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3 years ago

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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1 year ago

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2 years ago

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2 years ago

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2 years ago

Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s

Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

$N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where N is amount of radioactive nuclide after t time, $N_{0}$ is initial amount of radioactive nuclide and $t_{\frac{1}{2}}$ is half life of radioactive nuclide

Here $N_{0}=4.60 mg$ , $t=4.1\times 10^{9}yr$ and $t_{\frac{1}{2}}=4.5\times 10^{9}yr$

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so, N = 2.446 mg

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2 years ago