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+{

+ "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": {}

+  }

+ ]

+}
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