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.
Answer:
The chemical formula does not show how the atoms are connected to one another.
Explanation:
With a chemical formula, you can see the types of elements that make up the compound, the number of atoms of each element in a molecule, and the chemical symbols of the elements in the compound.
I think the answer might be nuclear physics
Answer:
composting scraps
recycling is the action or process of converting waste into reusable material.
Question:
A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.
Answer:
The answer to the question is as follows
The mass of solution the student should use is 23.42 g.
Explanation:
To solve the question we note the following
A solution containing 42.7 % w/w of isopropenylbenzene in acetone has 42.7 g of isopropenylbenzene in 100 grams of the solution
Therefore we have 10 g of isopropenylbenzene contained in
100 g * 10 g/ 42.7 g = 23.42 g of solution
Available solution = 120 g
Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.
