Answer:
14 moles AlCl3 x 3 moles PbCl2/2 moles AlCl3 = 21 moles of Pb(NO3)2
Explanation:
Using dimensional analysis and the mole ratios from the balanced equation, we can calculate moles of PbCl2/2 formed from 14 moles of AlCl3...
I hope this helps and if I am wrong I am sorry but i am positive that your answer.
Sc (neutral) [Ar] 3d1 4s2
Sc+ [Ar] 3d1 4s1
Sc2+ [Ar] 3d1
hope this helped!
Answer:
hydrogen oxygen carbon and nitrogen
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
