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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /858/CH2 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '858/CH2')
32 files changed, 617 insertions, 0 deletions
diff --git a/858/CH2/EX2.1/example_1.sce b/858/CH2/EX2.1/example_1.sce new file mode 100755 index 000000000..ac461f69c --- /dev/null +++ b/858/CH2/EX2.1/example_1.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 2.1 page number 71\n\n")
+
+//to find the volume of oxygen that can be obtained
+
+p1=15 //in bar
+p2=1.013 //in bar
+t1=283 //in K
+t2=273 //in K
+v1=10 //in l
+
+v2=p1*v1*t2/(t1*p2);
+
+printf("volume of oxygen = %f liters",v2)
diff --git a/858/CH2/EX2.10/example_10.sce b/858/CH2/EX2.10/example_10.sce new file mode 100755 index 000000000..89a458150 --- /dev/null +++ b/858/CH2/EX2.10/example_10.sce @@ -0,0 +1,14 @@ +clc
+clear
+printf("example 2.10 page number 74\n\n")
+
+//to find amount of KClO3 precipitated
+
+solubility_70=30.2 //in gm/100gm
+w_solute=solubility_70*350/130.2; //in gm
+
+w_water=350-w_solute;
+solubility_30=10.1 //in gm/100gm
+precipitate=(solubility_70-solubility_30)*w_water/100
+
+printf("amount precipitated = %f gm",precipitate)
diff --git a/858/CH2/EX2.11/example_11.sce b/858/CH2/EX2.11/example_11.sce new file mode 100755 index 000000000..70c5af3e9 --- /dev/null +++ b/858/CH2/EX2.11/example_11.sce @@ -0,0 +1,12 @@ +clc
+clear
+printf("example 2.11 page number 74\n\n")
+
+//to find the pressure for solubility of CO2
+
+absorbtion_coefficient=1.71 //in liters
+molar_mass=44;
+solubility=absorbtion_coefficient*molar_mass/22.4; //in gm
+pressure=8/solubility*101.3;
+
+printf("pressure required = %f kPa",pressure)
diff --git a/858/CH2/EX2.12/example_12.sce b/858/CH2/EX2.12/example_12.sce new file mode 100755 index 000000000..0108a5adc --- /dev/null +++ b/858/CH2/EX2.12/example_12.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 2.12 page number 74\n\n")
+
+//to find the vapor pressure of water
+
+w_water=540 //in gm
+w_glucose=36 //in gm
+m_water=18; //molar mass of water
+m_glucose=180; //molar mass of glucose
+
+x=(w_water/m_water)/(w_water/m_water+w_glucose/m_glucose);
+p=8.2*x;
+depression=8.2-p;
+
+printf("depression in vapor pressure = %f Pa",depression*1000)
diff --git a/858/CH2/EX2.13/example_13.sce b/858/CH2/EX2.13/example_13.sce new file mode 100755 index 000000000..83470bdc8 --- /dev/null +++ b/858/CH2/EX2.13/example_13.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 2.13 page number 75\n\n")
+
+//to find the boiling point of solution
+
+w_glucose=9 //in gm
+w_water=100 //in gm
+E=0.52;
+m=90/180; //moles/1000gm water
+
+delta_t=E*m;
+boiling_point=100+delta_t;
+
+printf("boiling_point of water = %f degreeC",boiling_point)
diff --git a/858/CH2/EX2.14/example_14.sce b/858/CH2/EX2.14/example_14.sce new file mode 100755 index 000000000..87413d016 --- /dev/null +++ b/858/CH2/EX2.14/example_14.sce @@ -0,0 +1,20 @@ +clc
+clear
+printf("example 2.14 page number 75\n\n")
+
+//to find the molar mass and osmotic pressure
+
+K=1.86;
+c=15 //concentration of alcohol
+delta_t=10.26;
+
+m=delta_t/K; //molality
+M=c/(m*85); //molar mass
+printf("molar mass = %f gm\n\n",M*1000)
+
+density=0.97 //g/ml
+cm=c*density/(M*100);
+printf("molar concentration of alcohol = %f moles/l\n\n",cm)
+
+p=cm*8.314*293 //osmotic pressure
+printf("osmotic pressure = %f Mpa\n\n",p/1000)
diff --git a/858/CH2/EX2.15/example_15.sce b/858/CH2/EX2.15/example_15.sce new file mode 100755 index 000000000..668dd839c --- /dev/null +++ b/858/CH2/EX2.15/example_15.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 2.15 page number 75\n\n")
+
+//to find u_in, M_v, k'
+
+u_in = 0.575 //from the graph
+u_s = 0.295 //in mPa-s
+
+M_v = (u_in/(5.80*10^-5))^(1/0.72);
+u_red = 0.628; //in dl/g
+
+c = 0.40 //in g/dl
+k = (u_red-u_in)/((u_in^2)*c);
+
+printf("k = %f \nMv = %f\nu_in = %f dl/gm",k,M_v,u_in)
diff --git a/858/CH2/EX2.16/example_16.sce b/858/CH2/EX2.16/example_16.sce new file mode 100755 index 000000000..18944e881 --- /dev/null +++ b/858/CH2/EX2.16/example_16.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 2.16 page number 76\n\n")
+
+//to find the molecular formula
+
+C=54.5 //% of carbon
+H2=9.1 //% of hydrogen
+O2=36.4 //% of oxygen
+x=C/12; //number of carbon molecules
+y=O2/16; //number of oxygen molecules
+z=H2/2 //number of hydrogen molecules
+molar_mass=88;
+density=44;
+
+ratio=molar_mass/density;
+x=ratio*2;
+y=ratio*1;
+z=ratio*4;
+
+printf("x = %f, y = %f, z = %f",x,y,z)
+printf("\n\nformula of butyric acid is = C4H8O2")
diff --git a/858/CH2/EX2.17/example_17.sce b/858/CH2/EX2.17/example_17.sce new file mode 100755 index 000000000..9a7ffc6a8 --- /dev/null +++ b/858/CH2/EX2.17/example_17.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 2.17 page number 77\n\n")
+
+//to find molecular foemula
+C=93.75 //% of carbon
+H2=6.25 //% of hydrogen
+x=C/12 //number of carbon atoms
+y=H2/2 //number of hydrogen atoms
+molar_mass=64
+density=4.41*29;
+
+ratio=density/molar_mass;
+
+x=ratio*5;
+y=ratio*4;
+
+
+printf("x = %f, y = %f",x,y)
+printf("\n\nformula of butyric acid is = C10H8")
+
diff --git a/858/CH2/EX2.18/example_18.sce b/858/CH2/EX2.18/example_18.sce new file mode 100755 index 000000000..b4fda753a --- /dev/null +++ b/858/CH2/EX2.18/example_18.sce @@ -0,0 +1,31 @@ +clc
+clear
+printf("example 2.18 page number 77\n\n")
+
+//to find molecular formula
+C=50.69 //% of carbon
+H2=4.23 //% of hydrogen
+O2=45.08 //% of oxygen
+a=C/12; //number of carbon molecules
+c=O2/16; //number of oxygen molecules
+b=H2/2; //number of hydrogen molecules
+molar_mass=71;
+
+function M=f(m)
+ M=(2.09*1000)/(60*m);
+
+endfunction
+
+M=f((1.25/5.1));
+
+printf("actual molecular mass = %f\n\n",M)
+
+ratio=M/molar_mass;
+a=ratio*3;
+b=ratio*3;
+c=ratio*2;
+
+
+printf("a = %f, b = %f, c = %f",a,b,c)
+printf("\n\nformula of butyric acid is = C6H6O4")
+
diff --git a/858/CH2/EX2.19/example_19.sce b/858/CH2/EX2.19/example_19.sce new file mode 100755 index 000000000..74672fa19 --- /dev/null +++ b/858/CH2/EX2.19/example_19.sce @@ -0,0 +1,24 @@ +clc
+clear
+printf("example 2.19 page number 78\n\n")
+
+//to find the molecular formula
+C=64.6 //% of carbon
+H2=5.2 //% of hydrogen
+O2=12.6 //% of oxygen
+N2=8.8 //% of nitrogen
+Fe=8.8 //% of iron
+
+a=C/12; //number of carbon molecules
+c=8.8/14; //number of nitrogen molecules
+b=H2/2; //number of hydrogen molecules
+d=O2/16; //number of oxygen molecules
+e=Fe/56 //number of iron atoms
+
+cm=243.4/(8.31*293) //concentration
+
+ molar_mass=63.3/cm;
+
+ printf("a = %f, b = %f, c = %f, d = %f, e = %f",a*6.5,b*6.5,c*6.5,d*6.5,e*6.5)
+printf("\n\nformula of butyric acid is = C34H33N4O5Fe")
+
diff --git a/858/CH2/EX2.2/example_2.sce b/858/CH2/EX2.2/example_2.sce new file mode 100755 index 000000000..0c2176d66 --- /dev/null +++ b/858/CH2/EX2.2/example_2.sce @@ -0,0 +1,25 @@ +clc
+clear
+printf("example 2.2 page number 71\n\n")
+
+//to find volumetric composition,partial pressue of each gas and total pressure of mixture
+
+nCO2=2/44; //moles of CO2
+nO2=4/32; //moles of O2
+nCH4=1.5/16; //moles of CH4
+
+total_moles=nCO2+nO2+nCH4;
+yCO2=nCO2/total_moles;
+yO2=nO2/total_moles;
+yCH4=nCH4/total_moles;
+
+printf (" Composition of mixture = \nCH4 = %f \nO2 = %f \n CO2 = %f \n\n",yCH4,yO2,yCO2)
+
+pCO2=nCO2*8.314*273/(6*10^-3);
+pO2=nO2*8.314*273/(6*10^-3);
+pCH4=nCH4*8.314*273/(6*10^-3);
+
+printf ("pressure of CH4 = %f kPa \npressure of O2 = %f kPa\n pressure of CO2 =%f kPa\n\n",pCH4*10^-3,pO2*10^-3,pCO2*10^-3)
+
+total_pressure=pCO2+pCH4+pO2;
+printf ("total pressure = %f Kpa",total_pressure*10^-3)
diff --git a/858/CH2/EX2.20/example_20.sce b/858/CH2/EX2.20/example_20.sce new file mode 100755 index 000000000..aaf4f41d0 --- /dev/null +++ b/858/CH2/EX2.20/example_20.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 2.20 page number 78\n\n")
+
+//to find sequence of deposition
+E1=-0.25;
+E2=0.80;
+E3=0.34;
+
+a=[E1;E2;E3];
+b=gsort(a);
+
+printf("sorted potential in volts =")
+disp (b)
+disp ("E2>E3>E1")
+disp ("silver>copper>nickel")
+
+
diff --git a/858/CH2/EX2.21/example_21.sce b/858/CH2/EX2.21/example_21.sce new file mode 100755 index 000000000..d7da98cf4 --- /dev/null +++ b/858/CH2/EX2.21/example_21.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 2.21 page number 79\n\n")
+
+//to find the emf of cell
+
+E0_Zn=-0.76;
+E0_Pb=-0.13;
+c_Zn=0.1;
+c_Pb=0.02;
+
+E_Zn=E0_Zn+(0.059/2)*log10(c_Zn);
+E_Pb=E0_Pb+(0.059/2)*log10(c_Pb);
+E=E_Pb-E_Zn;
+
+printf("emf of cell = %f V",E)
+printf("\n\nSince potential of lead is greater than that of zinc thus reduction will occur at lead electrode and oxidation will occur at zinc electrode")
diff --git a/858/CH2/EX2.22/example_22.sce b/858/CH2/EX2.22/example_22.sce new file mode 100755 index 000000000..4e0ac0209 --- /dev/null +++ b/858/CH2/EX2.22/example_22.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 2.22 page number 79\n\n")
+
+//to find the emf of cell
+E0_Ag=0.80;
+E0_AgNO3=0.80;
+c_Ag=0.001;
+c_AgNO3=0.1;
+
+E_Ag=E0_Ag+(0.059)*log10(c_Ag);
+E_AgNO3=E0_AgNO3+(0.059)*log10(c_AgNO3);
+E=E_AgNO3-E_Ag;
+
+printf("emf of cell = %f V" ,E)
+printf("\n\nsince E is positive, the left hand electrode will be anode and the electron will travel in the external circuit from the left hand to the right hand electrode")
diff --git a/858/CH2/EX2.23/example_23.sce b/858/CH2/EX2.23/example_23.sce new file mode 100755 index 000000000..c3ea8b288 --- /dev/null +++ b/858/CH2/EX2.23/example_23.sce @@ -0,0 +1,10 @@ +clc
+clear
+printf("example 2.23 page number 79\n\n")
+
+//to find emf of cell
+pH=12; //pH of solution
+E_H2=0;
+E2=-0.059*pH;
+E=E_H2-E2;
+printf("EMF of cell = %f V",E)
diff --git a/858/CH2/EX2.24/example_24.sce b/858/CH2/EX2.24/example_24.sce new file mode 100755 index 000000000..d279c1848 --- /dev/null +++ b/858/CH2/EX2.24/example_24.sce @@ -0,0 +1,12 @@ +clc
+clear
+printf("example 2.24 page number 80\n\n")
+
+//to find amount of silver deposited
+I=3 //in Ampere
+t=900 //in s
+m_eq=107.9 //in gm/mol
+F=96500;
+
+m=(I*t*m_eq)/F;
+printf("mass = %f gm",m)
diff --git a/858/CH2/EX2.25/example_25.sce b/858/CH2/EX2.25/example_25.sce new file mode 100755 index 000000000..508af8f6c --- /dev/null +++ b/858/CH2/EX2.25/example_25.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 2.25 page number 80\n\n")
+
+//to find the time for electroplating
+volume=10*10*0.005; //in cm3
+mass=volume*8.9;
+F=96500;
+atomic_mass=58.7 //in amu
+current=2.5 //in Ampere
+
+charge=(8.9*F*2)/atomic_mass;
+yield=0.95;
+actual_charge=charge/(yield*3600);
+t=actual_charge/current;
+
+printf("time required = %f hours",t)
diff --git a/858/CH2/EX2.26/example_26.sce b/858/CH2/EX2.26/example_26.sce new file mode 100755 index 000000000..f763adc9c --- /dev/null +++ b/858/CH2/EX2.26/example_26.sce @@ -0,0 +1,12 @@ +clc
+clear
+printf("example 2.26 page number 80\n\n")
+
+//to find hardness of water
+m_MgSO4=90 //in ppm
+MgSO4_parts=120;
+CaCO3_parts=100;
+
+hardness=(CaCO3_parts/MgSO4_parts)*m_MgSO4;
+
+printf("hardness of water = %f mg/l",hardness)
diff --git a/858/CH2/EX2.27/example_27.sce b/858/CH2/EX2.27/example_27.sce new file mode 100755 index 000000000..ff5328e3c --- /dev/null +++ b/858/CH2/EX2.27/example_27.sce @@ -0,0 +1,30 @@ +clc
+clear
+printf("example 2.26 page number 80\n\n")
+
+m1 = 162 //mass of calcium bi carbonate in mg
+m2 = 73 //mass of magnesium bi carbonate in mg
+m3 = 136 // mass of calsium sulfate in mg
+m4 = 95 // mass of magnesium cloride
+m5 = 500 //mass of sodium cloride in mg
+m6 = 50 // mass of potassium cloride in mg
+
+content_1 = m1*100/m1; //content of calcium bi carbonate in mg
+content_2 = m2*100/(2*m2); //content of magnesium bi carbonate in mg
+content_3 = m3*100/m3; // content of calsium sufate in mg
+content_4 = m4*100/m4; // content of magnesium cloride
+
+//part_1
+
+temp_hardness = content_1 + content_2; //depends on bicarbonate only
+total_hardness = content_1+content_2+content_3+content_4;
+printf("total hardness = %f\n temporary hardness = %f \n",temp_hardness,total_hardness)
+
+//part 2
+wt_lime = (74/100)*(content_1+2*content_2+content_4);
+actual_lime = wt_lime/0.85;
+printf("amount of lime required = %f \n",actual_lime)
+
+soda_required = (106/100)*(content_1+content_4);
+actual_soda = soda_required/0.98;
+printf("amount of soda required = %f \n",actual_soda)
diff --git a/858/CH2/EX2.28/example_28.sce b/858/CH2/EX2.28/example_28.sce new file mode 100755 index 000000000..95249c066 --- /dev/null +++ b/858/CH2/EX2.28/example_28.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 2.28 page number 82\n\n")
+
+//to find hardness of water
+
+volume_NaCl=50 //in l
+c_NaCl=5000 //in mg/l
+
+m=volume_NaCl*c_NaCl;
+equivalent_NaCl=50/58.5;
+
+hardness=equivalent_NaCl*m;
+
+printf("hardness of water = %f mg/l",hardness/1000)
diff --git a/858/CH2/EX2.29/example_29.sce b/858/CH2/EX2.29/example_29.sce new file mode 100755 index 000000000..8eb909d7a --- /dev/null +++ b/858/CH2/EX2.29/example_29.sce @@ -0,0 +1,27 @@ +clc
+clear
+printf("example 2.29 page number 82\n\n")
+
+//to find the total vapor pressure and molar compositions
+
+m_benzene = 55 //in kg
+m_toluene = 28 //in kg
+m_xylene = 17 // in kg
+
+mole_benzene = m_benzene/78;
+mole_toluene = m_toluene/92;
+mole_xylene = m_xylene/106;
+
+mole_total = mole_benzene+mole_toluene+mole_xylene;
+x_benzene = mole_benzene/mole_total;
+x_toluene = mole_toluene/mole_total;
+x_xylene = mole_xylene/mole_total;
+
+P = x_benzene*178.6+x_toluene*74.6+x_xylene*28;
+printf("total pressure = %f kPa\n",P)
+
+benzene = (x_benzene*178.6*100)/P;
+toluene = (x_toluene*74.6*100)/P;
+xylene = (x_xylene*28*100)/P;
+
+printf("xylene = %f \n toluene = %f \n benzene = %f",xylene,toluene,benzene)
diff --git a/858/CH2/EX2.3/example_3.sce b/858/CH2/EX2.3/example_3.sce new file mode 100755 index 000000000..d4eaa19c6 --- /dev/null +++ b/858/CH2/EX2.3/example_3.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 2.3 page number 72\n\n")
+
+//to find equivalent mass of metal
+
+P=104.3 //total pressure in KPa
+pH2O=2.3 //in KPa
+pH2=P-pH2O; //in KPa
+
+VH2=209*pH2*273/(293*101.3)
+
+printf("volume of hydrogen obtained = %f ml\n\n",VH2)
+
+//calculating amount of metal having 11.2l of hydrogen
+
+m=350/196.08*11.2 //mass of metal in grams
+printf("mass of metal equivalent to 11.2 litre/mol of hydrogen = %f gm",m)
diff --git a/858/CH2/EX2.30/example_30.sce b/858/CH2/EX2.30/example_30.sce new file mode 100755 index 000000000..8f659237c --- /dev/null +++ b/858/CH2/EX2.30/example_30.sce @@ -0,0 +1,36 @@ +clc
+clear
+printf("example 2.30 page number 83\n\n")
+
+//to find the mixture composition
+
+vapor_pressure=8 //in kPa
+pressure=100 //in kPa
+
+//part 1
+volume=1 //in m3
+volume_ethanol=volume*(vapor_pressure/pressure);
+volume_air=1-volume_ethanol;
+printf("volumetric composition:- \nair composition = %f\n ethanol compostion = %f",volume_air*100,volume_ethanol*100)
+
+//part 2
+molar_mass_ethanol=46;
+molar_mass_air=28.9;
+mass_ethanol=0.08*molar_mass_ethanol; //in kg
+mass_air=0.92*molar_mass_air; //in kg
+fraction_ethanol=(mass_ethanol*100)/(mass_air+mass_ethanol);
+fraction_air=(mass_air*100)/(mass_air+mass_ethanol);
+printf("\n\ncomposition by weight:-\nAir = %f Ethanol vapor = %f",fraction_air,fraction_ethanol)
+
+//part 3
+mixture_volume=22.3*(101.3/100)*(299/273); //in m3
+weight_ethanol=mass_ethanol/mixture_volume;
+printf("\n\nweight of ethanol/cubic meter = %f Kg",weight_ethanol)
+
+//part 4
+w_ethanol=mass_ethanol/mass_air;
+printf("\n\nweight of ethanol/kg vapor free air = %f Kg",w_ethanol)
+
+//part 5
+moles_ethanol=0.08/0.92;
+printf("\n\nkmol of ethanol per kmol of vapor free air = %f",moles_ethanol)
diff --git a/858/CH2/EX2.31/example_31.sce b/858/CH2/EX2.31/example_31.sce new file mode 100755 index 000000000..93b25e8b9 --- /dev/null +++ b/858/CH2/EX2.31/example_31.sce @@ -0,0 +1,27 @@ +clc
+clear
+printf("example 2.31 page number 84\n\n")
+
+//to find relative saturation and dew point
+
+vapor_pressure=8 //in kPa
+volume_ethanol=0.05;
+
+//basis 1kmol of mixture
+
+partial_pressure=volume_ethanol*100;
+relative_saturation=partial_pressure/vapor_pressure;
+mole_ratio=volume_ethanol/(1-volume_ethanol);
+printf("mole ratio = %f \nrelative saturation = %f",mole_ratio,relative_saturation*100)
+
+//basis 1kmol saturated gas mixture at 100kPa
+volume_vapor=(8/100)*100;
+ethanol_vapor=volume_vapor/100;
+air_vapor=1-ethanol_vapor;
+saturation_ratio=ethanol_vapor/air_vapor;
+percentage_saturation=mole_ratio/saturation_ratio;
+
+printf("\n\npercentage saturation = %f",percentage_saturation)
+
+//dew point
+printf("\n\ncorresponding to partial pressure of 5kPa we get a dew point of 17.3 degree celcius")
diff --git a/858/CH2/EX2.32/example_32.sce b/858/CH2/EX2.32/example_32.sce new file mode 100755 index 000000000..459b9b9a4 --- /dev/null +++ b/858/CH2/EX2.32/example_32.sce @@ -0,0 +1,31 @@ +clc
+clear
+printf("example 2.32 page number 84\n\n")
+
+//to find the properties of humid air
+
+p = 4.24 //in kPa
+H_rel = 0.8;
+p_partial = p*H_rel;
+molal_H = p_partial/(100-p_partial);
+printf("initial molal humidity = %f\n\n",molal_H)
+
+//part 2
+P = 200 //in kPa
+p_partial = 1.70 //in kPa
+final_H = p_partial/(P-p_partial);
+printf("final molal humidity = %f\n\n",final_H)
+
+//part 3
+p_dryair = 100 - 3.39;
+v = 100*(p_dryair/101.3)*(273/303);
+moles_dryair = v/22.4;
+vapor_initial = molal_H*moles_dryair;
+vapor_final = final_H*moles_dryair;
+water_condensed = (vapor_initial-vapor_final)*18;
+printf("amount of water condensed = %f \n\n",water_condensed)
+
+//part 4
+total_air = moles_dryair+vapor_final;
+final_v = 22.4*(101.3/200)*(288/273)*total_air;
+printf("final volume of wety air = %f \n\n",final_v)
diff --git a/858/CH2/EX2.4/example_4.sce b/858/CH2/EX2.4/example_4.sce new file mode 100755 index 000000000..c4a8d2f30 --- /dev/null +++ b/858/CH2/EX2.4/example_4.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 2.4 page number 72\n\n")
+
+//to find NaCl content in NaOH solution
+
+w=2 //in gm
+m=0.287 //in gm
+
+//precipitate from 58.5gm of NaCl=143.4gm
+
+mNaCl=58.5/143.4*m;
+
+printf("mass of NaCl = %f gm\n",mNaCl )
+
+percentage_NaCl=mNaCl/w*100;
+printf("amount of NaCl = %f",percentage_NaCl)
diff --git a/858/CH2/EX2.5/example_5.sce b/858/CH2/EX2.5/example_5.sce new file mode 100755 index 000000000..b4989d530 --- /dev/null +++ b/858/CH2/EX2.5/example_5.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 2.5 page number 72\n\n")
+
+//to find the carbon content in sample
+
+w=4.73 //in gm5
+VCO2=5.30 //in liters
+
+weight_CO2=44/22.4*VCO2;
+carbon_content=12/44*weight_CO2;
+
+percentage_content=(carbon_content/w)*100;
+
+printf("percentage amount of carbon in sample = %f",percentage_content)
diff --git a/858/CH2/EX2.6/example_6.sce b/858/CH2/EX2.6/example_6.sce new file mode 100755 index 000000000..a9d65e5dd --- /dev/null +++ b/858/CH2/EX2.6/example_6.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 2.6 page number 73\n\n")
+//to find the volume of air
+
+volume_H2=0.5 //in m3
+volume_CH4=0.35 //in m3
+volume_CO=0.08 //in m3
+volume_C2H4=0.02 //in m3
+volume_oxygen=0.21 //in m3 in air
+
+//required oxygen for various gases
+H2=0.5*volume_H2;
+CH4=2*volume_CH4;
+CO=0.5*volume_CO;
+C2H4=3*volume_C2H4;
+
+total_O2=H2+CH4+CO+C2H4;
+oxygen_required=total_O2/volume_oxygen;
+
+printf("amount of oxygen required = %f cubic meter",oxygen_required)
diff --git a/858/CH2/EX2.7/example_7.sce b/858/CH2/EX2.7/example_7.sce new file mode 100755 index 000000000..11020ae1e --- /dev/null +++ b/858/CH2/EX2.7/example_7.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 2.7 page number 73\n\n")
+
+//to find the volume of sulphuric acid and mass of water consumed
+
+density_H2SO4 = 1.10 //in g/ml
+mass_1 = 100*density_H2SO4; //mass of 100ml of 15% solution
+mass_H2SO4 = 0.15*mass_1;
+density_std = 1.84 //density of 96% sulphuric acid
+mass_std = 0.96*density_std; //mass of H2SO4 in 1ml 96% H2SO4
+
+volume_std = mass_H2SO4/mass_std; //volume of 96%H2SO4
+mass_water = mass_1 - mass_H2SO4;
+
+printf("volume of 0.96 H2SO4 required = %f ml",volume_std)
+printf("\nmass of water required = %f g",mass_water)
diff --git a/858/CH2/EX2.8/example_8.sce b/858/CH2/EX2.8/example_8.sce new file mode 100755 index 000000000..08e4c91ff --- /dev/null +++ b/858/CH2/EX2.8/example_8.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 2.8 page number 73\n\n")
+
+//to find molarity,molality and normality
+
+w_H2SO4=0.15 //in gm/1gm solution
+density=1.10 //in gm/ml
+m=density*1000; //mass per liter
+weight=m*w_H2SO4; //H2SO4 per liter solution
+molar_mass=98;
+
+Molarity=weight/molar_mass;
+printf("Molarity = %f mol/l\n\n",Molarity)
+
+equivalent_mass=49;
+normality=weight/equivalent_mass;
+printf("Normality = %f N\n\n",normality)
+
+molality=176.5/molar_mass;
+printf("Molality = %f",molality)
diff --git a/858/CH2/EX2.9/example_9.sce b/858/CH2/EX2.9/example_9.sce new file mode 100755 index 000000000..025b0e232 --- /dev/null +++ b/858/CH2/EX2.9/example_9.sce @@ -0,0 +1,9 @@ +clc
+clear
+printf("example 2.9 page number 74\n\n")
+
+molar_mass_BaCl2=208.3; //in gm
+equivalent_H2SO4=0.144;
+normality=equivalent_H2SO4*1000/28.8;
+
+printf("Normality = %f N",normality)
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