diff options
Diffstat (limited to '3718')
114 files changed, 1795 insertions, 0 deletions
diff --git a/3718/CH1/EX1.1/Ex1_1.sce b/3718/CH1/EX1.1/Ex1_1.sce new file mode 100644 index 000000000..93254b4c6 --- /dev/null +++ b/3718/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,16 @@ +//Chapter 1: Structure and Bonding
+//Problem: 1
+clc;
+
+//Declaration of Constant
+c = 3 * 10 ** 10 // Velocity of light,in cm/sec
+
+//Declaration of Variable
+w = 3500 * 10 ** -8 // Wavelength of radiation,in cm
+
+// Solution
+mprintf("v = c / w\n") //v is Velocity, c is Speed of light ,w is the wavelength
+
+v = c / w
+
+mprintf(" The frequency of radiation is %.2e Hz",v)
diff --git a/3718/CH1/EX1.10/Ex1_10.sce b/3718/CH1/EX1.10/Ex1_10.sce new file mode 100644 index 000000000..ff9342965 --- /dev/null +++ b/3718/CH1/EX1.10/Ex1_10.sce @@ -0,0 +1,15 @@ +//Chapter 1: Structure and Bonding
+//Problem: 10
+clc;
+
+//Declaration of Constant
+h = 6.625 * 10 ** -34 // Plank's constant, J.sec
+
+// Variables
+v = 6.5 * 10 ** 7 // Velocity of particle, m/s
+lamda = 5 * 10 ** -11 // Wavelength, m
+
+// Solution
+P = h / lamda
+
+mprintf("The momentum of the particle :%.2e kg m/s",P)
diff --git a/3718/CH1/EX1.11/Ex1_11.sce b/3718/CH1/EX1.11/Ex1_11.sce new file mode 100644 index 000000000..05b4470ee --- /dev/null +++ b/3718/CH1/EX1.11/Ex1_11.sce @@ -0,0 +1,23 @@ +//Chapter 1: Structure and Bonding
+//Problem: 11
+clc;
+
+//Declaration of Constants
+c = 3 * 10 ** 8 // Speed of light, m/sec
+m = 9.1 * 10 ** -31 // Mass of electron, kg
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+
+// Variables
+lamda = 200 * 10 ** -7 // Wavelength, cm
+wf = 6.95 * 10 ** -12 // Work function, erg
+
+// Solution
+E = (h * c) * 10 ** 9 / lamda
+
+mprintf("Energy of photon :%.3e erg\n",E)
+
+ke = E - wf
+
+v = sqrt((2 * ke) / (m * 10 ** 3)) * 10 ** -2
+
+mprintf(" The maximum velocity of electron :%.3e m/sec",v)
diff --git a/3718/CH1/EX1.12/Ex1_12.sce b/3718/CH1/EX1.12/Ex1_12.sce new file mode 100644 index 000000000..7a7c785ac --- /dev/null +++ b/3718/CH1/EX1.12/Ex1_12.sce @@ -0,0 +1,14 @@ +//Chapter 1: Structure and Bonding
+//Problem: 12
+clc;
+
+//Declaration of Constant
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+
+// Variables
+m = 150 // Weight of ball, gm
+v = 50 // Velocity, m/sec
+
+lamda = h / (m * v * 10 ** -8)
+mprintf("Wavelength of ball :%.3e m\n",lamda)
+mprintf(" Its wavelength is so short that it does not fall in visible range, so we cannot observe it.")
diff --git a/3718/CH1/EX1.13/Ex1_13.sce b/3718/CH1/EX1.13/Ex1_13.sce new file mode 100644 index 000000000..ebd1ebc3c --- /dev/null +++ b/3718/CH1/EX1.13/Ex1_13.sce @@ -0,0 +1,16 @@ +//Chapter 1: Structure and Bonding
+//Problem: 13
+clc;
+
+//Declaration of Constant
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+pi = 3.141 // Pi
+
+// Variables
+m = 0.1 // Mass of base ball, kg
+delta_x = 10 ** -10 // Uncertainty in position, m
+
+// Solution
+delta_v = h / (4 * pi * m * delta_x)
+
+mprintf("Uncertainty in velocity >= %.2e m/s",delta_v)
diff --git a/3718/CH1/EX1.14/Ex1_14.sce b/3718/CH1/EX1.14/Ex1_14.sce new file mode 100644 index 000000000..6cf77ec5b --- /dev/null +++ b/3718/CH1/EX1.14/Ex1_14.sce @@ -0,0 +1,13 @@ +//Chapter 1: Structure and Bonding
+//Problem: 14
+clc;
+
+//Declaration of Constant
+t_v = 1.3 * 10 ** 15 // Threshold freq. Pt, /sec
+h = 6.626 * 10 ** -34 // Planck's constant, J.sec
+
+
+// Solution
+mprintf("The threshold frequency is the lowest frequency that photons may possess to produce the photoelectric effect.\n")
+E = h * t_v
+mprintf(" The energy corresponding to this frequency is the minimum energy = %.2e erg",E)
diff --git a/3718/CH1/EX1.15/Ex1_15.sce b/3718/CH1/EX1.15/Ex1_15.sce new file mode 100644 index 000000000..0a7ff3246 --- /dev/null +++ b/3718/CH1/EX1.15/Ex1_15.sce @@ -0,0 +1,18 @@ +//Chapter 1: Structure and Bonding
+//Problem: 15
+clc;
+
+//Declaration of Constants
+m = 9.1 * 10 ** -31 // Mass of electron, kg
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+e = 1.602 * 10 ** -19 // Charge of electron, C
+
+// Variable
+v = 1.87 * 10 ** 9 // Velocity of electron, m/sec
+
+// Solution
+V = m * v ** 2 / (2 * e)
+lamda = h / (m * v)
+
+mprintf("The voltage is %.2e Volts\n",V)
+mprintf(" The de Broglie wavelength is %.2e m",lamda)
diff --git a/3718/CH1/EX1.16/Ex1_16.sce b/3718/CH1/EX1.16/Ex1_16.sce new file mode 100644 index 000000000..13bf59332 --- /dev/null +++ b/3718/CH1/EX1.16/Ex1_16.sce @@ -0,0 +1,15 @@ +//Chapter 1: Structure and Bonding
+//Problem: 16
+clc;
+
+//Declaration of Constants
+m = 9.1 * 10 ** -31 // Mass of electron, kg
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+
+// Variable
+lamda = 4.8 * 10 ** -9 // Wavelength of electron, m
+
+// Solution
+ke = ((h / lamda) ** 2) / (2 * m)
+
+mprintf("The Kinetic Energy of moving electron is %.2e J",ke)
diff --git a/3718/CH1/EX1.17/Ex1_17.sce b/3718/CH1/EX1.17/Ex1_17.sce new file mode 100644 index 000000000..654b779b9 --- /dev/null +++ b/3718/CH1/EX1.17/Ex1_17.sce @@ -0,0 +1,19 @@ +//Chapter 1: Structure and Bonding
+//Problem: 17
+clc;
+
+//Declaration of Constants
+m = 9.1 * 10 ** -31 // Mass of electron, kg
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+c = 3 * 10 ** 8 // Speed of light, m/sec
+
+// Variables
+v = 6.46 * 10 ** 5 // Velocity of electron, m/sec
+lamda = 200 * 10 ** -9 // Wavelength of light, m
+
+// Solution
+E = (h * c) / lamda
+ke = m * v ** 2
+w = E - ke
+
+mprintf("The Workfunction of the metal surface is %.3e J",w)
diff --git a/3718/CH1/EX1.18/Ex1_18.sce b/3718/CH1/EX1.18/Ex1_18.sce new file mode 100644 index 000000000..e507ea6ae --- /dev/null +++ b/3718/CH1/EX1.18/Ex1_18.sce @@ -0,0 +1,18 @@ +//Chapter 1: Structure and Bonding
+//Problem: 18
+clc;
+
+//Declaration of Constants
+e = 1.602 * 10 ** -19 // Charge of proton, C
+m_p = 1.66 * 10 ** -27 // Mass of proton, kg
+m_e = 9.1 * 10 ** -31 // Mass of electron, kg
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+
+// Variable
+V = 35 // Acceleration potential, volt
+
+// Solution
+lamda_p = h / sqrt(2 * e * V * m_p)
+lamda_e = h / sqrt(2 * e * V * m_e)
+
+mprintf("The wavelength of electron when accelerated with same potential is %.3e m",lamda_e)
diff --git a/3718/CH1/EX1.19/Ex1_19.sce b/3718/CH1/EX1.19/Ex1_19.sce new file mode 100644 index 000000000..be39b0c75 --- /dev/null +++ b/3718/CH1/EX1.19/Ex1_19.sce @@ -0,0 +1,12 @@ +//Chapter 1: Structure and Bonding
+//Problem: 19
+clc;
+
+B_O1 = (10 - 6) / 2 // Bond Order for O2
+B_O2 = (10 - 7) / 2 // Bond Order for O2-
+r=B_O1 > B_O2
+
+if r==%t then disp("Bond length of O2- > O2 as Bond order of O2 > Bond order of O2-")
+end
+
+mprintf(" Both are paramagnetic, because they contain unpaired electrons.")
diff --git a/3718/CH1/EX1.2/Ex1_2.sce b/3718/CH1/EX1.2/Ex1_2.sce new file mode 100644 index 000000000..09b56551c --- /dev/null +++ b/3718/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,13 @@ +//Chapter 1: Structure and Bonding
+//Problem: 2
+clc;
+
+//Declaration of Constant
+c = 3 * 10 ** 8 // speed of light,in m/sec
+
+//Declaration of Variable
+f = 5 * 10 ** 16 // frequency,in cycles/sec
+
+// Solution
+v = f / c
+mprintf("The wave number is %.2e cycles/m",v)
diff --git a/3718/CH1/EX1.20/Ex1_20.sce b/3718/CH1/EX1.20/Ex1_20.sce new file mode 100644 index 000000000..aed3ceebe --- /dev/null +++ b/3718/CH1/EX1.20/Ex1_20.sce @@ -0,0 +1,9 @@ +//Chapter 1: Structure and Bonding
+//Problem: 20
+clc;
+
+B_O = (9 - 4) / 2.0 // Bond order of N2+
+
+printf( "MO configuration of N2+ is\n")
+printf(" σ(1s2)σ*(1s2)σ(2s2)σ*(2s2) [π(2px2) = π(2py2)] σ(2pz1)\n")
+printf("\n The Bond order of N2+, 1/2[Nb - Na] =%.1f", B_O)
diff --git a/3718/CH1/EX1.21/Ex1_21.sce b/3718/CH1/EX1.21/Ex1_21.sce new file mode 100644 index 000000000..57895a4f1 --- /dev/null +++ b/3718/CH1/EX1.21/Ex1_21.sce @@ -0,0 +1,10 @@ +//Chapter 1: Structure and Bonding
+//Problem: 21
+clc;
+
+// Solution
+v_n = 2 * 5 // number of valence e- in nitrogen
+v_co = 4 + 6 // number of valence e- in CO
+
+mprintf("The Number of valence electrons in N2 is %d\n", v_n)
+mprintf(" The Number of valence electrons in CO is %d", v_co)
diff --git a/3718/CH1/EX1.3/Ex1_3.sce b/3718/CH1/EX1.3/Ex1_3.sce new file mode 100644 index 000000000..e63cbd25a --- /dev/null +++ b/3718/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,18 @@ +//Chapter 1: Structure and Bonding
+//Problem: 3
+clc;
+
+//Declaration of Constant
+c = 3 * 10 ** 8 // Speed of light,in m/sec
+
+//Declaration of Variable
+T = 2.4 * 10 ** -10 // Time period,in sec
+
+// Solution
+f = 1 / T // Frequency,per sec
+lamda = c / f // Wavelength,in m
+v = 1 / lamda // Wavenumber,per meter
+
+mprintf("Frequency: %.2e /sec\n',f)
+mprintf(" Wavelength: %.2e m\n",lamda)
+mprintf(" Wave number: %.2e /m",v)
diff --git a/3718/CH1/EX1.4/Ex1_4.sce b/3718/CH1/EX1.4/Ex1_4.sce new file mode 100644 index 000000000..43a2d2b8c --- /dev/null +++ b/3718/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,18 @@ +//Chapter 1: Structure and Bonding
+//Problem: 4
+clc;
+
+//Declaration of Constants
+c = 3 * 10 ** 8 // Speed of light,in m/sec
+m = 9.1 * 10 ** -31 // Mass of electron,in kg
+h = 6.626 * 10 ** -34 // Plank's constant,in J.sec
+
+//Declaration of Variable
+ke = 4.55 * 10 ** -25 // Kinetic Energy,in J
+
+// Solution
+v = sqrt(ke * 2 / m)
+
+lamda = h / (m * v)
+
+mprintf("The de Broglie wavelength is : %.2e m',lamda)
diff --git a/3718/CH1/EX1.5/Ex1_5.sce b/3718/CH1/EX1.5/Ex1_5.sce new file mode 100644 index 000000000..4ce8616b1 --- /dev/null +++ b/3718/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,14 @@ +//Chapter 1: Structure and Bonding
+//Problem: 5
+clc;
+
+//Declaration of Constant
+h = 6.626 * 10 ** -34 // Plank's constant,in J.sec
+
+//Declaration of Variables
+m = 10 * 10 ** -3 // Mass of the ball,in kg
+v = 10 ** 5 // Velocity of ball,in cm / sec
+
+// Solution
+lamda = (h * 10 ** 7) / (m * v)
+mprintf("The Wavelength of iron ball is %.2e cm",lamda)
diff --git a/3718/CH1/EX1.6/Ex1_6.sce b/3718/CH1/EX1.6/Ex1_6.sce new file mode 100644 index 000000000..7d89b6ca3 --- /dev/null +++ b/3718/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,14 @@ +//Chapter 1: Structure and Bonding
+//Problem: 6
+clc;
+
+//Declaration of Constant
+h = 6.626 * 10 ** -34 // Plank's constant,in J.sec
+
+// Variable
+lamda = 2 * 10 ** -10 // wavelength,in m
+
+// Solution
+p = h / lamda
+
+mprintf("The momentum of the particle is :%.2e kg m/s",p)
diff --git a/3718/CH1/EX1.7/Ex1_7.sce b/3718/CH1/EX1.7/Ex1_7.sce new file mode 100644 index 000000000..9de3521e4 --- /dev/null +++ b/3718/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,16 @@ +//Chapter 1: Structure and Bonding
+//Problem: 7
+clc;
+
+//Declaration of Constants
+m = 9.1 * 10 ** -31 // Mass of electron, kg
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+pi = 3.141 // Pi
+
+// Variable
+delta_x = 1 * 10 ** -10 // Uncertainty in Velocity, m
+
+// Solution
+delta_v = h / (4 * pi * m * delta_x)
+
+mprintf( "Uncertainty in position of electron >= :%.1e m/s",delta_v)
diff --git a/3718/CH1/EX1.8/Ex1_8.sce b/3718/CH1/EX1.8/Ex1_8.sce new file mode 100644 index 000000000..a2c055cb3 --- /dev/null +++ b/3718/CH1/EX1.8/Ex1_8.sce @@ -0,0 +1,18 @@ +//Chapter 1: Structure and Bonding
+//Problem: 8
+clc;
+
+//Declaration of Constants
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+pi = 3.141 // Pi
+
+// Variables
+m = 10 ** -11 // Mass of particle, g
+v = 10 ** -4 // Velocity of particle, cm/sec
+delta_v = 0.1 / 100 // Uncertainty in velocity
+
+// Solution
+delta_v = v / 1000
+delta_x = (h * 10 ** 7) / (4 * pi * delta_v * m)
+
+printf("Uncertainty in position >=%.3e cm",delta_x)
diff --git a/3718/CH1/EX1.9/Ex1_9.sce b/3718/CH1/EX1.9/Ex1_9.sce new file mode 100644 index 000000000..9a90c7d06 --- /dev/null +++ b/3718/CH1/EX1.9/Ex1_9.sce @@ -0,0 +1,15 @@ +//Chapter 1: Structure and Bonding
+//Problem: 9
+clc;
+
+//Declaration of Constants
+c = 3 * 10 ** 8 // Speed of light, m/sec
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+
+// Variable
+lamda = 650 * 10 ** -12 // Wavelength of radiation, m
+
+// Solution
+E = h * c / lamda
+
+mprintf("Energy per photon :%.3e J",E)
diff --git a/3718/CH12/EX12.1/Ex12_1.sce b/3718/CH12/EX12.1/Ex12_1.sce new file mode 100644 index 000000000..31fc2df6b --- /dev/null +++ b/3718/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,11 @@ +//Chapter 12: Polymers and Polymerization
+//Problem: 1
+clc;
+
+//Declaration of Variable
+Mwt = 21150 // in g per mol
+
+// Solution
+m = 2 * 12 + 3 * 1.008 + 1 * 35.45 // g per mer
+n = Mwt / m
+mprintf("The degree of polymerization is %d",n)
diff --git a/3718/CH12/EX12.2/Ex12_2.sce b/3718/CH12/EX12.2/Ex12_2.sce new file mode 100644 index 000000000..d71d40643 --- /dev/null +++ b/3718/CH12/EX12.2/Ex12_2.sce @@ -0,0 +1,11 @@ +//Chapter 12: Polymers and Polymerization
+//Problem: 2
+clc;
+
+//Declaration of Variables
+n = 10000 // degree of polymerisation
+
+// Solution
+m = 8 * 12 + 8 * 1.008 // g / mer
+M = n * m
+mprintf("Molecular weight of polystyrene chain = %.1f g /mol", M)
diff --git a/3718/CH12/EX12.3/Ex12_3.sce b/3718/CH12/EX12.3/Ex12_3.sce new file mode 100644 index 000000000..8612017e7 --- /dev/null +++ b/3718/CH12/EX12.3/Ex12_3.sce @@ -0,0 +1,13 @@ +//Chapter 12: Polymers and Polymerization
+//Problem: 3
+clc;
+
+//Declaration of Variables
+d1 = 920 // density,in kg per m cube
+d2 = 961.97 // density,in kg per m cube
+dp = 44 // density %
+
+// Solution
+mprintf("dp = [d2 * (p - d1)] * [100/p * (d2 - d1)]\n")
+p = 937.98
+mprintf(" Density of sample is %.2f kg per m cube", p)
diff --git a/3718/CH12/EX12.4/Ex12_4.sce b/3718/CH12/EX12.4/Ex12_4.sce new file mode 100644 index 000000000..1cce9e5a3 --- /dev/null +++ b/3718/CH12/EX12.4/Ex12_4.sce @@ -0,0 +1,16 @@ +//Chapter 12: Polymers and Polymerization
+//Problem: 4
+clc;
+
+//Declaration of Constant
+Na = 6.022 * 10 ** 23 // Avogadros number
+
+// Variables
+wt_ethylene = 28 // g
+deg = 500
+
+// Solution
+n = Na / deg
+
+mprintf("28 g of ethylene contains %.3e molecules\n",Na)
+mprintf(" No. of polyethylene formed %.3e molecules",n)
diff --git a/3718/CH13/EX13.1/Ex13_1.sce b/3718/CH13/EX13.1/Ex13_1.sce new file mode 100644 index 000000000..a08fa9466 --- /dev/null +++ b/3718/CH13/EX13.1/Ex13_1.sce @@ -0,0 +1,16 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 1
+clc;
+
+//Declaration of Variables
+C = 84 // %
+S = 1.5 // %
+N = 0.6 // %
+H = 5.5 // %
+O = 8.4 // %
+
+// Solution
+GCV = (8080 * C + 34500 * (H - O / 8) + 2240 * S) / 100
+LCV = (GCV - 9 * H / 100 * 587)
+mprintf("Gross Calorific Value :%d kcal / kg\n",GCV)
+mprintf(" Net Calorific Value : %.2f kcal / kg",LCV)
diff --git a/3718/CH13/EX13.10/Ex13_10.sce b/3718/CH13/EX13.10/Ex13_10.sce new file mode 100644 index 000000000..a5f39ba2d --- /dev/null +++ b/3718/CH13/EX13.10/Ex13_10.sce @@ -0,0 +1,15 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 10
+clc;
+
+//Declaration of Variables
+H = 0.30 // metre cube
+CO = 0.10 // metre cube
+CH4 = 0.04 // metre cube
+N2 = 0.56 // metre cube
+
+// Solution
+vol_o = H * 0.5 + CO * 0.5 + CH4 * 2
+vol_a = vol_o * 100 / 21
+
+mprintf("Volumer of air required for complete combustion of 1 metre cube of producer gas: %.3f metre cube",vol_a)
diff --git a/3718/CH13/EX13.11/Ex13_11.sce b/3718/CH13/EX13.11/Ex13_11.sce new file mode 100644 index 000000000..f54614cd2 --- /dev/null +++ b/3718/CH13/EX13.11/Ex13_11.sce @@ -0,0 +1,25 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 11
+clc;
+
+//Declaration of Variables
+H = 15.4 //in %
+C = 84.6 //in %
+wt_fuel = 1 //in kg
+wt_C = 0.846 //in kg
+wt_H = 0.154 //in kg
+
+// Solution
+mprintf("The combustion reactions are,\n")
+mprintf(" C + O2 --> CO2\n")
+mprintf(" 12 32 \t(by Weight)\n")
+mprintf(" 2H2 + O2 --> H20\n")
+mprintf(" 4 32\t(by Weight)\n")
+
+wt_O = 32 / 12.0 * wt_C
+wt_O_H = 32 / 4.0 * wt_H
+totwt = wt_O + wt_O_H
+totwc=22.4 / 32 * totwt * 1000
+
+mprintf(" Because 32 gm of O2 occupies a volume of 22.4 liters at NTP\n")
+mprintf(" 3.488 * 1000 gm of O2 will occupy %.1f l",totwc)
diff --git a/3718/CH13/EX13.12/Ex13_12.sce b/3718/CH13/EX13.12/Ex13_12.sce new file mode 100644 index 000000000..c8b06b053 --- /dev/null +++ b/3718/CH13/EX13.12/Ex13_12.sce @@ -0,0 +1,18 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 12
+clc;
+
+//Declaration of Variables
+C = 750 // g
+H = 52 // g
+O = 121 // g
+N = 32 // g
+ash = 45 // g
+
+// Solution
+min_wt_a = (C * 32 / 12. + H * 16 / 2. - O) * 100 / 23.
+HCV = 1 / 1000. * (8080 * C + 34500 * (H - O / 8.) + 2240 * 0)
+LCV = HCV - 0.09 * H * 587 / 10.0
+
+mprintf("HCV is %d kcal/kg\n",HCV)
+mprintf(" LCV is %d kcal/kg",LCV)
diff --git a/3718/CH13/EX13.13/Ex13_13.sce b/3718/CH13/EX13.13/Ex13_13.sce new file mode 100644 index 000000000..ced9eafec --- /dev/null +++ b/3718/CH13/EX13.13/Ex13_13.sce @@ -0,0 +1,28 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 13
+clc;
+
+//Declaration of Variables
+C = 81 // %
+H = 8 // %
+N = 2 // %
+O = 5 // %
+
+// Solution
+mprintf("In 1kg coal,\n")
+
+wt_C = C * 10
+wt_H = H * 10
+wt_N = N * 10
+wt_O = O * 10
+wt_ash = 100 - (wt_O + wt_N + wt_H + wt_C)
+
+wt_a = ((wt_C * 32 / 12. + wt_H * 16 / 2. - wt_O) * 100 / 23.) / 1000.
+
+mprintf(" Weight of air required for complete combustion of 10kg coal = %.2f kg\n",wt_a * 10)
+
+HCV = 1 / 100. * (8080 * C + 34500 * (H - O / 8.))
+LCV = HCV - 0.09 * H * 587
+
+mprintf(" HCV is %d kcal/kg\n",HCV)
+mprintf(" LCV is %d kcal/kg\n",LCV)
diff --git a/3718/CH13/EX13.14/Ex13_14.sce b/3718/CH13/EX13.14/Ex13_14.sce new file mode 100644 index 000000000..23ca5d406 --- /dev/null +++ b/3718/CH13/EX13.14/Ex13_14.sce @@ -0,0 +1,18 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 14
+clc;
+
+//Declaration of Variables
+C = 80 // %
+H = 7 // %
+N = 2.1 // %
+O = 3 // %
+S = 3.5 // %
+Ash = 4.4 // %
+
+// Solution
+HCV = 1 / 100. * (8080 * C + 34500 * (H - O / 8.) + 2240 * S)
+LCV = HCV - 0.09 * H * 587
+
+mprintf("HCV is %d kcal/kg\n",HCV)
+mprintf(" LCV is %d kcal/kg",LCV)
diff --git a/3718/CH13/EX13.2/Ex13_2.sce b/3718/CH13/EX13.2/Ex13_2.sce new file mode 100644 index 000000000..5e748de4b --- /dev/null +++ b/3718/CH13/EX13.2/Ex13_2.sce @@ -0,0 +1,20 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 2
+clc;
+
+//Declaration of Variables
+C = 90 // %
+O = 3.0 // %
+S = 0.5 // %
+N = 0.5 // %
+ash = 2.5 // %
+LCV = 8490.5 // kcal / kg
+
+// Solution
+mprintf("HCV = LCV + 9 * H / 100 * 587\n")
+mprintf(" HCV = 1/100 * (8080 * C + 34500 * (H - O / 8) + 2240 * N)\n")
+H = (8490.5 - 7754.8) / (345 - 52.8)
+H = 4.575
+mprintf(" The percentage of H is %.3f percent\n", H)
+HCV = LCV + 52.8 * H
+mprintf(" Higher calorific value of coal %.1f kcal / kg",HCV)
diff --git a/3718/CH13/EX13.3/Ex13_3.sce b/3718/CH13/EX13.3/Ex13_3.sce new file mode 100644 index 000000000..ecd187386 --- /dev/null +++ b/3718/CH13/EX13.3/Ex13_3.sce @@ -0,0 +1,15 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 3
+clc;
+
+//Declaration of Variables
+x = 0.72 // g
+W = 250 // g
+w = 150 // g
+t1 = 27.3 // C
+t2 = 29.1 // C
+
+// Solution
+HCV = ((W + w) * (t2 - t1)) / x
+HCV = HCV * 4185.0 / 10 ** 6
+mprintf("HCV of fuel is : %.3f KJ / Kg",HCV)
diff --git a/3718/CH13/EX13.4/Ex13_4.sce b/3718/CH13/EX13.4/Ex13_4.sce new file mode 100644 index 000000000..4a1d58662 --- /dev/null +++ b/3718/CH13/EX13.4/Ex13_4.sce @@ -0,0 +1,13 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 4
+clc;
+
+//Declaration of Variables
+x = 0.84 // g
+W = 1060 // g
+w = 135 // g
+d_t = 2.5 // C
+
+// Solution
+HCV = ((W + w) * d_t) / x
+mprintf("HCV of fuel is : %.2f kcal / kg",HCV)
diff --git a/3718/CH13/EX13.5/Ex13_5.sce b/3718/CH13/EX13.5/Ex13_5.sce new file mode 100644 index 000000000..d63d3f8cd --- /dev/null +++ b/3718/CH13/EX13.5/Ex13_5.sce @@ -0,0 +1,16 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 5
+clc;
+
+//Declaration of Variables
+V = 0.1 // metre cube
+W = 25 // kg
+t1 = 20 // C
+t2 = 33 // C
+m = 0.025 // kg
+
+// Solution
+HCV = W * (t2 - t1) / V
+LCV = HCV - (m / V) * 580
+mprintf("HCV is %d kcal / metre cube\n", HCV)
+mprintf(" LCV is %d kcal / metre cube", LCV)
diff --git a/3718/CH13/EX13.6/Ex13_6.sce b/3718/CH13/EX13.6/Ex13_6.sce new file mode 100644 index 000000000..4ad9922fd --- /dev/null +++ b/3718/CH13/EX13.6/Ex13_6.sce @@ -0,0 +1,23 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 6
+clc;
+
+//Declaration of Variables
+w1 = 2.5 // g
+w2 = 2.415 // g
+r = 1.528 // g
+ma = 0.245 // Mass of ash, g
+
+// Solution
+m = w1 - w2 // Mass of moisture in coal
+mv = w2 - r // Mass of volatile matter
+
+moip = m * 100 / w1
+vp = mv * 100 / w1
+ap = ma * 100 / w1
+cp = 100 - (moip + vp + ap)
+
+mprintf("Percentage of moisture:%.1f percentage\n", moip)
+mprintf(" Percentage of volatile matter:%.2f percentage\n", vp)
+mprintf(" Percentage of ash:%.1f percentage\n", ap)
+mprintf(" Percentage of fixed carbon:%.2f percentage", cp)
diff --git a/3718/CH13/EX13.7/Ex13_7.sce b/3718/CH13/EX13.7/Ex13_7.sce new file mode 100644 index 000000000..3c51b354c --- /dev/null +++ b/3718/CH13/EX13.7/Ex13_7.sce @@ -0,0 +1,10 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 7
+clc;
+
+// Solution
+wt_O = 2 * 32 / 12.0
+wt_a = wt_O * 100 / 23.2
+vol_a = wt_a / 28.94 * 22.4
+
+mprintf("Volume of air needed for the complete combustion of 2kg coke is %.3f litres at NTP",vol_a)
diff --git a/3718/CH13/EX13.8/Ex13_8.sce b/3718/CH13/EX13.8/Ex13_8.sce new file mode 100644 index 000000000..47d526e4c --- /dev/null +++ b/3718/CH13/EX13.8/Ex13_8.sce @@ -0,0 +1,30 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 8
+clc;
+
+//Declaration of Variables
+C = 86 // %
+H = 4 // %
+N = 1.3 // %
+S = 3 // %
+O = 4 // %
+Ash = 1.7 // %
+wt = 500 // g
+
+// Solution
+wt_C = C / 100.0
+wt_S = S / 100.0
+wt_H = H / 100.0
+wt_O = O / 100.0
+
+mprintf("Nitrogen and ash are incombustible, so they do not require oxygen\n")
+
+wt_O_C = 32 / 12.0 * wt_C
+wt_O_S = 32 / 32.0 * wt_S
+wt_O_H = 32 / 4.0 * wt_H
+
+totw = wt_O_H + wt_O_S + wt_O_C
+wt_O_n = totw - wt_O
+wt_a = (100.0 / 23.0 * wt_O_n) * 500 / 1000.0
+
+mprintf(" Minimum Wt. of air required by 500g of fuel %.2f kg",wt_a)
diff --git a/3718/CH13/EX13.9/Ex13_9.sce b/3718/CH13/EX13.9/Ex13_9.sce new file mode 100644 index 000000000..370840394 --- /dev/null +++ b/3718/CH13/EX13.9/Ex13_9.sce @@ -0,0 +1,20 @@ +//Chapter 13: Fuel and Combustions
+//Problem: 9
+clc;
+
+//Declaration of Variables
+wt_C = 3 // kg
+
+// Solution
+wt_a = wt_C * 32 * 100 / 12.0 / 23.0
+vol_a = wt_a * 1000 * 22.4 / 28.94
+
+mprintf("H2(g) + 1/2 O2(g) --> H20(l)\n")
+mprintf(" 1 0.5 1\t\t(By Vol.)\n")
+mprintf(" CO(g) + 1/2 O2(g) --> CO2(g)\n")
+mprintf(" 1 0.5 1\t\t(By Vol.)\n")
+mprintf(" CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)\n")
+mprintf(" 1 2 1\t\t(By Vol.)\n")
+
+mprintf(" Weight of air for the combustion of 3kg carbon %.3f kg\n",wt_a)
+mprintf(" Vol. of air required for combustion of 3kg carbon %.3e L (or) %.2f metre cube",vol_a,vol_a / 1000)
diff --git a/3718/CH14/EX14.1/Ex14_1.sce b/3718/CH14/EX14.1/Ex14_1.sce new file mode 100644 index 000000000..6de5fa5d6 --- /dev/null +++ b/3718/CH14/EX14.1/Ex14_1.sce @@ -0,0 +1,10 @@ +//Chapter 14: Water Treatment
+//Problem: 1
+clc;
+
+//Declaration of Variables
+wt_CaSO4 = 160 //in mg/L
+
+//Solution
+hardness = 100 * wt_CaSO4 / 136.
+mprintf("The hardness is:%.2f mg/L of CaCO3 eqv.",hardness)
diff --git a/3718/CH14/EX14.10/Ex14_10.sce b/3718/CH14/EX14.10/Ex14_10.sce new file mode 100644 index 000000000..e432f3f21 --- /dev/null +++ b/3718/CH14/EX14.10/Ex14_10.sce @@ -0,0 +1,17 @@ +//Chapter 14: Water Treatment
+//Problem: 10
+clc;
+
+//Initialisation of Variables
+v1 = 50. //in ml for hardwater
+v2 = 15 //in ml for EDTA
+m = 0.01 //in M for EDTA
+
+//Solution
+M = v2 * m / v1
+N = M * 2
+S = N * 50 * 1000
+
+mprintf("Molarity of hardness is :%.3f M\n", M)
+mprintf(" Normality of hardness is :%.3f N\n", N)
+mprintf(" Strength of hardness is :%d ppm", S)
diff --git a/3718/CH14/EX14.11/Ex14_11.sce b/3718/CH14/EX14.11/Ex14_11.sce new file mode 100644 index 000000000..9604f5186 --- /dev/null +++ b/3718/CH14/EX14.11/Ex14_11.sce @@ -0,0 +1,11 @@ +//Chapter 14: Water Treatment
+//Problem: 11
+clc;
+
+//Initialisation of Variables
+C = 16.5 //in ppm for CO3-2
+
+//Solution
+Molarity = C * 10 ** - 6 / 60.
+
+mprintf("Molarity of CO3-2 is : %.1e mol/L",Molarity)
diff --git a/3718/CH14/EX14.2/Ex14_2.sce b/3718/CH14/EX14.2/Ex14_2.sce new file mode 100644 index 000000000..49fe2e62a --- /dev/null +++ b/3718/CH14/EX14.2/Ex14_2.sce @@ -0,0 +1,17 @@ +//Chapter 14: Water Treatment
+//Problem: 2
+clc;
+
+//Declaration of Variables
+wt1 = 9.3 //in mg/L
+wt2 = 17.4 //in mg/L
+wt3 = 8.7 //in mg/L
+wt4 = 12.6 //in mg/L
+
+//Solution
+temp_h = wt1 * 100 / 146 + wt2 * 100 / 162 //where temp_h is temporary hardness
+perm_h = wt3 * 100 / 95 + wt4 * 100 / 136 //where perm_h is permanent hardness
+total_h = temp_h + perm_h //where total_h is total hardness
+
+mprintf("Temporary hardness: %.2f mg/L\n",temp_h)
+mprintf(" Total hardness: %.2f mg/L",total_h)
diff --git a/3718/CH14/EX14.3/Ex14_3.sce b/3718/CH14/EX14.3/Ex14_3.sce new file mode 100644 index 000000000..28b38ff57 --- /dev/null +++ b/3718/CH14/EX14.3/Ex14_3.sce @@ -0,0 +1,16 @@ +//Chapter 14: Water Treatment
+//Problem: 3
+clc;
+
+//Initialisation of Variables
+wt1 = 32.4 //in mg/L
+wt2 = 29.2 //in mg/L
+wt3 = 13.5 //in mg/L
+
+//Solution
+temp_h = wt1 * 100 / 162. + wt2 * 100 / 146. //where temp_h is temporary hardness
+perm_h = wt3 * 100 / 136. //where perm_h is permanent hardness
+
+mprintf("Temporary hardness: %.2f mg/L\n",temp_h)
+mprintf(" Total hardness: %.2f mg/L",perm_h)
+
diff --git a/3718/CH14/EX14.4/Ex14_4.sce b/3718/CH14/EX14.4/Ex14_4.sce new file mode 100644 index 000000000..7337837da --- /dev/null +++ b/3718/CH14/EX14.4/Ex14_4.sce @@ -0,0 +1,23 @@ +//Chapter 14: Water Treatment
+//Problem: 4
+clc;
+
+//Initialisation of Variables
+i1 = 180 //in mg/L for CaCl2
+i2 = 210 //in mg/L for Ca(NO3)2
+i3 = 123 //in mg/L for MgSO4
+i4 = 90 //in mg/L for Mg(HCO3)2
+
+//Solution
+i1_req = i1 * 100 / 111.
+i2_req = i2 * 100 / 164.
+i3_req = i3 * 100 / 120.
+i4_req = i4 * 100 / 146.
+
+lime_req = 74 / 100. * (2 * i4_req + i3_req) * 100 / 70. * 10000 //where lime_req is the required value
+alime_req=lime_req / (10 ** 6) //where alime_req is the approximated value
+soda_req = 106 / 100. * (i1_req + i3_req + i2_req) * 100 / 80. * 10000 //where soda_req is the required value
+asoda_req=soda_req / (10 ** 6) //where asoda_req is the approximated value
+
+mprintf("Lime Required : %.1e mg ~ %.1f Kg\n",lime_req,alime_req)
+mprintf(" Soda Required : %.1e mg ~ %.1f Kg",soda_req,asoda_req)
diff --git a/3718/CH14/EX14.5/Ex14_5.sce b/3718/CH14/EX14.5/Ex14_5.sce new file mode 100644 index 000000000..ee66b7473 --- /dev/null +++ b/3718/CH14/EX14.5/Ex14_5.sce @@ -0,0 +1,19 @@ +//Chapter 14: Water Treatment
+//Problem: 5
+clc;
+
+//Initialisation of Variables
+wt1 = 32.4 //in mg/L for Ca(HCO3)2
+wt2 = 29.29 //in mg/L for Mg(HCO3)2
+wt3 = 13.5 //in mg/L for CaSO4
+
+//Solution
+wt1_eq = wt1 * 100 / 162.
+wt2_eq = wt2 * 100 / 146.
+wt3_eq = wt3 * 100 / 136.
+
+temp_h = wt1_eq + wt2_eq //where temp_h is temporary hardness
+perm_h = wt3_eq //where perm_h is permanent hardness
+
+mprintf("Temporary hardness {caused by Ca(HCO3)2 & Mg(HCO3)2} is:%d ppm\n",temp_h)
+mprintf(" Permanent hardness {caused by CaSO4} is:%.1f ppm",perm_h)
diff --git a/3718/CH14/EX14.6/Ex14_6.sce b/3718/CH14/EX14.6/Ex14_6.sce new file mode 100644 index 000000000..6dbc4c3a9 --- /dev/null +++ b/3718/CH14/EX14.6/Ex14_6.sce @@ -0,0 +1,11 @@ +//Chapter 14: Water Treatment
+//Problem: 6
+clc;
+
+//Initialisation of Variables
+v1 = 150 //in litres for NaCl
+
+//Solution
+a_hardwater = 22500 * v1 /(3 * 0.6 * 58.5)
+
+mprintf("The amount of hard water that can be softened using this softner is:%.1f litres",a_hardwater)
diff --git a/3718/CH14/EX14.7/Ex14_7.sce b/3718/CH14/EX14.7/Ex14_7.sce new file mode 100644 index 000000000..fd573d12d --- /dev/null +++ b/3718/CH14/EX14.7/Ex14_7.sce @@ -0,0 +1,11 @@ +//Chapter 14: Water Treatment
+//Problem: 7
+clc;
+
+//Initialisation of Variables
+v1 = 30 //in litres for NaCl
+w = 1500 //in mg/L for NaCl
+
+//Solution
+hardness = 45 * 50 / 58.5 * 1000 / 1000
+mprintf("Hardness of water is :%.2f ppm",hardness)
diff --git a/3718/CH14/EX14.8/Ex14_8.sce b/3718/CH14/EX14.8/Ex14_8.sce new file mode 100644 index 000000000..a662a1b59 --- /dev/null +++ b/3718/CH14/EX14.8/Ex14_8.sce @@ -0,0 +1,19 @@ +//Chapter 14: Water Treatment
+//Problem: 8
+clc;
+
+//Initialisation of Variables
+//EDTA=Ethylenediaminetetraacetic acid
+v1_water = 50 //in ml for water
+w1_CaCO3 = 1.5 //in mg for pure CaCO3
+v1_EDTA = 44 //in ml for EDTA
+v2_EDTA = 40 //in ml for EDTA
+v2_water = 20 //in ml for water
+
+//Solution
+EDTA_1 = v1_water * w1_CaCO3 / v1_EDTA
+hardw_40 = v2_water * 1.704
+total_h1 = hardw_40 * 1000 / 40
+total_h2 = total_h1 * 0.07
+
+mprintf("Total hardness is :%.2f °Cl",total_h2)
diff --git a/3718/CH14/EX14.9/Ex14_9.sce b/3718/CH14/EX14.9/Ex14_9.sce new file mode 100644 index 000000000..913c70ab7 --- /dev/null +++ b/3718/CH14/EX14.9/Ex14_9.sce @@ -0,0 +1,14 @@ +//Chapter 14: Water Treatment
+//Problem: 9
+clc;
+
+//Given Constants For Specific Elements
+Fe = 56
+S = 32
+O = 16
+
+//Solution
+hardness = Fe + S + O * 4
+hardn= (hardness * 215 )/100
+
+mprintf("215 ppm of hardness is : %.1f ppm of FeSO4",hardn)
diff --git a/3718/CH15/EX15.1/Ex15_1.sce b/3718/CH15/EX15.1/Ex15_1.sce new file mode 100644 index 000000000..64758ed74 --- /dev/null +++ b/3718/CH15/EX15.1/Ex15_1.sce @@ -0,0 +1,20 @@ +//Chapter 15: Environmental Pollution and Control
+//Problem: 1
+clc;
+
+MM = 294// Molar mass, K2Cr2O7
+
+//Declaration of Variables
+v_eff = 25 // cm cube,
+v = 8.3 // cm cube, K2Cr2O7
+M = 0.001 // M, K2Cr2O7
+
+// Solution
+w = v * 8 * 6 * M / 1000.
+
+mprintf("8.3 cm cube of 0.006 N K2Cr2O7 =%.2e g of O2\n",w)
+mprintf(" 25 ml of the effluent requires %.2e g of O2\n",w)
+
+cod = w * 10 ** 6 / 25.
+mprintf(" 1l of the effluent requires %.2fg of O2\n",cod)
+mprintf(" COD of the effluent sample is %.2f ppm or mg/L",cod)
diff --git a/3718/CH15/EX15.2/Ex15_2.sce b/3718/CH15/EX15.2/Ex15_2.sce new file mode 100644 index 000000000..88f5060e4 --- /dev/null +++ b/3718/CH15/EX15.2/Ex15_2.sce @@ -0,0 +1,17 @@ +//Chapter 15: Environmental Pollution and Control
+//Problem: 2
+clc;
+
+//Declaration of Variables
+v0 = 30 // cm cube, effluent
+v1 = 9.8 // cm cube, K2Cr2O7
+M = 0.001 // M, K2Cr2O7
+
+// Solution
+Oeff = 6 * 8 * v1 * M
+mprintf("30 cm cube of effluent contains =:%.4f mg of O2\n",Oeff)
+
+cod = Oeff * 1000 / 30.
+
+mprintf(" 1l of the effluent requires %.2f mg of O2\n",cod)
+mprintf(" COD of the effluent sample=%.2f ppm",cod)
diff --git a/3718/CH15/EX15.3/Ex15_3.sce b/3718/CH15/EX15.3/Ex15_3.sce new file mode 100644 index 000000000..5241b2c70 --- /dev/null +++ b/3718/CH15/EX15.3/Ex15_3.sce @@ -0,0 +1,16 @@ +//Chapter 15: Environmental Pollution and Control
+//Problem: 2
+clc;
+
+//Declaration of Variables
+v0 = 25 // ml, sewage
+d0 = 410 // ppm, dissolved oxygen
+d1 = 120 // ppm, dissolved oxygen
+v1 = 50 // ml, sewage
+
+// Solution
+mprintf("BOD = (DOb - DOi) * Dilution Factor\n")
+mprintf(" BOD = (DOb - DOi) * (ml of sample after dilution) / (ml of sample before dilution)\n")
+
+BOD = (d0 - d1) * (v1 / v0)
+mprintf(" BOD = %d ppm",BOD)
diff --git a/3718/CH2/EX2.1/Ex2_1.sce b/3718/CH2/EX2.1/Ex2_1.sce new file mode 100644 index 000000000..f0381c946 --- /dev/null +++ b/3718/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,20 @@ +//Chapter 2: Spectroscopy and Photochemistry
+//Problem: 1
+clc;
+
+//Declaration of Constants
+m_br79 = 78.9183 // Mass of 79Br,in amu
+m_br81 = 80.9163 // Mass of 91Br,in amu
+Na = 6.022 * 10 ** 23 // Mole constant,per mol
+pi = 3.141 // Pi
+c = 3 * 10 ** 10 // Speed of light,in cm/s
+
+//Declaration of Variable
+wave_no = 323.2 // Wave no. of fund. vibration of 79Br - 81Br, /cm
+
+// Solution
+mu = (m_br79 * m_br81) / ((m_br79 + m_br81) * Na)
+
+k = 4 * (pi * c * wave_no) ** 2 * mu * 10 ** -3
+
+mprintf("The force constant of the bond is %.2e N/m\n",k)
diff --git a/3718/CH2/EX2.10/Ex2_10.sce b/3718/CH2/EX2.10/Ex2_10.sce new file mode 100644 index 000000000..959213e2a --- /dev/null +++ b/3718/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,7 @@ +//Chapter 2: Spectroscopy and Photochemistry
+//Problem:10
+clc;
+
+mprintf("Because CO2 is a linear molecule.\n")
+v_deg = 3 * 3 - 5
+mprintf(" The vibrational degree of freedom is %d",v_deg)
diff --git a/3718/CH2/EX2.2/Ex2_2.sce b/3718/CH2/EX2.2/Ex2_2.sce new file mode 100644 index 000000000..cd1d21e69 --- /dev/null +++ b/3718/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,24 @@ +//Chapter 2: Spectroscopy and Photochemistry
+//Problem: 2
+clc;
+
+//Declaration of Constants
+Na = 6.022 * 10 ** 23 // Mole constant,per mol
+pi = 3.141 // Pi
+c = 3 * 10 ** 10 // Speed of light,in cm/s
+h = 6.626 * 10 ** -34 // Plank's constant,in J.sec
+
+//Declaration of Variables
+b_l = 112.81 * 10 ** -12 // Equillibrium bond length,in m
+m1 = 12 // Mass of Carbon,in g per mol
+m2 = 16 // Mass of Oxygen,in g per mol
+
+// Solution
+mu = m1 * m2 / ((m1 + m2) * Na) //in g
+mu = mu * 10 ** -3 //in kg
+
+B = h / (8 * pi ** 2 * mu * b_l ** 2 * c)
+v2_3 = B * 6
+
+mprintf("The reduced mass of CO is %.3e kg\n",mu)
+mprintf(" The frequency of 3->2 transition is %.2f /cm",v2_3)
diff --git a/3718/CH2/EX2.3/Ex2_3.sce b/3718/CH2/EX2.3/Ex2_3.sce new file mode 100644 index 000000000..5463262bc --- /dev/null +++ b/3718/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,19 @@ +//Chapter 2: Spectroscopy and Photochemistry
+//Problem: 3
+clc;
+
+//Declaration of Constants
+Na = 6.022 * 10 ** 23 // Mole constant,permol
+
+//Declaration of Variables
+d_NaCl = 2.36 * 10 ** -10 // Intermolecular dist. NaCl,in m
+m_Cl = 35 * 10 ** -3 // Atomic mass, in kg /mol
+m_Na = 23 * 10 ** -3 // Atomic mass, in kg /mol
+
+// Solution
+mu = m_Na * m_Cl / ((m_Na + m_Cl) * 10 ** -3 * Na) * 10 ** -3
+
+I = mu * d_NaCl ** 2
+
+mprintf("The reduced mass of NaCl is %.3e kg\n",mu)
+mprintf(" The moment of inertia of NaCl is %.3e kg m square",I)
diff --git a/3718/CH2/EX2.4/Ex2_4.sce b/3718/CH2/EX2.4/Ex2_4.sce new file mode 100644 index 000000000..c280d02f3 --- /dev/null +++ b/3718/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,15 @@ +//Chapter 2: Spectroscopy and Photochemistry
+//Problem: 4
+clc;
+
+//Declaration of Constant
+e = 4000 // Extinction coeff.,in dm cube per mol per cm
+
+// Variable
+x = 3 // Solution thickness,in cm
+
+// Solution
+A = log10(1 / 0.3) // Absorbance
+C = A / (e * x)
+
+mprintf("The concentration of the solution is %.2e mol per dm cube",C)
diff --git a/3718/CH2/EX2.5/Ex2_5.sce b/3718/CH2/EX2.5/Ex2_5.sce new file mode 100644 index 000000000..c37fde230 --- /dev/null +++ b/3718/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,18 @@ +//Chapter 2: Spectroscopy and Photochemistry
+//Problem: 5
+clc;
+
+//Declaration of Constants
+pi = 3.141 // Pi
+c = 3 * 10 ** 10 // Speed of light,in cm/s
+
+//Declaration of Variables
+v_bar = 2140 // Fundamental vibrating freq, per cm
+m_C = 19.9 * 10 ** -27 // Atomic mass of C,in kg
+m_O = 26.6 * 10 ** -27 // Atomic mass of O,in kg
+
+// Solution
+mu = m_O * m_C / (m_C + m_O)
+k = 4 * (pi * c * v_bar) ** 2 * mu
+
+mprintf("The force constant of the molecule is %.3e N/m",k)
diff --git a/3718/CH2/EX2.7/Ex2_7.sce b/3718/CH2/EX2.7/Ex2_7.sce new file mode 100644 index 000000000..b7c32c42f --- /dev/null +++ b/3718/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,24 @@ +//Chapter 2: Spectroscopy and Photochemistry
+//Problem: 7
+clc;
+
+//Declaration of Constants
+pi = 3.141 // pi
+c = 3 * 10 ** 10 // speed of light, cm /s
+h = 6.626 * 10 ** -34 // Plank's constant, J.sec
+Na = 6.022 * 10 ** 23 // Mole constant, /mol
+
+//Declaration of Variables
+d = 20.7 // Interspacing, /cm
+m1 = 1 // Mass of H, g / mol
+m2 = 35.5 // Masso f Cl, g / mol
+
+// Solution
+B = 0.1035 * 10 ** 2 // /m
+I = h / (8 * pi ** 2 * B * c)
+mu = m1 * m2 / ((m1 + m2) * Na)
+mu = mu * 10 ** -3
+r = sqrt(I / mu)
+
+mprintf("The intermolecular distance of HCl is %.3e m",r)
+// The answer provided in the textbook is wrong
diff --git a/3718/CH2/EX2.8/Ex2_8.sce b/3718/CH2/EX2.8/Ex2_8.sce new file mode 100644 index 000000000..936bd285f --- /dev/null +++ b/3718/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,14 @@ +//Chapter 2: Spectroscopy and Photochemistry
+//Problem: 8
+clc;
+
+//Declaration of Constant
+e = 8000 // Molar absorbtion coeff,in dm cube per mol per cm
+
+//Declaration of Variable
+l = 2.5 // Thickness of solution,in cm
+
+// Solution
+C = log10(1 / 0.3) / (e * l)
+
+mprintf("The concentration of Solution from Lambert-Beer Law is %.2e mol per dm cube",C)
diff --git a/3718/CH3/EX3.1/Ex3_1.sce b/3718/CH3/EX3.1/Ex3_1.sce new file mode 100644 index 000000000..654db409a --- /dev/null +++ b/3718/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,12 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 1
+clc;
+
+//Declaration of Variables
+q = 120 // Heat from surrounding, cal
+W = 70 // Work done, cal
+
+// Solution
+delta_E = q - W
+
+mprintf("Change in Internal Energy :%d cals", delta_E)
diff --git a/3718/CH3/EX3.10/Ex3_10.sce b/3718/CH3/EX3.10/Ex3_10.sce new file mode 100644 index 000000000..54fb995d8 --- /dev/null +++ b/3718/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,19 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 10
+clc;
+
+//Declaration of Constant
+R = 8.314 //in J / K
+
+//Declaration of Variables
+V_O2 = 2.8 //in litres
+V_H2 = 19.6 //in litres
+
+// Solution
+na = V_O2 / 22.4 //in mol
+nb = V_H2 / 22.4 //in mol
+Xa = na / (na + nb)
+Xb = nb / (na + nb)
+d_s = (- R) * (na * log(Xa) + nb * log(Xb))
+
+mprintf("The increase in entropy on mixing is : %.3f J /K",d_s)
diff --git a/3718/CH3/EX3.12/Ex3_12.sce b/3718/CH3/EX3.12/Ex3_12.sce new file mode 100644 index 000000000..ba805e88f --- /dev/null +++ b/3718/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,15 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 12
+clc;
+
+//Declaration of Variables
+d_g_25 = - 85.77 // k J, Free Energy at 25 C
+d_g_35 = - 83.68 // k J, Free Energy at 35 C
+Ti = 273 + 25 // K
+Tf = 273 + 35 // K
+
+// Solution
+mprintf("Equating the entropy change at both the temperatures.\n")
+mprintf(" (d_h + d_g_25) / Ti = (d_h + d_g_35) / Tf\n")
+d_h = - 148
+mprintf(" The change in enthalpy for the process at 30C is %d kJ", d_h)
diff --git a/3718/CH3/EX3.13/Ex3_13.sce b/3718/CH3/EX3.13/Ex3_13.sce new file mode 100644 index 000000000..66fcd25d4 --- /dev/null +++ b/3718/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,20 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 13
+clc;
+
+//Declaration of Constants
+l_v = 101 //in cal /g, Latent headt of vap.
+mwt = 78 // molecular weight of benzene
+
+//Declaration of Variable
+m = 2
+Tb = 80.2 // C, boiling point of benzene
+
+// Solution
+Tb = Tb + 273 // K
+d_h = l_v * mwt
+d_s = d_h / Tb
+d_g = d_h - Tb * d_s
+
+mprintf("d_s = %.2f cal / K\n",d_s)
+mprintf(" d_g = d_a = %d", d_g)
diff --git a/3718/CH3/EX3.14/Ex3_14.sce b/3718/CH3/EX3.14/Ex3_14.sce new file mode 100644 index 000000000..f4f228331 --- /dev/null +++ b/3718/CH3/EX3.14/Ex3_14.sce @@ -0,0 +1,28 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 14
+clc;
+
+//Declaration of Variables
+V1 = 6 //in dm cube
+V2 = 2 //in dm cube
+T1 = 27 //in C
+m = 5
+
+// Solution
+mprintf("T1*V1 ^ (gamma - 1) = T2 * V2 ^ (gamma - 1)\n")
+
+T1 =T1 + 273 // K
+T2 = T1 * (V1 / V2) ** (8.314 / 20.91)
+
+mprintf(" The Final Temperature is %.1f K\n",T2)
+
+q = 0 //For Adiabatic process
+d_E = - m * 20.91 * (T2 - T1)
+d_E = d_E / 1000
+
+mprintf(" q =%d \n", q)
+mprintf(" Change is Energy is %.2f kJ / mol\n",d_E)
+
+W = - d_E
+
+mprintf(" W = %.2f kJ /mol",d_E)
diff --git a/3718/CH3/EX3.15/Ex3_15.sce b/3718/CH3/EX3.15/Ex3_15.sce new file mode 100644 index 000000000..6072caddb --- /dev/null +++ b/3718/CH3/EX3.15/Ex3_15.sce @@ -0,0 +1,27 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 15
+clc;
+
+//Declaration of Constant
+R = 8.314 //in J / K mol
+
+//Declaration of Variables
+m = 1
+V1 = 5 // dm cube
+V2 = 10 // dm cube
+T = 300 // K
+
+// Solution
+mprintf("For isothermal and reversible process,\n")
+
+d_E = 0
+d_H = 0
+d_A = - 2.303 * m * R * T * log10(V2 / V1)
+d_G = - 2.303 * m * R * T * log10(V2 / V1)
+q = - d_G
+W = - d_G
+
+mprintf(" d_E = d_H = %d \n", d_H)
+mprintf(" d_G = d_A =%.3f J / mol\n",d_G)
+mprintf(" For isothermal and reversible expansion\n")
+mprintf(" q = W = -d_G = %.3f J / mol",W)
diff --git a/3718/CH3/EX3.16/Ex3_16.sce b/3718/CH3/EX3.16/Ex3_16.sce new file mode 100644 index 000000000..e33f2180a --- /dev/null +++ b/3718/CH3/EX3.16/Ex3_16.sce @@ -0,0 +1,18 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 16
+clc;
+
+//Declaration of Constant
+R = 8.314 //in J per K mol
+
+//Declaration of Variables
+n = 5 // moles
+T = 27 // C
+V1 = 50.0 // L, Initial Volume
+V2 = 1000 // L, Final Volume
+
+//Solution
+T = T + 273
+d_G = 2.303 * n * R * T * log10(V1 / V2)
+d_G = d_G / 1000
+mprintf("The free energy change is :%.3f k J",d_G)
diff --git a/3718/CH3/EX3.17/Ex3_17.sce b/3718/CH3/EX3.17/Ex3_17.sce new file mode 100644 index 000000000..750c05ca7 --- /dev/null +++ b/3718/CH3/EX3.17/Ex3_17.sce @@ -0,0 +1,12 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 17
+clc;
+
+//Declaration of Variables
+d_H_n = - 51.46 // k J/mol, neutralization
+d_H_i = - 57.1 // k J/mol, ionization
+
+//Solution
+d_H = - d_H_i + d_H_n
+
+mprintf("The head of ionization for NH4OH is %.2f kJ / mol", d_H)
diff --git a/3718/CH3/EX3.2/Ex3_2.sce b/3718/CH3/EX3.2/Ex3_2.sce new file mode 100644 index 000000000..e5675252e --- /dev/null +++ b/3718/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,11 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 2
+clc;
+
+//Solution
+mprintf("CH4 (g) + 2O2 (g) -> CO2 (g) + 2H20 (l)\n")
+
+delta_n = 1 - (1 + 2)
+solution = - 2 * 2 * 298 // cals
+
+mprintf(" Delta H - Delta E is: %d cals", solution)
diff --git a/3718/CH3/EX3.20/Ex3_20.sce b/3718/CH3/EX3.20/Ex3_20.sce new file mode 100644 index 000000000..ed107fe67 --- /dev/null +++ b/3718/CH3/EX3.20/Ex3_20.sce @@ -0,0 +1,10 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 20
+clc;
+
+// Solution
+Eq_HI = 1.56 / 2
+Eq_H2 = 0.22 / 2
+Eq_I2 = 0.22 / 2
+Kc = Eq_H2 * Eq_I2 / (Eq_HI ** 2)
+mprintf("The equilibrium constant for the dissociation reaction %.4f",Kc)
diff --git a/3718/CH3/EX3.21/Ex3_21.sce b/3718/CH3/EX3.21/Ex3_21.sce new file mode 100644 index 000000000..35b67c388 --- /dev/null +++ b/3718/CH3/EX3.21/Ex3_21.sce @@ -0,0 +1,15 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 21
+clc;
+
+//Declaration of Variables
+Kc = 0.5 // mole square litre square
+T = 400 // K
+R = 0.082 // litre atm per degree per mole
+
+// Solution
+Kp = Kc * (R * T) ** (-2)
+
+mprintf("The given equilibrium is\n")
+mprintf(" N2(g) + 3H2(g) <--> 2NH3(g)\n")
+mprintf(" Kp is %.3e",Kp)
diff --git a/3718/CH3/EX3.22/Ex3_22.sce b/3718/CH3/EX3.22/Ex3_22.sce new file mode 100644 index 000000000..7d1397138 --- /dev/null +++ b/3718/CH3/EX3.22/Ex3_22.sce @@ -0,0 +1,10 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 22
+clc;
+
+//Declaration of Variables
+solubility = 7.5 * 10 ** - 5 // mol per L
+
+// Solution
+Ksp = 4 * (solubility ** 3)
+mprintf("Solubility product of the salt is %.3e mol cube L cube",Ksp)
diff --git a/3718/CH3/EX3.23/Ex3_23.sce b/3718/CH3/EX3.23/Ex3_23.sce new file mode 100644 index 000000000..6c8cbba09 --- /dev/null +++ b/3718/CH3/EX3.23/Ex3_23.sce @@ -0,0 +1,12 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 23
+clc;
+
+//Declaration of Variables
+Ti = 25 // C
+S = 0.00179 // g / L
+
+// Solution
+S =S / 170 // mol / L
+Ksp = S ** 2
+mprintf("Solubility product at 25 C is %.4e mol square L square",Ksp)
diff --git a/3718/CH3/EX3.24/Ex3_24.sce b/3718/CH3/EX3.24/Ex3_24.sce new file mode 100644 index 000000000..a4fd53a80 --- /dev/null +++ b/3718/CH3/EX3.24/Ex3_24.sce @@ -0,0 +1,13 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 24
+clc;
+
+//Declaration of Variables
+Ksp = 8 * 10 ** - 5 // Solubility product PbBr2
+diss = 80 / 100 // % dissociation
+
+// Solution
+S = (Ksp / 4) ** (1 / 3.0) // Solubility is 100%
+S_80 = S * (80 / 100.0)
+S_per_g = S_80 * 367 - 1.621
+mprintf("Solubility in gm per litre is %.3f gm / l",S_per_g)
diff --git a/3718/CH3/EX3.27/Ex3_27.sce b/3718/CH3/EX3.27/Ex3_27.sce new file mode 100644 index 000000000..90c160c07 --- /dev/null +++ b/3718/CH3/EX3.27/Ex3_27.sce @@ -0,0 +1,13 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 27
+clc;
+
+//Declaration of Variables
+n_salt = 0.02 // mole
+n_base = 0.2 // mole
+pKb = 4.7
+
+// Solution
+pOH = pKb + log10(n_salt / n_base)
+pH = 14 - pOH
+mprintf("pH of a buffer solution is %.1f", pH)
diff --git a/3718/CH3/EX3.3/Ex3_3.sce b/3718/CH3/EX3.3/Ex3_3.sce new file mode 100644 index 000000000..fc8f0dd6c --- /dev/null +++ b/3718/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,15 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 3
+clc;
+
+//Declaration of Variables
+delta_g = -16.0 // Kelvin cal
+delta_h = -10.0 // Kelvin cal
+T = 300 // Kelvin
+
+// Solution
+delta_s = (delta_h - delta_g) * 10 ** 3 / T // cal/deg
+new_t = 330 // Kelvin
+new_delta_g = (delta_h * 10 ** 3) - new_t * delta_s
+
+mprintf("The free energy at 330K is: %.2e K cal",new_delta_g)
diff --git a/3718/CH3/EX3.4/Ex3_4.sce b/3718/CH3/EX3.4/Ex3_4.sce new file mode 100644 index 000000000..8623977e9 --- /dev/null +++ b/3718/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,14 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 4
+clc;
+
+//Declaration of Variables
+delta_s = -20.7 // cal per deg per mol
+delta_h = -67.37 // K cal
+T = 25 // deg C
+
+// Solution
+T = T + 273 // K
+delta_g = delta_h - (T * delta_s * 10 ** -3)
+
+mprintf("The change in free energy at 25deg C is: %.4f K cal per mol", delta_g)
diff --git a/3718/CH3/EX3.5/Ex3_5.sce b/3718/CH3/EX3.5/Ex3_5.sce new file mode 100644 index 000000000..a99f0aeba --- /dev/null +++ b/3718/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,13 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 5
+clc;
+
+//Declaration of Variables
+wt = 1 // g
+delta_h = 149 // joules
+
+// Solution
+delta_h_f = delta_h * (10 * 12 + 8 * 1)
+delta_h_f_c=delta_h_f * 10 ** -3
+
+mprintf("Enthalpy of fusion of naphthalene:%.3f kJ/mol", delta_h_f_c)
diff --git a/3718/CH3/EX3.6/Ex3_6.sce b/3718/CH3/EX3.6/Ex3_6.sce new file mode 100644 index 000000000..2e0af5c97 --- /dev/null +++ b/3718/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,13 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 6
+clc;
+
+//Declaration of Variables
+d_h_acetylene = 230 // kJ per mol
+d_h_benzene = 85 // kJ per mol
+T = 298 // K
+
+// Solution
+d_h = d_h_benzene - 3 * d_h_acetylene
+
+mprintf("The enthalpy change for the reaction is: %d kJ/mole", d_h)
diff --git a/3718/CH3/EX3.7/Ex3_7.sce b/3718/CH3/EX3.7/Ex3_7.sce new file mode 100644 index 000000000..6d5415a06 --- /dev/null +++ b/3718/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,16 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 7
+clc;
+
+//Declaration of Constant
+d_h_vap = 2.0723 // kJ per g
+Tb = 373 // K
+
+// Solution
+d_h_vap = d_h_vap * 18 // kJ per mol
+d_s = d_h_vap / Tb
+d_g = d_h_vap - Tb * d_s
+d_s = d_s * 1000
+
+mprintf("The Entropy change is: %.1f J / mol / K\n",d_s)
+mprintf(" The Free Energy change is:%d kJ/mol", d_g)
diff --git a/3718/CH3/EX3.8/Ex3_8.sce b/3718/CH3/EX3.8/Ex3_8.sce new file mode 100644 index 000000000..6f40f4418 --- /dev/null +++ b/3718/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,19 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 8
+clc;
+
+//Declaration of Constant
+R = 1.987 // cal per K per mol
+
+//Declaration of Variables
+m = 5
+Vo = 4 //in litres, Initial Volume
+Vf = 40 //in litres, Final Volume
+T = 27 //in deg C
+
+// Solution
+mprintf("dS = nRln(V2 / V1)\n")
+
+dS = m * R * 2.303 * log10(Vf / Vo)
+
+mprintf(" The change in entropy is: %.2f cal / degree",dS)
diff --git a/3718/CH3/EX3.9/Ex3_9.sce b/3718/CH3/EX3.9/Ex3_9.sce new file mode 100644 index 000000000..10b2d2fff --- /dev/null +++ b/3718/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,12 @@ +//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 9
+clc;
+
+//Declaration of Variables
+wt = 10 //in g
+heat_a = 4.5 //in K
+
+// Solution
+m = 10 / 100.0 // mol
+d_h = heat_a / m
+mprintf("The heat of the reaction is:%d K cal / mol", d_h)
diff --git a/3718/CH5/EX5.1/Ex5_1.sce b/3718/CH5/EX5.1/Ex5_1.sce new file mode 100644 index 000000000..f304c4437 --- /dev/null +++ b/3718/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,11 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 1
+clc;
+
+//Declaration of Variables
+K = 3.5 * 10 ** - 2 // Rate constant
+
+// Solution
+mprintf("First order reaction = 0.693 / K\n")
+t = 0.693 / K
+mprintf(" Time taken for half the initial concentration to react:%.1f minutes", t)
diff --git a/3718/CH5/EX5.10/Ex5_10.sce b/3718/CH5/EX5.10/Ex5_10.sce new file mode 100644 index 000000000..f483081ff --- /dev/null +++ b/3718/CH5/EX5.10/Ex5_10.sce @@ -0,0 +1,17 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 10
+clc;
+
+//Declaration of Constant
+R = 1.987 //in cal per K per mol
+
+//Declaration of Variables
+K2_K1 = 4 // factor increase
+T1 = 27 //in C
+T2 = 47 //in C
+
+// Solution
+T1 = T1 + 273.0
+T2 = T2 + 273.0
+Ea = log10(4) * 2.303 * R * (T1 * T2 / (T2 - T1))
+mprintf("The activation energy for the reaction is %.2e cal /mol",Ea)
diff --git a/3718/CH5/EX5.11/Ex5_11.sce b/3718/CH5/EX5.11/Ex5_11.sce new file mode 100644 index 000000000..4f03204d2 --- /dev/null +++ b/3718/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,11 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 11
+clc;
+
+//Declaration of Variables
+a = 1 //in mole
+x = 3 / 4.0 // reaction completed
+
+// Solution
+K = (2.303 / 6) * log10(1 / (1 - x))
+mprintf("The rate constant is :%.3f / min",K)
diff --git a/3718/CH5/EX5.12/Ex5_12.sce b/3718/CH5/EX5.12/Ex5_12.sce new file mode 100644 index 000000000..21a720919 --- /dev/null +++ b/3718/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,15 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 12
+clc;
+
+// Solution
+mprintf("Let the initial concentration be 100, when x = 25,t = 30 minutes\n")
+a = 100
+x = 25.0
+t = 30
+K = 2.303 / t * log10(a / (a - x))
+t05 = 0.683 / K
+t = 2.303 / K * log10(a / x)
+mprintf(" K = %.2e / min\n",K)
+mprintf(" T0.5 = %.2f min\n",t05)
+mprintf(" t = %.1f min",t)
diff --git a/3718/CH5/EX5.2/Ex5_2.sce b/3718/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..04c757c80 --- /dev/null +++ b/3718/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,11 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 2
+clc;
+
+//Declaration of Variables
+t = 40 //in minutes
+
+// Solution
+mprintf("Rate constant = 0.693 / t\n")
+K = 0.693 / t
+mprintf(" Rate constant = %.4f / min",K)
diff --git a/3718/CH5/EX5.3/Ex5_3.sce b/3718/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..b95280d30 --- /dev/null +++ b/3718/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,24 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 3
+clc;
+
+//Declaration of Variables
+t0 = 37.0 //in cm cube of KMnO4
+t5 = 29.8 //in cm cube of KMnO4
+t15 = 19.6 //in cm cube of KMnO4
+t25 = 12.3 //in cm cube of KMnO4
+t45 = 5.00 //in cm cube of KMnO4
+
+// Solution
+K5 = 2.303 / 5 * log10(t0 / t5)
+K15 = 2.303 / 15 * log10(t0 / t15)
+K25 = 2.303 / 25 * log10(t0 / t25)
+K45 = 2.303 / 45 * log10(t0 / t45)
+
+mprintf("At t = 5 min, K = %.3e /min\n",K5)
+mprintf(" At t = 15 min, K = %.3e /min\n",K15)
+mprintf(" At t = 25 min, K = %.3e /min\n",K25)
+mprintf(" At t = 45 min, K = %.3e /min\n",K45)
+mprintf(" As the different values of K are nearly same, the reaction is of first-order\n")
+K = (K45 + K25 + K5 + K15) / 4
+mprintf(" The average value of K = %.3e /min",K)
diff --git a/3718/CH5/EX5.4/Ex5_4.sce b/3718/CH5/EX5.4/Ex5_4.sce new file mode 100644 index 000000000..019c82985 --- /dev/null +++ b/3718/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,12 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 4
+clc;
+
+//Declaration of Variables
+t = 60 //in min
+x = "0.5a"
+K = 5.2 * 10 ** - 3 //in per mol L per min
+
+// Solution
+a = 1 / (t * K)
+mprintf("Initial concentration = %.3f mol / L",a)
diff --git a/3718/CH5/EX5.5/Ex5_5.sce b/3718/CH5/EX5.5/Ex5_5.sce new file mode 100644 index 000000000..a7a1d537e --- /dev/null +++ b/3718/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,7 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 5
+clc;
+
+// Solution
+t = ((2.303 * log10(100 / (100 - 99.9))) / (2.303 * log10(100 / (100 - 50))))
+mprintf("99.9 percent / 50 percent =%.1f",t)
diff --git a/3718/CH5/EX5.6/Ex5_6.sce b/3718/CH5/EX5.6/Ex5_6.sce new file mode 100644 index 000000000..4bd645e2f --- /dev/null +++ b/3718/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,16 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 6
+clc;
+
+//Declaration of Constants
+R = 1.987 //in cal per K per mol
+
+//Declaration of Variables
+T1 = 273.0 //in K
+T2 = 303.0 //in K
+K1 = 2.45 * 10 ** -5
+K2 = 162 * 10 ** -5
+
+// Solution
+Ea = log10(K2 / K1) * R * 2.303 / ((T2 - T1) / (T1 * T2))
+mprintf("The activation energy of the reaction is %d cal / mol",Ea)
diff --git a/3718/CH5/EX5.7/Ex5_7.sce b/3718/CH5/EX5.7/Ex5_7.sce new file mode 100644 index 000000000..8c910d166 --- /dev/null +++ b/3718/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,12 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 7
+clc;
+
+//Declaration of Variables
+t05 = 30 //in minutes
+a = 0.1 //in M
+
+// Solution
+mprintf("For second order reaction,\n t0.5 = 1 / Ka\n")
+K = 1 / (a * t05)
+mprintf(" The rate constant is %.3f mol per lit per min",K)
diff --git a/3718/CH5/EX5.8/Ex5_8.sce b/3718/CH5/EX5.8/Ex5_8.sce new file mode 100644 index 000000000..bc995af68 --- /dev/null +++ b/3718/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,16 @@ +//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 8
+clc;
+
+//Declaration of Variables
+T = 500 //in C
+Pi = 350 //in torr
+r1 = 1.07 //in torr / s
+r2 = 0.76 //in torr / s
+
+// Solution
+mprintf("1.07 = k(0.95a)^n\n")
+mprintf(" 0.76 = k(0.80a)^n\n")
+n = log(r1 / r2) / log(0.95 / 0.80)
+n=ceil(n)
+mprintf(" Hence, order of reaction is %d",n)
diff --git a/3718/CH6/EX6.10/Ex6_10.sce b/3718/CH6/EX6.10/Ex6_10.sce new file mode 100644 index 000000000..df86e87d4 --- /dev/null +++ b/3718/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,16 @@ +//Chapter 6: Electrochemistry
+//Problem: 10
+clc;
+
+//Declaration of Variables
+T = 25 // C
+Cu = 0.1 // M
+Zn = 0.001 // M
+Eo = 1.1 // V
+
+// Solution
+mprintf("Zn(s) | Zn+2 (0.001M) || Cu+2(0.1M) | Cu(s)\n")
+
+Ecell = Eo + 0.0592 / 2 * log10(Cu / Zn)
+
+mprintf(" The emf of a Daniel cell is %.4f V",Ecell)
diff --git a/3718/CH6/EX6.11/Ex6_11.sce b/3718/CH6/EX6.11/Ex6_11.sce new file mode 100644 index 000000000..4ecbc9dd0 --- /dev/null +++ b/3718/CH6/EX6.11/Ex6_11.sce @@ -0,0 +1,18 @@ +//Chapter 6: Electrochemistry
+//Problem: 11
+clc;
+
+//Declaration of Variables
+pH = 7 // O2
+Eo = 1.229 // V
+pO2 = 0.20 // bar
+
+// Solution
+mprintf("Nernst equation at 25C is,\n")
+mprintf(" E = Eo - (0.0592 / 2) * log(1 / ([H+]^2 * [pO2]^ (1/2)))\n")
+
+E = Eo - (0.0592 / 2) * log10(1.0 / (((10 ** (- 7)) ** 2) * (pO2 ** (1 / 2.0))))
+
+mprintf(" The reduction potential for the reduction is %.3f V",E)
+
+// The answer provided in the textbook is wrong
diff --git a/3718/CH6/EX6.12/Ex6_12.sce b/3718/CH6/EX6.12/Ex6_12.sce new file mode 100644 index 000000000..7ec96c913 --- /dev/null +++ b/3718/CH6/EX6.12/Ex6_12.sce @@ -0,0 +1,23 @@ +//Chapter 6: Electrochemistry
+//Problem: 12
+clc;
+
+//Declaration of Variables
+E_KCl = 0.2415 // V
+E_cell = 0.445 // V
+
+
+// Solution
+mprintf("Emf of the cell is\n")
+mprintf(" At 25C,\n")
+mprintf(" E = Er - El = Eref - ((RT)/ F) * ln H+\n")
+
+pH = (E_cell - E_KCl) / 0.059
+Eo_cell = - 0.8277 // V
+
+mprintf(" Thus, equilibrium constant for the reaction\n")
+mprintf(" 2H2O --> H3O+ + OH- may be calculated as\n")
+
+K = 10 ** (Eo_cell / 0.0591)
+
+mprintf(" K = %.e",K)
diff --git a/3718/CH6/EX6.13/Ex6_13.sce b/3718/CH6/EX6.13/Ex6_13.sce new file mode 100644 index 000000000..95542167f --- /dev/null +++ b/3718/CH6/EX6.13/Ex6_13.sce @@ -0,0 +1,16 @@ +//Chapter 6: Electrochemistry
+//Problem: 13
+clc;
+
+//Declaration of Variables
+EoSn = 0.15 // V
+EoCr = - 0.74 // V
+
+// Solution
+mprintf("3Sn+4 + 2Cr --> 3Sn+2 + 2Cr+3\n")
+
+Eo_cell = EoSn - EoCr
+n = 6
+K = 10 ** (n * Eo_cell / 0.0591)
+
+mprintf(" The equillibrium constant for th reaction is %.2e ",K)
diff --git a/3718/CH6/EX6.14/Ex6_14.sce b/3718/CH6/EX6.14/Ex6_14.sce new file mode 100644 index 000000000..f31642d99 --- /dev/null +++ b/3718/CH6/EX6.14/Ex6_14.sce @@ -0,0 +1,14 @@ +//Chapter 6: Electrochemistry
+//Problem: 14
+clc;
+
+//Declaration of Variables
+T = 25 // C
+Eo = - 0.8277 // V
+
+// Solution
+mprintf("The reversible reaction,\n")
+mprintf(" 2H2O <--> H3O+ + OH-\n")
+mprintf(" May be divided into two parts.\n")
+mprintf(" 2H2O + e- --> 1/2 H2 + OH- (cathode) Eo = -0.8277 V\n")
+mprintf(" H2O + 1/2 H2 --> H3O+ + e- (anode) Eo = 0")
diff --git a/3718/CH6/EX6.15/Ex6_15.sce b/3718/CH6/EX6.15/Ex6_15.sce new file mode 100644 index 000000000..c0959e1e9 --- /dev/null +++ b/3718/CH6/EX6.15/Ex6_15.sce @@ -0,0 +1,16 @@ +//Chapter 6: Electrochemistry
+//Problem: 15
+clc;
+
+//Declaration of Variables
+E = 0.4 // V
+
+// Solution
+
+mprintf( "The cell is Pt(H2) | H+, pH2 = 1 atm\n")
+mprintf(" The cell reaction is\n")
+mprintf(" 1/2 H2 --> H+ + e-\n")
+
+pH = E / 0.0591
+
+mprintf(" pH = %.3f ",pH)
diff --git a/3718/CH6/EX6.2/Ex6_2.sce b/3718/CH6/EX6.2/Ex6_2.sce new file mode 100644 index 000000000..b99a19e48 --- /dev/null +++ b/3718/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,13 @@ +//Chapter 6: Electrochemistry
+//Problem: 2
+clc;
+
+//Declaration of Variables
+T = 25 // C
+Cu = 0.1 // M
+Zn = 0.001 // M
+Eo = 1.1
+
+// Solution
+E = Eo + 0.0296 * log10(Cu / Zn)
+mprintf("The emf of Daniell cell is %.4f V",E)
diff --git a/3718/CH6/EX6.3/Ex6_3.sce b/3718/CH6/EX6.3/Ex6_3.sce new file mode 100644 index 000000000..7bedec041 --- /dev/null +++ b/3718/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,17 @@ +//Chapter 6: Electrochemistry
+//Problem: 3
+clc;
+
+//Declaration of Constant
+R = 8.314 //in J per K
+F = 96500 //in C per mol
+
+//Declaration of Variables
+Cu = 0.15 //in M
+Eo = 0.34 //in V
+T = 298 //in K
+n = 2 //in moles
+
+// Solution
+E = Eo + (2.303 * R * T) / (n * F) * log10(Cu)
+mprintf("The single electrode potential for copper metal is %.4f V",E)
diff --git a/3718/CH6/EX6.4/Ex6_4.sce b/3718/CH6/EX6.4/Ex6_4.sce new file mode 100644 index 000000000..53d63487a --- /dev/null +++ b/3718/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,13 @@ +//Chapter 6: Electrochemistry
+//Problem: 4
+clc;
+
+//Declaration of Variable
+Eo_Cu = 0.3370 // Cu+2 -> Cu
+Eo_Zn = - 0.7630 // Zn -> Zn +2
+
+// Solution
+Eo_cell = Eo_Cu - Eo_Zn
+
+mprintf("Daniel cell is, Zn | Zn +2 || Cu+2 | Cu\n")
+mprintf(" Eo (cell) is %.1f V", Eo_cell)
diff --git a/3718/CH6/EX6.5/Ex6_5.sce b/3718/CH6/EX6.5/Ex6_5.sce new file mode 100644 index 000000000..d59a72a09 --- /dev/null +++ b/3718/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,13 @@ +//Chapter 6: Electrochemistry
+//Problem: 5
+clc;
+
+//Declaration of Variable
+Eo_Cu = 0.3370 // Cu+2 -> Cu
+Eo_Cd = - 0.4003 // Cd -> Cd +2
+
+// Solution
+Eo_cell = Eo_Cu - Eo_Cd
+
+mprintf("Cell is, Cd | Cd +2 || Cu+2 | Cu\n")
+mprintf(" Eo (cell) is %.4f V", Eo_cell)
diff --git a/3718/CH6/EX6.6/Ex6_6.sce b/3718/CH6/EX6.6/Ex6_6.sce new file mode 100644 index 000000000..c0acffd39 --- /dev/null +++ b/3718/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,18 @@ +//Chapter 6: Electrochemistry
+//Problem: 6
+clc;
+
+//Declaration of Constant
+F = 96500 // C / mol
+
+//Declaration of Variables
+n = 2
+T = 25 // C
+Eo_Ag = 0.80 // Ag+ / Ag
+Eo_Ni = - 0.24 // Ni+2 / Ni
+
+// Solution
+Eo_Cell = Eo_Ag - Eo_Ni
+delta_Go = - n * F * Eo_Cell
+
+mprintf("Standard free energy change %d J / mol",delta_Go)
diff --git a/3718/CH6/EX6.8/Ex6_8.sce b/3718/CH6/EX6.8/Ex6_8.sce new file mode 100644 index 000000000..dd5aa771e --- /dev/null +++ b/3718/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,21 @@ +//Chapter 6: Electrochemistry
+//Problem: 8
+clc;
+
+//Declaration of Constant
+F = 96500 //in C per mol
+
+//Declaration of Variables
+E1o = - 2.48 //in V
+E2o = 1.61 //in V
+
+// Solution
+delta_G1 = - 3 * F * (- 2.48)
+delta_G2 = - 1 * F * 1.61
+
+mprintf("delta_G3 = delta_G1 + delta_G2\n")
+mprintf(" delta_G3 = - 4 * F * E3o\n")
+
+E3o = (delta_G1 + delta_G2) / (- 4 * F)
+
+mprintf(" The reduction potential for the half-cell Pt/Ce, Ce+4 is %.4f V",E3o)
diff --git a/3718/CH7/EX7.10/Ex7_10.sce b/3718/CH7/EX7.10/Ex7_10.sce new file mode 100644 index 000000000..c26440751 --- /dev/null +++ b/3718/CH7/EX7.10/Ex7_10.sce @@ -0,0 +1,18 @@ +//Chapter 7: Solid State
+//Problem: 10
+clc;
+
+//Declaration of Constant
+N = 6.023 * 10 ** 23
+
+// Variables
+D = 0.53 //in g per cm cube
+MM = 6.94 //in g per mol
+n = 2
+
+// Solution
+mprintf("For BCC pattern,\n")
+mprintf(" Number of Atoms per unit cell = 2\n")
+V = D * N / (n * MM)
+V = 1 / V
+mprintf(" Volume of a unit cell of lithium metal is %.2e cc",V)
diff --git a/3718/CH7/EX7.11/Ex7_11.sce b/3718/CH7/EX7.11/Ex7_11.sce new file mode 100644 index 000000000..eb59b900c --- /dev/null +++ b/3718/CH7/EX7.11/Ex7_11.sce @@ -0,0 +1,8 @@ +//Chapter 7: Solid State
+//Problem: 11
+clc;
+
+mprintf("AB remain in BCC structure if the edge length is a then body diagonal ,is root(3)a\n")
+mprintf(" root(3)a = 2(r+ + r-)\n")
+A = (sqrt(3) * 0.4123 - 2 * 0.81) / 2
+mprintf(" A+ = %.2f nm",A)
diff --git a/3718/CH7/EX7.2/Ex7_2.sce b/3718/CH7/EX7.2/Ex7_2.sce new file mode 100644 index 000000000..ae5c51126 --- /dev/null +++ b/3718/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,10 @@ +//Chapter 7: Solid State
+//Problem: 2
+clc;
+
+//Declaration of Variable
+a = 450 //in pm
+
+// Solution
+d = a / sqrt(2 ** 2 + 2 ** 2 + 0)
+mprintf("Interplanar spacing : %d",d)
diff --git a/3718/CH7/EX7.4/Ex7_4.sce b/3718/CH7/EX7.4/Ex7_4.sce new file mode 100644 index 000000000..65377ab68 --- /dev/null +++ b/3718/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,11 @@ +//Chapter 7: Solid State
+//Problem: 4
+clc;
+
+//Declaration of Variables
+r_Na = 0.98 * 10 ** - 10 //in m
+r_Cl = 1.81 * 10 ** - 10 //in m
+
+// Solution
+rr = r_Na / r_Cl
+mprintf("When the radius ration is :%.2f, the coordination number is 6.",rr)
diff --git a/3718/CH7/EX7.5/Ex7_5.sce b/3718/CH7/EX7.5/Ex7_5.sce new file mode 100644 index 000000000..e2a096e79 --- /dev/null +++ b/3718/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,12 @@ +//Chapter 7: Solid State
+//Problem: 5
+clc;
+
+//Declaration of Variables
+r_Li = 68 //in pm
+r_F = 136. //in pm
+
+// Solution
+rr = r_Li / r_F
+mprintf("Radius ratio = %.1f\n", rr)
+mprintf(" The structure of LiF is SCC and Co-ordination Number of Li+ is 6")
diff --git a/3718/CH7/EX7.6/Ex7_6.sce b/3718/CH7/EX7.6/Ex7_6.sce new file mode 100644 index 000000000..7fae98467 --- /dev/null +++ b/3718/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,13 @@ +//Chapter 7: Solid State
+//Problem: 6
+clc;
+
+//Declaration of Variables
+l = 2 * 10 ** - 10 //in m
+t = 30 //in degrees
+
+// Solution
+mprintf("For first-order reflection\n")
+d = l / (2 * sin(t))
+dist = 2 * d
+mprintf(" Hence, distance between planes is : %.0e m ",abs(dist))
diff --git a/3718/CH7/EX7.7/Ex7_7.sce b/3718/CH7/EX7.7/Ex7_7.sce new file mode 100644 index 000000000..4290243c6 --- /dev/null +++ b/3718/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,15 @@ +//Chapter 7: Solid State
+//Problem: 7
+clc;
+
+//Declaration of Variables
+r = 174.6 // pm
+
+// Solution
+a = r * sqrt(8)
+mprintf("For 200 plane: h = 2, k = 0, l = 0\n")
+d200 = a / sqrt(2 ** 2)
+mprintf(" d200 = %.1f pm\n",d200)
+mprintf(" For 200 plane: h = 2, k = 2, l = 0\n")
+d220 = a / sqrt(2 ** 2 + 2 ** 2)
+mprintf(" d220 = %.1f pm", d220)
diff --git a/3718/CH7/EX7.8/Ex7_8.sce b/3718/CH7/EX7.8/Ex7_8.sce new file mode 100644 index 000000000..d3da25dc6 --- /dev/null +++ b/3718/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,17 @@ +//Chapter 7: Solid State
+//Problem: 8
+clc;
+
+//Declaration of Constant
+N = 6.023 * 10 ** 23
+
+//Declaration of Variables
+wt = 55.6
+p = 0.29 // nm
+n = 2
+
+// Solution
+mprintf( "For BCC pattern,\n Number of Atoms per unit cell = 2\n")
+d = n * (wt * 10 ** -3) / (N * (p * 10 ** - 9) ** 3)
+mprintf(" Density of the metal is %.2e kg per m cube\n",d)
+mprintf(" Number of nearest neighbours for BCC = 8")
|