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Diffstat (limited to 'Chemistry/Chapter_4.ipynb')
-rwxr-xr-x | Chemistry/Chapter_4.ipynb | 290 |
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diff --git a/Chemistry/Chapter_4.ipynb b/Chemistry/Chapter_4.ipynb deleted file mode 100755 index 13f747e8..00000000 --- a/Chemistry/Chapter_4.ipynb +++ /dev/null @@ -1,290 +0,0 @@ -{
- "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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