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Brrunno [24]
3 years ago
11

What is the equilibrium constant expression for the ksp of sr3(po4)2?

Chemistry
2 answers:
Eddi Din [679]3 years ago
7 0

The equilibrium constant expression for KSP of Sr3(PO4)2 is


KSP={(Sr^2+)^3 (PO4^3-)^2/ Sr3(PO4)2}


Explanation


write the ionic equation for Sr3(PO4)2


Sr3(PO4)2 → 3Sr^2+ + 2 PO4^3-


KSP is given by (concentration of the products raised to their coefficient /concentration of reactants raised to their coefficient)

Pie3 years ago
6 0

The equilibrium constant expression for KSP of Sr3(PO4)2 is


KSP={(Sr^2+)^3 (PO4^3-)^2/ Sr3(PO4)2}


Explanation


write the ionic equation for Sr3(PO4)2


Sr3(PO4)2 → 3Sr^2+ + 2 PO4^3-


KSP is given by (concentration of the products raised to their coefficient /concentration of reactants raised to their coefficient)

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Answer:

<em>so mass in gram=560grams</em>

Explanation:

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<em>i hope this will help you :)</em>

8 0
3 years ago
Read 2 more answers
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