PbI(ii) ionization in the solution of PBI(ii) into water is:
<span>PbI</span>₂(solution) <==> Pb₂⁺ + 2I⁻
If the conc. of PbI(ii) in the sol. is xM then the conc. of Lead(ii) will be x M and conc. of iodide will be 2 x M.
Therefore,
<span>Ksp=<span>[Pb</span></span>²⁺][I-]²
Plugging the values:
1.4×10⁻⁸ = x ⋅ (2x)²
1.4×10⁻⁸ = 4x³
x³ = {1.4×10⁻⁸}÷4
x³ = 0.35 x 10⁻⁸
or
x³ = 3.5 x 10⁻⁹
x = 1.51 x 10⁻³
Hence,
Concentration of iodide ions in the solution:
2x = 3.02 x 10⁻³
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I believe the complete
given is that the bucket of sand contains 9.5 billion grains of sand. That is:
bucket = 9,500,000,000
grains
and the concentration of
brown sand is 2 ppm:
concentration brown sand =
2 / 1,000,000
Therefore:
brown sand = 9,500,000,000
* (2 / 1,000,000)
<span>brown sand = 19,000 grains</span>
density of acetone in gm/cm^3 is 0.791
=28.80cm^3*0.791 g/cm63
=22.78 g
The answer is Joule. Kilograms and Grams is a unit for weight
