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/Ex5_1.sce b/2465/CH5/EX5.1/Ex5_1.sce deleted file mode 100644 index aa4ffc6d9..000000000 --- a/2465/CH5/EX5.1/Ex5_1.sce +++ /dev/null @@ -1,23 +0,0 @@ -//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/Ex5_10.sce b/2465/CH5/EX5.10/Ex5_10.sce deleted file mode 100644 index d2c27a4d5..000000000 --- a/2465/CH5/EX5.10/Ex5_10.sce +++ /dev/null @@ -1,29 +0,0 @@ -//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/Ex5_11.sce b/2465/CH5/EX5.11/Ex5_11.sce deleted file mode 100644 index 06b492323..000000000 --- a/2465/CH5/EX5.11/Ex5_11.sce +++ /dev/null @@ -1,23 +0,0 @@ -//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/Ex5_12.sce b/2465/CH5/EX5.12/Ex5_12.sce deleted file mode 100644 index 608882bf0..000000000 --- a/2465/CH5/EX5.12/Ex5_12.sce +++ /dev/null @@ -1,23 +0,0 @@ -//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/Ex5_15.sce b/2465/CH5/EX5.15/Ex5_15.sce deleted file mode 100644 index 0b9f3a800..000000000 --- a/2465/CH5/EX5.15/Ex5_15.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex5_16.sce b/2465/CH5/EX5.16/Ex5_16.sce deleted file mode 100644 index 90d6670a2..000000000 --- a/2465/CH5/EX5.16/Ex5_16.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex5_17.sce b/2465/CH5/EX5.17/Ex5_17.sce deleted file mode 100644 index 7156929ab..000000000 --- a/2465/CH5/EX5.17/Ex5_17.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex5_2.sce b/2465/CH5/EX5.2/Ex5_2.sce deleted file mode 100644 index e1d462398..000000000 --- a/2465/CH5/EX5.2/Ex5_2.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex5_3.sce b/2465/CH5/EX5.3/Ex5_3.sce deleted file mode 100644 index 08a5f99ec..000000000 --- a/2465/CH5/EX5.3/Ex5_3.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex5_4.sce b/2465/CH5/EX5.4/Ex5_4.sce deleted file mode 100644 index 9f7f895e6..000000000 --- a/2465/CH5/EX5.4/Ex5_4.sce +++ /dev/null @@ -1,15 +0,0 @@ -//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/Ex5_5.sce b/2465/CH5/EX5.5/Ex5_5.sce deleted file mode 100644 index 371a55fec..000000000 --- a/2465/CH5/EX5.5/Ex5_5.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex5_6.sce b/2465/CH5/EX5.6/Ex5_6.sce deleted file mode 100644 index f607293bc..000000000 --- a/2465/CH5/EX5.6/Ex5_6.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex5_7.sce b/2465/CH5/EX5.7/Ex5_7.sce deleted file mode 100644 index 728c0cf49..000000000 --- a/2465/CH5/EX5.7/Ex5_7.sce +++ /dev/null @@ -1,27 +0,0 @@ -//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/Ex5_9.sce b/2465/CH5/EX5.9/Ex5_9.sce deleted file mode 100644 index 9e528c241..000000000 --- a/2465/CH5/EX5.9/Ex5_9.sce +++ /dev/null @@ -1,15 +0,0 @@ -//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/Ex8_1.sce b/2465/CH8/EX8.1/Ex8_1.sce deleted file mode 100644 index 9917fc3f0..000000000 --- a/2465/CH8/EX8.1/Ex8_1.sce +++ /dev/null @@ -1,11 +0,0 @@ -//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/Ex8_2.sce b/2465/CH8/EX8.2/Ex8_2.sce deleted file mode 100644 index f6a1b0244..000000000 --- a/2465/CH8/EX8.2/Ex8_2.sce +++ /dev/null @@ -1,21 +0,0 @@ -//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/Ex8_3.sce b/2465/CH8/EX8.3/Ex8_3.sce deleted file mode 100644 index 6c3156976..000000000 --- a/2465/CH8/EX8.3/Ex8_3.sce +++ /dev/null @@ -1,32 +0,0 @@ -//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/Ex8_4.sce b/2465/CH8/EX8.4/Ex8_4.sce deleted file mode 100644 index a9d10cf1d..000000000 --- a/2465/CH8/EX8.4/Ex8_4.sce +++ /dev/null @@ -1,24 +0,0 @@ -//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/Ex8_5.sce b/2465/CH8/EX8.5/Ex8_5.sce deleted file mode 100644 index 90551f959..000000000 --- a/2465/CH8/EX8.5/Ex8_5.sce +++ /dev/null @@ -1,15 +0,0 @@ -//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/Ex8_6.sce b/2465/CH8/EX8.6/Ex8_6.sce deleted file mode 100644 index 984f0c31f..000000000 --- a/2465/CH8/EX8.6/Ex8_6.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex8_7.sce b/2465/CH8/EX8.7/Ex8_7.sce deleted file mode 100644 index 1d5b25cf9..000000000 --- a/2465/CH8/EX8.7/Ex8_7.sce +++ /dev/null @@ -1,14 +0,0 @@ -//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/Ex8_8.sce b/2465/CH8/EX8.8/Ex8_8.sce deleted file mode 100644 index 7e43b504c..000000000 --- a/2465/CH8/EX8.8/Ex8_8.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex9_1.sce b/2465/CH9/EX9.1/Ex9_1.sce deleted file mode 100644 index 8c983f2ae..000000000 --- a/2465/CH9/EX9.1/Ex9_1.sce +++ /dev/null @@ -1,19 +0,0 @@ -//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/Ex9_10.sce b/2465/CH9/EX9.10/Ex9_10.sce deleted file mode 100644 index dcd2b7092..000000000 --- a/2465/CH9/EX9.10/Ex9_10.sce +++ /dev/null @@ -1,22 +0,0 @@ -//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/Ex9_11.sce b/2465/CH9/EX9.11/Ex9_11.sce deleted file mode 100644 index 3c674c678..000000000 --- a/2465/CH9/EX9.11/Ex9_11.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex9_12.sce b/2465/CH9/EX9.12/Ex9_12.sce deleted file mode 100644 index 0af82950a..000000000 --- a/2465/CH9/EX9.12/Ex9_12.sce +++ /dev/null @@ -1,21 +0,0 @@ -//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/Ex9_2.sce b/2465/CH9/EX9.2/Ex9_2.sce deleted file mode 100644 index b2d50cde9..000000000 --- a/2465/CH9/EX9.2/Ex9_2.sce +++ /dev/null @@ -1,16 +0,0 @@ -//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/Ex9_3.sce b/2465/CH9/EX9.3/Ex9_3.sce deleted file mode 100644 index 3e51aedc3..000000000 --- a/2465/CH9/EX9.3/Ex9_3.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex9_4.sce b/2465/CH9/EX9.4/Ex9_4.sce deleted file mode 100644 index fbd6e46ab..000000000 --- a/2465/CH9/EX9.4/Ex9_4.sce +++ /dev/null @@ -1,25 +0,0 @@ -//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/Ex9_5.sce b/2465/CH9/EX9.5/Ex9_5.sce deleted file mode 100644 index eb9776f64..000000000 --- a/2465/CH9/EX9.5/Ex9_5.sce +++ /dev/null @@ -1,17 +0,0 @@ -//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/Ex9_6.sce b/2465/CH9/EX9.6/Ex9_6.sce deleted file mode 100644 index 6cec6eaf8..000000000 --- a/2465/CH9/EX9.6/Ex9_6.sce +++ /dev/null @@ -1,11 +0,0 @@ -//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/Ex9_7.sce b/2465/CH9/EX9.7/Ex9_7.sce deleted file mode 100644 index 88a5053e4..000000000 --- a/2465/CH9/EX9.7/Ex9_7.sce +++ /dev/null @@ -1,16 +0,0 @@ -//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/Ex9_8.sce b/2465/CH9/EX9.8/Ex9_8.sce deleted file mode 100644 index 522db0bde..000000000 --- a/2465/CH9/EX9.8/Ex9_8.sce +++ /dev/null @@ -1,15 +0,0 @@ -//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/Ex9_9.sce b/2465/CH9/EX9.9/Ex9_9.sce deleted file mode 100644 index b0181a822..000000000 --- a/2465/CH9/EX9.9/Ex9_9.sce +++ /dev/null @@ -1,18 +0,0 @@ -//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)
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