3.4 LOS MEDIOS DE COMUNICACIÓN Y LOS VALORES
3.4.2 La influencia de la televisión sobre el desarrollo socio
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APPENDICES APPENDIX I
Preparation of Acid solutions
For the acid solution; This is done possible by calculation method of the dilution system of the acid of 2M concentration using
For Sulphuric Acid
Method of Dilution:- mass concentration = Specific gravity of the acid * %purity / Molecularweight of the acid X dilution volume
So it is mathematically written as
M.Conc = SG x %Purity x DV = 1.84 x 98 x 10 = 18.4cm3 Mole Weight 98
Dilution formulae is done using the formulae of concentration determination i.e C1V1 = C2V2 i.e 18.4 x V1 = 2 x 1000 V1 = 108.7cm3
There for 108.7cm3 of conc H2SO4 is diluted in 1dm3 of beaker to make up the 2M sulphuric acid
For Hydrochloric Acid
Method of Dilution:- mass concentration = Specific gravity of the acid * %purity / Molecularweight of the acid X dilution volume
So it is mathematically written as
M.Conc = SG x %Purity x DV = 1.18 x 36 x 10 = 11.6cm3 Mole Weight 36.5
Dilution formulae is done using the formulae of concentration determination i.e C1V1 = C2V2 i.e 11.6 x V1 = 2 x 1000 V1 = 86.2cm3
There for 86.2cm3 of conc HCl is diluted in 1dm3 of beaker to make up the 2M Hydrochloric acid
APPENDIXX II PROXIMATE ANALYSIS
Moisture Content analysis Moisture content analysis result
Sample ID Weight of sample before drying (g) M2
Weight of sample after drying (g) M1
Percentage content
% (= m2/w1 *100)
Saw dust (SD) 2 1.8 10±0.20
Rice husk (RH) 2 1.9 5±0.50
G/Nut shells (GS) 2 1.9 5±0.20
Ash content determination result
Likely, this is thus proven successfull as discussed earlier in the methodology in which results are presented in percentage composition and is presented as seen below in the table.
Ash content analysis result
Sample ID Weight of sample before drying (g) M2
Weight of sample after drying (g) M1
Percentage content
% (= X2/X1 * 100)
Saw Dust (SD) 1.8 0.262 14.54±0.060
Rice husk (RH) 1.9 0.570 30.00±0.050
G/Nut Shells (GS)
1.9 0.294 15.47±0.055
Organic Matter Content and %Carbon Content Analysis Result
Sample ID Organic Matter composition% % carbon
Saw Dust (SD) 75.46±0.06 43.77±0.039
Rice husk (RH) 65.00±0.05 37.70±0.029
G/Nut Shells (GS) 79.53±0.06 46.13±0.030
Result summary of the analytical experiments of the Physical Parameters
Sample ID Moisture Ash Org Matter % Carbon
W1(g) W2(g) %MC X1(g) X2(g) %AC %OM %CC
SD 1 2 1.80 10 1.80 0.263 14.61 75.39 43.726
SD2 2 1.80 10 1.80 0.261 14.50 75.50 43.790
SD3 2 1.80 10 1.80 0.262 14.50 75.50 43.790
Mean 2 1.80 10±0.20 1.80 0.262 14.54±0.06 75.46±0.06 43.767±0.038
RH1 2 1.90 5 1.90 0.571 30.05 64.95 37.671
RH2 2 1.90 5 1.90 0.570 30.00 65.00 37.70
RH3 2 1.90 5 1.90 0.569 29.95 65.05 37.729
Mean 2 1.90 5±0.50 1.90 0.570 30.00±0.05 65.00±0.05 37.70±0.029
GS1 2 1.90 5 1.90 0.293 15.42 79.58 46.156
GS2 2 1.90 5 1.90 0.295 15.53 79.47 46.093
GS3 2 1.90 5 1.90 0.294 15.47 79.53 46.127
Mean 2 1.90 5±0.20 1.90 0.294 15.47±0.06 79.53±0.06 46.127±0.03
Proximate analysis result
0 10 20 30 40 50 60 70 80 90
moisture content ash content org matter content % carbon content
Proximate Analysis Result
Saw dust Rice husk G/nut cobs
APPENDIX III Reducing sugar quantification
This is achieved by calculating the amount of stock solution and a reference from the Standard solution graph from the reference line curve as seen from the absorption of the DNS solvent used which is summariesd as tabulated below the equationfocal point in all the calculations involved
But dilution factor is said to be of 1g of glucose = 100ml of distl H2O as such then 10g
= 10000ml of glucose. So therefore stock solution is said to be in 10,000mg ie ppm Standard dilutions are in 10000ppm as thus
Standard reducing sugar result table
Concentration in 2000ppm 0.00 2000 4000 6000 8000 Absorption nm 0.00 0.366 0.421 0.464 0.510
y = 2E-05x + 0.3215 R² = 0.9972
0 0.1 0.2 0.3 0.4 0.5 0.6
0 2000 4000 6000 8000 10000
AAbsorption in nm
Concentratio in ppm
Graph of Standard Glucose Result Curve
Series1 Linear (Series1)
Using X = y-0.321/2E-05 we get the sugar yield in ppm
Actual sugar conc is calculated by multiplyin the sugar yield in ppm by Dilution Factor (5/2 = 2.5) and the sample used in ppm (i.e 25g in 100ml is same as 250mg in 1000ppm
= 4) i.e X x 2.5 x 4 = Actual sugar concentration
Then the Sugar conc in g/100ml of sample is found by multiplying the actual conc by 1000 as in the case of (250g = Actual Sugar yield is same as 1000g = X yield) therefore Actual yield / 10000 = sugar (mg) which is conversion from ppm to g/100ml an
example is seen below.
First sample has 0.3636nm of absorption as our Y value then we say X = 0.3636 - 0.321 / 2E-5 = 2130ppm
Actual sugae yield = 2130 * 2.5 * 4 = 21300ppm Sugar value = 21300/10,000ppm = 1.23g/100ml%
So, same calculation method is applied to all tthe samples REDUCING SUGAR CONTENT RESULT
Result for the Sugar content Analysis for HCl Treatment (%dry weight)
Sample ID Concentration % Absorbance (nm)
HClSD2 6.1374±0.1507 0.484*
HClSD4 2.1000±0.1800 0.363
HClSD6 9.4000±0.0100 0.509*
HClSD8 6.5807±0.1299 0.473*
HClRH2 8.5500±0.0600 0.492
HClRH4 9.6167±1.2790 0.510*
HClRH6 2.6063±0.0126 0.720*
HClRH8 4.4053±0.0019 0.548*
HClGS2 3.4483±0.0026 0.611*
HClGS4 1.5577±0.0106 0.963*
HClGS6 4.9500±0.0600 0.420
HClGS8 3.9370±0.0015 0.575*
Result of Sugar content analysis of H2SO4 Treatment
Sample ID Concentration % Absorbance (nm)
HSSD2 1.2500±0.1500 0.346
HSSD4 2.8410±0.0242 0.673*
HSSD6 2.7143±0.1303 0.690*
HSSD8 2.8571±0.0008 0.671*
HSRH2 2.9442±0.0264 0.661*
HSRH4 10.6433±0.129 0.536
HSRH6 7.2000±0.0050 0.465
HSRH8 4.2017±0.0010 0.559*
HSGS2 2.2000±0.2500 0.356
HSGS4 3.2000±0.025 0.385
HSGS6 2.7859±0.0388 0.680*
HSGS8 1.0001±0.0500 0.341
n*= are all samples with a high wavelenght that are diluted with 10ml of distilled water. (So 5ml of test sample x 10ml of water = 50) thus are then multiplied by the dilution factors (50) to get the right concentrate
GRAPH OF REDUCING SUGAR CONTENT 0
2 4 6 8 10 12
HCLSD2 HCLSD4 HCLSD6 HCLSD8 HCLRD2 HCLRD4 HCLRD6 HCLRD8 HCLGC2 HCLGC4 HCLGC6 HCLGC8 HSSD2 HSSD4 HSSD6 HSSD8 HSRD2 HSRD4 HSRD6 HSRD8 HSGC2 HSGC4 HSGC6 HSGC8
Concentration (ppm)
Sample waste label
Reducing Sugar Concentration
stdev conc
REDUCING SUGAR TEST RESULT ANALYSIS
One-way Analysis of Variance (ANOVA)for reducing sugar content analysis The P value is < 0.0001, considered extremely significant. Variation among column means is significantly greater than expected by chance. Test for linear trend between column means and column number
Slope = 0.1960 r squared = 0.0729 There is a significant linear trend.
Is the nonlinear variation statistically significant? The P value is < 0.0001, considered extremely significant. After accounting for the linear trend, the remaining variation among column means is significant.
APPENDIX IV Ethanol quantification
Result from the UV-Vis spetrophotometer is utilised and thus the concentration is found by multiplying the wavelength from the spectra with the dilution factor and dilution factor is gotten by the 1ml of the cromium reagent used with the 4ml of the sample of ethanol produced i.e Conc = concentration x dilution factor A.C = conc x (4+1) as demonstrated below ( Dilution Factor = 4+1 = 5 so DF /ml of sample = 5/4
= 1.25)
From the equation of the Standard graph we’ve
Standard graph of ethanol produced
y = 1.283x + 0.1722 R² = 0.8959
0 0.2 0.4 0.6 0.8 1 1.2 1.4
0 0.2 0.4 0.6 0.8 1
Absorption in nm
Concentration in %
Standard Graph of Ethanol Produced
Series1 Linear (Series1)
Table 3.8A;- Table of Standard Readings of the sample i.e Ethanol
Concentration (ppm) 0.00 0.2 0.4 0.6 0.8
Absorption 0.00 0.595 0.799 0.905 1.128
From the equation of the graph which is Y = 1.283X + 0.172 We make X the subject as it becomes X = Y - 0.172/1.283
Where Y is the Absorption from the Spectral Readings and X is the rated ethanol concentration
Actual Concentration is calculated by Act Ethanol = X x Dilution Factor (1.25)
Since 25g of sample waste is hydrolysed in 100ml of dilute acid to get 98ml of fermentable substrate we’ve = 25 x 98/100 = 24.5g
We now measure the amount of ethanol distilled and denominate it from it by saying;
24.5g x 100 = ethanol amount in 1000g of sample as lebelled (Eth. conc/1000g)
To get it in 100g/sample we say the Amount gotten as Actual Ehanol divided/ethanol conc in 1000g/s x 100 (Ethanol Conc = Actual Ethanol/eth.Conc in 1000g/s x 100%) Sample calculation
X = Y - 0.172 / 1.283 is same as 1.215 – 0.172/1.283 = 1.080939 Actual Ethanol = 1.080939 x 1.25 = 1.51875
Amount of distilled ethanol is 18ml per 25g/100% so we’ve 24.5/18 *100 = 136.11g/1000ml
So to get it in 100ml/25g we say 1.51875/136.11 *100 = 1.1158g/100%
So same calculation is done for all sample that are listed averagely below in the table.
Table 3.9A: Result of the ethanol concentration for the Hydrochloric Acid treatment Sample ID Wavelenght (580) Eth. CONCENTRATION
(%/g)
HClSD2 1.295 0.9250±0.0094
HClSD4 1.306 0.9329±0.0047
HClSD6 1.346 0.9714±0.0125
HClSD8 1.271 1.1672±0.0047
HClRH2 1.342 1.2325±0.0000
HClRH4 1.285 1.7702±0.0082
HClRH6 1.296 1.0571±0.0000
HClRH8 1.349 0.9636±0.0047
HClGS2 1.398 0.9253±0.0047
HClGS4 1.315 1.0749±0.0047
HClGS6 1.314 1.4078±0.0047
HClGS8 1.371 0.9793±0.0058
Table 3.10A; Table of Ethanol concentration for the Hydrogen Sulphide treatment Sample ID Wavelenght (580) ACTUAL CONCENTRATION (%/g)
HSSD2 1.312 1.2813±0.0047
HSSD4 1.397 1.1404±0.0082
HSSD6 1.471 1.0507±0.0047
HSSD8 1.363 1.8081±0.0047
HSRH2 1.397 2.4234±0.0047
HSRH4 1.387 2.6893±0.0082
HSRH6 1.331 1.2297±0.0082
HSRH8 1.338 1.4335±0.0047
HSGS2 1.276 0.9115±0.0001
HSGS4 1.323 1.2150±0.0047
HSGS6 1.390 1.0638±0.0047
HSGS8 1.300 0.8633±0.0309
Graph Of Ethanol Concentration
One-way Analysis of Variance (ANOVA) for ethanol concentration The P value is < 0.0001, considered extremely significant. Variation among column means is significantly greater than expected by chance. Test for linear trend between column means and column number
Slope = 0.08029 r squared = 0.1564
Is the linear trend statistically significant? The P value is < 0.0001, considered extremely significant. There is a significant linear trend.
Is the nonlinear variation statistically significant? The P value is < 0.0001, considered extremely significant.
After accounting for the linear trend, the remaining variation among column means is significant.
This result was obtained from dividing variation among columns into linear and nonlinear components:
0 0.5 1 1.5 2 2.5 3
HClSD2 HClSD4 HClSD6 HClSD8 HClRD2 HClRD4 HClRD6 HClRD8 HClGC2 HClGC4 HClGC6 HClGC8 HSSD2 HSSD4 HSSD6 HSSD8 HSRD2 HSRD4 HSRD6 HSRD8 HSGC2 HSGC4 HSGC6 HSGC8
Concentration and Stand. dev
Sample ID
Ethanol Concentration g/100ml
ethanol concentration st dev ethanol concentration mean
Comprehensive tabulated result of practical done in general.
Sample ID
Quantity (ml)
Wavelenght (580)nm
Act.Eth. CONC (%/g)
Abs (491)nm
Sug.Conc
%/100g HClSD2 14 1.295 0.9250±0.0094 0.484 6.137±0.151 HClSD4 14 1.306 0.9329±0.0047 0.363 2.100±0.180 HClSD6 14 1.346 0.9714±0.0125 0.509 9.400±0.010 HClSD8 18 1.271 1.1672±0.0047 0.437 6.581±0.129 HClRH2 18 1.342 1.2325±0.0000 0.492 8.550±0.060 HClRH4 27 1.285 1.7702±0.0082 0.510 9.617±1.279 HClRH6 16 1.296 1.0571±0.0000 0.720 2.506±0.013 HClRH8 14 1.349 0.9636±0.0047 0.548 4.405±0.002 HClGS2 13 1.398 0.9253±0.0047 0.611 3.448±0.002 HClGS4 16 1.315 1.0749±0.0047 0.963 1.558±0.011 HClGS6 21 1.314 1.4078±0.0047 0.420 4.950±0.060 HClGS8 14 1.371 0.9793±0.0058 0.575 3.937±0.002 HSSD2 19 1.312 1.2813±0.0047 0.346 1.250±0.150 HSSD4 16 1.397 1.1404±0.0082 0.673 2.841±0.024 HSSD6 14 1.471 1.0507±0.0047 0.690 2.714±0.130 HSSD8 26 1.363 1.8081±0.0047 0.671 2.857±0.001 HSRH2 14 1.397 2.4234±0.0047 0.661 2.944±0.003 HSRH4 18 1.387 2.6893±0.0082 0.536 10.643±0.129 HSRH6 15 1.331 1.2297±0.0082 0.465 7.200±0.005 HSRH8 13 1.338 1.4335±0.0047 0.559 4.202±0.001 HSGS2 34 1.276 0.9115±0.0001 0.356 2.200±0.250 HSGS4 38 1.323 1.2150±0.0047 0.385 3.200±0.025 HSGS6 18 1.390 1.0638±0.0047 0.680 2.786±0.038 HSGS8 21 1.300 0.8633±0.0309 0.341 1.000±0.050
Data represented above are in triplicate of their mean ± Standard deviation for the Act Ethanol conc. And the Reducing Sugar content while the rest are the quantity of the ethanol distilled and the wavelengths of absorbance for both the ethanol content and the reducing sugar content respectively.
————— 9/17/2014 11:41:47 AM ————————————————————
Results for: Sugar Result Analysis.MTW
General Linear Model: obsv versus Acid, Sample, Percent
Factor Type Levels Values Acid fixed 2 H2, HC Sample fixed 3 GS, RH, SD Percent fixed 4 2, 4, 6, 8
Analysis of Variance for obsv, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P Acid 1 46.811 46.811 46.811 610.57 0.000 Sample 2 138.385 138.385 69.193 902.50 0.000 Percent 3 18.650 18.650 6.217 81.09 0.000 Acid*Sample 2 40.950 40.950 20.475 267.06 0.000 Acid*Percent 3 60.129 60.129 20.043 261.43 0.000 Sample*Percent 6 157.794 157.794 26.299 343.03 0.000 Acid*Sample*Percent 6 84.783 84.783 14.130 184.31 0.000 Error 48 3.680 3.680 0.077
Total 71 551.183
S = 0.276889 R-Sq = 99.33% R-Sq(adj) = 99.01%
Unusual Observations for obsv
Obs obsv Fit SE Fit Residual St Resid 16 8.4000 9.6167 0.1599 -1.2167 -5.38 R 17 10.9500 9.6167 0.1599 1.3333 5.90 R
R denotes an observation with a large Standardized residual.
Grouping Information Using Tukey Method and 95.0% Confidence
Acid Sample Percent N Mean Grouping H2 RH 4 3 10.643 A HC RH 4 3 9.617 B HC SD 6 3 9.400 B C HC RH 2 3 8.550 C H2 RH 6 3 7.200 D HC SD 8 3 6.581 D E HC SD 2 3 6.137 E HC GS 6 3 4.950 F HC RH 8 3 4.405 F G H2 RH 8 3 4.202 F G H HC GS 8 3 3.937 G H I HC GS 2 3 3.448 H I J H2 GS 4 3 3.200 I J K H2 RH 2 3 2.944 J K L H2 SD 8 3 2.857 J K L H2 SD 4 3 2.841 J K L H2 GS 6 3 2.786 J K L H2 SD 6 3 2.714 J K L HC RH 6 3 2.506 K L H2 GS 2 3 2.200 L M HC SD 4 3 2.100 L M N HC GS 4 3 1.558 M N O H2 SD 2 3 1.250 N O H2 GS 8 3 1.000 O Means that do not share a letter are significantly different.