| 1 |
What is the primary role of transition metal ion catalysts in the catalytic ozonation process for nanoplastic removal?
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To decrease the reaction time without enhancing removal efficiency |
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because transition metal can easily give and accept electrons this would enchancing the chance of chemical reaction to occurs
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chemical reactions encompass changes that only involve the positions of electrons in the forming and breaking of chemical bonds between atoms
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| 2 |
According to the article, what was the observed effect of using Co2+ at 1 mM on the mineralization rate of polystyrene nanoplastics during ozonation?
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Increased mineralization rate by 70% |
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transition metal can easily give and accept electrons this would enchancing the chance of chemical reaction to occurs thats why the rate is changed
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the highest PSNPs ozonation performance was achieved, decreasing the turbidity up to 65 % and achieving 70 % of mineralization in the same ozonation time.
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| 3 |
In the context of nanoplastics removal, what does the scavenger experiment with methanol demonstrate about the catalytic ozonation process?
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Methanol acts as an alternative catalyst |
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ozonation at disinfection doses reduced the particle size of PSNPs ,the catalytic process demonstrated greater efficacy in the total removal of PSNPs.
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the catalytic process demonstrated greater efficacy in the total removal of PSNPs.
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| 4 |
If the initial concentration of nanoplastics is 20 mg/L and the catalytic ozonation achieves a 70% mineralization rate, what is the concentration of remaining nanoplastics?
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6 mg/L |
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70/100 x20 = 14
20-14 = 6
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70/100 x20 = 14
20-14 = 6
The question asks for remaing nanoplastic
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| 5 |
Given an ozone flow rate of 0.5 NL/min and an ozonation time of 120 minutes, how much ozone (in grams) has been used if the ozone concentration is 10 mg/NL?
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600 grams |
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v=s/t
0.5x120
s=60
c=m/v
10x60=600
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in this point
i give s as a v (volume)
accoding to concentration formula after my calculation, i got the answer m=600
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| 6 |
If reducing the turbidity of water by catalytic ozonation with Co2+ from 100 NTU to 35 NTU represents a 65% reduction, what was the original turbidity?
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100 NTU |
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because it reduces by 65% and 65% of 100 is 65
100-65 = 35 then the original turbidity will be 100
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because it reduce by 65% and 65% of 100 is 65
100-65 = 35 then the original turbidity will be 100
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| 7 |
What is a major benefit of catalytic ozonation over single ozonation in water treatment?
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It provides higher mineralization rates |
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according to the experiment the catalytic ozonation has higher rate of mineralization than single ozonation
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The single ozonation of PSNPs, in the absence of catalyst, led to low turbidity reduction (33 %) and low mineralization rate (16 %) even after 2 h reaction time. Nevertheless, the rapid size reduction of PSNPs by >99 % in less than 5 min of ozonation (TOD: 30 mg L−1) was confirmed. This fact could imply a new issue under the ozone disinfection conditions by the formation of smaller-size particles. However, in the presence of Co2+ (1 mM), the highest PSNPs ozonation performance was achieved, decreasing the turbidity up to 65 % and achieving 70 %
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| 8 |
Which of the following is NOT a transition metal ion used as a catalyst in the study?
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Ca2+ |
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ca2+ is not a transition metal
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because ca2+ is located at the left side of the table that means ca2+ is a alkane earth metal
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| 9 |
What environmental issue does the removal of nanoplastics address?
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Reduction of water pollution |
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because PSNPs cause the water to be contaminated
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the PNSPs molecule is very small even after the single ozonation so it can easily cause water to contaminate
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| 10 |
What analytical technique is NOT mentioned as used for monitoring the degradation of nanoplastics?
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Mass Spectrometry |
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because the GPC is use for finding molecule weight
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because the GPC is use for finding molecule weight so Mass Spectrometry is not neccessary
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| 11 |
What effect does the melt fiber spinning system have on the crystallinity of PET fibers?
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It decreases the crystallinity to less than 10%. |
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test with a one time use water bottle extruded,drawn and spooled as thin fibers that cool by passive heat dissipation rapidly enough to quench the polymer to low crystallinity (<10%)
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https://www.sciencedirect.com/science/article/pii/S2666821124000425
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| 12 |
What role does the spooling speed play in the fiber spinning system described in the article?
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Higher spooling speeds lead to lower crystallinity and smaller fiber diameters. |
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according to article it said that" we estimate the fiber spinning also increases the feedstock surface-area-to-volume ratio by up to 15-fold"
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we estimate the fiber spinning also increases the feedstock surface-area-to-volume ratio by up to 15-fold
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| 13 |
According to the article, what was the impact of using the LCC-ICCG enzyme on PET depolymerization?
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It significantly increased the monomer release from PET. |
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because LCC-ICCG enzyme use as a seperator so it increase the monomer release from PET.
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This enzyme is an engineered version of the leaf branch compost cutinase
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| 14 |
If the initial mass of PET before enzymatic treatment in a bioreactor was 500 grams and 96.9% mass was lost due to depolymerization, what is the final mass of PET?
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484.5 grams |
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96.9/100 x 500 = 484.5
ans 484.5 grams
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96.9/100 x 500 = 484.5
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| 15 |
Considering the average crystallinity of PET fibers after treatment is 9.7%, what would be the crystallinity if the process conditions were unaltered but the drop distance doubled?
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19.4% |
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double distance means double crystallinity
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Small scale PET depolymerization testing was performed at 10 mL volumes in test tubes. Unless otherwise noted, all reactions were performed by incubating select PET melt spin fibers (19 g/L buffer) with purified LCC-ICCG enzyme (1 mg/g PET) in 100 mM potassium phosphate buffer, pH 8 at 65 °C with agitation. All melt spin fibers were spooled from an 18″ drop, except the 0 rpm condition, which was from a 9″ drop. Bench reactions were performed in triplicate in 50 mL centrifuge tubes and agitated at an angle (∼45° tilt) at 200 rpm
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| 16 |
If the surface area to volume ratio of PET increased 15-fold due to processing, and the initial ratio was 0.1 mm²/mm³, what is the new ratio?
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1.5 mm²/mm³ |
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0.1x15
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0.1x15
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| 17 |
What is a major advantage of the melt fiber spinning system over traditional recycling methods?
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Reduces the need for high capital investment |
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according to the article , In conclusion, this pre-processing approach was demonstrated to be effective for real-world PET products and has advantages over other approaches in its low capital cost requirement,
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In conclusion, this pre-processing approach was demonstrated to be effective for real-world PET products and has advantages over other approaches in its low capital cost requirement,
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| 18 |
What does the term 'amorphous content' refer to in the context of PET recycling?
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The unstructured, non-crystalline state of PET |
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amorphus is about the structured of the object
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https://www.sciencedirect.com/science/article/pii/S2666821124000425
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| 19 |
What is the primary benefit of reducing the crystallinity of PET in recycling processes?
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It increases the mechanical strength of the recycled PET. |
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so after the recycling process PET can still be use
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https://www.sciencedirect.com/science/article/pii/S2666821124000425
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| 20 |
Which measurement technique was used to assess the polymer spectra to confirm the presence of PET?
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Fourier Transform Infrared Spectroscopy |
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To see polymer
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https://www.sciencedirect.com/science/article/pii/S2666821124000425
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