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.
STP stands for standard temperature and pressure. Standard pressure is equivalent to 1 atm, and standard temperature is equivalent to 273.15 K. Therefore, your answer is A. the temperature is 273.15 kelvin.
Hope this helps!
Answer:
covalent
Explanation:
The carbon and the nitrogen very often form bonds in nature, carbon-nitrogen bonds, which are covalent types of bonds. In fact, the bonds between the carbon and nitrogen are one of the most abundant in the biochemistry and the organic chemistry. The bonds between these two can be double bonds, as well as triple bonds. The carbon-nitrogen bonds have the tendency to be strongly polarized toward the nitrogen.
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
The answer would be 0.25 g/mL.
I determined the density by dividing the mass by the volume which gives you the density. D = mass/volume.
<span>6 g / 24 mL = 0.25 g/mL
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