For lead fluoride Ksp = s³
For Aluminum hydroxide Ksp = s⁴
<u>Explanation:</u>
The expression for the solubility product can be written as the product of the concentration of the ions present in the solution.
PbF₂ → Pb²⁺ + 2 F⁻
Then the Ksp can be written as,
Ksp = [ Pb²⁺] [F⁻]²
If the concentration can be considered as s, then the Ksp can be written as,
Ksp = s× s² = s³
For Aluminum hydroxide on dissociation it gives 4 ions as,
Al(OH)₃ → Al³⁺ + 3 OH⁻
Ksp = [Al³⁺] [OH⁻]³
= s × s³ = s⁴
Thus,For lead fluoride Ksp = s³
For Aluminum hydroxide Ksp = s⁴
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Sirve para considerar la cantidad de espacio que un objeto puede llenar (densidad), medir la energía de los átomos (estado de la materia), analizar los resultados de la combinación de sustancias (reacciones), entre otras muchas cosas interesantes.
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It could be either A or B hope this helps
Answer:
"2.48 mole" of H₂ are formed. A further explanation is provided below.
Explanation:
The given values are:
Mole of Al,
= 3.22 mole
Mole of HBr,
= 4.96 mole
Now,
(a)
The number of mole of H₂ are:
⇒ 
or,
⇒ 
⇒ 
⇒ 
(b)
The limiting reactant is:
= 
(c)
The excess reactant is:
= 
