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postnew [5]
3 years ago
10

How plastic polymer types of compositions, surface hydrophobicity, crystallinity, porosity, size, shapes, aging or degree of wea

thering of plastics affect adsorption of water pollutants on microplastics?
Chemistry
1 answer:
Andreas93 [3]3 years ago
4 0

Explanation:

Scientific evidences abound of the occurrence of plastic pollution, from mega- to nano-sized plastics, in virtually all matrixes of the environment. Apart from the direct effects of plastics and microplastics pollution such as entanglement, inflammation of cells and gut blockage due to ingestion, plastics are also able to act as vectors of various chemical contaminants in the aquatic environment. This paper provides a review of the association of plastic additives with environmental microplastics, how the structure and composition of polymers influence sorption capacities and highlights some of the models that have been employed to interpret experimental data from recent sorption studies. The factors that influence the sorption of chemical contaminants such as the degree of crystallinity, surface weathering, and chemical properties of contaminants. and the implications of chemical sorption by plastics for the marine food web and human health are also discussed. It was however observed that most studies relied on pristine or artificially aged plastics rather than field plastic samples for studies on chemical sorption by plastics.

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Ozone has a molecular formula of O3. If 7.92*1024 atoms of oxygen react to form ozone, how many
prisoha [69]

337.92 moles of Ozone will be produced

1 Oxygen atom is 8 g

1 mole of ozone, O3 = 8 * 3 = 24 g

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? = 337.92 moles


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brainly.com/question/15356425

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2 years ago
A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heatin
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Answer:

12.371 g

Explanation:

Given :

m_{evaporating\ dish}=1.135\ g

m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g

m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g

m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g

Mass of salt hydrate:

m_{evaporating\ dish}=1.135\ g

m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g

m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g

Mass of salt anhydrous:

m_{evaporating\ dish}=1.135\ g

m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g

m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g

Mass of water:

m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g

m_{water}=12.371\ g

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4 years ago
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Three ways (out of the many) are:

Change of temperature
Color change
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3 years ago
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