adding book

author: hardythe1 2014-08-13 11:41:01 +0530
committer: hardythe1 2014-08-13 11:41:01 +0530
commit: 728bf707ac994b2cf05a32d8985d5ea27536cf34 (patch)
tree: 5530b1509900ca8d6d21384e33036f50734de927 /Chemistry
parent: f3e94078a83634b4353ab0cd2de3b0e204a48ac7 (diff)
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diff --git a/Chemistry/Chapter_1.ipynb b/Chemistry/Chapter_1.ipynb
new file mode 100755
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--- /dev/null
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@@ -0,0 +1,310 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chpater 1:Chemistry-The Study of Change"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:1.1,Page no:19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "m=301;#mass of gold, g\n",
+      "v=15.6;# volume of gold, cm**3\n",
+      "\n",
+      "#Calculation\n",
+      "d=m/v;#density of gold, g/cm**3\n",
+      "\n",
+      "#Result\n",
+      "print\"The density of gold is : \",round(d,1),\" g/cm**3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density of gold is :  19.3  g/cm**3\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:1.2,Page no:19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=13.6;#density of mercury, g/ml\n",
+      "v=5.50;# volume of mercury, ml\n",
+      "\n",
+      "#Calculation\n",
+      "m=d*v;#mass of mercury, g\n",
+      "\n",
+      "#Result\n",
+      "print\"The mass of mercury is : \",round(m,1),\" g\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of mercury is :  74.8  g\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:1.3,Page no:20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "C1=224.0#melting point of solder, C\n",
+      "F=-452.0#boiling point of helium, F\n",
+      "C3=-38.9#meltiing point of mercury, C\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "F1=(C1*(9.0/5.0)+32.0)       #melting point of solder, F\n",
+      "#(b)\n",
+      "C2=(F-32.0)*5.0/9.0       #boiling point of helium, C\n",
+      "#(c)\n",
+      "K=C3+273.15#meltiing point of mercury, K\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The melting point of solder is : \",round(F1),\" F\"\n",
+      "print\"(b) The boiling point of helium is :\",round(C2),\"C\"\n",
+      "print\"(c) The meltiing point of mercury is :\",round(K,1),\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The melting point of solder is :  435.0  F\n",
+        "\n",
+        "(b) The boiling point of helium is : -269.0 C\n",
+        "\n",
+        "(c) The meltiing point of mercury is : 234.2 K\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:1.5,Page no:25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "A1=11254.1;#g\n",
+      "B1=0.1983;#g\n",
+      "A2=66.59;#L\n",
+      "B2=3.113;#L\n",
+      "A3=8.16;#m\n",
+      "B3=5.1355;\n",
+      "A4=0.0154;#kg\n",
+      "B4=88.3;#mL\n",
+      "A5=2.64*10**3;#cm\n",
+      "B5=3.27*10**2;#cm\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "C1=A1+B1;#g\n",
+      "#(b)\n",
+      "C2=A2-B2;#L\n",
+      "#(c)\n",
+      "C3=A3*B3;#m\n",
+      "#(d)\n",
+      "C4=A4/B4;#kg/mL\n",
+      "#(e)\n",
+      "C5=A5+B5;#cm\n",
+      "print\"(a) (\",round(A1,1),\"+\",round(B1,4),\")g=\",round(C1,1),\"g\\n\"\n",
+      "print\"(b) (\",A2,\" -\",B2,\")L=\",round(C2,2),\"L\\n\"\n",
+      "print\"(c) \",A3,\"m*\",B3,\"=\",C3,\"m =\",round(C3,1),\"m\\n\"\n",
+      "print\"(d) \",A4,\" kg /\",B4,\" mL =\",C4,\"=\",round(C4,6),\"kg/mL=%.2e\"%round(C4,6),\"kg/mL\\n\"\n",
+      "print\"(e) (\",(A5*10**-3),\"*10**3 + %.2f\"%(B5*10**-2),\"*10**2 )cm =\",(C5*10**-3),\"*10**3 cm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) ( 11254.1 + 0.1983 )g= 11254.3 g\n",
+        "\n",
+        "(b) ( 66.59  - 3.113 )L= 63.48 L\n",
+        "\n",
+        "(c)  8.16 m* 5.1355 = 41.90568 m = 41.9 m\n",
+        "\n",
+        "(d)  0.0154  kg / 88.3  mL = 0.000174405436014 = 0.000174 kg/mL=1.74e-04 kg/mL\n",
+        "\n",
+        "(e) ( 2.64 *10**3 + 3.27 *10**2 )cm = 2.967 *10**3 cm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:1.6,Page no:29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "lb=0.0833;#pound mass, lb\n",
+      "\n",
+      "#Calculation\n",
+      "g=lb*453.6;#pound mass to gram mass, 1lb=453.6g\n",
+      "mg=1000*g;#gram to milligram\n",
+      "\n",
+      "#Result\n",
+      "print\"The mass of glucose is :%.2e\"%mg,\"mg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of glucose is :3.78e+04 mg\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:1.7,Page no:30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "L=5.2;#volume in litres\n",
+      "\n",
+      "#Calculation\n",
+      "cc=1000*L;#litre to cm**3\n",
+      "mc=cc/10**6;#cm**3 to m**3\n",
+      "\n",
+      "#Result\n",
+      "print\"The volume of blood is %.1e\"%mc,\"m**3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume of blood is 5.2e-03 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:1.8,Page no:30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "gpcc=0.808;#density in gram per cm**3\n",
+      "\n",
+      "#Calculation\n",
+      "kgpmc=1000*gpcc;#g/cm**3 to kg/m**3, as 1000g=1kg and 1cm=10**-2 m\n",
+      "\n",
+      "#Result\n",
+      "print\"The density of liquid nitrogen is :\",kgpmc,\"kg/m**3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density of liquid nitrogen is : 808.0 kg/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_11.ipynb b/Chemistry/Chapter_11.ipynb
new file mode 100755
index 00000000..6d364ba0
--- /dev/null
+++ b/Chemistry/Chapter_11.ipynb
@@ -0,0 +1,226 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11:Intermolecular Forces and Liquids and Solids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:11.3,Page no:479"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Variable declaration\n",
+      "atoms=8*1.0/8.0+6*1.0/2.0 #atoms in a cell\n",
+      "d=19.3 #density, g/cc\n",
+      "Au=197.0 #mol mass of Au, g\n",
+      "NA=6.022*10**23 #avogadro no.\n",
+      "\n",
+      "#Calculation\n",
+      "m=atoms*Au/NA #mass of 1 cell, g\n",
+      "V=m/d #volume, cc\n",
+      "a=V**(1/3.0) #edge length, cm\n",
+      "r=a/math.sqrt(8.0)/100.0 #radius in m\n",
+      "\n",
+      "#Result\n",
+      "print\"The atomic radius of Au is :\",round(r*10**12),\"pm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The atomic radius of Au is : 144.0 pm\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:11.4,Page no:481"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Variable declaration\n",
+      "n=1.0 \n",
+      "lamda=154 #wavelength, pm\n",
+      "theta=19.3 #angle of reflection, degree\n",
+      "\n",
+      "#Calculation\n",
+      "d=n*lamda/(2*math.sin(theta*math.pi/180.0)) #spacing between the planes\n",
+      "\n",
+      "#Result\n",
+      "print\"The spacing between planes is :\",round(d),\"pm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The spacing between planes is : 233.0 pm\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:11.6,Page no:483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Na=22.99 #mass of one atom of Na, amu\n",
+      "Cl=35.45 #mass of one atom of Cl, amu\n",
+      "NA=6.022*10**23 #avogadro no.\n",
+      "\n",
+      "#Calculation\n",
+      "mass=4*(Na+Cl)/NA #mass in a unit cell in grams\n",
+      "a=564*10**-10 #edge length, cm\n",
+      "V=a**3 #volume of unit cell, cc\n",
+      "d=mass/V #density in g/cc\n",
+      "\n",
+      "#Result\n",
+      "print\"The density of NaCl is :\",round(d,2),\"g/cm**3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density of NaCl is : 2.16 g/cm**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:11.7,Page no:492"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Variable declaration\n",
+      "P1=401.0 #vapor pressure at 18C, mm Hg\n",
+      "T1=18.0+273.0 #temperature, K\n",
+      "T2=32.0+273.0 #temperature, K\n",
+      "deltaH=26000.0 #heat of vaporisation, J/mol\n",
+      "R=8.314 #gas constant, J/K.mol\n",
+      "\n",
+      "#Calculation\n",
+      "X=deltaH/R*(T1-T2)/(T1*T2) \n",
+      "P2=401*math.exp(-X) #vapor pressure at 32C, mmHg(from ln(P1/P1)=deltaH/R*((T1-T2)/(T1*T2)))\n",
+      "\n",
+      "#Result\n",
+      "print\"The pressure at 32 C is \",round(P2,1),\"mm Hg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pressure at 32 C is  656.7 mm Hg\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:11.8,Page no:497"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "m=346 #mass of H2O in g\n",
+      "s=4.184 #specific heat of H2O, J/g C\n",
+      "deltaH=40.79 #heat of vaporisation in kJ\n",
+      "H2O=18.02 #mol mass of H2O, g\n",
+      "s2=1.99 #specific heat of steam, J/g C\n",
+      "\n",
+      "#Calculation\n",
+      "#from 0 to 100 C\n",
+      "deltaT=100.0-0.0 #change in Temp, C\n",
+      "q1=round((m*s*deltaT)/1000.0) #heating H2O, kJ\n",
+      "#for evaporation at 100 C\n",
+      "deltaT2=182-100 #change in temp of steam, kJ\n",
+      "q2=round(m*deltaH/H2O) #heat of vaporising water, kJ\n",
+      "#for steam from 100 to 182 C\n",
+      "q3=round(m*s2*deltaT2/1000.0,1) #heating steam, kJ\n",
+      "q=round(q1)+round(q2)+round(q3,1) #overall energy required, kJ\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print\"The overall energy required is :\",round(q),\"kJ\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The overall energy required is : 985.0 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_12.ipynb b/Chemistry/Chapter_12.ipynb
new file mode 100755
index 00000000..c565711d
--- /dev/null
+++ b/Chemistry/Chapter_12.ipynb
@@ -0,0 +1,446 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12:Physical Properties of Solutions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.2,Page no:518"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "msolute=0.892 #mass of solute, g\n",
+      "msolvent=54.6 #mass of solvent, g\n",
+      "\n",
+      "#Calculation\n",
+      "percent=msolute/(msolute+msolvent)*100 #concentration, percent by mass\n",
+      "\n",
+      "#Result\n",
+      "print\"The concentration of KCl solution by mass is :\",round(percent,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The concentration of KCl solution by mass is : 1.61 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.3,Page no:518"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration \n",
+      "mass=24.4 #mass of H2SO4, g\n",
+      "M=98.09 #mol maass of H2SO4, g\n",
+      "\n",
+      "#Calculation\n",
+      "n=mass/M #moles of H2SO4\n",
+      "massH2O=0.198 #mass of H2O, kg\n",
+      "m=n/massH2O #molality of H2SO4, molal\n",
+      "\n",
+      "#Result\n",
+      "print\"The molality of sulfuric acid solution is :\",round(m,2),\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molality of sulfuric acid solution is : 1.26 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.4,Page no:520"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#considering 1L solution\n",
+      "msolution=976 #mass of solution, g\n",
+      "n=2.45 #moles\n",
+      "CH3OH=32.04 #mol. mass of CH3OH, g\n",
+      "\n",
+      "#Calculation\n",
+      "msolute=n*CH3OH #mass of solute, g\n",
+      "msolvent=(msolution-msolute)/1000 #mass of solvent, kg\n",
+      "m=n/msolvent #molality, molal\n",
+      "\n",
+      "#Result\n",
+      "print\"The molality of CH3OH solution is :\",round(m,2),\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molality of CH3OH solution is : 2.73 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.5,Page no:520"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#considering 100g of solution\n",
+      "percent=35.4 #mass percent of H3PO4\n",
+      "H3PO4=97.99 #mol mass of H3PO4\n",
+      "\n",
+      "#Calculation\n",
+      "n=percent/H3PO4 #moles of H3PO4\n",
+      "mH2O=(100-percent)/1000 #mass of solvent\n",
+      "m=n/mH2O #molality of H3PO4, molal\n",
+      "\n",
+      "#Result\n",
+      "print\"the molality of H3PO4 solution is :\",round(m,2),\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the molality of H3PO4 solution is : 5.59 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.6,Page no:525"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "c=6.8*10**-4 #solubility of N2 in water, M\n",
+      "P=1 #pressure, atm\n",
+      "\n",
+      "#Calculation\n",
+      "k=c/P #henry's constant\n",
+      "#for partial pressure of N2=0.78atm\n",
+      "P=0.78 #partial pressure of N2, atm\n",
+      "c=k*P #solubility of N2, M\n",
+      "\n",
+      "#Result\n",
+      "print\"The solubility of N2 gas in water is :%.1e\"%c,\"M\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The solubility of N2 gas in water is :5.3e-04 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.7,Page no:527"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "H2O=18.02 #mol mass of H2O, g\n",
+      "V=460 #volume of water, mL\n",
+      "glucose=180.2 #mol. mass of glucose, g\n",
+      "mass=218 #mass of gllucose, g\n",
+      "\n",
+      "#Calculation\n",
+      "n1=V/H2O #moles of water\n",
+      "n2=mass/glucose #moles of glucose\n",
+      "x1=n1/(n1+n2) #mole fraction of water\n",
+      "P=31.82 #vapor pressure of pure water, mmHg\n",
+      "P1=x1*P #vapor pressure afteraddition of glucose, mmHg\n",
+      "#Result\n",
+      "print\"Vapor pressure is:\",round(P1,1),\"mm Hg\"\n",
+      "print\"The vapor pressure lowering is :\",round(P-P1,1),\"mmHg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vapor pressure is: 30.4 mm Hg\n",
+        "The vapor pressure lowering is : 1.4 mmHg\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.8,Page no:532"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "mH2O=2.505 #mass of H2O, kg\n",
+      "mEG=651 #mass of EG, g\n",
+      "EG=62.07 #mol mass of EG, g\n",
+      "\n",
+      "#Calculation\n",
+      "n=mEG/EG #moles of EG\n",
+      "m=n/mH2O #molality of EG\n",
+      "Kf=1.86 #molal freezing point depression constant, C/m\n",
+      "deltaTf=Kf*m #depression in freezing point, C\n",
+      "Kb=0.52 #molal boiling point elevation constant, C/m\n",
+      "deltaTb=Kb*m #elevation in boiling point, C\n",
+      "\n",
+      "#Result\n",
+      "print\"The depression in freezing point is\",round(deltaTf,2),\"C\"\n",
+      "print\"Elevation in boiling point is :\",round(deltaTb,1),\"C\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The depression in freezing point is 7.79 C\n",
+        "Elevation in boiling point is : 2.2 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.9,Page no:536"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "pie=30 #osmotic pressure, atm\n",
+      "R=0.0821 #gas constant, L atm/K mol\n",
+      "T=298 #temp., K\n",
+      "\n",
+      "#Calculation\n",
+      "M=pie/(R*T) #molar concentration, M\n",
+      "\n",
+      "#Result\n",
+      "print\"The molar concentration is :\",round(M,2),\"M\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molar concentration is : 1.23 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.10,Page no:537"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "deltaTf=1.05 #depression in freezing point, C\n",
+      "Kf=5.12 #molal freezing point depression constant\n",
+      "\n",
+      "#Calculation\n",
+      "m=deltaTf/Kf #molality of solution, molal\n",
+      "mbenzene=301/1000.0 #mass of benzene, kg\n",
+      "n=m*mbenzene #moles of sapmle\n",
+      "msample=7.85 #mass of sample, g\n",
+      "molarmass=msample/n #molar mass of sample, g/mol\n",
+      "\n",
+      "#Result\n",
+      "print\"The molar mass of the sample is :\",round(molarmass),\"g/mol \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molar mass of the sample is : 127.0 g/mol \n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.11,Page no:538"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "R=0.0821 #gas constant, L atm/K mol\n",
+      "T=298 #temp, K\n",
+      "pie=10/760.0 #osmotic pressure, atm\n",
+      "\n",
+      "#Calculation\n",
+      "M=pie/(R*T) #molarity of the solution, M\n",
+      "#taking 1L of solution\n",
+      "mass=35 #mass of Hg, g\n",
+      "n=M #moles\n",
+      "molarmass=mass/n #molar mass of hemoglobin, g/mol\n",
+      "\n",
+      "#Result\n",
+      "print\"The molar mass of the hemoglobin is :%.2e\"%molarmass,\"g/mol\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molar mass of the hemoglobin is :6.51e+04 g/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.12,Page no:540"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "R=0.0821 #gas constant, L atm/K mol\n",
+      "T=298 #temp, K\n",
+      "pie=0.465 #osmotic pressure, atm\n",
+      "M=0.01 #molarity of the solution, M\n",
+      "\n",
+      "#Calculation\n",
+      "i=pie/(M*R*T) #vant hoff factor of KI\n",
+      "\n",
+      "#Result\n",
+      "print\"The vant hoff factor of KI at 25 C is :\",round(i,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The vant hoff factor of KI at 25 C is : 1.9\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_13.ipynb b/Chemistry/Chapter_13.ipynb
new file mode 100755
index 00000000..72db9250
--- /dev/null
+++ b/Chemistry/Chapter_13.ipynb
@@ -0,0 +1,441 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13:Chemical Kinetics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.2,Page no:564"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "dO2=-0.024 #rate of reaction of O2, M/s\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "dN2O5=-2*dO2 #rate of formation of N2O5, M/s\n",
+      "#(b)\n",
+      "dNO2=4*dO2 #rate of reaction of NO2, M/s\n",
+      "\n",
+      "#Result\n",
+      "print\"(a).The rate of formation of N2O5 is :\",dN2O5,\"M/s\"\n",
+      "print\"(b).The rate of reaction of NO2 is :\",dNO2,\"M/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a).The rate of formation of N2O5 is : 0.048 M/s\n",
+        "(b).The rate of reaction of NO2 is : -0.096 M/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.3,Page no:568"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "NO1=5*10**-3 #conc of NO from 1st experiment, M\n",
+      "H21=2*10**-3 #conc of H2 from 1st experiment, M\n",
+      "r1=1.3*10**-5 #initial rate from 1st experiment, M/s\n",
+      "NO2=10*10**-3 #conc of NO from 2nd experiment, M\n",
+      "H22=2*10**-3 #conc of H2 from 2nd experiment, M\n",
+      "r2=5*10**-5 #initial rate from 1st experiment, M/s\n",
+      "NO3=10*10**-3 #conc of NO from 3rd experiment, M\n",
+      "H23=4*10**-3 #conc of H2 from 3rd experiment, M\n",
+      "r3=10*10**-5 #initial rate from 3rd experiment, M/s\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "#(a)\n",
+      "#r=k*NO**x*H2**y, dividing r2/r1 and r3/r2\n",
+      "x=math.log(r2/r1)/math.log(NO2/NO1) #since H21=H22\n",
+      "y=math.log(r3/r2)/math.log(H23/H22) #since NO3=NO2\n",
+      "x=round(x) \n",
+      "y=round(y) \n",
+      "#(b)\n",
+      "k=r2/((NO2)**x*H22**y) #rate constant, /M**2 s\n",
+      "#(c)\n",
+      "NO=12*10**-3 #conc of NO, M\n",
+      "H2=6*10**-3 #conc of H2, M\n",
+      "rate=k*(NO**x)*H2**y #rate, M/s\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) the rate of reaction is : r=k[NO]**\",x,\"*[H2]**\",y\n",
+      "print\"(b) the rate constant of the reaction is :%.1e\"%k,\" /M**2 s\" \n",
+      "print\"(c) the rate of reaction at given concentration is :%.1e\"%rate,\" M/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) the rate of reaction is : r=k[NO]** 2.0 *[H2]** 1.0\n",
+        "(b) the rate constant of the reaction is :2.5e+02  /M**2 s\n",
+        "(c) the rate of reaction at given concentration is :2.2e-04  M/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 72
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.4,Page no:571"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "t=8.8*60 #time, s\n",
+      "k=6.7*10**-4 #rate constant, s-1\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "#(a)\n",
+      "Ao1=0.25 #initial conc, M\n",
+      "A1=math.exp(-k*t+math.log(Ao1)) #final conc, M\n",
+      "#(b)\n",
+      "A2=0.15 #initial conc, M\n",
+      "Ao2=0.25 #final conc, M\n",
+      "t2=-math.log(A2/Ao2)/(60*k) #time, min\n",
+      "#(c)\n",
+      "percent=74.0 \n",
+      "#let initial conc be 1\n",
+      "Ao3=1.0 #initial conc, M\n",
+      "A3=1.0-percent/100.0 #final conc, M\n",
+      "t3=-math.log(A3/Ao3)/(k*60.0) #time, min\n",
+      "print\"(a) the concentration of cyclopropane at given time is :\",round(A1,2),\"M\"\n",
+      "print\"(b) the time required is :\",round(t2),\"min\"\n",
+      "print\"(c) the time required for required conversion is :%.1f\"%t3,\"min(approx)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) the concentration of cyclopropane at given time is : 0.18 M\n",
+        "(b) the time required is : 13.0 min\n",
+        "(c) the time required for required conversion is :33.5 min(approx)\n"
+       ]
+      }
+     ],
+     "prompt_number": 75
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.5,Page no:573"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "t=[0,100,150,200,250,300] #time(data given), s\n",
+      "P=[284.0,220.0,193.0,170.0,150.0,132.0] #pressure(data given) in mmHg corresponding to time values\n",
+      "lnP=[0,0,0,0,0,0]\n",
+      "\n",
+      "#Calculation\n",
+      "import matplotlib.pyplot as plt\n",
+      "import numpy as np\n",
+      "from pylab import *\n",
+      "%pylab inline\n",
+      "lnP=numpy.log(P) #lnP values corresponding to P\n",
+      "A=plot(t,lnP)\n",
+      "plt.xlim(0,350)\n",
+      "plt.ylim(4.8,5.8)\n",
+      "plt.ylabel('$ln Pt$')\n",
+      "plt.xlabel('$t(s)$')\n",
+      "plt.title('Plot of ln Pt versus time for the decomposition of azomethane\\n')\n",
+      "slope=np.polyfit(t,lnP,1)\n",
+      "k=-slope[0]\n",
+      "\n",
+      "#Result\n",
+      "show(A)\n",
+      "print\"The rate constant for the decomposition is :%.2e\"%k,\"s**-1\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['power', 'draw_if_interactive', 'random', 'fft', 'linalg', 'info']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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oKBEBgImJtLP+7FngwQNp4MlvvmEbjIiqTmdbKN26dUNAQABWr15dZvrNmzex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\nbACx9wAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x856d978>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rate constant for the decomposition is :2.55e-03 s**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 76
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.6,Page no:576"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "k=5.36*10**-4 #rate constant, s-1\n",
+      "\n",
+      "#Calculation\n",
+      "t_half=0.693/(60*k) #half life of the reaction, min\n",
+      "\n",
+      "#Result\n",
+      "print\"The half life for the decomposition of ethane is :\",round(t_half,1),\"min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The half life for the decomposition of ethane is : 21.5 min\n"
+       ]
+      }
+     ],
+     "prompt_number": 77
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.7,Page no:578"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "k=7*10**9 #rate constant, M s\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "t=2*60 #half life of the reaction, s\n",
+      "Ao1=0.086 \n",
+      "A1=(k*t+1/Ao1)**-1 \n",
+      "#(b)\n",
+      "Ao2=0.6 \n",
+      "t_half2=1/(Ao2*k) #half life of the reaction, s\n",
+      "Ao3=0.42 \n",
+      "t_half3=1/(Ao3*k) #half life of the reaction, s\n",
+      "\n",
+      "#Result\n",
+      "print\"(a.)The concentration of I is :%.1e\"%A1,\"M\"\n",
+      "print\"(b.)The half life for Io=0.6 is :%.1e\"%t_half2,\"s\"\n",
+      "print\"    The half life for Io=0.42 is :%.1e\"%t_half3,\"s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a.)The concentration of I is :1.2e-12 M\n",
+        "(b.)The half life for Io=0.6 is :2.4e-10 s\n",
+        "    The half life for Io=0.42 is :3.4e-10 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 79
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.8,Page no:584"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "T=[700.0,730.0,760.0,790.0,810.0] #temperature(data given), K\n",
+      "R=8.314 #gas constant, kJ/mol\n",
+      "k=[0.011,0.035,0.105,0.343,0.789] #rate constant(data given) in 1/M**1/2 s corresponding to temperature values\n",
+      "\n",
+      "#Calculation\n",
+      "x=numpy.reciprocal(T) #1/T values corresponding to Temp values above, K-1\n",
+      "x=numpy.round(x,5)\n",
+      "import math\n",
+      "import numpy\n",
+      "from pylab import *\n",
+      "%pylab inline\n",
+      "lnk=numpy.log(k) #lnk values corresponding to k\n",
+      "lnk[0]=numpy.round(lnk[0],2)\n",
+      "lnk[1]=numpy.round(lnk[1],2)\n",
+      "lnk[2]=numpy.round(lnk[2],3)\n",
+      "lnk[3]=numpy.round(lnk[3],3)\n",
+      "lnk[4]=numpy.round(lnk[4],3)\n",
+      "A=plot(x,lnk,marker='o', linestyle='-')\n",
+      "plt.xlim(1.2*10**-3,1.45*10**-3)\n",
+      "plt.ylim(-5.0,0.0)\n",
+      "plt.ylabel('$ln k$')\n",
+      "plt.xlabel('$1/T(K^-1)$')\n",
+      "plt.title('Plot of ln k versus 1/T\\n')\n",
+      "slope=np.polyfit(x,lnk,1)\n",
+      "Ea=-slope[0]*R #activation energy, kJ/mol\n",
+      "\n",
+      "#Result\n",
+      "show(A)\n",
+      "print\"The activation energy for the decomposition is :%.3e\"%(Ea/1000),\"kJ/mol\"\n",
+      "print\"NOTE:Slope is approximately calculated in book,that's why wrong answer\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['power', 'draw_if_interactive', 'random', 'fft', 'linalg', 'info']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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CdjZMnWoubz9lijlHxcfy9hjnycjIICMjw5LPtmzS4vbt2+nevTuBgYEA5OXl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\ngieeAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x8535d30>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The activation energy for the decomposition is :1.797e+02 kJ/mol\n",
+        "NOTE:Slope is approximately calculated in book,that's why wrong answer\n"
+       ]
+      }
+     ],
+     "prompt_number": 80
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.9,Page no:586"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Variable declaration\n",
+      "k1=3.46*10**-2 #rate constant at T1\n",
+      "T1=298 #temp K\n",
+      "T2=350 #temp K\n",
+      "R=8.314 #gas constant,J/K mol\n",
+      "Ea=50.2*1000 #activation energy, J/mol\n",
+      "\n",
+      "#Calculation\n",
+      "k2=k1/math.exp(Ea/R*(T1-T2)/(T1*T2)) #from equation ln(k1/k2)=Ea*(T1-T2)/(T1*T2*R), S-1\n",
+      "\n",
+      "#Result\n",
+      "print\"The rate constant at temp 350 is :\",round(k2,3),\"s**-1\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rate constant at temp 350 is : 0.702 s**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 81
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_14.ipynb b/Chemistry/Chapter_14.ipynb
new file mode 100755
index 00000000..29fa23a9
--- /dev/null
+++ b/Chemistry/Chapter_14.ipynb
@@ -0,0 +1,359 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14:Chemical Equilibrium"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:14.2,Page no:622"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "NO=0.0542 #equilibrium conc of NO, M\n",
+      "O2=0.127 #equilibrium conc of O2, M\n",
+      "NO2=15.5 #equilibrium conc of NO2, M\n",
+      "\n",
+      "#Calculation\n",
+      "Kc=NO2**2/(O2*NO**2) #equilibrium constant for given reaction\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of the equilibrium constant of the reaction is %.2e\"%Kc\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of the equilibrium constant of the reaction is 6.44e+05\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:14.3,Page no:623"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "PCl3=0.463 #equilibrium pressure of PCl3, atm\n",
+      "PCl5=0.875 #equilibrium pressure of PCl5, atm\n",
+      "Kp=1.05 #equilibrium constant of the reaction\n",
+      "\n",
+      "#Calculation\n",
+      "Cl2=Kp*PCl5/PCl3 #equilibrium pressure of Cl2 in atm, formula from the definition of equilibrium constant\n",
+      "\n",
+      "#Result\n",
+      "print\"The value of the equilibrium pressure of the Cl2 gas is :\",round(Cl2,2),\"atm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of the equilibrium pressure of the Cl2 gas is : 1.98 atm\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:14.4,Page no:623"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Kc=10.5 \n",
+      "delta_n=1-3 \n",
+      "T=273+220 \n",
+      "\n",
+      "#Calculation\n",
+      "Kp=Kc*(0.0821*T)**delta_n \n",
+      "\n",
+      "#Result\n",
+      "print\"The value of the equilibrium constant of the reaction is :%.2e\"%Kp"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of the equilibrium constant of the reaction is :6.41e-03\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:14.6,Page no:626"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "CO2=0.236 #pressure of CO2 gas, atm\n",
+      "T=273+800 \n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "Kp=CO2 \n",
+      "#(b)\n",
+      "delta_n=1 \n",
+      "Kc=Kp*(0.0821*T)**-delta_n \n",
+      "\n",
+      "#Result\n",
+      "print\"(a) the value of Kp of the reaction is :\",Kp\n",
+      "print\"(b) the value of Kc of the reaction is %.2e\"%Kc"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) the value of Kp of the reaction is : 0.236\n",
+        "(b) the value of Kc of the reaction is 2.68e-03\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:14.8,Page no:633"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Kc=1.2 #equilibrium constant for the reaction\n",
+      "N2=.249/3.5 #conc of N2, M\n",
+      "H2=(3.21*10**-2)/3.5 #conc of H2, M\n",
+      "NH3=(6.42*10**-4)/3.5 #conc of NH3, M\n",
+      "\n",
+      "#Calculation\n",
+      "Qc=NH3**2/(N2*H2**3) #reaction quotient initial\n",
+      "print\"Qc=\",round(Qc,3),\"(approx)\"\n",
+      "\n",
+      "#Result\n",
+      "if(Qc==Kc):\n",
+      "    d=\"the system is in equilibrium\" \n",
+      "elif(Qc<Kc):\n",
+      "    d=\"the system is not in equilibrium and the reaction will move from left to right\" \n",
+      "else:\n",
+      "    d=\"the system is not in equilibrium and the reaction will move from right to left\" \n",
+      "print d"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Qc= 0.613 (approx)\n",
+        "the system is not in equilibrium and the reaction will move from left to right\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:14.9,Page no:635"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Kc=54.3 \n",
+      "H2i=0.5 #initial moles of H2\n",
+      "I2i=0.5 #initial moles of I2\n",
+      "\n",
+      "#Calculation\n",
+      "#Let us assume that x moles have reacted, so, HI=2x, H2=0.5-x, I2=0.5-x, when we substitute in Kc=(HI)**2/(H2)*(I2) we get 54.3=(2x)**2/((0.5-x)*(0.5-x)) taking root we get 7.37=2*x/0.5-x\n",
+      "x=0.393 #from the above equation\n",
+      "H2=0.5-x \n",
+      "I2=0.5-x \n",
+      "HI=2*x \n",
+      "\n",
+      "#Result\n",
+      "print\"The equilibrium concentration of H2 is :\",H2,\"M\" \n",
+      "print\"The equilibrium concentration of I2 is :\",I2,\"M\"\n",
+      "print\"The equilibrium concentration of HI is :\",HI,\"M\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The equilibrium concentration of H2 is : 0.107 M\n",
+        "The equilibrium concentration of I2 is : 0.107 M\n",
+        "The equilibrium concentration of HI is : 0.786 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:14.10,Page no:636"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Variable declaration\n",
+      "Kc=54.3 \n",
+      "HIo=0.0224 \n",
+      "H2o=0.00623 \n",
+      "I2o=0.00414 \n",
+      "#let us assume that x moles have reacted, so, HI=HIo+2x, H2=0.00623-x, I2=0.00414-x, when we substitute in Kc=(HI)**2/(H2)*(I2) we get 54.3=(2x+0.0224)**2/((0.00623-x)*(0.00414-x)) simplifying we get 50.3x**2-0.654x+8.98*10**-4=0\n",
+      "a=50.3 \n",
+      "b=-0.654 \n",
+      "c=8.98*10**-4 \n",
+      "\n",
+      "#Calculation\n",
+      "x1=(-b+math.sqrt(b**2-4*a*c))/(2*a) \n",
+      "x2=(-b-math.sqrt(b**2-4*a*c))/(2*a) \n",
+      "if(x1>I2o):\n",
+      "    x=x2 \n",
+      "else:\n",
+      "    x=x1 \n",
+      "H2=0.00623-x \n",
+      "I2=0.00414-x \n",
+      "HI=2*x+0.0224 \n",
+      "\n",
+      "#Result\n",
+      "print\"The equilibrium concentration of H2 is :\",round(H2,5),\"M\"\n",
+      "print\"The equilibrium concentration of I2 is :\",round(I2,5),\"M\"\n",
+      "print\"The equilibrium concentration of HI is :\",round(HI,4),\"M\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The equilibrium concentration of H2 is : 0.00467 M\n",
+        "The equilibrium concentration of I2 is : 0.00258 M\n",
+        "The equilibrium concentration of HI is : 0.0255 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:14.11,Page no:639"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Kc=2.37*10**-3 #equilibrium constant for the reaction\n",
+      "N2=0.683 #conc of N2, M\n",
+      "H2=8.8 #conc of H2, M\n",
+      "NH3=3.65 #conc of NH3, M\n",
+      "\n",
+      "#Calculation\n",
+      "Qc=NH3**2/(N2*H2**3) #reaction quotient initial\n",
+      "print\"Qc=%.2e\"%Qc\n",
+      "\n",
+      "#Result\n",
+      "if(Qc==Kc):\n",
+      "    d=\"the system is in equilibrium\" \n",
+      "elif(Qc<Kc):\n",
+      "    d=\"the system is not in equilibrium and the reaction will move from left to right\" \n",
+      "else:\n",
+      "    d=\"the system is not in equilibrium and the reaction will move from right to left\" \n",
+      "print d\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Qc=2.86e-02\n",
+        "the system is not in equilibrium and the reaction will move from right to left\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_15.ipynb b/Chemistry/Chapter_15.ipynb
new file mode 100755
index 00000000..fe8c12e3
--- /dev/null
+++ b/Chemistry/Chapter_15.ipynb
@@ -0,0 +1,472 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 15:Acids and Bases"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:15.2,Page no:662"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "OH=0.0025  # [OH-] ion concentration, M\n",
+      "Kw=1*10**-14  # ionic product of water, M**2\n",
+      "#Calculation\n",
+      "H=Kw/OH  # From the formula (ionic product)Kw=[H+]*[OH-]\n",
+      "#Result\n",
+      "print\"The [H+] ion concentration of the solution is :\",H,\"M\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The [H+] ion concentration of the solution is : 4e-12 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:15.3,Page no:665"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "H1=3.2*10**-4  #Concentration of [H+] ion on first occasion,\n",
+      "H2=1*10**-3  #Concentration of [H+] ion on second occasion, M\n",
+      "\n",
+      "#Calculation\n",
+      "pH1=-math.log10(H1) #from the definition of pH\n",
+      "pH2=-math.log10(H2) #from the definition of pH\n",
+      "\n",
+      "#Result\n",
+      "print\"pH of the solution on first occasion is:\",round(pH1,2)\n",
+      "print\"pH of the solution on second occasion is :\",pH2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pH of the solution on first occasion is: 3.49\n",
+        "pH of the solution on second occasion is : 3.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:15.4,Page no:665"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "pH=4.82 #Given\n",
+      "\n",
+      "#Calculation\n",
+      "H=10**(-pH) #Concentration of [H+] ion, M, formula from the definition of pH\n",
+      "\n",
+      "#Result\n",
+      "print\"The [H+] ion concentration of the solution is :%.1e\"%H,\"M\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The [H+] ion concentration of the solution is :1.5e-05 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:15.5,Page no:666"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Variable declaration\n",
+      "OH=2.9*10**-4 #Concentration of [OH-] ion, M\n",
+      "#Calculation\n",
+      "pOH=-math.log10(OH) #by definition of p(OH)\n",
+      "pH=14-pOH \n",
+      "#Result\n",
+      "print\"\\t the pH of the solution is :\",round(pH,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t the pH of the solution is : 10.46\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:15.6,Page no:669"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "ConcHCl=1*10**-3 #Concentration of HCl solution, M\n",
+      "ConcBaOH2=0.02 #Concentration of Ba(OH)2 solution, M\n",
+      "\n",
+      "#Calculation\n",
+      "#for HCL solution\n",
+      "H=ConcHCl #Concentration of [H+] ion after ionisation of HCl\n",
+      "pH=-math.log10(H) \n",
+      "#for Ba(OH)2 solution\n",
+      "OH=ConcBaOH2*2 #Concentration of [OH-] ion after ionisation of Ba(OH)2 as two ions are generated per one molecule of Ba(OH)2\n",
+      "pOH=-math.log10(OH) \n",
+      "pH2=14-pOH \n",
+      "\n",
+      "#Result\n",
+      "print\"(a). The pH of the HCl solution is :\",pH\n",
+      "print\"(b). The pH of the Ba(OH)2 solution is :\",round(pH2,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a). The pH of the HCl solution is : 3.0\n",
+        "(b). The pH of the Ba(OH)2 solution is : 12.6\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:15.8,Page no:675"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Variable declaration\n",
+      "InitHNO2=0.036 #Initial concentration of HNO2 solution, M\n",
+      "#Let 'x' be the equilibrium concentration of the [H+] and [NO2-] ions, M\n",
+      "Ka=4.5*10**-4 #ionisation constant of HNO2, M\n",
+      "\n",
+      "#Calculation\n",
+      "x=math.sqrt(Ka*InitHNO2) #from the definition of ionisation constant Ka=[H+]*[NO2-]/[HNO2]=x*x/(0.036-x), which reduces to x*x/0.036, as x<<InitHNO2 (approximation)\n",
+      "approx=x/InitHNO2*100 #this is the percentage of approximation taken. if it is more than 5%, we will be having higher deviation from correct value\n",
+      "if(approx>5):\n",
+      "     x1=(-Ka+math.sqrt((Ka**2)-(-4*1*Ka*InitHNO2)))/(2*1) \n",
+      "     x2=(-Ka-math.sqrt((Ka**2)-(-4*1*Ka*InitHNO2)))/(2*1) \n",
+      "     if(x1>0):\n",
+      "         x=x1 \n",
+      "     else :\n",
+      "          x=x2 \n",
+      "pH=-math.log10(x) #since x is the conc. of [H+] ions\n",
+      "\n",
+      "#Result\n",
+      "print\"The pH of the HNO2 solution is :\",round(pH,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pH of the HNO2 solution is : 2.42\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:15.9,Page no:676"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "pH=2.39 # pH of the HCOOH acid solution\n",
+      "InitHCOOH=0.1 #initial concentration of the solution\n",
+      "\n",
+      "#Calculation\n",
+      "H=10**(-pH) #[H+] ion concentration from the definition of pH, M\n",
+      "HCOO_=H     #[HCOO-] ion concentration,M\n",
+      "HCOOH=InitHCOOH-H   #HCOOH concentration in M\n",
+      "Ka=(H*HCOO_)/(HCOOH) #ionisation constant of the acid, M, Ka=[H+]*[HCOO-]/[HCOOH]\n",
+      "\n",
+      "#Result\n",
+      "print\"The ionisation constant of the given solution is :%.2e\"%Ka,\"M(approx)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ionisation constant of the given solution is :1.73e-04 M(approx)\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:15.10,Page no:678"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "InitNH3=0.4 #Initial concentration of NH3 solution, M\n",
+      "#Let 'x' be the equilibrium concentration of the [OH-] and [NH4+] ions, M\n",
+      "Kb=1.8*10**-5 #ionisation constant of NH3, M\n",
+      "\n",
+      "#Calculation\n",
+      "x=math.sqrt(Kb*InitNH3) #from the definition of ionisation constant Kb=[OH-]*[NH4+]/[NH3]=x*x/(InitNH3-x), which reduces to x*x/InitNH3, as x<<InitNH3 (approximation)\n",
+      "approx=x/InitNH3*100 #this is the percentage of approximation taken. if it is more than 5%, we will be having higher deviation from correct value\n",
+      "if(approx>5):\n",
+      "     x1=(-Kb+math.sqrt((Kb**2)-(-4*1*Kb*InitNH3)))/(2*1) \n",
+      "     x2=(-Kb-math.sqrt((Kb**2)-(-4*1*Kb*InitNH3)))/(2*1) \n",
+      "     if(x1>0):#as only one root is positive\n",
+      "          x=x1 \n",
+      "     else:\n",
+      "          x=x2 \n",
+      "pOH=-math.log10(x) #since x is the conc. of [H+] ions\n",
+      "pH=14-pOH \n",
+      "\n",
+      "#Result\n",
+      "print\"The pH of the NH3 solution is :\",round(pH,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pH of the NH3 solution is : 11.43\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:15.11,Page no:682"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "InitC2H2O4=0.1 #Initial concentration of C2H2O4 solution, M\n",
+      "#Let 'x1' be the equilibrium concentration of the [H+] and [C2HO4-] ions, M\n",
+      "#First stage of ionisation\n",
+      "import math\n",
+      "Kw=10**-14 #ionic product of water, M**2\n",
+      "Ka1=6.5*10**-2 #ionisation constant of C2H2O4, M\n",
+      "\n",
+      "#Calculation\n",
+      "x=math.sqrt(Ka1*InitC2H2O4) #from the definition of ionisation constant Ka1=[H+]*[C2HO4-]/[C2H2O4]=x*x/(InitC2H2O4-x), which reduces to x*x/InitC2H2O4, as x<<InitC2H2O4 (approximation)\n",
+      "approx=x/InitC2H2O4*100 #this is the percentage of approximation taken. if it is more than 5%, we will be having higher deviation from correct value\n",
+      "if(approx>5):\n",
+      "     x1=(-Ka1+math.sqrt((Ka1**2)-(-4*1*Ka1*InitC2H2O4)))/(2*1) \n",
+      "     x2=(-Ka1-math.sqrt((Ka1**2)-(-4*1*Ka1*InitC2H2O4)))/(2*1) \n",
+      "     if(x1>0):#as only one root is positive\n",
+      "          x=x1 \n",
+      "     else: \n",
+      "          x=x2 \n",
+      "C2H2O4=InitC2H2O4-x #equilibrium value\n",
+      "\n",
+      "#Result\n",
+      "print\"The concentration of the [H2C2O4] in the solution is :\",round(C2H2O4,3),\"M\"\n",
+      "\n",
+      "\n",
+      "#Second stage of ionisation\n",
+      "\n",
+      "#Variable declaration\n",
+      "InitC2HO4=x #concentration of C2HO4 from first stage of ionisation\n",
+      "Ka2=6.1*10**-5 #ionisation constant of C2HO4-, M\n",
+      "\n",
+      "#Calculation\n",
+      "#Let 'y' be the concentration of the [C2HO4-] dissociated to form [H+] and [C2HO4-] ions, M\n",
+      "y=Ka2 #from the definition of ionisation constant Ka2=[H+]*[C2O4-2]/[C2HO4-]=(0.054+y)*y/(0.054-y), which reduces to y, as y<<InitC2HO4 (approximation)\n",
+      "approx=y/InitC2HO4*100 #this is the percentage of approximation taken. if it is more than 5%, we will be having higher deviation from correct value\n",
+      "if(approx>5):\n",
+      "     x1=(-Ka2+math.sqrt((Ka2**2)-(-4*1*Ka2*InitC2HO4)))/(2*1) \n",
+      "     x2=(-Ka2-math.sqrt((Ka2**2)-(-4*1*Ka2*InitC2HO4)))/(2*1) \n",
+      "     if(x1>0):#as only one root is positive\n",
+      "          y=x1 \n",
+      "     else: \n",
+      "          y=x2 \n",
+      "C2HO4=InitC2HO4-y #from first and second stages of ionisation\n",
+      "H=x+y #from first and second stages of ionisation\n",
+      "C2O4=y #from the assumption\n",
+      "OH=Kw/H # From the formula (ionic product)Kw=[H+]*[OH-]\n",
+      "\n",
+      "#Result\n",
+      "print\"The concentration of the [HC2O4-] ion in the solution is :\",round(C2HO4,3),\"M\"\n",
+      "print\"The concentration of the [H+] ion in the solution is :\",round(H,3),\"M\" \n",
+      "print\"The concentration of the [C2O4^2-] ion in the solution is :\",C2O4,\"M\"\n",
+      "print\"The concentration of the [OH-] ion in the solution is :%.1e\"%OH,\"M\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The concentration of the [H2C2O4] in the solution is : 0.046 M\n",
+        "The concentration of the [HC2O4-] ion in the solution is : 0.054 M\n",
+        "The concentration of the [H+] ion in the solution is : 0.054 M\n",
+        "The concentration of the [C2O4^2-] ion in the solution is : 6.1e-05 M\n",
+        "The concentration of the [OH-] ion in the solution is :1.8e-13 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:15.13,Page no:690"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "InitCH3COONa=0.15 #Initial concentration of CH3COONa solution, M\n",
+      "InitCH3COO=InitCH3COONa #concentration of [CH3COO-] ion after dissociation of CH3COONa solution, M\n",
+      "#Let 'x' be the equilibrium concentration of the [OH-] and [CH3COOH] ions after hydrolysis of [CH3COO-], M\n",
+      "Kb=5.6*10**-10 #equilibrium constant of hydrolysis, M\n",
+      "import math\n",
+      "\n",
+      "#Calculation\n",
+      "x=math.sqrt(Kb*InitCH3COO) #from the definition of ionisation constant Kb=[OH-]*[CH3COOH]/[CH3COO-]=x*x/(0.15-x), which reduces to x*x/0.15, as x<<0.15 (approximation)\n",
+      "approx=x/InitCH3COO*100 #this is the percentage of approximation taken. if it is more than 5%, we will be having higher deviation from correct value\n",
+      "if(approx>5):\n",
+      "     x1=(-Kb+math.sqrt((Kb**2)-(-4*1*Kb*InitCH3COO)))/(2*1) \n",
+      "     x2=(-Kb-math.sqrt((Kb**2)-(-4*1*Kb*InitCH3COO)))/(2*1) \n",
+      "     if(x1>0):#as only one root is positive\n",
+      "          x=x1 \n",
+      "     else: \n",
+      "          x=x2 \n",
+      "pOH=-math.log10(x) #since x is the conc. of [OH-] ions\n",
+      "pH=14-pOH \n",
+      "\n",
+      "#Result\n",
+      "print\"The pH of the salt solution is :\",round(pH,2)\n",
+      "percenthydrolysis=x/InitCH3COO*100 \n",
+      "print\"The percentage of hydrolysis of the salt solution is :\",round(percenthydrolysis,4),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pH of the salt solution is : 8.96\n",
+        "The percentage of hydrolysis of the salt solution is : 0.0061 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_16.ipynb b/Chemistry/Chapter_16.ipynb
new file mode 100755
index 00000000..96bc5bbb
--- /dev/null
+++ b/Chemistry/Chapter_16.ipynb
@@ -0,0 +1,584 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16:Acid-Base Equilibria and Solubility Equilibria"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:16.1,Page no:715"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "InitCH3COOH1=0.2                #Initial concentration of CH3COOH solution, M\n",
+      "#Let 'x' be the equilibrium concentration of the [H+] and [CH3COO-] ions after dissociation of [CH3COOH], M\n",
+      "Ka=1.8*10**-5                   #equilibrium constant of acid, M\n",
+      "InitCH3COONa=0.3                #Initial concentration of CH3COONa solution and is equal to conc of Na+ and CH3COO- as it completely dissociates, M\n",
+      "InitCH3COOH2=0.2                #Initial concentration of CH3COOH solution, M\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "import math\n",
+      "x1=math.sqrt(Ka*InitCH3COOH1)   #from the definition of ionisation constant Ka=[H+]*[CH3COO-]/[CH3COOH]=x*x/(0.2-x), which reduces to x*x/0.2, as x<<0.2 (approximation)\n",
+      "pH1=-math.log10(x1)             #since x is the conc. of [H+] ions\n",
+      "\n",
+      "#(b)\n",
+      "#Let 'x' be the equilibrium concentration of the [H+] and hence conc of [CH3COO-] ions is '0.3 + x', M\n",
+      "x2=Ka*InitCH3COOH2/InitCH3COONa           #from the definition of ionisation constant Ka=[H+]*[CH3COO-]/[CH3COOH]=x*(0.3+x)/(0.2-x), which reduces to x*0.3/0.2(approximation)\n",
+      "pH2=-math.log10(x2)                       #since x is the conc. of [H+] ions\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) the pH of CH3COOH solution is \",round(pH1,2)\n",
+      "print\"(b) the pH of CH3COOH and CH3COONa solution is :\",round(pH2,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) the pH of CH3COOH solution is  2.72\n",
+        "(b) the pH of CH3COOH and CH3COONa solution is : 4.92\n"
+       ]
+      }
+     ],
+     "prompt_number": 92
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.3,Page no:719"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Ka=1.8*10**-5                     #ionisation constant of acid\n",
+      "InitCH3COONa=1                    #Initial concentration of CH3COONa solution and is equal to conc of Na+ and CH3COO- as it completely dissociates, M\n",
+      "InitCH3COOH=1 #Initial concentration of CH3COOH solution, M\n",
+      "HCl=0.1 #moles of HCl added to 1L solution\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "#(a)\n",
+      "x=Ka*InitCH3COOH/InitCH3COONa #from the definition of ionisation constant Ka=[H+]*[CH3COO-]/[CH3COOH]=x*(1+x)/(1-x), which reduces to x(approximation)\n",
+      "pH=-math.log10(x) #since x is the conc. of [H+] ions\n",
+      "#(b)\n",
+      "CH3COO=InitCH3COONa-HCl #conc of CH3COO- ions, M\n",
+      "CH3COOH=InitCH3COOH+HCl #conc of CH3COOH, M\n",
+      "x2=Ka*CH3COOH/CH3COO #from the definition of ionisation constant Ka=[H+]*[CH3COO-]/[CH3COOH]=x*(0.9+x)/(1.1-x), which reduces to x*0.9/1.1(approximation)\n",
+      "pH2=-math.log10(x2) #since x is the conc. of [H+] ions\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) the pH of CH3COOH and CH3COONa solution is :\",round(pH,2)\n",
+      "print\"(b) the pH of solution after adding HCl is :\",round(pH2,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) the pH of CH3COOH and CH3COONa solution is : 4.74\n",
+        "(b) the pH of solution after adding HCl is : 4.66\n"
+       ]
+      }
+     ],
+     "prompt_number": 93
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:16.5,Page no:728"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "InitCH3COOH=0.1 #Initial concentration of CH3COOH solution, M\n",
+      "VCH3COOH=25 #volumeof CH3COOH, mL\n",
+      "nCH3COOH=InitCH3COOH*VCH3COOH/1000 \n",
+      "Ka=1.8*10**-5 #equilibrium constant of acid, M\n",
+      "Kb=5.6*10**-10 #equilibrium constant of base, M\n",
+      "N=0.1 #Initial concentration of NaOH solution, M\n",
+      "V=10 #Initial volume of NaOH solution, mL\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "n=N*V/1000 #Initial moles of NaOH solution\n",
+      "import math\n",
+      "nCH3COOH_tit=nCH3COOH-n #moles of CH3COOH after titration\n",
+      "nCH3COO=n #moles of CH3COO after titration\n",
+      "H=nCH3COOH_tit*Ka/nCH3COO #conc of H+ ions, M\n",
+      "pH=-math.log10(H) #since H is the conc. of [H+] ions\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) the pH of the solution is :\",round(pH,2)\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "#(b)\n",
+      "V2=25.0 #Initial volume of NaOH solution, mL\n",
+      "n2=N*V2/1000.0 #Initial moles of NaOH solution\n",
+      "nCH3COOH_tit2=nCH3COOH-n2 #moles of CH3COOH after titration\n",
+      "nCH3COO2=n2 #moles of CH3COO- ions after titration\n",
+      "V_total=V2+VCH3COOH #total volume after titration\n",
+      "CH3COO=nCH3COO2/V_total*1000 #conc of CH3COO- ions, M\n",
+      "x=math.sqrt(Kb*CH3COO) #from the definition of ionisation constant Kb=[OH-]*[CH3COOH]/[CH3COO-]=x*x/(0.05-x), which reduces to x*x/0.05, as x<<0.05 (approximation)\n",
+      "pOH=-math.log10(x) #since x is the conc. of [OH-] ions\n",
+      "pH2=14.0-pOH \n",
+      "\n",
+      "#Result\n",
+      "print\"(b) the pH of the solution is :\",round(pH2,2)\n",
+      "\n",
+      "#Calculation\n",
+      "#(c)\n",
+      "N=0.1 #Initial concentration of NaOH solution, M\n",
+      "V=35 #Initial volume of NaOH solution, mL\n",
+      "n=N*V/1000 #Initial moles of NaOH solution\n",
+      "n_tit=n-nCH3COOH #moles of NaOH after titration\n",
+      "nCH3COO=nCH3COOH #moles of CH3COO- ions after titration\n",
+      "V_total=V+VCH3COOH #total volume\n",
+      "OH=n_tit/V_total*1000 #conc of OH- ions, M\n",
+      "pOH=-math.log10(OH) #since OH is the conc. of [OH-] ions\n",
+      "pH=14-pOH \n",
+      "\n",
+      "#Result\n",
+      "print\"(c) the pH of the solution is :\",round(pH,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) the pH of the solution is : 4.57\n",
+        "(b) the pH of the solution is : 8.72\n",
+        "(c) the pH of the solution is : 12.22\n"
+       ]
+      }
+     ],
+     "prompt_number": 94
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:16.6,Page no:730"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "InitNH3=0.1 #Initial concentration of NH3 solution, M\n",
+      "VNH3=25 #volume of NH3, mL\n",
+      "nNH3=InitNH3*VNH3/1000 \n",
+      "Ka=5.6*10**-10 #equilibrium constant of acid, M\n",
+      "N=0.1 #Initial concentration, M\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "V=VNH3/InitNH3*N #Initial volume, mL\n",
+      "V_total=V+VNH3 #total volume of the mixture, mL\n",
+      "n_NH4Cl=nNH3 #moles of NH4Cl\n",
+      "NH4Cl=n_NH4Cl/V_total*1000 #conc of NH4+ ions formed, M\n",
+      "x=math.sqrt(Ka*NH4Cl) #from the definition of ionisation constant Ka=[H+]*[NH3]/[NH4+]=x*x/(NH4+-x), which reduces to x*x/NH4+, as x<<NH4+ (approximation)\n",
+      "pH=-math.log10(x) #since x is the conc. of [H+] ions\n",
+      "\n",
+      "#Result\n",
+      "print\"The pH of the solution at equivalent point is :\",round(pH,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pH of the solution at equivalent point is : 5.28\n"
+       ]
+      }
+     ],
+     "prompt_number": 95
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.8,Page no:738"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "solubility=0.67 #solubility of CaSO4, g/L\n",
+      "\n",
+      "#Calculation\n",
+      "M=136.2 #mol mass of CaSO4, g\n",
+      "s=solubility/M #concentration, M\n",
+      "Ksp=s**2 #solubility product\n",
+      "\n",
+      "#Result\n",
+      "print\"The Ksp of CaSO4 is :%.1e\"%Ksp"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Ksp of CaSO4 is :2.4e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 96
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:16.9,Page no:739"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Ksp=2.2*10**-20 #solubility product\n",
+      "M=97.57 #mol mass of Cu(OH)2, g\n",
+      "\n",
+      "#Calculation\n",
+      "s=(Ksp/4.0)**(1.0/3.0) #concentration, M\n",
+      "solubility=s*M #solubility of Cu(OH)2, g/L\n",
+      "\n",
+      "#Result\n",
+      "print\"The solubility of Cu(OH)2 is :%.2e\"%solubility,\"g/L\"\n",
+      "\n",
+      "#ALTERNATIVE METHOD:\n",
+      "\n",
+      "#Variable declaration\n",
+      "Kspt=2.2*10**-20             #Ksp value from table 16.2\n",
+      "M=97.75          #Molar mass of Cu(OH)2\n",
+      "\n",
+      "#Calculation\n",
+      "from scipy.optimize import fsolve\n",
+      "def f(s):\n",
+      "    Cu2=s\n",
+      "    OH_=2*s\n",
+      "    return(Kspt-(Cu2*(OH_**2)))\n",
+      "s=fsolve(f,1)\n",
+      "solubility=round(s,8)*M\n",
+      "\n",
+      "#Result\n",
+      "print \"Alternative method: Solubility of Cu(OH)2=%.1e\"%solubility,\"g/L\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The solubility of Cu(OH)2 is :1.72e-05 g/L\n",
+        "Alternative method: Solubility of Cu(OH)2=1.8e-05"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " g/L\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:16.10,Page no:741"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Ksp=1.1*10**-10 #solubility product of BaSO4\n",
+      "#for Ba2+ ion\n",
+      "N=0.004 #normality, M\n",
+      "V=200 #vol in mL\n",
+      "n=N*V/1000 #moles\n",
+      "#for K2SO4sol\n",
+      "N1=0.008 #normality, M\n",
+      "V1=600 #vol in mL\n",
+      "\n",
+      "#Calculation\n",
+      "n1=N1*V1/1000 #moles\n",
+      "Nnew=n*1000/(V+V1) #conc of Ba2+ ions in final sol\n",
+      "N1new=n1*1000/(V+V1) #conc of SO4 2- ions in final sol\n",
+      "Q=Nnew*N1new #as Q=[Ba2+][SO4 2-]\n",
+      "\n",
+      "#Result\n",
+      "print\"Q=\",Q\n",
+      "if(Q>Ksp): #determination of precipitation\n",
+      "    print\"Q>Ksp, The solution is supersaturated and hence a precipitate will form\"\n",
+      "else: \n",
+      "    print\"Q<Ksp, The solution is not supersaturated and hence a precipitate will not form\"  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q= 6e-06\n",
+        "Q>Ksp, The solution is supersaturated and hence a precipitate will form\n"
+       ]
+      }
+     ],
+     "prompt_number": 99
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:16.11,Page no:743"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Br=0.02 #conc of Ag+ ions, M\n",
+      "Ksp1=7.7*10**-13 #solubility product of AgBr\n",
+      "Ksp2=1.6*10**-10 #solubility product of AgCl\n",
+      "Cl=0.02 #conc of Cl- ions, M\n",
+      "\n",
+      "#Calculation\n",
+      "#for Br\n",
+      "Ag1=Ksp1/Br #conc of Ag+ ions at saturated state, M\n",
+      "#for Cl\n",
+      "Ag2=Ksp2/Cl #conc of Ag+ ions at saturated state, M\n",
+      "\n",
+      "#Result\n",
+      "print \"[Ag+]=\",Ag2,\"M\"\n",
+      "print\"To precipitate Br- without precipitating Cl- the concentration of Ag must be greater than %.1e\"%Ag1,\"M but less than\",Ag2,\"M\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "[Ag+]= 8e-09 M\n",
+        "To precipitate Br- without precipitating Cl- the concentration of Ag must be greater than 3.9e-11 M but less than 8e-09 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 100
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:16.12,Page no: 745"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "N_AgNO3=6.5*10**-3 #normality of AgNO3, M\n",
+      "AgCl=143.4 #mol mass of AgCl, g\n",
+      "Ksp=1.6*10**-10 #solubility product of AgCl\n",
+      "\n",
+      "#Calculation\n",
+      "Ag=N_AgNO3 #conc of Ag+ ions as 's' is negligible, M\n",
+      "s=Ksp/Ag #as Ksp=[Ag+][Cl-], molar solubility of AgCl, M\n",
+      "solubility=s*AgCl #solubility of AgCl in AgBr solution, g/L\n",
+      "\n",
+      "#Result\n",
+      "print\"The solubility of AgCl in AgBr solution is :%.2e\"%solubility,\"g/L\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The solubility of AgCl in AgBr solution is :3.53e-06 g/L\n"
+       ]
+      }
+     ],
+     "prompt_number": 101
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:16.14,Page no:748"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "FeCl2=0.003 #normality of FeCl2, M\n",
+      "Fe=FeCl2 #as Fe2+ is strong electrolyte, conc of Fe2+=conc of FeCl2, M\n",
+      "Ksp=1.6*10**-14 #solubility product of FeCl2\n",
+      "Kb=1.8*10**-5 #ionisation constant of base\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "OH=math.sqrt(Ksp/Fe) #as Ksp=[Fe2+][OH-]**2, conc of OH- ions, M\n",
+      "x=(OH**2)/Kb+OH #as Kb=[NH4+][OH-]/[NH3]\n",
+      "\n",
+      "#Result\n",
+      "print\"To initiate precipitation the conc of NH3 must be slightly greater than :%.1e\"%x,\"M\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "To initiate precipitation the conc of NH3 must be slightly greater than :2.6e-06 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 102
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:16.15,Page no:750"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "CuSO4=0.2 #normality of CuSO4, M\n",
+      "NH3=1.2 #initial conc of NH3, M\n",
+      "VNH3=1 #volume of NH3, L\n",
+      "Kf=5*10**13 #formation constant\n",
+      "\n",
+      "#Calculation\n",
+      "CuNH34=CuSO4 #conc of Cu(NH3)4 2+, M\n",
+      "NH3=NH3-4*CuNH34 #conc of NH3 after formation of complex, as 4 moles of NH3 react to form 1 mole complex, M\n",
+      "x=CuNH34/(NH3**4*Kf) #as Kf=[Cu(SO4)3 2+]/[Cu2+][NH3]**4\n",
+      "\n",
+      "#Result\n",
+      "print\"The conc of Cu2+ ions in equilibrium is :%.1e\"%x,\"M\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conc of Cu2+ ions in equilibrium is :1.6e-13 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 103
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:16.16,Page no:751"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "InitNH3=1 #initial conc of NH3, M\n",
+      "Ksp=1.6*10**-10 #solubility product of AgCl\n",
+      "Kf=1.5*10**7 #formation constant of complex\n",
+      "\n",
+      "#Calculation\n",
+      "K=Ksp*Kf #overall equilibrium constant\n",
+      "import math\n",
+      "s=math.sqrt(K)/(1+2*InitNH3*math.sqrt(K)) #molar solubility of AgCl, M\n",
+      "\n",
+      "#Result\n",
+      "print\"Amount of AgCl which can be dissolved in 1 L of 1 M NH3 sol in equilibrium is :\",round(s,3),\"M\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Amount of AgCl which can be dissolved in 1 L of 1 M NH3 sol in equilibrium is : 0.045 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 91
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_17.ipynb b/Chemistry/Chapter_17.ipynb
new file mode 100755
index 00000000..8debeae7
--- /dev/null
+++ b/Chemistry/Chapter_17.ipynb
@@ -0,0 +1,92 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17:Chemistry in the Atmosphere"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:17.1,Page no:774"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "E=498.7*10**3/(6.022*10**23) #energy in J/molecule\n",
+      "h=6.63*10**-34               #plancks constant, J s\n",
+      "\n",
+      "#Calculation\n",
+      "v=E/h #frequency of the photon, s**-1\n",
+      "lamda=3*10**8/v #wavelength in m, since v*lambda=speed of light in vacuum\n",
+      "\n",
+      "#Variable declaration\n",
+      "print\"The maximum wavelength of the photon which can dissociate an O2 molecule is :\",round(lamda*10**9),\"nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum wavelength of the photon which can dissociate an O2 molecule is : 240.0 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:17.3,Page no:792"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Rninitial=1 #initial mass of Rn, g\n",
+      "\n",
+      "#Calculation\n",
+      "Rnfinal=Rninitial*0.5**10 #final mass of Rn, g\n",
+      "\n",
+      "#Result\n",
+      "print\"The amount of Rn left after 10 half lives is :%.1e\"%Rnfinal,\"g\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of Rn left after 10 half lives is :9.8e-04 g\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_18.ipynb b/Chemistry/Chapter_18.ipynb
new file mode 100755
index 00000000..aabe2f15
--- /dev/null
+++ b/Chemistry/Chapter_18.ipynb
@@ -0,0 +1,284 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 18:Entropy, Free Energy,and Equilibrium"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:18.2,Page no:809"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#(a)\n",
+      "SCaO=39.8 #standard entropy of CaO, J/K mol\n",
+      "SCO2=213.6 #standard entropy of CO2, J/K mol\n",
+      "SCaCO3=92.9 #standard entropy of CaCO3, J/K mol\n",
+      "#(b)\n",
+      "SNH3=193 #standard entropy of NH3, J/K mol\n",
+      "SN2=192 #standard entropy of N2, J/K mol\n",
+      "SH2=131 #standard entropy of H2, J/K mol\n",
+      "#(c)\n",
+      "SHCl=187 #standard entropy of HCl, J/K mol\n",
+      "SH2=131 #standard entropy of H2, J/K mol\n",
+      "SCl2=223 #standard entropy of Cl2, J/K mol\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "deltaSrxn1=SCaO+SCO2-SCaCO3 #standard entropy change of the reaction, J/K mol\n",
+      "#(b)\n",
+      "deltaSrxn2=2*SNH3-(SN2+3*SH2) #standard entropy change of the reaction, J/K mol\n",
+      "#(c)\n",
+      "deltaSrxn3=2*SHCl-SH2-SCl2 #standard entropy change of the reaction, J/K mol\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) the standard entropy of reaction is :\",deltaSrxn1,\"J/K mol\"\n",
+      "print\"(b) the standard entropy of reaction is :\",deltaSrxn2,\"J/K mol\"\n",
+      "print\"(c) the standard entropy of reaction is :\",deltaSrxn3,\"J/K mol\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) the standard entropy of reaction is : 160.5 J/K mol\n",
+        "(b) the standard entropy of reaction is : -199 J/K mol\n",
+        "(c) the standard entropy of reaction is : 20 J/K mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:18.4,Page no:817"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#(a)\n",
+      "GCO2=-394.4 #free energy of formation of CO2, kJ/mol\n",
+      "GH2O=-237.2 #free energy of formation of H2O, kJ/mol\n",
+      "GCH4=-50.8 #free energy of formation of CH4, kJ/mol\n",
+      "GO2=0 #free energy of formation of O2, kJ/mol\n",
+      "#(b)\n",
+      "GMg=0 #free energy of formation of Mg, kJ/mol\n",
+      "GMgO=-569.6 #free energy of formation of MgO, kJ/mol\n",
+      "GO2=0 #free energy of formation of O2, kJ/mol\n",
+      "\n",
+      "#Calculation\n",
+      "deltaGrxn1=(GCO2+GH2O*2)-(GCH4+2*GO2) #standard free energy change of the reaction, kJ/mol\n",
+      "deltaGrxn=(GO2+GMg*2)-(2*GMgO) #standard free energy change of the reaction, kJ/mol\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The standard free energy change of reaction is :\",deltaGrxn1,\"kJ/mol\"\n",
+      "print\"(b) The standard free energy change of reaction is :\",round(deltaGrxn),\"kJ/mol\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The standard free energy change of reaction is : -818.0 kJ/mol\n",
+        "(b) The standard free energy change of reaction is : 1139.0 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:18.5,Page no:820"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#for fusion\n",
+      "T1=5.5+273 #temperature of fusion, K\n",
+      "deltaH1=10.9*1000 #change in enthalpy, J/mol\n",
+      "#for vaporisation\n",
+      "T2=80.1+273 #temperature of vaporisation, K\n",
+      "deltaH2=31*1000 #change in enthalpy, J/mol\n",
+      "\n",
+      "#Calculation\n",
+      "deltaSf=deltaH1/T1 #since in fusion deltaG=0, J/ K mol\n",
+      "deltaSv=deltaH2/T2 #since in vaporisation deltaG=0, J/ K mol\n",
+      "\n",
+      "#Result\n",
+      "print\"The entropy change for fusion is\",round(deltaSf,1),\"J/K.mol\" \n",
+      "print\"The entropy change for condensation is : \",round(deltaSv,1),\"J/K mol\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The entropy change for fusion is 39.1 J/K.mol\n",
+        "The entropy change for condensation is :  87.8 J/K mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:18.6,Page no:823"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "T=298 #temperature, K\n",
+      "R=8.314 #gas constant, J/K mol\n",
+      "GH2=0 #free energy of formation of H2, kJ/mol\n",
+      "GH2O=-237.2 #free energy of formation of H2O, kJ/mol\n",
+      "GO2=0 #free energy of formation of O2, kJ/mol\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "deltaG=1000*(2*GH2+GO2-2*GH2O) #free energy of rxn, J/mol\n",
+      "Kp=math.exp(-deltaG/(R*T)) #equilibrium constant for rxn\n",
+      "\n",
+      "#Result\n",
+      "print\"deltaG_rxn=\",deltaG/1000,\"kJ/mol\"\n",
+      "print\"The equilibrium constant for the given reaction is :%.e\"%Kp"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "deltaG_rxn= 474.4 kJ/mol\n",
+        "The equilibrium constant for the given reaction is :7e-84\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:18.7,Page no:824"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "T=298 #temperature, K\n",
+      "R=8.314 #gas constant, J/K mol\n",
+      "Ksp=1.6*10**-10 #solubility constant\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "deltaG=-R*T*math.log(Ksp) #here solubility product is equal to equilibrium constant\n",
+      "\n",
+      "#Result\n",
+      "print\"The free energy for the given reaction is :\",round(deltaG*10**-3),\"kJ/mol\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The free energy for the given reaction is : 56.0 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:18.8,Page no:825"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "T=298 #temperature, K\n",
+      "R=8.314 #gas constant, J/K mol\n",
+      "deltaG0=5.4*10**3 #standard free energy, kJ/mol\n",
+      "pNO2=0.122 #pressure of NO2, atm\n",
+      "pN2O4=0.453 #pressure of N2O4, atm\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "deltaG=deltaG0+R*T*math.log(pNO2**2/pN2O4) #here solubility product is equal to equilibrium constant\n",
+      "\n",
+      "#Result\n",
+      "if(deltaG<0):#equilibrium determination\n",
+      "    d=\"net reaction proceeds from left to right to reach equilibrium\" \n",
+      "else:\n",
+      "    d=\"net reaction proceeds from right to left to reach equilibrium\" \n",
+      "print\"The free energy for the given reaction is :\",round(deltaG*10**-3,2),\"kJ/mol and\",d\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The free energy for the given reaction is : -3.06 kJ/mol and net reaction proceeds from left to right to reach equilibrium\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_19.ipynb b/Chemistry/Chapter_19.ipynb
new file mode 100755
index 00000000..c904a842
--- /dev/null
+++ b/Chemistry/Chapter_19.ipynb
@@ -0,0 +1,267 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 19:Electrochemistry"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:19.3,Page no:848"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "E0cathode=0.8 #standard electrode potential of cathode(Ag+/Ag), V\n",
+      "E0anode=-2.37 #standard electrode potential of anode(Mg2+/Mg), V\n",
+      "\n",
+      "#Calculation\n",
+      "E0cell=E0cathode-E0anode #standard emf of the cell, V\n",
+      "\n",
+      "#Result\n",
+      "print\"The standard emf of the cell is :\",E0cell,\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The standard emf of the cell is : 3.17 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:19.4,Page no:850"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "n=2.0 \n",
+      "E0cathode=0.15 #standard electrode potential of cathode(Cu2+/Cu+), V\n",
+      "E0anode=-0.14 #standard electrode potential of anode(Sn2+/Sn), V\n",
+      "\n",
+      "#Calculation\n",
+      "E0cell=E0cathode-E0anode #standard emf of the cell, V\n",
+      "import math\n",
+      "K=math.exp(round(n*E0cell/0.0257,1)) #equilibrium constant, from the formula E0cell=0.0257*lnK/n\n",
+      "\n",
+      "#Result\n",
+      "print\"The equilibrium constant for the given reaction is :%.e\"%K "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0.29\n",
+        "The equilibrium constant for the given reaction is :7e+09\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:19.5,Page no:851"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "n=6 \n",
+      "F=96500 #faraday constant, J/V mol\n",
+      "E0cathode=-2.87 #standard electrode potential of cathode(Ca2+/Ca), V\n",
+      "E0anode=1.5 #standard electrode potential of anode(Au3+/Au), V\n",
+      "\n",
+      "#Calculation\n",
+      "E0cell=E0cathode-E0anode #standard emf of the cell, V\n",
+      "deltaG0=-n*F*E0cell #standard free energy change for the reaction, kJ/mol\n",
+      "\n",
+      "#Result\n",
+      "print\"The standard free energy change for the reaction is :%.2e\"%(deltaG0/1000),\"kJ/mol\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The standard free energy change for the reaction is :2.53e+03 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:19.6,Page no:853"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "n=2 \n",
+      "F=96500 #faraday constant, J/V mol\n",
+      "Co2=0.15 #conc of Co2+ ions, M\n",
+      "Fe2=0.68 #conc of Fe2+ ions, M\n",
+      "E0cathode=-0.44 #standard electrode potential of cathode(Fe2+/Fe), V\n",
+      "E0anode=-0.28 #standard electrode potential of anode(Co2+/Co), V\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "E0cell=E0cathode-E0anode #standard emf of the cell, V\n",
+      "Ecell=E0cell-0.0257/2*math.log(Co2/Fe2) #calculation of cell potential at non standard conditions, V\n",
+      "\n",
+      "#Result\n",
+      "print \"E=\",round(Ecell,2),\"V\"\n",
+      "if(Ecell>0):\n",
+      "    print\"The reaction would proceed spontaneously in the direction written\"\n",
+      "else:\n",
+      "    print\"The reaction is not spontaneously in the direction written\"\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "E= -0.14 V\n",
+        "The reaction is not spontaneously in the direction written\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:19.7,Page no:855"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "n=2 \n",
+      "Zn=1 #conc of Zn2+ ions, M\n",
+      "pH2=1 #pressure of H2 gas, atm\n",
+      "Ecell=0.54 #emf of the cell, V\n",
+      "E0cell=0.76 #standard emf of the cell, V\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Q=math.exp(-(Ecell-E0cell)*2/0.0257) #since Ecell=E0cell-0.0257/2*log(Q) where Q=(Zn2+)*pH2/(H+)**2\n",
+      "H=math.sqrt(Zn*pH2/Q) #the conc of H+ ions, M\n",
+      "\n",
+      "#Result\n",
+      "print\"The molar concentration of H+ ion is :%.e\"%H,\"M\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molar concentration of H+ ion is :2e-04 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:19.9,Page no:870"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "t=7.44*3600 #time, sec\n",
+      "A=1.26 #current in ampere\n",
+      "F=96500.0 #faraday constant, J/V mol\n",
+      "R=0.0821 #gas constant, L atm/K\n",
+      "T=273.0 #temperature in Kelvin\n",
+      "P=1.0 #pressure in atm\n",
+      "\n",
+      "#Calculation\n",
+      "q=A*t #charge passed, coulomb\n",
+      "ne=q/F #moles of electrons\n",
+      "nO2=ne/4.0 #moles of oxygen\n",
+      "nH2=ne/2.0 #moles of H2\n",
+      "VO2=nO2*R*T/P #volume of oxygen gas generated\n",
+      "VH2=nH2*R*T/P #volume of H2 gas generated\n",
+      "\n",
+      "#Result\n",
+      "print\"The volume of O2 gas is:\",round(VO2,2),\"L\"\n",
+      "print\"Volume of H2 gas generated is:\",round(VH2,2),\"L\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume of O2 gas is: 1.96 L\n",
+        "Volume of H2 gas generated is: 3.92 L\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_23.ipynb b/Chemistry/Chapter_23.ipynb
new file mode 100755
index 00000000..c36a0d3c
--- /dev/null
+++ b/Chemistry/Chapter_23.ipynb
@@ -0,0 +1,66 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 23:Nuclear Chemistry"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:23.2,Page no:994"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "NA=6.022*10**23 #avogadro number\n",
+      "c=3*10**8 #speed of light, m/s\n",
+      "p=1.007825 #mass of proton, amu\n",
+      "n=1.008665 #mass of neutron, amu\n",
+      "mI=126.9004 #atomic mass of I, amu\n",
+      "\n",
+      "#Calculation\n",
+      "pI=53*p+74*n #estimated mass of I, amu\n",
+      "deltam=mI-pI #mass defect, amu\n",
+      "deltaE=-deltam*c**2 #energy released, amu m**2/s**2\n",
+      "deltaE=deltaE/(NA*1000) #energy released in J\n",
+      "print\"Nuclear binding energy is %.2e\"%deltaE,\"J\"\n",
+      "deltaE=deltaE/127 #binding energy per nucleon, J\n",
+      "\n",
+      "#Result\n",
+      "print\"The nuclear binding energy per nucleon is :%.2e\"%deltaE,\"J/nucleon\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Nuclear binding energy is 1.73e-10 J\n",
+        "The nuclear binding energy per nucleon is :1.36e-12 J/nucleon\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_3.ipynb b/Chemistry/Chapter_3.ipynb
new file mode 100755
index 00000000..96753167
--- /dev/null
+++ b/Chemistry/Chapter_3.ipynb
@@ -0,0 +1,676 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3:Mass Relationships in Chemical Reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.1,Page no:81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Cu63=69.09 #percent of Cu 62.93 amu\n",
+      "Cu65=30.91 #percent of Cu 64.9278 amu\n",
+      "\n",
+      "#Calculation\n",
+      "AverageAMU=62.93*Cu63/100+64.9278*Cu65/100 # average amu\n",
+      "\n",
+      "#Result\n",
+      "print\"The average atomic mass of Copper is :\",round(AverageAMU,2),\"amu\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average atomic mass of Copper is : 63.55 amu\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.2,Page no:84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "mass=6.46 #mass of He, g\n",
+      "\n",
+      "#Calculation\n",
+      "moles=mass/4.003 #no. of moles of He, as mol. mass of He is 4.003 amu\n",
+      "\n",
+      "#Result\n",
+      "print\"The no. of moles of He is :\",round(moles,2),\"mol He\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The no. of moles of He is : 1.61 mol He\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.3,Page no:84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "moles=0.356 #moles of Zn\n",
+      "\n",
+      "#Calculation\n",
+      "mass=moles*65.39 #mass of Zn, g, 1 mole=65.39 g\n",
+      "\n",
+      "#Result\n",
+      "print\"The mass of Zn is :\",round(mass,1),\"g Zn\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of Zn is : 23.3 g Zn\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.4,Page no:84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Na=6.022*10**23 # Avogadro number, atoms/mol\n",
+      "mass=16.3 #mass of sulfur, g\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "moles=mass/32.07 #moles of S\n",
+      "atoms=moles*Na #number of atoms of S\n",
+      "\n",
+      "#Result\n",
+      "print\"The no. of atoms of S is :%.2e\"%atoms,\"S atoms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The no. of atoms of S is :3.06e+23 S atoms\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.5,Page no:86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "MassO=16.0 #mass of O, amu\n",
+      "MassS=32.07 #mass of S, amu\n",
+      "MassN=14.01 #mass of N, amu\n",
+      "MassH=1.008 #mass of H, amu\n",
+      "MassC=12.01 #mass of C, amu\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "MassSO2=MassS+MassO*2 #mass of SO2, amu\n",
+      "#(b)\n",
+      "MassC8H10N4O2=8*MassC+10*MassH+4*MassN+2*MassO \n",
+      "\n",
+      "#Result\n",
+      "print\"(a).The molecular mass of SO2 is :\",MassSO2,\"amu\\n\"\n",
+      "print\"(b).The molecular mass of C8H10N4O2 is :\",MassC8H10N4O2,\"amu\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a).The molecular mass of SO2 is : 64.07 amu\n",
+        "\n",
+        "(b).The molecular mass of C8H10N4O2 is : 194.2 amu\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.6,Page no:86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Mass=6.07 #mass of CH4, g\n",
+      "MassC=12.01 #mol. mass of C, amu\n",
+      "MassH=1.008 #mol. mass of H, amu\n",
+      "\n",
+      "#Calculation\n",
+      "MassCH4=MassC+4*MassH #mol. mass of CH4, amu\n",
+      "Moles=Mass/MassCH4 #no. of moles of CH4\n",
+      "\n",
+      "#Result\n",
+      "print\"The no. of moles of CH4 is :\",round(Moles,3),\"mol CH4\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The no. of moles of CH4 is : 0.378 mol CH4\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.7,Page no:87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Na=6.07*10**23 # Avogadro number, atoms/mol\n",
+      "Mass=25.6 #mass of Urea, g\n",
+      "MolMass=60.06 #mol. mass of Urea, g\n",
+      "\n",
+      "#Calculation\n",
+      "moles=Mass/MolMass #moles of Urea, mol\n",
+      "Atoms=moles*Na*4 #No. of atoms of Hydrogen\n",
+      "\n",
+      "#Result\n",
+      "print\"The no. of atoms of hydrogen are :%.2e\"%Atoms,\"H atoms\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The no. of atoms of hydrogen are :1.03e+24 H atoms\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.8,Page no:89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "H=1.008 #molar mass of H, g\n",
+      "P=30.97 #molar mass of P, g\n",
+      "O=16 #molar mass of O, g\n",
+      "MolMass=97.99 #mol. mass of H3PO4, g\n",
+      "\n",
+      "#Calculation\n",
+      "percentH=3*H/MolMass*100 #percent of H\n",
+      "percentP=P/MolMass*100 #percent of P\n",
+      "percentO=4*O/MolMass*100 #percent of O\n",
+      "\n",
+      "#Result\n",
+      "print\"The percent by mass of Hydrogen is :\",round(percentH,3),\"%\\n\"\n",
+      "print\"The percent by mass of Phosphorus is :\",round(percentP,2),\"%\\n\"\n",
+      "print\"The percent by mass of Oxygen is :\",round(percentO,2),\"%\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percent by mass of Hydrogen is : 3.086 %\n",
+        "\n",
+        "The percent by mass of Phosphorus is : 31.61 %\n",
+        "\n",
+        "The percent by mass of Oxygen is : 65.31 %\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.9,Page no:90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "H=1.008 #molar mass of H, g\n",
+      "C=12.01 #molar mass of C, g\n",
+      "O=16.0 #molar mass of O, g\n",
+      "percentC=40.92 #percent of C\n",
+      "\n",
+      "#Calculation\n",
+      "nC=percentC/C \n",
+      "percentH=4.58 #percent of H\n",
+      "nH=percentH/H \n",
+      "percentO=54.5 #percent of O\n",
+      "nO=percentO/O \n",
+      "if(nC>nH):# determining the smallest subscript\n",
+      "    small=nH\n",
+      "else:\n",
+      "    small=nC \n",
+      "    if(small>nO):\n",
+      "        small=nO \n",
+      "\n",
+      "nC=nC/small #dividing by the smallest subscript\n",
+      "nH=nH/small \n",
+      "nO=nO/small \n",
+      "#the approximate values of these variables are to be multiplied by appropriate number to make it an integer by trial and error method\n",
+      "#in this case we need to multiply with 3 to get integer values\n",
+      "nC=nC*3 \n",
+      "nH=nH*3 \n",
+      "nO=nO*3 \n",
+      "\n",
+      "#Result\n",
+      "print\"The empirical formula of ascorbic acid is : C%.f\"%nC,\"H%.f\"%nH,\"O%.f\"%nO"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The empirical formula of ascorbic acid is : C3 H4 O3\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.10,Page no:91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "massCuFeS2=3.71*10**3 #given mass of CuFeS2, kg\n",
+      "CuFeS2=183.5 #mol. mass of CuFeS2, g\n",
+      "Cu=63.55 #mol. mass of Cu, g\n",
+      "\n",
+      "#Calculation\n",
+      "percentCu=Cu/CuFeS2*100 #percent Cu in CuFeS2\n",
+      "massCu=percentCu*massCuFeS2/100#mass of Cu in given CuFeS2, kg\n",
+      "\n",
+      "#Result\n",
+      "print\"The mass of Cu in CuFeS2 is :%.2e\"%massCu,\"kg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of Cu in CuFeS2 is :1.28e+03 kg\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.11,Page no:93"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "M_N=1.52      #Mass of nitrogen in g\n",
+      "M_O=3.47      #Mass of oxygen in g\n",
+      "N=14.01       #Atomic mass of N\n",
+      "Molar1=95     #Molar mass in g \n",
+      "O=16          #Atomic mass of O\n",
+      "\n",
+      "#Calculation\n",
+      "n_N=M_N/N       #No of moles of N\n",
+      "n_O=M_O/O       #No of moles of O\n",
+      "emp=N+2*(O)     #Empirical molar massof NO2\n",
+      "ratio=Molar1/emp\n",
+      "ratio=round(ratio)\n",
+      "actual=ratio*emp\n",
+      "\n",
+      "#Result\n",
+      "print\"Molecular formula is N%.3g\"%n_N,\"O%.3g\"%n_O\n",
+      "print \"Actual molar mass is\",actual,\"g,which is between 90 g and 95 g\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Molecular formula is N0.108 O0.217\n",
+        "Actual molar mass is 92.02 g,which is between 90 g and 95 g\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.13,Page no:101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "CO2=44.01 #mol. mass of CO2, g\n",
+      "Glucose=180.2 #mol. mass of Glucose, g\n",
+      "massGlucose=856 #given mass of Glucose, g\n",
+      "\n",
+      "#Calculation\n",
+      "moleGlucose=massGlucose/Glucose # moles of glucose\n",
+      "moleCO2=moleGlucose*6 #1 mole glucose gives 6 moles of CO2\n",
+      "massCO2=moleCO2*CO2 # mass of CO2, g\n",
+      "\n",
+      "#Result\n",
+      "print\"The mass of CO2 is :%.2e\"%massCO2,\"g CO2\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of CO2 is :1.25e+03 g CO2\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.14,Page no:102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "H2=2.016 #mol. mass of H2, g\n",
+      "Li=6.941 #mol. mass of Li, g\n",
+      "mH2=9.89 #mass of H2, g\n",
+      "\n",
+      "#Calculation\n",
+      "nH2=mH2/H2 #moles of H2\n",
+      "nLi=2*nH2 #moles of Li, 1mol H2 given by 2mol Li\n",
+      "mLi=Li*nLi ##mass of Li, g\n",
+      "\n",
+      "#Result\n",
+      "print\"The mass of Li is :\",round(mLi,1),\"g Li\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of Li is : 68.1 g Li\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.15,Page no:104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Urea=60.06 #mol. mass of Urea, g\n",
+      "NH3=17.03 #mol. mass of NH3, g\n",
+      "CO2=44.01 #mol. mass of CO2, g\n",
+      "\n",
+      "#(a)\n",
+      "#for NH3\n",
+      "mNH3=637.2 #mass of NH3, g\n",
+      "\n",
+      "#Calculation\n",
+      "nNH3=mNH3/NH3 #moles of NH3\n",
+      "nUrea1=nNH3/2#moles of Urea\n",
+      "#for CO2\n",
+      "mCO2=1142#mol. mass of CO2, g\n",
+      "nCO2=mCO2/CO2 #moles of CO2\n",
+      "nUrea2=nCO2 #moles of Urea\n",
+      "if(nUrea1>nUrea2): #finding limiting reagent\n",
+      "    nUrea=nUrea2 \n",
+      "    limiting=\"CO2\" \n",
+      "else:\n",
+      "    limiting=\"NH3\" \n",
+      "    nUrea=nUrea1 \n",
+      "    \n",
+      "#Result\n",
+      "print\"(a).The limiting reagent is :\",limiting\n",
+      "#(b)\n",
+      "mUrea=nUrea*Urea #mass of urea produced\n",
+      "print\"(b).The mass of the Urea produced is :\",round(mUrea),\"g (NH2)2CO\"\n",
+      "\n",
+      "#(c)\n",
+      "if(limiting==\"NH3\") :#finding excess reagent\n",
+      "    nCO2excess=nCO2-nNH3/2 \n",
+      "    mCO2excess=nCO2excess*CO2 \n",
+      "    print\"(c).The mass of excess CO2 is :\",round(mCO2excess),\"g \\n\"\n",
+      "else: \n",
+      "    nNH3excess=nNH3-2*nCO2 \n",
+      "    mNH3excess=nNH3excess*NH3 \n",
+      "    print\"The mass of excess NH3 is :\",mNH3excess,\"g\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a).The limiting reagent is : NH3\n",
+        "(b).The mass of the Urea produced is : 1124.0 g (NH2)2CO\n",
+        "(c).The mass of excess CO2 is : 319.0 g \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.16,Page no:106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)\n",
+      "#for TiCl4\n",
+      "\n",
+      "#Variable declaration\n",
+      "mTiCl4=3.54*10**7 #mass of TiCl4, g\n",
+      "nTiCl4=mTiCl4/189.7 #moles of TiCl4\n",
+      "nTi1=nTiCl4*1.0 #moles of Ti\n",
+      "\n",
+      "#for Mg\n",
+      "mMg=1.13*10**7 #mass of Mg, g\n",
+      "nMg=mMg/24.31 #moles of Mg\n",
+      "nTi2=nMg/2.0 #moles of Ti\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "if(nTi1>nTi2):   #finding imiting reagent\n",
+      "     nTi=nTi2 \n",
+      "else:\n",
+      "    nTi=nTi1 \n",
+      "    mTi=nTi*47.88 \n",
+      "    print\"(a).The theoretical yield is :%.2e\"%mTi,\"g\\n\"\n",
+      "    print\"\\nNOTE:Variation from answer due to approx value of mTi taken in book\\n\"\n",
+      "\n",
+      "#(b)\n",
+      "\n",
+      "mTiactual=7.91*10**6 #given, actual Ti produced\n",
+      "p_yield=mTiactual/mTi*100.0 \n",
+      "print\"(b).The percent yield is :\",round(p_yield,1),\"%\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a).The theoretical yield is :8.93e+06 g\n",
+        "\n",
+        "\n",
+        "NOTE:Variation from answer due to approx value of mTi taken in book\n",
+        "\n",
+        "(b).The percent yield is : 88.5 %\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_4.ipynb b/Chemistry/Chapter_4.ipynb
new file mode 100755
index 00000000..13f747e8
--- /dev/null
+++ b/Chemistry/Chapter_4.ipynb
@@ -0,0 +1,290 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4:Reactions in Aqueous Solutions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.6,Page no:148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "K2Cr2O7=294.2 #mol mass of K2Cr2O7, g\n",
+      "M=2.16 #Concentration of K2Cr2O7, M\n",
+      "V=0.250 #volume of K2Cr2O7, L\n",
+      "\n",
+      "#Calculation\n",
+      "moles=M*V #moles of K2Cr2O7\n",
+      "mass=moles*K2Cr2O7 \n",
+      "\n",
+      "#Result\n",
+      "print\"The mass of the K2Cr2O7 needed is :\",round(mass),\"g K2Cr2O7\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of the K2Cr2O7 needed is : 159.0 g K2Cr2O7\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.7,Page no:149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "mGlucose=3.81 #mass of Glucose, g\n",
+      "Glucose=180.2 #mol mass of Glucose, g\n",
+      "M=2.53 #Concentration of Glucose, M\n",
+      "\n",
+      "#Calculation\n",
+      "moles=mGlucose/Glucose #moles of Glucose\n",
+      "V=moles/M #volume of Glucose, L\n",
+      "\n",
+      "#Result\n",
+      "print\"The volume of the Glucose needed is :\",round(V*1000,2),\"mL soln\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume of the Glucose needed is : 8.36 mL soln\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.8,Page no:150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "M2=1.75 #final Concentration of H2SO4, M\n",
+      "V2=500 #final volume of H2SO4, mL\n",
+      "M1=8.61 #initial Concentration of H2SO4, M\n",
+      "\n",
+      "#Calculation\n",
+      "V1=M2*V2/M1 #initail volume of H2SO4, mL\n",
+      "\n",
+      "#Result\n",
+      "print\"The volume of the H2SO4 needed to dilute the solution is :\",round(V1),\"mL\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume of the H2SO4 needed to dilute the solution is : 102.0 mL\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.9,Page no:152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "mSample=0.5662 #mass of sample, g\n",
+      "Cl=35.5 #mol mass of Cl, g\n",
+      "AgCl=143.4 #mol mass of AgCl, g\n",
+      "mAgCl=1.0882 #mass of AgCl formed, g\n",
+      "\n",
+      "#Calculation\n",
+      "p_Cl_AgCl=Cl/AgCl*100.0 #percent Cl in AgCl\n",
+      "mCl=p_Cl_AgCl*mAgCl/100.0 #mass of Cl in AgCl, g\n",
+      "mCl=round(mCl,3)\n",
+      "#the same amount of Cl is present in initial sample\n",
+      "p_Cl=mCl/mSample*100.0 #percent Cl in initial sample\n",
+      "\n",
+      "#Result\n",
+      "print\"The percentage of Cl in sample is :\",round(p_Cl,2),\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage of Cl in sample is : 47.51 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.10,Page no:154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "mKHP=0.5468 #mass of KHP, g\n",
+      "KHP=204.2 #mol mass of KHP, g\n",
+      "\n",
+      "#Calculation\n",
+      "nKHP=mKHP/KHP #moles of KHP\n",
+      "VNaOH=23.48 #volume of NaOH, mL\n",
+      "MNaOH=nKHP/VNaOH*1000 #molarity of NaOH sol, M\n",
+      "\n",
+      "#Result\n",
+      "print\"The molarity of NaOH solution is :\",round(MNaOH,4),\"M\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molarity of NaOH solution is : 0.114 M\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.11,Page no:155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "MNaOH=0.610 #molarity of NaOH, M\n",
+      "VH2SO4=20 #volume of H2SO4, mL\n",
+      "MH2SO4=0.245 #molarity of H2SO4, M\n",
+      "\n",
+      "#Calculation\n",
+      "nH2SO4=MH2SO4*VH2SO4/1000 #moles of H2SO4\n",
+      "VNaOH=2*nH2SO4/MNaOH #Volume of NaOH, L\n",
+      "\n",
+      "#Result\n",
+      "print\"The volume of NaOH solution is :\",round(VNaOH*1000,1),\"mL\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume of NaOH solution is : 16.1 mL\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.12,Page no:157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "MKMnO4=0.1327 #molarity of KMnO4, M\n",
+      "VKMnO4=16.42 #volume of KMnO4, mL\n",
+      "\n",
+      "#Calculation\n",
+      "nKMnO4=MKMnO4*VKMnO4/1000 \n",
+      "nFeSO4=5*nKMnO4 \n",
+      "VFeSO4=25 #volume of FeSO4, mL\n",
+      "MFeSO4=nFeSO4/VFeSO4*1000 \n",
+      "\n",
+      "#Result\n",
+      "print\"The molarity of FeSO4 solution is :\",round(MFeSO4,3),\"M\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molarity of FeSO4 solution is : 0.436 M\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_5.ipynb b/Chemistry/Chapter_5.ipynb
new file mode 100755
index 00000000..9649fe13
--- /dev/null
+++ b/Chemistry/Chapter_5.ipynb
@@ -0,0 +1,768 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5:Gases"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.1,Page no:177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Pbaro=688 #pressure in mm Hg\n",
+      "\n",
+      "#Calculation\n",
+      "Patm=Pbaro/760.0 #pressure in atm\n",
+      "\n",
+      "#Result\n",
+      "print\"The presuure in atmospheres is :\",round(Patm,3),\"atm\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The presuure in atmospheres is : 0.905 atm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.2,Page no:178"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Pbaro=732 #pressure in mm Hg\n",
+      "\n",
+      "#Calculation\n",
+      "Patm=Pbaro/760.0 #pressure in atm\n",
+      "P=Patm*1.01325*10**2 #pressure in kilo Pascal\n",
+      "\n",
+      "#Result\n",
+      "print\"The presuure in kilo pascals is :\",round(P,1),\"kPa\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The presuure in kilo pascals is : 97.6 kPa\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.3,Page no:186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "V=5.43 #volume, L\n",
+      "t=69.5 #temperature, C\n",
+      "\n",
+      "#Calculation\n",
+      "T=t+273 #temperature, K\n",
+      "n=1.82 #moles\n",
+      "R=0.0821 #universal gas constant, L.atm/(K.mol)\n",
+      "P=n*R*T/V #pressure, atm\n",
+      "\n",
+      "#Result\n",
+      "print\"The presuure in atmospheres is :\",round(P,2),\"atm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The presuure in atmospheres is : 9.42 atm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.4,Page no:187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "m=7.4 #mass of NH3, g\n",
+      "#at STP for NH3 for 1mole of NH3\n",
+      "V1=22.41 # volume, L\n",
+      "NH3=17.03 #molar mass of NH3, g\n",
+      "\n",
+      "#Calculation\n",
+      "n=m/NH3 #moles of NH3\n",
+      "V=n*V1 #volume, L\n",
+      "\n",
+      "#Result\n",
+      "print\"The volume of NH3 under given conditions is :\",round(V,2),\"L\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume of NH3 under given conditions is : 9.74 L\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.5,Page no:188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "V1=0.55 #volume, L\n",
+      "P1=1 #pressure at sea level, atm\n",
+      "P2=0.4 #pressurea at 6.5km height, atm\n",
+      "\n",
+      "#n1=n2 and T1=T2 given hence P1V1=P2V2\n",
+      "#Calculation\n",
+      "V2=P1*V1/P2 \n",
+      "\n",
+      "#Result\n",
+      "print\"The volume of He balloon at height 6.5km above sea level is :\",round(V2,1),\"L\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume of He balloon at height 6.5km above sea level is : 1.4 L\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.6,Page no:189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "P1=1.2 # pressure initial, atm\n",
+      "T1=18+273 #temperature initial, K\n",
+      "T2=85+273 #temperature final, K\n",
+      "#volume is constant\n",
+      "\n",
+      "#Calculation\n",
+      "P2=P1*T2/T1 # pressure final,atm\n",
+      "\n",
+      "#Result\n",
+      "print\"The final pressure is :\",round(P2,2),\"atm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final pressure is : 1.48 atm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.7,Page no:189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "P1=6.4 # pressure initial, atm\n",
+      "P2=1.0 # pressure final, atm\n",
+      "T1=8+273 #temperature initial, K\n",
+      "T2=25+273 #temperature final, K\n",
+      "V1=2.1 #volume initial, mL\n",
+      "\n",
+      "#Calculation\n",
+      "V2=P1*V1*T2/(T1*P2) # volume final, mL\n",
+      "\n",
+      "#Result\n",
+      "print\"The final volume is :\",round(V2),\"mL\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final volume is : 14.0 mL\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.8,Page no:191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "\n",
+      "#taking 1 mole of CO2\n",
+      "n=1 \n",
+      "R=0.0821 #universal gas constant, L. atm/K.mol\n",
+      "t=55 #temperature, C\n",
+      "T=t+273 #temperature, K\n",
+      "P=0.99 #.pressure, atm\n",
+      "M=44.01 #molar mass of CO2, g\n",
+      "\n",
+      "#Calculation\n",
+      "d1=P*M/(R*T) #density of CO2, g/L\n",
+      "#altenate method\n",
+      "#taking 1 mole of CO2\n",
+      "mass=M #mass of CO2 in g =mol mass since we are considering 1 mole of CO2\n",
+      "V=n*R*T/P #volume, L\n",
+      "d2=mass/V #density=mass/volume, g/L\n",
+      "\n",
+      "#Result\n",
+      "print\"The density of CO2 is :\",round(d1,2),\"g/L\\n\"\n",
+      "print\"(Alternate Method)the density of CO2 is :\",round(d2,2),\"g/L\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density of CO2 is : 1.62 g/L\n",
+        "\n",
+        "(Alternate Method)the density of CO2 is : 1.62 g/L\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.9,Page no:192"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=7.71 # density, g/mL(given)\n",
+      "R=0.0821 #universal gas constant, L. atm/K.mol\n",
+      "T=36+273 # temp, K\n",
+      "P=2.88 #pressure, atm\n",
+      "\n",
+      "#Calculation\n",
+      "M1=d*R*T/P # mol. mass, g/mol\n",
+      "#alternate method\n",
+      "#considering 1 L of compound\n",
+      "V=1 #volume, L\n",
+      "n=P*V/(R*T) #no of moles\n",
+      "m=7.71 #mass per 1 L, g\n",
+      "M2=m/n # mol. mass, g/mol\n",
+      "\n",
+      "#Result\n",
+      "print\"The molecular mass of given compound is :\",round(M1,1),\"g/mol\\n\"\n",
+      "print\"{alternate method} the molecular mass of given compound is :\",round(M2,1),\" g/mol\\n\" \n",
+      "print\"The molecular formula can be only found by trial and error method as given in the book \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molecular mass of given compound is : 67.9 g/mol\n",
+        "\n",
+        "{alternate method} the molecular mass of given compound is : 67.9  g/mol\n",
+        "\n",
+        "The molecular formula can be only found by trial and error method as given in the book \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.10,Page no:193"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "percentSi=33 #percent of Si in compound\n",
+      "percentF=67 #percent of F in compound\n",
+      "P=1.7 #pressure, atm\n",
+      "T=35+273 #temp. in K\n",
+      "m=2.38 #mass, g\n",
+      "V=0.21 #volume, L\n",
+      "R=0.0821 #universal Gas constant, L.atm/K.mol\n",
+      "m_sif3=85.09\n",
+      "\n",
+      "#Calculation\n",
+      "nSi=percentSi/28.01 #moles of Si in 100g compound\n",
+      "nF=percentF/19 #moles of F in 100g compound\n",
+      "n=P*V/(R*T) #moles\n",
+      "M=m/n #mol. mass=mass/moles, g/mol\n",
+      "e=M/m_sif3\n",
+      "\n",
+      "#Result\n",
+      "print\"The molecular mass of given compound is :\",round(M),\"g/mol\\n\"\n",
+      "print\"Therefore,molecular formula is (SiF3)%.f\"%round(e),\"OR Si2F6\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molecular mass of given compound is : 169.0 g/mol\n",
+        "\n",
+        "Therefore,molecular formula is (SiF3)2 OR Si2F6\n"
+       ]
+      }
+     ],
+     "prompt_number": 57
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.11,Page no:194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "VC2H2=7.64 #volume of acetylene, L\n",
+      "\n",
+      "#Calculation\n",
+      "VO2=VC2H2*5/2.0 #volume of O2 required for complete combustion as 5mol O2 react with 2mol acetylene for complete combustion\n",
+      "\n",
+      "#Result\n",
+      "print\"The volume of O2 required for complete combustion of acetylene is :\",VO2,\"L\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume of O2 required for complete combustion of acetylene is : 19.1 L\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.12,Page no:195"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "R=0.0821 #universal Gas constant, L.atm/K.mol\n",
+      "T=80+273 #temp in K\n",
+      "P=823/760.0 #pressure in atm\n",
+      "m=60.0 #mass of NaN3, g\n",
+      "NaN3=65.02 #mol. mass of NaN3, g\n",
+      "\n",
+      "#Calculation\n",
+      "nN2=m*3.0/(2.0*NaN3) #moles of N2\n",
+      "nN2=round(nN2,2)\n",
+      "VN2=nN2*R*T/P #from ideal gas law\n",
+      "#Result\n",
+      "print\"\\t the volume of N2 generated is :\",round(VN2,1),\"L\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t the volume of N2 generated is : 36.9 L\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 61
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.13,Page no:196"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "R=0.0821 #universal Gas constant, L.atm/K.mol\n",
+      "T=312 #temp in K\n",
+      "V=2.4*10**5 #volume, L\n",
+      "P1=7.9*10**-3 #pressure initial in atm\n",
+      "P2=1.2*10**-4 #pressure final in atm\n",
+      "\n",
+      "#Calculation\n",
+      "Pdrop=P1-P2 #pressure drop, atm\n",
+      "n=Pdrop*V/(R*T) #moles of Co2 reacted\n",
+      "Li2CO3=73.89 #mol. mass of Li2CO3, g\n",
+      "mLi2CO3=n*Li2CO3 #mass of Li2CO3, g\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print\"The mass of Li2CO3 formed is :%.1e\"%mLi2CO3,\"g Li2CO3\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of Li2CO3 formed is :5.4e+03 g Li2CO3\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 64
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.14,Page no:199"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "nNe=4.46 #moles of Ne\n",
+      "nXe=2.15 #moles of Xe\n",
+      "nAr=0.74 #moles of Ar\n",
+      "PT=2 #total pressure in atm\n",
+      "\n",
+      "#Calculation\n",
+      "XNe=nNe/(nNe+nAr+nXe) #mole fraction of Ne\n",
+      "XAr=nAr/(nNe+nAr+nXe) #mole fraction of Ar\n",
+      "XXe=nXe/(nNe+nAr+nXe) #mole fraction of Xe\n",
+      "PNe=XNe*PT #partial pressure of Ne\n",
+      "PAr=XAr*PT #partial pressure of Ar\n",
+      "PXe=XXe*PT #partial pressure of Xe\n",
+      "\n",
+      "#Result\n",
+      "print\"The partial pressures of Ne, Ar and Xe are:\"\n",
+      "print\"PNe=\",round(PNe,2),\"atm\\n\",\"PAr=\",round(PAr,2),\"atm\\nPXe=\",round(PXe,3),\"atm \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The partial pressures of Ne, Ar and Xe are:\n",
+        "PNe= 1.21 atm\n",
+        "PAr= 0.2 atm\n",
+        "PXe= 0.585 atm \n"
+       ]
+      }
+     ],
+     "prompt_number": 70
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.15,Page no:200"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "PT=762 #pressure total, mmHg\n",
+      "PH2O=22.4 #pressure of water vapor, mmHg\n",
+      "PO2=PT-PH2O #pressure of O2, frm Dalton's law, mmHg\n",
+      "M=32 #mol mass of O2, g\n",
+      "R=0.0821 #universal Gas constant, L.atm/K.mol\n",
+      "T=24+273 #temp in K\n",
+      "V=0.128 #volume in L\n",
+      "\n",
+      "#Calculation\n",
+      "m=(PO2/760)*V*M/(R*T) #mass of mass of O2 collected, g\n",
+      "\n",
+      "#Result\n",
+      "print\"The mass of O2 collected is :\",round(m,3),\"g\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of O2 collected is : 0.163 g\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 71
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.16,Page no:207"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "R=8.314 #universal Gas constant, J/K mol\n",
+      "T=25+273 #temp in K\n",
+      "import math\n",
+      "#Calculation\n",
+      "#for O2\n",
+      "M=4.003*10**-3 #mol mass in kg\n",
+      "Urms=math.sqrt(3*R*T/M) #rms velocity, m/s\n",
+      "print\"\\t the rms velocity of O2 collected is :%.2e\"%Urms,\"m/s\\n\"\n",
+      "#for N2\n",
+      "M=28.02*10**-3 #mol mass in kg\n",
+      "Urms=math.sqrt(3*R*T/M) #rms velocity, m/s\n",
+      "\n",
+      "#Result\n",
+      "print\"\\t the rms velocity of N2 collected is :\",round(Urms),\" m/s\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t the rms velocity of O2 collected is :1.36e+03 m/s\n",
+        "\n",
+        "\t the rms velocity of N2 collected is : 515.0  m/s\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.17,Page no:209"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "t2=1.5 #diffusion time of compound, min\n",
+      "t1=4.73 #diffusion time of Br, min\n",
+      "M2=159.8 #mol mass of Br gas, g\n",
+      "\n",
+      "#Calculation\n",
+      "M=(t2/t1)**2*M2 #molar gas of unknown gas, g(from Graham's Law of Diffusion)\n",
+      "\n",
+      "#Result\n",
+      "print\"\\t the molar mass of unknown gas is :\",round(M,1),\"g/mol\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t the molar mass of unknown gas is : 16.1 g/mol\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 76
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.18,Page no:213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#(a)\n",
+      "V=5.2 #volume, L\n",
+      "T=47+273 \n",
+      "n=3.5 \n",
+      "R=0.0821 #universal Gas constant, L.atm/K.mol\n",
+      "\n",
+      "#Calculation\n",
+      "P=n*R*T/V \n",
+      "print\"The pressure of NH3 gas from ideal gas equation is :\",round(P,1),\"atm\\n\"\n",
+      "#(b)\n",
+      "a=4.17 #constant, atm.L2/mol2\n",
+      "b=0.0371 #constant, L/mol\n",
+      "Pc=a*n**2/V**2 #pressure correction term, atm\n",
+      "Vc=n*b #volume correction term, L\n",
+      "P=n*R*T/(V-Vc)-Pc #from van der waals equation, pressure, atm\n",
+      "\n",
+      "#Result\n",
+      "print\"The pressure of NH3 gas from van der waals equation is :\",round(P,1),\"atm\\n\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pressure of NH3 gas from ideal gas equation is : 17.7 atm\n",
+        "\n",
+        "The pressure of NH3 gas from van der waals equation is : 16.2 atm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 80
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_6.ipynb b/Chemistry/Chapter_6.ipynb
new file mode 100755
index 00000000..c09bdeb7
--- /dev/null
+++ b/Chemistry/Chapter_6.ipynb
@@ -0,0 +1,422 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6:Thermochemistry"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.1,Page no:237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "P1=0 #external pressure, atm\n",
+      "Vf1=6 #final volume, L\n",
+      "Vi1=2 #initial volume, L\n",
+      "P2=1.2 #external pressure, atm\n",
+      "Vf2=6 #final volume, L\n",
+      "Vi2=2 #initial volume, L\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "W1=-P1*(Vf1-Vi1) #work in atm.L\n",
+      "#(b)\n",
+      "W2=-P2*(Vf2-Vi2) #work in atm.L\n",
+      "W2=W2*101.3 #work in J\n",
+      "\n",
+      "#Result\n",
+      "print\"(a).The work done in expansion against vacuum is :\",W1,\"J\\n\"\n",
+      "print\"(b).The work done in expansion against 1.2 atm pressure is :%.1e\"%W2,\"J\\n\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a).The work done in expansion against vacuum is : 0 J\n",
+        "\n",
+        "(b).The work done in expansion against 1.2 atm pressure is :-4.9e+02 J\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.2,Page no:238"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "q=-128 #heat transfer from the gas, J\n",
+      "w=462 #work done in compressing the gas, J\n",
+      "\n",
+      "#Calculation\n",
+      "deltaE=q+w #change in energy of the gas, J\n",
+      "\n",
+      "#Result\n",
+      "print\"The change in energy for the process is :\",deltaE,\"J\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The change in energy for the process is : 334 J\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.3,Page no:243"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "mSO2=87.9 #mass in g\n",
+      "SO2=64.07 #molar mass in g\n",
+      "\n",
+      "#Calculation\n",
+      "nSO2=mSO2/SO2 #moles of SO2\n",
+      "deltaH=-198.2 #heat produced for 2 mol, in kJ/mol\n",
+      "deltaH=deltaH/2  #for one mole SO2,in kJ/mol\n",
+      "Hprod=deltaH*nSO2 #heat produced in this case, in kJ/mol\n",
+      "\n",
+      "#Result\n",
+      "print\"The heat produced in a reaction is :\",round(Hprod),\"kJ\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat produced in a reaction is : -136.0 kJ\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.4,Page no:245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "R=8.314 #gas constant, J/K. mol\n",
+      "T=298 #temp in K\n",
+      "deltaH=-566 #enthalpy change, kJ/mol\n",
+      "deltan=2-3 #change in gas moles\n",
+      "\n",
+      "#Calculation\n",
+      "deltaE=deltaH-R*T*deltan/1000.0 #change in internal energy, kJ/mol\n",
+      "\n",
+      "#Result\n",
+      "print\"The change in internal energy in the reaction is :\",round(deltaE,1),\"kJ/mol\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The change in internal energy in the reaction is : -563.5 kJ/mol\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.5,Page no:246"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "m=466 #mass in g\n",
+      "s=4.184 #specific heat in J/g C\n",
+      "\n",
+      "#Calculation\n",
+      "deltaT=74.6-8.5 #change in temp, C/K\n",
+      "q=m*s*deltaT/1000 #amount of heat absorbed, kJ\n",
+      "\n",
+      "#Result\n",
+      "print\"\\t the amount of heat absorbed is :\",round(q),\"kJ\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\t the amount of heat absorbed is : 129.0 kJ\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.6,Page no:248"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Ccal=10.17 #heat capacity, kJ/C\n",
+      "deltaT=25.95-20.28 #change in temp, C\n",
+      "m=1.435 #mass of naphthalene, g\n",
+      "molm=128.2 #mol mass of naphthalene, g\n",
+      "\n",
+      "#Calculation\n",
+      "qcal=Ccal*deltaT \n",
+      "q=-qcal*molm/m #molar heat of combustion of naphthalene, kJ\n",
+      "\n",
+      "#Result\n",
+      "print\"The molar heat of combustion of naphthalene is :%.3g\"%q,\"kJ/mol\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The molar heat of combustion of naphthalene is :-5.15e+03 kJ/mol\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.7,Page no:249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#for water\n",
+      "m=100 #mass, g\n",
+      "s=4.184 #specific heat, J/g C\n",
+      "deltaT=23.17-22.5 #change in temp., C\n",
+      "qH2O=m*s*deltaT #heat gained by water, J\n",
+      "\n",
+      "#for lead\n",
+      "qPb=-qH2O #heat lost by lead, J\n",
+      "m=26.47 #mass, g\n",
+      "\n",
+      "#Calculation\n",
+      "deltaT=23.17-89.98 #change in temp., C\n",
+      "s=qPb/(m*deltaT) #specific heat, J/g C\n",
+      "\n",
+      "#Result\n",
+      "print\"The specific heat of lead is :\",round(s,3),\"J/g C\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The specific heat of lead is : 0.159 J/g C\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.8,Page no:249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#for water\n",
+      "m=100+100 #mass, g\n",
+      "s=4.184 #specific heat, J/g C\n",
+      "deltaT=25.86-22.5 #change in temp., C\n",
+      "\n",
+      "#Calculation\n",
+      "qsoln=m*s*deltaT/1000 #heat gained by water, kJ\n",
+      "qrxn=-qsoln \n",
+      "Hneut=qrxn/(0.5*0.1) \n",
+      "\n",
+      "#Result\n",
+      "print\"The heat of neutralization is :\",round(Hneut,1),\"kJ/mol\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat of neutralization is : -56.2 kJ/mol\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.9,Page no:256"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "deltaH1=-393.5        #kJ/mol\n",
+      "deltaH2=-285.8        #kJ/mol\n",
+      "deltaH3=-2598.8       #kJ/mol\n",
+      "\n",
+      "#Calculation\n",
+      "deltaH4=2*(deltaH1)\n",
+      "deltaH5=(-1.0/2.0)*deltaH3\n",
+      "std_H=deltaH4+deltaH2+deltaH5\n",
+      "#Result\n",
+      "print\"Standard enthalpy of formation of acetylene is:\",std_H,\"kJ/mol\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1299.4\n",
+        "Standard enthalpy of formation of acetylene is: 226.6 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.10,Page no:258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "deltaH_Fel=12.4        #Heat of frmtn of Fe(l) in kJ/mol\n",
+      "deltaH_Al2O3=-1669.8   #Heat of formation of Al2O3 in kJ/mol\n",
+      "deltaH_Al=0            #Heat of formation of Al\n",
+      "deltaH_Fe2O3=-822.2     #Heat of formtion of Fe2O3 kJ/mol\n",
+      "M_Al=26.98           #Molar mass of Al\n",
+      "\n",
+      "#Calculation\n",
+      "deltaH_rxn=(deltaH_Al2O3+2*deltaH_Fel)-(2*deltaH_Al+deltaH_Fe2O3)\n",
+      "ratio=deltaH_rxn/2.0\n",
+      "heat_released=ratio/M_Al\n",
+      "\n",
+      "#Result\n",
+      "print\"Heat released per gram of Al is\",round(heat_released,2),\"kJ/g\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat released per gram of Al is -15.25 kJ/g\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_7.ipynb b/Chemistry/Chapter_7.ipynb
new file mode 100755
index 00000000..ebd9568b
--- /dev/null
+++ b/Chemistry/Chapter_7.ipynb
@@ -0,0 +1,312 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7:Quantum theorem and electronic structure of Atoms"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:7.1,Page no:27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "lamda=522*10**-9 #wavelength, m\n",
+      "c=3*10**8 #speed of light in vacuum, m/s\n",
+      "\n",
+      "#Calculation\n",
+      "v=c/lamda #frequency, Hz\n",
+      "\n",
+      "#Result\n",
+      "print\"The frequency of the wave is :%.2e\"%v,\"Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The frequency of the wave is :5.75e+14 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:7.2,Page no:280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "c=3*10**8 #speed of light in vacuum, m/s\n",
+      "h=6.63*10**-34 #planck's constant, J s\n",
+      "lamda1=5*10**-5 #wavelength, m\n",
+      "lamda2=5*10**-11 #wavelength, m\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)\n",
+      "E1=h*c/lamda1 #energy, J\n",
+      "#(b)\n",
+      "E2=h*c/lamda2 #energy, J\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) the energy of the photon is :%.2e\"%E1,\"J\"\n",
+      "print\"(b) the energy of the photon is :%.2e\"%E2,\"J\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) the energy of the photon is :3.98e-21 J\n",
+        "(b) the energy of the photon is :3.98e-15 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 59
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:7.3,Page no:281"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "W=3.42*10**-19      #Work function of cesium in J\n",
+      "h=6.63*10**-34      #planck's constant in J.s\n",
+      "v2=10**15     #Freq for irridation in s-1\n",
+      "\n",
+      "#Calculation\n",
+      "#part a\n",
+      "KE=0\n",
+      "#hv=KE+w\n",
+      "v=W/h\n",
+      "#part b\n",
+      "KE=h*v2-W\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)Minimum frequency of light is %.2e\"%v,\"s**-1\"\n",
+      "print\"(b).Kinetic energy of ejected electron is\",KE,\"J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Minimum frequency of light is 5.16e+14 s**-1\n",
+        "(b).Kinetic energy of ejected electron is 3.21e-19 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:7.4,Page no:287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "c=3*10**8 #speed of light in vacuum, m/s\n",
+      "h=6.63*10**-34 #planck's constant, J s\n",
+      "Rh=2.18*10**-18 #rydberg's constant, J\n",
+      "ni=5.0 #initial orbit\n",
+      "nf=2.0 #final orbit\n",
+      "\n",
+      "#Calculation\n",
+      "deltaE=Rh*(1/ni**2-1/nf**2) \n",
+      "lamda=c*h/-deltaE \n",
+      "\n",
+      "#Result\n",
+      "print\"The wavelength of the photon is :\",round(lamda*10**9),\"nm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength of the photon is : 434.0 nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 61
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:7.5,Page no:291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "h=6.63*10**-34 #planck's constant, J s\n",
+      "m1=0.06 #mass, kg\n",
+      "u1=68.0 #speed, m/s\n",
+      "m2=9.1094*10**-31 #mass, kg\n",
+      "u2=68 #speed, m/s\n",
+      "\n",
+      "#Calculation\n",
+      "# (a)\n",
+      "lamda1=h/(m1*u1) #wavelength, m\n",
+      "#(b)\n",
+      "lamda2=h/(m2*u2) #wavelength, m\n",
+      "\n",
+      "#Result\n",
+      "print\"The wavelength of the tennis ball is :\",lamda1,\"m\"\n",
+      "print\"The wavelength of the electron is :%.1e\"%lamda2,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength of the tennis ball is : 1.625e-34 m\n",
+        "The wavelength of the electron is :1.1e-05 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 62
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:,7.7,Page no:299"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "n=3    #Principal quantum no\n",
+      "\n",
+      "#Calculation\n",
+      "#The possible values of l are 0,1,2\n",
+      "s=1    #One 3s orbital,n=3,l=0,ml=0\n",
+      "p=3    #Three p obitals\n",
+      "d=5    #five 'd' orbital\n",
+      "total=s+p+d\n",
+      "\n",
+      "#Result\n",
+      "print\"The total no of orbitals is:\",total"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total no of orbitals is: 9\n"
+       ]
+      }
+     ],
+     "prompt_number": 63
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:7.9,Page no:306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "n=3    #Principal quantum number\n",
+      "l=[0,1,2]\n",
+      "n=[0,0,0]\n",
+      "\n",
+      "#Calculation\n",
+      "print\"Value of l \\t\\t No of orbitals(2l+1)\"\n",
+      "print\"--------------------------------------------\"\n",
+      "total=0\n",
+      "for i in range(0,3):\n",
+      "    n[i]=2*l[i]+1\n",
+      "    print l[i],\"\\t\\t\\t\\t\",n[i]\n",
+      "    total=n[i]+total\n",
+      "max=2*total\n",
+      "\n",
+      "#Result\n",
+      "print\"The maximum number of electrons that can reside is\",max\n",
+      "   \n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of l \t\t No of orbitals(2l+1)\n",
+        "--------------------------------------------\n",
+        "0 \t\t\t\t1\n",
+        "1 \t\t\t\t3\n",
+        "2 \t\t\t\t5\n",
+        "The maximum number of electrons that can reside is 18\n"
+       ]
+      }
+     ],
+     "prompt_number": 64
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/Chapter_9.ipynb b/Chemistry/Chapter_9.ipynb
new file mode 100755
index 00000000..a6cb1698
--- /dev/null
+++ b/Chemistry/Chapter_9.ipynb
@@ -0,0 +1,130 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9:Chemical Bonding I\n",
+      "Basic Concepts"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.13,Page no:397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#Type of bond broken:\n",
+      "hh_no=1         #No.of bonds broken\n",
+      "hh_be=436.4     #Bond enthalpy in kJ/mol\n",
+      "hh_ec=436.4     #Energy Change in kJ/mol\n",
+      "cl_no=1         #No.of bonds broken\n",
+      "cl_be=242.7     #Bond enthalpy in kJ/mol\n",
+      "cl_ec=242.7     #Energy Change in kJ/mol\n",
+      "#Type of bonds formed\n",
+      "hcl_no=2         #No.of bonds formed\n",
+      "hcl_be=431.9     #Bond enthalpy in kJ/mol\n",
+      "hcl_ec=863.8     #Energy Change in kJ/mol\n",
+      "#data:\n",
+      "Hf_hcl=-92.3     #Heat of formation of Hcl in kJ/mol\n",
+      "Hf_H2=0          #Heat of formation of H2 in kJ/mol\n",
+      "Hf_Cl2=0         #Heat of formation of Cl2 in kJ/mol \n",
+      "\n",
+      "#Calculation\n",
+      "energy_in=hh_ec+cl_ec       #Total energy input in kJ/mol\n",
+      "energy_rlsd=hcl_ec          #Total energy released\n",
+      "delta_H=energy_in-energy_rlsd      #enthalpy change in kJ/mol\n",
+      "delta_Ho=2*(-92.3)-(Hf_H2+Hf_Cl2)    #Ethalpy change from eq 6.18\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print \"Using equation 9.3,delta_H=\",delta_H,\"kJ/mol\"\n",
+      "print\"Alternatively:\"\n",
+      "print\"Enthalpy change is:\",delta_Ho,\"kJ/mol\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Using equation 9.3,delta_H= -184.7 kJ/mol\n",
+        "Alternatively:\n",
+        "Enthalpy change is: -184.6 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.14,Page no:398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#Type of bond broken:\n",
+      "hh_no=2          #No.of bonds broken\n",
+      "hh_be=436.4      #Bond enthalpy in kJ/mol\n",
+      "hh_Ec=872.8      #Energy Change in kJ/mol\n",
+      "oo_no=1           #No.of bonds broken\n",
+      "oo_be=498.7       #Bond enthalpy in kJ/mol\n",
+      "oo_Ec=498.7       #Energy Change in kJ/mol\n",
+      "#Types of bond formed\n",
+      "nn_oh=4          #No.of bonds formed\n",
+      "nn_be=460        #Bond enthalpy in kJ/mol\n",
+      "bb_Ec=1840       #Energy Change in kJ/mol\n",
+      "#data:\n",
+      "Hf_h2o=-241.8     #Heat of formation of H2O in kJ/mol\n",
+      "Hf_h2=0           #Heat of formation of H2 in kJ/mol\n",
+      "Hf_O2=0           #Heat of formation of O2 in kJ/mol\n",
+      "\n",
+      "#Calculation\n",
+      "energy_in=hh_Ec+oo_Ec       #Total energy input in kJ/mol\n",
+      "energy_rlsd=bb_Ec           #Total energy released\n",
+      "delta_h=energy_in-energy_rlsd            #enthalpy change in kJ/mol\n",
+      "delta_Ho=2*(Hf_h2o)-(2*Hf_h2+Hf_O2)      #Ethalpy change from eq 6.18\n",
+      "\n",
+      "#Result\n",
+      "print\"Enthalpy change is\",delta_Ho,\"kJ/mol\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enthalpy change is -483.6 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemistry/README.txt b/Chemistry/README.txt
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--- /dev/null
+++ b/Chemistry/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Kiran Bala
+Course: btech
+College/Institute/Organization: Ideal Institute of Technology,Ghaziabad
+Department/Designation: Electrical & Electronics Engg
+Book Title: Chemistry
+Author: Raymond Chang
+Publisher: McGraw Hill
+Year of publication: 2010
+Isbn: 9780073511092
+Edition: 10
+\ No newline at end of file
diff --git a/Chemistry/screenshots/screen1.png b/Chemistry/screenshots/screen1.png
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diff --git a/Chemistry/screenshots/screen2.png b/Chemistry/screenshots/screen2.png
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diff --git a/Chemistry/screenshots/screen3.png b/Chemistry/screenshots/screen3.png
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author	hardythe1	2014-08-13 11:41:01 +0530
committer	hardythe1	2014-08-13 11:41:01 +0530
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tree	5530b1509900ca8d6d21384e33036f50734de927 /Chemistry
parent	f3e94078a83634b4353ab0cd2de3b0e204a48ac7 (diff)
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