Answer:
there is 55.44 gr of silver ion in the pool
Explanation:
assuming that the pool has an rectangular area , the volume covered by the pool is
volume of pool = area*depth = 15m* 12m * 2.2 m = 396 m³
assuming that all the pool is covered with the solution that has the silver ion , then
mass = concentration in ppm * density of solution * volume of pool = 0.14*10⁻⁶ gr of silver/gr of solution * 1 gr of solution/1 ml * 396 m³ * 10⁶L/m³ = 0.14*396 gr = 55.44 gr
therefore there is 55.44 gr of silver ion in the pool
The mass of silver ion will be "55 g".
Given:
Length,
Breadth,
Height,
Concentration of Ag ion,
→ The volume covered by the pool will be:
= [tex]l\times b\times h[/tex]
= [tex]15\times 12\times 2.2[/tex]
= [tex]396 \ m^3[/tex]
hence,
→ In reservoir, the mass of Ag ion will be:
= [tex]Concentration\times density\times volume[/tex]
= [tex]0.14\times 10^{-6}\times 1\times 396\times 10^6[/tex]
= [tex]55 \ g[/tex]
Thus the solution above is right.
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