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RideAnS [48]
3 years ago
11

Which statement is true about this equation?

Chemistry
2 answers:
lara [203]3 years ago
3 0
The appropriate response is the fourth one. The announcement is valid about this condition beneath is in spite of the fact that it is unequal, it can be adjusted by specifically utilizing observer particles.

I hope the answer will help you. 

bulgar [2K]3 years ago
3 0

Answer:

D

Explanation:

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Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g s
Nina [5.8K]

Answer:

D) Ca(OH)₂ will not precipitate because Q <  Ksp

Explanation:

Here we have first a chemical reaction in which Ca(OH)₂  is produced:

CaC₂(s)  + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂  is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product  constant for Ca(OH)₂  is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M  = 0.002 M  ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q=  [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is option D

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Answer:

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