Answer:
The answer is B, to convert kelvin to temperature you :
Subtract 273 (k - 273)
And what are the compounds?
The answer is E.
Hope it helps.
Water is an essential part of life and its availability is important for all living creatures. On the other side, the world is suffering from a major problem of drinking water. There are several gases, microorganisms and other toxins (chemicals and heavy metals) added into water during rain, flowing water, etc. which is responsible for water pollution. This review article describes various applications of nanomaterial in removing different types of impurities from polluted water. There are various kinds of nanomaterials, which carried huge potential to treat polluted water (containing metal toxin substance, different organic and inorganic impurities) very effectively due to their unique properties like greater surface area, able to work at low concentration, etc. The nanostructured catalytic membranes, nanosorbents and nanophotocatalyst based approaches to remove pollutants from wastewater are eco-friendly and efficient, but they require more energy, more investment in order to purify the wastewater. There are many challenges and issues of wastewater treatment. Some precautions are also required to keep away from ecological and health issues. New modern equipment for wastewater treatment should be flexible, low cost and efficient for the commercialization purpose.
Explanation:
(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.
1) 100 g of 0.500% (w/w) NaI
Mass of solution = 100 g
Mass of solute = x
Required w/w % of solution = 0.500%
0.500 grams of solute needed to make 100 g of 0.500% (w/w) NaI.
2) 250 g of 0.500% (w/w) NaBr
Mass of solution = 250 g
Mass of solute = x
Required w/w % of solution = 0.500%
1.25 grams of solute needed to make 250 g of 0.500% (w/w) NaBr
3) 500 g of 1.25% (w/w) glucose
Mass of solution = 500 g
Mass of solute = x
Required w/w % of solution = 1.25%
6.25 grams of solute needed to make 500 g of 1.25% (w/w) (glucose)
4) 750 g of 2.00% (w/w) sulfuric acid.
Mass of solution = 750 g
Mass of solute = x
Required w/w % of solution = 2.00%
15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.
