Answer:it is wrong answer
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The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.
Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O ) = 0
2× ( Oxidation number of Al) +3(−2)=0
2× ( oxidation number of Al) +6
∴ Oxidation number of Al =+3
<u>Answer:</u> The solubility product of silver (I) phosphate is 
<u>Explanation:</u>
We are given:
Solubility of silver (I) phosphate = 1.02 g/L
To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:
Molar mass of silver (I) phosphate = 418.6 g/mol

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of silver (I) phosphate follows:
3s s
The expression of
for above equation follows:

We are given:

Putting values in above expression, we get:

Hence, the solubility product of silver (I) phosphate is 