Answer:
Amount of heat required = 153.62 J
Explanation:
Given:
Mass = 163.45 g
∆Hvap = 43.3 kJ/mol
Molar mass C₂H₅OH = 46.07 g/mol
Find:
Amount of heat required
Computation:
Amount of heat required = Number of moles x Molar mass C₂H₅OH
Amount of heat required = [163.45/46.07][43.3]
Amount of heat required = 153.62 J
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.
Answer:just breath its ok just use brainly
Explanation:
Answer:
pt1
The half-life of the element is 1,600 years.
pt2
k = ln(2)/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1,600 years) = 4.33 x 10⁻⁴ year⁻¹.
kt = ln([A₀]/[A]),
where, k is the rate constant of the reaction (k = 4.33 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 312.0 g).
[A] is the remaining concentration of the sample ([A] = 9.75 g).
∴ t = (1/k) ln([A₀]/[A]) = (1/4.33 x 10⁻⁴ year⁻¹) ln(312.0 g/9.75 g) = [8,000 year].
-Hops
