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:
1.089%
Explanation:
From;
ν =1/2πc(k/meff)^1/2
Where;
ν = wave number
meff = reduced mass or effective mass
k = force constant
c= speed of light
Let
ν =1/2πc (k/meff)^1/2 vibrational wave number for 23Na35 Cl
ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl
The between the two is obtained from;
ν' - ν /ν = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2
Therefore;
ν' - ν /ν = [meff/m'eff]^1/2 - 1
Substituting values, we have;
ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2 -1
ν' - ν /ν = -0.01089
percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;
ν' - ν /ν * 100
|(-0.01089)| × 100 = 1.089%
Given :
A 10.99 g sample of NaBr contains 22.34% Na by mass.
To Find :
How many grams of sodium does a 9.77g sample of sodium bromine contain.
Solution :
By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.
Therefore , percentage of Na by mass in NaBr will be same for every amount .
Percentage of Na in 9.77 g NaBr is 22.34 % too .
Gram of Na = 
.
Hence , this is the required solution .
Answer:
NH₄⁺
H₂PO₄⁻
H₃O⁺
Explanation:
- An ion is an atom or molecule with a net electric charge due to the loss or gain of one or more electrons.
- Ion may be positively charged "cation" or negatively charged "anion".
- Neutral molecule has a net charge of zero.
<em>So, the species that are ions are: </em>
NH₄⁺
H₂PO₄⁻
H₃O⁺
