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raketka [301]
3 years ago
11

The solubility of CaF2 is 0.00021 mole per liter. What is the solubility product constant for CaF2? A) 7.3 x 10-12 B) 3.7 x 10-1

1 C) 8.5 x 10-8 D) 4.4 x 10-8 E) 1.9 x 10-11
Chemistry
1 answer:
Naddika [18.5K]3 years ago
3 0

Answer:

B) Ksp = 3.7 E-11

Explanation:

CaF2 ↔ Ca2+  +  2F-

  S            S          2S,,,,,,,,,,,,,,,in the equilibrium

⇒ Ksp = [ Ca2+ ] * [ F- ]²

⇒ Ksp = S * ( 2S )²

⇒ Ksp = 4S³

⇒ Ksp = 4 * ( 0.00021 )³

⇒ Ksp = 4 * 9.261 E-12

⇒ Ksp = 3.704 E-11

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Two solutions are created by mixing one solution containing lithium nitrate with one containing sodium phosphate.
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Answer:

Solution A that will form a precipitate with Ksp = 2.3 x 10−4

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                                  Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

                                                     3S               S

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<u>Solution A </u>

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0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole

                                     3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄

(Mole/Stoichiometry)    \frac{0.15}{3}                \frac{0.24}{1}

                                   = 0.05            = 0.24

Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.

So concentration of Li₃PO₄ is equal to 0.05.

                       

6 0
2 years ago
Which is easier to remove: a valence electron from, Li or Na? Explain why.
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The atomic number of Li is 3

Electron configuration of Li : 1s² 2s¹

The atomic number of Na is 11

Electron configuration of Na : 1s²2s²2p⁶3s¹

Thus there is one electron in the valence shell of Li (2s¹) and that of Na (3s¹). However, the valence electron in Na is in a shell that is farther away from the nucleus compared to that of Li. As a result, the Na valence electron will be held less tightly by the nucleus i.e. it will experience a reduced nuclear attraction and can be removed easily than the Li 2s electron.

7 0
3 years ago
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