A plant would not be able to complete photosynthesis without
2 answers:
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Solution :
Given :
The steady state flow = 8000
The concentration of the particulate matter = 18 mg/L
Therefore, the total quantity of a particulate matter in fluid
If 60 mg of alum required for one litre of the water treatment.
So Alum required for
or 480 kg/d
Therefore the alum required is 480 kg/d
1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give
mg of alum ppt. per litre
480 kg of alum will give = 480 x 0.234 kg/d
= 112.32 kg/d ppt of alum
Daily total solid load is
= 256.32 kg/d
So, the total concentration of the suspended solid after alum addition
= 32.04 mg/L
Therefore total alum requirement = 480 kg/d
b). Initial pH = 7.4
The dissociation reaction of aluminium hydroxide as follows :
After addition, the aluminium hydroxide pH of water will increase due to increase in ions.
Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.
c). The reaction of and water as follows :
For the atmospheric pressure :
And the pH is reduced into the range of 5.9 to 6.4
Answer:
The density of Lithium β is 0.5798 g/cm³
Explanation:
For a face centered cubic (FCC) structure, there are total number of 4 atoms in the unit cell.
we need to calculate the mass of these atoms because density is mass per unit volume.
Atomic mass of Lithium is 6.94 g/mol
Then we calculate the mass of four atoms;
⇒next, we estimate the volume of the unit cell in cubic centimeter
given the edge length or lattice constant a = 0.43nm
a = 0.43nm = 0.43 X 10⁻⁹ m = 0.43 X 10⁻⁹ X 10² cm = 4.3 X 10⁻⁸cm
Volume of the unit cell = a³ = (4.3 X 10⁻⁸cm)³ = 7.9507 X 10⁻²³ cm³
⇒Finally, we calculate the density of Lithium β
Density = mass/volume
Density = (4.6097 X 10⁻²³ g)/(7.9507 X 10⁻²³ cm³)
Density = 0.5798 g/cm³