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Mashcka [7]
3 years ago
7

BRAINLIEST IF ANSWERED IN THE NEXT 5M

Chemistry
1 answer:
belka [17]3 years ago
5 0
(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻
                      k=2        k=1
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Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.
madreJ [45]

Answer:

Molar \ solubility=3.12x10^{-5}M

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-

The equilibrium expression is:

Ksp=[Ca^{2+}][F^-]^2

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent x is computed as follows:

3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M

Thus, the molar solubility equals the reaction extent x, therefore:

Molar \ solubility=3.12x10^{-5}M

Regards.

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Explanation:

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Pls Brainliest! It would mean a lot! ;)

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