Indoor plants - Free Essay Example

Sample details Pages: 23 Words: 6839 Downloads: 7 Date added: 2017/06/26 Category Statistics Essay Did you like this example? Are indoor plants adapted to get rid formaldehyde, Sipin, Elly Lorreta one of the noxious wastes commonly found at home 002348-019 nowadays? 1.0 Introduction I did further research and found out that indoor air pollution phenomenon has urged the NASA (National Aeronautics and Space Administration) scientists to study the functions of plants to provide clean indoor air. NASA has become the pioneer towards this research and recently has been widened by many other associations like the Wolverton Environmental Services, Inc. endorsed by the Plants for Clean Air Council in Mitchellville, Maryland[1]. Don’t waste time! Our writers will create an original "Indoor plants" essay for you Create order Research done by NASA has found out that there are certain plants that have the function to purify the air in a building[2]. They detoxify the existing toxins and pollutants which originate from the things used in daily activities nowadays; fabrics, detergents and also furniture. These pollutants can be classified into three common indoor pollutants according to the list of indoor contaminant that are currently present. There are benzene, formaldehyde and trichloroethylene. (TCE)[3] Plants use the concept of transpiration to work onto this problem[4]. As the vaporized chemical enters the stomatal opening on the leaves of the indoor plants, they are either broken down directly or be sent downwards; down to the root system of the plants.[5] The presence of colonies of microbes at the root system breaks down various kinds of unhealthy compounds; in this case the indoor pollutants, and absorbs them as their source of food[6]. As for the mechanism of transpiration to remove the pollutant, water vapour that is liberated by the leaves of the plants will mix with the air in the atmosphere. Convection of air leads to the movement of the atmospheric air that is contaminated with the vaporized chemical downwards to the base of the plants. I chose 6 types of plants to be experimented by one fixed type of pollutant; formaldehyde. It is normally used in the production of grocery bags, facial tissues, waxed paper, waxed paper[7] and produced by tobacco products, gas cookers and open fireplaces.[8] In the experiment, this chemical is predicted to be absorbed by each plant. Plant that absorbs the chemical the most would be the efficient plant to be included in places mentioned before. 2.0 Aim To study the effect of plants transpiration towards the acidity and mass of formaldehyde in a transparent chamber. 3.0 Planning and method development Firstly, a chamber must be set up to place plants chosen. A pot of selected plant is placed into each chamber. 6 types of plants were chosen, therefore 6 chambers must be created. To make sure that air, sunlight and water could be continuously supplied, I decided that the chamber must be transparent, and there are holes to let air enters. The material that I chose is transparent plastic so that holes can be poked, the wall of the chambers can be flipped to water the plants everyday and plants get sufficient sunlight. I selected formaldehyde as the pollutant to the plants. In each of the chamber, I included formalin of the same amount in a beaker and let it evaporate in the chamber. As formalin CH2O, is a reducing agent[9], therefore it has the ability to release its hydrogen.[10] The more hydrogen ions present in it, the greater the strength of the acid. When evaporation of formalin happens continuously, there will be less in quantity of hydrogen atoms in the aqueous solution. Thus, the acidity of formaldehyde could decrease through evaporation; pH of the formalin increases. So, the pH of the formalin is ought to be checked for every interval of two days. Because concept of evaporation is used, it is for sure the volume of the formalin will reduce. The most effective method to measure this is by getting the mass decrease. I took the reading of the mass of formalin for every interval of two days. I decided to take note on the external condition of all the plants so that analysis on that can be don e to find its relativity with formalin. 4.0 Hypothesis My prediction is that indoor plants have the ability to get rid of formaldehyde, one of the noxious wastes commonly found at home nowadays by absorbing the chemicals through their microscopic openings perforated on their leaves; the stomata[11]. As the chemical evaporates, the molecules of the chemical are absorbed by the plants by gaining entrance through the stomata. These plants transport the absorbed chemical to their root system along the xylem of the plants to be broken down by the microbes present at the roots.[12] As formalin acts as a reducing agent, release of hydrogen could occur. Through evaporation of formalin, there will be less hydrogen atoms could remain in the aqueous solution. Thus, it is possible for the decrease in mass and increase in the pH of the formalin to occur when indoor plants are available. 5.0 Methodology 5.1 Variables a) Independent: * Types of plants chosen to be experimented There are variety types of plants chosen in order to know whether the hypothesis could be accepted. They are Boston fern (Nephrolepis exaltata Bostoniensis), Janet Craig(Dracaena deremensis), Florists mum(Chrysanthemum morifolium), Kimberly queen fern (Nephrolepis obliterata), Snake plant or mother-in-laws tongue (Sansevieria trifasciata Laurentii), Himalayan Balsam (Impatiens glandulifera) altogether. Himalayan Balsam (Impatiens glandulifera) acts as the control of the experiment to show its less in efficiency to absorb the toxin. Some plants have no ability to absorb the chosen toxin as good as in some indoor plants. b) Dependent: * The rate of absorption of formaldehyde The rate of absorption of formaldehyde is taken as the decrease in mass of formalin over time. This is documented for every interval of two days. Other than that, the acidity of formaldehyde in each chamber is also noted. This is done by using pH paper and pH meter to indicate the change in pH. The pH of the formalin in the chamber is recorded to see the pattern of change in acidity. c) Fixed: * The type of toxin chosen; formaldehyde Liquid formalin is selected to be one of the fixed variables in this experiment so that the analysis of the change in acidity can be done easily. More than one type of pollutant will promote confusion while conducting the experiment as the characteristic of one pollutant differ from one to another. Formalin is the aqueous state of the chemical formaldehyde and the concentration of the liquid formalin is 100%. I made the volume and the concentration of liquid formalin the same in every small beaker included in every transparent chamber. It is important to do so because the pH of the chemical and its mass are to be checked every 2 days throughout the duration of the experiment. The initial pH of the chemical is 3.510 while the initial volume of the chemical is 10 0.5 ml making its mass to be 10.19 0.01 g * The estimated size of the plants chosen The chosen plants are of the same size. There is no specific measurement for the plants sizes so therefore, the size is depending on the experimenters justification by fixing the number of leaves present in every plant chosen. This is due to the mechanism of the absorption of the chemical formalin happens through the microscopic opening present on the leaves; the stomata. It is therefore can be predicted that more tiny opening present on the leaves, the more effective would the rate of absorption be. I decided that the total number of leaves is approximately 15-20 leaves depending on the how broad the surface of the leaves is. * The size of the pyramidal transparent chamber The size of the pyramidal transparent chamber is to be made constant by using the same size and number of transparent plastic bags. The size of the plastic bags is 23cm x 38cm and they are cut into same shapes to fit it with the skeleton of the chamber. The base of the chamber is triangular in shape and constant with the area of (50cm x 50cm). 5.2 Materials MATERIALS QUANTITY JUSTIFICATION Formalin 120ml Formalin acts as the toxin in the experiment. Tap Water 5 litres This is used to water the plants everyday for 2 weeks duration. 5.3 Apparatus APPARATUS QUANTITY JUSTIFICATION Boston fern (N. exaltata) 1 pot These are the plants chosen to determine their effectiveness to absorb the formalin. Janet Craig (D. deremensis) 1 pot Florists mum (C. morifolium) 1 pot Kimberly queen fern (N. obliterata) 1 pot Snake plant (S. trifasciata) 1 pot Himalayan Balsam (I. glandulifera) 1 pot pH paper 1 box To check the acidity of formalin every 2 days. pH meter 1 To determine the pH of the formalin every 2 days. Disposable plastic cups 24 To be the base of the pyramidal transparent chamber. Plastic and bamboo chopsticks 54 To be the poles of the pyramidal transparent chamber. Electronic balance 1 To measure the decrease in mass of the liquid formalin for every 2 days. 50ml beaker 6 To place the liquid formalin in each chamber. 50ml measuring cylinder 1 To measure the amount of formalin in each 50ml beaker. Transparent plastics for packaging (23cm x 38cm) 1 pack To become the cover of the chamber. 5.4 Methodology to prepare a chamber for the plant A chamber has to be invented to place the chosen plants, considering the needs of those plants to get sufficient sunlight, air and water. I chose transparent plastics and attach them together to create a pyramidal transparent chamber. Holes were also poked to allow air move into the chamber. I included nine chopsticks to be the poles of chamber. A pole comprised of 3 combined chopsticks. To increase its stability, I poked a hole onto the bases of three disposable plastic cups and inserted the chopsticks into the holes. 5.5 Methodology to determine the change in acidity of formaldehyde After the chamber was set up, I prepared the solution of the toxin chosen; formalin.in a 50ml beaker. 10 0.5 ml of the chemical in each beaker was measured using 50ml measuring cylinder. 6 transparent chambers were set up to place 6 types of plants which were the Boston fern (N. exaltata), Janet Craig (D. deremensis), Florists mum (C. morifolium), Kimberly queen fern (N. obliterata), Snake plant (S. trifasciata), and Himalayan Balsam (I. glandulifera). All the 6 chambers contained different pots of plants and 10ml of formalin in a 50ml beaker. At intervals of 2 days, the mass of the formalin was recorded. The procedure to get the mass of formalin in each chamber was as follows; * Take the reading of the mass of 50ml beaker before filling in the formalin by using electronic balance. Repeat the steps 3 times in order to get the average reading. * Weigh the 50ml beaker containing formalin by using electronic balance. Repeat the procedure 3 times in order to get the average reading. The reading of the mass of the formalin + 50ml beaker at intervals of 2 days was recorded. The mass of the formalin was determined by subtracting the average value of the mass of formalin + 50ml beaker with the average mass of the 50ml beaker. The pH was again checked by using pH paper and also pH meter for 2 weeks. The change in colour of the pH paper and the reading of the pH meter were noted and documented. Each of the plants in the chamber was watered once a day using tap water. The amount of tap water must was 20ml per watering and watering time was at 10.30 a.m and 4.00 p.m. every day. Condition for each of the plants was observed for interval time of 2 days. All of results were recorded in a table. 5.5.1 Precaution 1. Beware while handling formalin because it is a dangerous chemical. Since a high concentration of formaldehyde will be used in the experiment, [13]it may cause burning sensation to the eyes, nose and lungs. Thus it could result in allergic reaction because of formalin. 2. Be cautious when building the pyramidal transparent chamber especially when dealing with the bamboo sticks. Avoid any sharp splinter of the bamboo stick from piercing into the skin. 6.0 Data collection TABLE 1: THE pH of FORMALIN IN EACH TRANSPARENT CHAMBER WITH DIFFERENT PLANTS IN 14 DAYS Transparent chamber containing plants Value of Ph of formalin in each transparent chamber according to number of days 2 days 4 days 6 days 8 days 10 days 12 days 14 days Boston fern (N. exaltata Bostoniensis) 3.510 3.550 3.570 4.020 4.130 4.260 4.310 Janet Craig (D. deremensis) 3.510 3.570 3.580 4.020 4.070 4.210 4.430 Florists mum (C. morifolium) 3.510 3.570 3.590 4.120 4.200 4.320 4.620 Kimberly queen fern (N. obliterate) 3.510 3.510 3.520 4.010 4.030 4.050 4.110 Snake plant (S. trifasciata Laurentii) 3.510 3.370 3.360 4.030 4.030 4.030 4.030 Himalayan Balsam (I. glandulifera) 3.510 3.370 3.370 3.350 3.350 3.350 3.350 Note: The pH of formalin in each beaker was checked at the same interval to ensure that none of the formalin being absorbed more by their respective plants. The time that they were checked was at a range of 4.00 p.m. until 4.45 p.m. 10 Are indoor plants adapted to get rid formaldehyde, Sipin, Elly Lorreta one of the noxious wastes commonly found at home 002348-019 nowadays? TABLE 2: MASS OF FORMALIN + 50ml BEAKER IN EACH CHAMBER CONTAINING DIFFERENT PLANTS IN 14 DAYS Transparent chamber containing plants Mass of formalin + 50ml beaker in each transparent chamber 0.01g 2 days 4 days 6 days 1st 2nd 3rd 1st 2nd 3rd 1st 2nd 3rd Boston fern (N. exaltata) 46.950 46.960 46.960 46.530 46.540 46.550 46.230 46.220 46.220 Janet Craig (D. deremensis) 46.910 46.910 46.910 46.520 46.520 46.510 46.310 46.310 46.310 Florists mum (C. morifolium) 46.940 46.940 46.950 46.610 46.600 46.610 46.350 46.340 46.350 Kimberly queen fern (N. obliterata) 46.970 46.970 46.970 46.620 46.620 46.640 46.430 46.410 46.410 Snake plant (S. trifasciata) 46.920 46.910 46.910 46.620 46.630 46.610 46.420 46.410 46.430 Himalayan Balsam(I. glandulifera) 46.940 46.940 46.930 46.780 46.790 46.790 46.720 46.710 46.720 Note: The mass of the formalin was measured at intervals of 2 days and it was at a range of time from 4.00 p.m. until 4.45 p.m. 10 Are indoor plants adapted to get rid formaldehyde, Sipin, Elly Lorreta one of the noxious wastes commonly found at home 002348-019 nowadays? Transparent chamber containing plants Mass of formalin + 50ml beaker in each transparent chamber 0.01g 8 days 10 days 12 days 14 days 1st 2nd 3rd 1st 2nd 3rd 1st 2nd 3rd 1st 2nd 3rd Boston fern (N. exaltata) 46.010 46.030 46.040 45.480 45.480 45.470 45.210 45.220 45.220 44.950 44.960 44.980 Janet Craig (D. deremensis) 45.520 45.530 45.530 45.030 45.030 45.020 44.960 44.960 44.920 44.580 44.590 44.580 Florists mum (C. morifolium) 45.550 45.550 45.560 45.220 45.210 45.220 44.940 44.940 44.950 44.130 44.130 44.140 Kimberly queen fern (N. obliterata) 45.500 45.510 45.510 45.320 45.350 45.350 44.980 44.980 44.990 44.220 44.230 44.230 Snake plant (S. trifasciata) 45.890 45.900 45.890 45.530 45.530 45.530 45.140 45.140 45.120 44.970 44.960 44.970 Himalayan Balsam(I. glandulifera) 46.680 46.680 46.680 46.340 46.340 46.320 46.290 46.290 47.300 46.250 46.240 46.250 10 Are indoor plants adapted to get rid formaldehyde, Sipin, Elly Lorreta one of the noxious wastes commonly found at home 002348-019 nowadays? Transparent chamber containing plants Change in colour of pH paper 2 days 4 days 6 days 8 days 10 days 12 days 14 days Boston fern (N. exaltata) Green leaves Green leaves Green leaves Green leaves Green leaves Green leaves Green leaves Janet Craig (D. deremensis) Green leaves Green leaves Green leaves Green leaves Green leaves Yellow leaves Brown Leaves Florists mum (C.morifolium) Green leaves Green leaves Green leaves Wilted flowers Wilted flowers Yellow leaves Yellow leaves K. queen fern (N. obliterata) Green leaves Green leaves Green leaves Green leaves Yellow leaves Yellow leaves Yellow leaves Snake plant (S. trifasciata) Green leaves Green leaves Green leaves Green leaves Green leaves Green leaves Green leaves H. Balsam (I. glandulifera) Green leaves Green leaves Yellow leaves Yellow leaves Yellow leaves Brown leaves Brown leaves TABLE 3: DAILY CONDITION OF PLANTS IN THE TRANSPARENT CHAMBERS IN 14 DAYS Note: Only Florists mum (C.morifolium) in this experiment has flowers. When the edges of the leaves becoming brown or yellow, it is indicated as having brown leaves or yellow leaves. The font in italic form indicates the adverse change onto the plants. 10 Are indoor plants adapted to get rid formaldehyde, Sipin, Elly Lorreta one of the noxious wastes commonly found at home 002348-019 nowadays? TABLE 4: CHANGE IN COLOUR OF pH PAPER WHEN pH OF FORMALIN FOR A DURATION OF TWO WEEKS Transparent chamber containing plants Change in colour of pH paper 2 days 4 days 6 days 8 days 10 days 12 days 14 days Boston fern (N. exaltata ) Light orange Light orange Light orange Light orange Light orange Light orange Light orange Janet Craig (D. deremensis) Light orange Light orange Light orange Light orange Light orange Light orange Light orange Florists mum (C. morifolium) Light orange Light orange Light orange Light orange Light orange Light orange Light orange K. queen fern (N. obliterata) Light orange Light orange Light orange Light orange Light orange Light orange Light orange Snake plant (S. trifasciata) Light orange Light orange Light orange Light orange Light orange Light orange Light orange H. Balsam (I. glandulifera) Light orange Light orange Light orange Light orange Light orange Light orange Light orange Note: The original colour of the pH paper is light yellow in colour 10 Are indoor plants adapted to get rid formaldehyde, Sipin, Elly Lorreta one of the noxious wastes commonly found at home 002348-019 nowadays? 7.0 Data processing 7.1 pH difference of formalin I discover that there are some changes in pH of the formalin in the transparent chamber. The following table shows the total difference in the final and the initial pH of the formalin in each transparent chamber. TABLE 5: DIFFERENCE IN pH OF FORMALIN IN EACH TRANSPARENT CHAMBER Transparent chamber containing plants Final pH Initial pH Difference in pH Boston fern (N. exaltata) 4.310 3.510 0.800 Janet Craig (D. deremensis) 4.430 3.510 0.920 Florists mum (C. morifolium) 4.620 3.510 1.110 Kimberly queen fern (N. obliterate) 4.110 3.510 0.600 Snake plant (S. trifasciata) 4.030 3.510 0.520 Himalayan Balsam (I. glandulifera) 3.350 3.510 0.160 Note: The method to calculate the pH of formalin in chamber containing Himalayan Balsam is inverted, since the pH value decreased so that negative value can be ignored. 7.2 Data for mean mass of formalin The following table shows the average mass of formalin + 50ml beaker for 14 days TABLE 6: AVERAGE MASS OF FORMALIN + 50ml BEAKER IN EACH CHAMBER CONTAINING DIFFERENT PLANTS IN 14 DAYS Transparent chamber containing plants Average mass of formalin+50ml beaker in each chamber 0.01g Day 2 Day 4 Day 6 Day 8 Day 10 Day 12 Day 14 Boston fern (N. exaltata) 46.960 46.540 46.220 46.030 45.480 45.220 44.960 Janet Craig (D. deremensis) 46.910 46.520 46.310 45.530 45.030 44.950 44.580 Florists mum (C. morifolium) 46.940 46.610 46.350 45.550 45.220 44.540 44.130 K. queen fern (N. obliterate) 46.970 46.630 46.420 45.510 45.340 44.980 44.240 Snake plant (S. trifasciata) 46.910 46.620 46.420 45.890 45.330 45.130 44.970 H. Balsam (I. glandulifera 46.940 46.790 46.720 46.680 46.330 46.290 44.250 Note: The average masses were obtained by totaling up the three mass values in three trials, and divide it into three. 7.3 Graph for the decreasing mass of formalin In order to get a graph of decrease in mass of formalin from day 0 to day 14, the real mass of formalin is required. Therefore, the table of mass of formalin for a duration of 14 days is made as follows. The formulation to calculate the mass of formalin in each beaker would be; Mass of formalin= [(Average mass of formalin+50ml beaker)- Average mass of 50ml beaker] TABLE 7: MASS OF FORMALIN IN EVERY 50ml BEAKER CONTAINED IN TRANSPARENT CHAMBER WITH DIFFERENT TYPES OF PLANTS Transparent chamber containing plants Mass of formalin 0.01g [(Average mass of formalin+50ml beaker) Average mass of 50ml beaker] Day 2 Day 4 Day 6 Day 8 Day 10 Day 12 Day 14 Boston fern (N. exaltata) 10.170 9.750 9.430 9.240 8.690 8.430 8.170 Janet Craig (D. deremensis) 10.120 9.730 9.520 8.740 8.240 8.160 7.790 Florists mum (C. morifolium) 10.150 9.820 9.560 8.760 8.430 8.150 7.340 K. queen fern (N. obliterate) 10.180 9.840 9.630 8.760 8.430 8.150 7.450 Snake plant (S. trifasciata) 10.120 9.830 9.630 9.100 8.540 8.340 8.180 H. Balsam (I. glandulifera 10.150 10.000 9.930 9.890 9.540 9.500 9.460 Note: The average mass of one 50ml beaker is 36.79 0.1g. This value was used to calculate the mass above. The bar graph of decrease in mass of the formalin against number of days for each beaker containing formalin in every transparent chamber is as follows; graph 1: decrease in mass of the formalin against number of days for each beaker containing formalin in every transparent chamber Note: The graph shows quite obvious inclination of mass of formalin in all chambers except for the H. Balsam (I. glandulifera) 7.4 Mass and percentage of formalin absorbed The initial average mass of the 10ml formalin in the 50ml beaker is 46.980 0.01g and the average mass of the 50ml beaker alone is 36.790 0.01g making the mass of the 10.000 0.1 ml formalin poured in to be 10.190 0.01g. From the data, there is a decreasing pattern of the mass of the formalin in the 50ml beaker. The percentage of decrease in mass of the 10.000 0.1 ml formalin in 14 days of time in respective transparent chamber of plants can be determined. Before that, the mass of formalin absorbed in all the 6 transparent chambers must be d up. Calculation is as follows; TABLE 8: MASS OF FORMALIN ABSORBED BY PLANTS IN EACH CHAMBER Name of plants in each chamber Mass of formalin absorbed [Initial mass (10.190)- Mass on the14th day] 0.01g Boston fern (N. exaltata) 2.020 Janet Craig (D. deremensis) 2.400 Florists mum (C. morifolium) 2.850 Kimberly queen fern (N. obliterate) 2.740 Snake plant (S. trifasciata) 2.010 H. Balsam (I. glandulifera 0.730 Note: The mass of formalin absorbed by plants in each chamber is referring to the decrease in mass of formalin throughout the 12 days duration. It is possible to calculate the percentage of decrease in mass of formalin absorbed by using the formulation below. The table below shows the percentage in respective 50ml beaker of formalin in all 6 chambers; Percentage of decrease in = Mass of formalin absorbed x 100% mass of formalin Initial mass of formalin TABLE 9: PERCENTAGE DECREASE IN MASS OF FORMALIN IN THE 50ml BEAKER IN RESPECTIVE TRANSPARENT CHAMBER Transparent chamber containing plants Percentage of decrease in mass of formalin absorbed Percentage of decrease in mass of formalin (%) Boston fern (N. exaltata) 2.020/10.190 x 100 19.820 Janet Craig (D. deremensis) 2.400/10.190 x 100 23.550 Florists mum (C. morifolium) 2.850/10.190 x 100 27.970 Kimberly queen fern (N. obliterate) 2.740/10.190 x 100 26.890 Snake plant (S. trifasciata) 2.010/10.190 x 100 19.730 Himalayan Balsam (I. glandulifera) 0.730/10.190 x 100 7.160 Note: The comparison of decrease in mass of formalin in beaker is based on the initial mass of formalin in the beaker. The greater the percentage of decrease in masses of formalin, the better the quality of air in the chamber, the better formalin absorber would the plant be. The following diagram shows the ascending order of the quality of plant as formalin absorber. Himalayan Balsam (I. glandulifera) Snake plant (S. trifasciata) Boston fern (N. exaltata) Janet Craig (D. deremensis) Kimberly queen fern (N. obliterate) Florists mum (C. morifolium) 7.5 Calculation for mean, standard deviation and T-test TABLE 10 : TABLE OF MEAN AND STANDARD DEVIATION FOR EVERY PLANTS CHOSEN Mass 0.01g Plants Boston fern (N. exaltata) Janet Craig (D. deremensis) Florists mum (C. morifolium) Kimberly queen fern (N. obliterata) Snake plant (S. trifasciata) Himalayan Balsam (I. glandulifera) 1st trial 2.000 2.330 2.810 2.000 1.950 0.690 2nd trial 2.000 2.320 2.810 2.740 1.950 0.700 3rd trial 1.980 2.330 2.810 2.740 1.940 0.680 Mean 1.993 2.327 2.810 2.493 1.947 0.690 Std. Dev 0.009 0.005 0.000 0.349 0.005 0.008 Note: The mean was determined by getting the difference of mass of formalin between 14th day with the 0 day; initial mass. The formulation to calculate t-test is as follows; t-value =_____difference in mean___ difference of standard error TABLE 11: TABLE OF T-VALUE FOR THE COMPARISON OF MASS DECREASE MEAN BETWEEN BOSTON FERN (N. exaltata) AND JANET CRAIG (D. deremensis) Mass 0.01g Plants Boston fern (N. exaltata) Janet Craig (D. deremensis) Difference between Boston fern and Janet Craig 1 trial 2.000 2.330 0.330 2 trial 2.000 2.320 0.320 3 trial 1.980 2.330 0.340 Mean 1.993 2.327 0.330 Std. Dev 0.009 0.005 0.008 Std. Error 1.151 1.343 0.191 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.728 Null Hypothesis: There is no significance difference for decrease in mass between Boston fern (N. exaltata) and Janet Craig (D. deremensis) | t | = 1.728 4.300 Thus, null hypothesis is rejected. The mean difference is not significant TABLE 12: TABLE OF T-VALUE FOR THE COMPARISON OF MASS DECREASE MEAN BETWEEN BOSTON FERN (N. exaltata) AND FLORISTS MUM (C. morifolium) Null Hypothesis: There is no significance difference for decrease in mass between Boston fern (N. exaltata) and Florists mum (C. morifolium) Mass 0.01g Plants Boston fern (N. exaltata) Florists mum (C. morifolium) Difference between Boston fern and Florists mum 1 trial 2.000 2.810 0.810 2 trial 2.000 2.810 0.810 3 trial 1.980 2.810 0.810 Mean 1.993 2.810 0.810 Std. Dev 0.009 0.000 0.000 Std. Error 1.151 1.622 0.468 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.731 | t | = 1.731 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 13: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN BOSTON FERN (N. exaltata) AND KIMBERLY QUEEN FERN (N. obliterata) Null Hypothesis: There is no significance difference for decrease in mass between Boston fern (N. exaltata) and Kimberly queen fern (N. obliterata) Mass 0.01g Plants Boston fern (N. exaltata) Kimberly queen fern (N. obliterata) Difference between Boston fern and Kimberly queen fern 1 trial 2.000 2.000 0.810 2 trial 2.000 2.740 0.810 3 trial 1.980 2.740 0.810 Mean 1.993 2.493 0.810 Std. Dev 0.009 0.349 0.000 Std. Error 1.151 1.439 0.468 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.730 | t | = 1.730 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 14: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN BOSTON FERN (N. exaltata) AND SNAKE PLANT (S. trifasciata) Null Hypothesis: There is no significance difference for decrease in mass between Boston fern (N. exaltata) and Snake plant (S. trifasciata) Mass 0.01g Plants Boston fern (N. exaltata) Snake plant (S. trifasciata) Difference between Boston fern and Snake plant 1 trial 2.000 1.950 0.050 2 trial 2.000 1.950 0.050 3 trial 1.980 1.940 0.040 Mean 1.993 1.950 0.050 Std. Dev 0.009 0.005 0.005 Std. Error 1.151 1.126 0.029 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.724 | t | = 1.724 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 15: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN BOSTON FERN (N. exaltata) AND HIMALAYAN BALSAM (I. glandulifera) Null Hypothesis: There is no significance difference for decrease in mass between Boston fern (N. exaltata) and Himalayan Balsam(I. glandulifera) Mass 0.01g Plants Boston fern (N. exaltata) Himalayan Balsam(I. glandulifera) Difference between Boston fern and Himalayan Balsam 1 trial 2.000 0.690 1.310 2 trial 2.000 0.700 1.300 3 trial 1.980 0.680 1.300 Mean 1.993 0.690 1.303 Std. Dev 0.009 0.008 0.005 Std. Error 1.151 0.398 0.752 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.733 | t | = 1.733 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 16: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN JANET CRAIG (D. deremensis) AND FLORISTS MUM (C. morifolium) Null Hypothesis: There is no significance difference for decrease in mass between Janet Craig (D. deremensis) and Florists mum (C. morifolium) Mass 0.01g Plants Janet Craig (D. deremensis) Florists mum (C. morifolium) Difference between Janet Craig and Florists mum 1 trial 2.330 2.810 0.480 2 trial 2.320 2.810 0.490 3 trial 2.330 2.810 0.480 Mean 2.327 2.810 0.483 Std. Dev 0.005 0.000 0.005 Std. Error 1.343 1.622 0.279 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.732 | t | = 1.732 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 17: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN JANET CRAIG (D. deremensis) AND KIMBERLY QUEEN FERN (N. obilterata) Null Hypothesis: There is no significance difference for decrease in mass between Janet Craig (D. deremensis) and Kimberly queen fern (N. obliterata) Mass 0.01g Plants Janet Craig (D. deremensis) Kimberly queen fern (N. obliterata) Difference between Janet Craig and Kimberly queen fern 1 trial 2.330 2.000 0.330 2 trial 2.320 2.740 0.420 3 trial 2.330 2.740 0.410 Mean 2.327 2.493 0.387 Std. Dev 0.005 0.349 0.040 Std. Error 1.343 1.440 0.223 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.734 | t | = 1.734 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 18: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN JANET CRAIG (D. deremensis) AND SNAKE PLANT (S. trifasciata) Null Hypothesis: There is no significance difference for decrease in mass between Janet Craig (D. deremensis) and Snake plant (S. trifasciata) Mass 0.01g Plants Janet Craig (D. deremensis) Snake plant (S. trifasciata) Difference between Janet Craig and Snake plant 1 trial 2.330 1.950 0.380 2 trial 2.320 1.950 0.370 3 trial 2.330 1.940 0.390 Mean 2.327 1.950 0.380 Std. Dev 0.005 0.005 0.008 Std. Error 1.343 1.126 0.219 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.735 | t | = 1.735 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 19: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN JANET CRAIG (D. deremensis) AND HIMALAYAN BALSAM (I. glandulifera) Null Hypothesis: There is no significance difference for decrease in mass between Janet Craig (D. deremensis) and Himalayan Balsam(I. glandulifera) Mass 0.01g Plants Janet Craig (D. deremensis) Himalayan Balsam(I. glandulifera) Difference between Janet Craig and Himalayan Balsam 1 trial 2.330 0.690 1.640 2 trial 2.320 0.700 1.620 3 trial 2.330 0.680 1.650 Mean 2.327 0.690 1.640 Std. Dev 0.005 0.008 0.013 Std. Error 1.343 0.398 0.947 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.732 | t | = 1.732 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 20: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN FLORISTS MUM (C. morifolium) AND KIMBERLY QUEEN FERN (N. obliterata) Null Hypothesis: There is no significance difference for decrease in mass between Florists mum (C. morifolium) and Kimberly queen fern (N. obliterata) Mass 0.01g Plants Florists mum (C. morifolium) Kimberly queen fern (N. obliterata) Difference between Florists mum and Kimberly queen fern 1 trial 2.810 2.000 0.810 2 trial 2.810 2.740 0.070 3 trial 2.810 2.740 0.070 Mean 2.810 2.493 0.327 Std. Dev 0.000 0.349 0.349 Std. Error 1.622 1.439 0.189 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.730 | t | = 1.730 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 21: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN FLORISTS MUM (C. morifolium) AND SNAKE PLANT (S. trifasciata) Null Hypothesis: There is no significance difference for decrease in mass between Florists mum (C. morifolium) and Snake plant (S. trifasciata) Mass 0.01g Plants Florists mum (C. morifolium) Snake plant (S. trifasciata) Difference between Florists mum and Snake plant 1 trial 2.810 1.950 0.860 2 trial 2.810 1.950 0.860 3 trial 2.810 1.940 0.870 Mean 2.810 1.950 0.860 Std. Dev 0.000 0.005 0.005 Std. Error 1.622 1.126 0.497 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.730 | t | = 1.730 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 22: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN FLORISTS MUM (C. morifolium) AND HIMALAYAN BALSAM (I. glandulifera) Null Hypothesis: There is no significance difference for decrease in mass between Florists mum (C. morifolium) and Himalayan Balsam (I. glandulifera) Mass 0.01g Plants Florists mum (C. morifolium) Himalayan Balsam(I. glandulifera) Difference between Florists mum and Himalayan Balsam 1 trial 2.810 0.690 2.120 2 trial 2.810 0.700 2.110 3 trial 2.810 0.680 2.130 Mean 2.810 0.690 2.120 Std. Dev 0.000 0.008 0.008 Std. Error 1.622 0.398 1.223 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.733 | t | = 1.733 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 23: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN KIMBERLY QUEEN FERN (N. obliterata) AND SNAKE PLANT (S. trifasciata) Null Hypothesis: There is no significance difference for decrease in mass between Kimberly queen fern (N. obliterata) and Snake plant (S. trifasciata) Mass 0.01g Plants Kimberly queen fern (N. obliterata) Snake plant (S. trifasciata) Difference between Kimberly queen fern (N. obliterate) 1 trial 2.000 1.950 0.050 2 trial 2.740 1.950 0.790 3 trial 2.740 1.940 0.800 Mean 2.493 1.950 0.547 Std. Dev 0.349 0.005 0.351 Std. Error 1.439 1.126 0.316 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.731 | t | = 1.731 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 24: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN KIMBERLY QUEEN FERN (N. obliterata) AND HIMALAYAN BALSAM (I. glandulifera) Null Hypothesis: There is no significance difference for decrease in mass between Kimberly queen fern (N. obliterata) and Himalayan Balsam(I. glandulifera) Mass 0.01g Plants Kimberly queen fern (N. obliterata) Himalayan Balsam(I. glandulifera) Difference between Kimberly queen fern and Himalayan Balsam 1 trial 2.000 0.690 1.310 2 trial 2.740 0.700 2.040 3 trial 2.740 0.680 2.060 Mean 2.493 0.690 1.803 Std. Dev 0.349 0.008 0.349 Std. Error 1.439 0.398 1.041 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.732 | t | = 1.732 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. TABLE 25: TABLE OF T-VALUE FOR THE COMPARISON OF MEAN BETWEEN SNAKE PLANT (S. trifasciata) AND HIMALAYAN BALSAM (I. glandulifera) Null Hypothesis: There is no significance difference for decrease in mass between Snake plant (S. trifasciata) and Himalayan Balsam(I. glandulifera) Mass 0.01g Plants Snake plant (S. trifasciata) Himalayan Balsam(I. glandulifera) Difference between Snake plant and Himalayan Balsam 1 trial 1.950 0.690 1.260 2 trial 1.950 0.700 1.250 3 trial 1.940 0.680 2.620 Mean 1.950 0.690 1.710 Std. Dev 0.005 0.008 0.643 Std. Error 1.126 0.398 0.987 Degree of freedom 2.000 Critical value at 5% level 4.300 t-value 1.733 | t | = 1.732 4.300 Thus, null hypothesis is rejected. The mean difference is not significant. 7.6 Analysis on mass of formalin From my prediction, only the indoor plants could eliminate indoor pollutants; as in this case the chemical formalin. While the outdoor plants are unable to do so. But I found out that decrease in mass of formalin in the entire 50ml beakers that contained the chemical is influenced by the ineffectiveness of the transparent chamber, which therefore rejects the theory that indoor plants transpiration alone could remove the pollutant. The ineffectiveness is referring to the external air movement that causes evaporation of formalin. According to the data processed, the percentage of formalin absorbed by each of the plants shows a very close difference to one another and it is irrelevant to assume that all the formalin that was lost was via transpiration. Further research was made to explain these big differences. I concluded that the difference in rate of transpiration in plants affect the rate at which volume of formalin is decreased. Thus, the greater the transpiration rates of a plant, the better quality of air it produces. My assumption on this is because of the availability amount of water vapor that could be emitted out by the leaves of the plants is great when the rate of transpiration of a plant is high. This enables more mixing of the water vapor in the atmosphere with the vaporized chemical. This means, there would be more food that is available to be broken down by the microbes at the root system of a plant. Of all the six plants chosen to be experimented, only 4 of them have high transpiration rates. They are Janet Craig (D. deremensis), Boston fern (N. exaltata), Kimberly queen fern (N. obliterate), and Florists mum (C. morifolium)[14]. The other two plants which are Snake plant (S. trifasciata) and Himalayan Balsam (I. glandulifera) have a much lower rate of transpiration when compared with the 4 plants. 7.7 Analysis on pH of formalin Formalin is acidic in nature and there should not be any change in pH of the formalin in the beaker because generally, a buffer; carboxylic acid is present in formalin.[15] With the buffer, the solution would be able to resist any change in pH even though there is any external factor that could alter the pH of the solution. Hence, the pH should remain constant. But since there is a change in the value of pH of the formalin contained in the 50ml beaker in all of the 6 transparent chambers, I then make an early assumption that the carboxylic acid that is contained in the solution does not buffer.[16] That was why the pH value of the formalin in the beaker in all of the 6 transparent chambers increased throughout the 14 days duration of experiment. There must be an external factor present in the chamber that affected the acidity of the formalin. Throughout my findings, I found an explanation for this. Formalin acts as a reducing agent.[17] Thus it can undergo oxidation which could relea se its hydrogen.[18] As formalin evaporates, it is being oxidized to become CHO+. The hydrogen would then combine with the water vapour emitted by the plants via transpiration; from H2O, becoming H3O+. These ions which are available in the air would be fixed by the microbes at the roots of the plants, becoming the source of food for the plants. 7.8 Analysis on the external condition of plants The external condition of the plants becoming worse as the experiment was carried out. Alteration of the colour of the edge of the leaves form green to yellow[19] and brown[20] and wilting of flowers[21] was due to the insufficient of water. Thus, it can be said that when the plant is lack of water, it would not be efficient in removing pollutants. The plants got dried up; changing colour from green to brown thus there was no stomata opening because the guard cells die. Eventually, the transfer of chemical downwards to the roots of the plants will not happen. It can be assumed that for plants that got dried up towards the end of experiment that the colour of the leaves started to become brown, the rate at which the mass of formalin decreases was not mainly supported by the process of vaporized chemical being absorbed through the stomata. Perhaps the colonies of bacteria are still present at the roots of the plants with this condition but the rate of decrease of formalin mass is reduc ed, not as rapid as the initial rate. 8.0 Conclusion From the data obtained, I conclude that one of the observable changes in the quantity of the formalin is the mass. This is influenced by the evaporation of the formalin. The percentage of formalin that could evaporate is minimized by having the transparent chamber to cover the plant and 10ml of formalin which means the evaporation of formalin was not greatly affected by wind movement. It is therefore possible for chosen indoor plants; Janet Craig (D. deremensis), Boston fern (N. exaltata), Kimberly queen fern (N. obliterate), Florists mum (C. morifolium), Snake plant (S. trifasciata) and Boston fern (N. exaltata) to remove the toxin, formaldehyde as there is quite large decrease in the mass of the formalin. Though Snake plant (S. trifasciata) does not have a high transpiration rate, it can still remove formalin quite large in quantity. I believe that this is affected by the factors such as number of holes poked onto the chamber, the external wind movement and some more mentioned in t he evaluation part. (Please refer to 9.0) As for Himalayan Balsam (I. glandulifera), an example of outdoor plant, it is probably able to remove indoor pollutant but just in small percentage as seen in the ranking done in the data processing, this plant provides the lowest quality of air in the chamber due to contamination by the pollutant formalin. It could just remove formalin for about 7.160% in 14 days duration. Besides, the only plant that experienced a rapid external change was Himalayan Balsam (I. glandulifera). The edges of the leaves become brown on the twelfth day. One of the closest possibilities is due to the failure of the experimenter to follow the period of watering everyday that it received less water.[22] It might be due to a very small surface area of this plant that it was deficient for it to cope with the concentrated amount of formalin in the chamber added as before the experiment was conducted properly, I have tried including the same amount of formalin in a chamber containing less than 10 leaves and the same result occurred during the 8 day interval. Referring to the T-test in table 11 to table 25, all of the t- values were rejected because it lied in the critical region. The null hypothesis selected suggested that there was no significance between the differences of mass decrease between the plants. For the data to correspond with the predicted result, null hypothesis should be accepted. But because the pattern of formalin mass decreasing was too small from one interval to another, some of the values of standard deviation obtained are zero. That was why the mass differences between the plants are not significant when the t-values are all less than the critical value at 5% level which is 4.300. Thus, the null hypothesis stated that the mass differences are not significant to each other. This did not indicate the unreliability of the data but it showed that limitations and weaknesses were present in the experiment. There is an increase in pH of the formalin and the pattern somehow shows that the acidity of the chemical has already decreased. The assumption made through the research made for this experiment was indoor plants have the ability to get rid of formaldehyde, one of the noxious wastes commonly found at home nowadays. 9.0 Evaluation There were several weaknesses in the procedure throughout the experiment. I have come out with some suggestions on what to be improved on the aspect of methodology and the aspect of apparatus and materials used so that the experiment would give a reliable result when it is repeated in the future. Regarding the technique to measure the acidity of chemical formalin, I found out that it was unreliable. The pH meter sometimes failed to function well. Sometimes, pH meter detect acidic chemical to be having a pH value more than 7. It is highly suggested to use colourimeter because this instrument could give the exact concentration of hydrogen, H+ in the solution[23] enabling the calculation of its pH by using the following formulation; pH= -log10 (concentration of H+ ions) Next, there was failure to exactly follow the watering time for all plants in the chamber which eventually affected the external condition of plants. Thus, a time table has to be prepared in the future so that the watering time is made standard each day. Other than that, there was a pot of plant with one of its leaves had fallen into the beaker containing formalin. This should not happen because it could have left effect onto the acidity of the formalin. Place the beaker of formalin further from the pot of plant in the chamber so that neither leaves nor flowers would fall into the beaker. Lastly, number of holes poked onto some of the transparent chambers which were not fixed. In my opinion, this is one of the causes of inefficiency of the chamber. The more holes present, the more rapid would the evaporation of chemical be. Therefore, fix the number of holes poked. For a better result, use a square plastic aquarium being inverted with 2 square polystyrene as its base. This would allow less movement of air but able to provide oxygen for the plants in the camber. All in all, the hypothesis of the experiment is accepted. It is proven that the indoor plants are able to remove indoor pollutants while plants that are not indoor plants are able to remove indoor pollutants with a lower rate. Thus the public can now use this concept to provide good air quality at homes. 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