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 sample of CaC2 (s) (molar mass 64 g/mol) is used instead and all of it reacts, which of the following will occur and why? (the value of Ksp for Ca(OH)2 is 8.0 x 10-8)(A) Ca(OH)2 will precipitate because Q >K sp.(B) Ca(OH)2will precipitate because Q K sp.(D) Ca(OH)2 will not precipitate because Q

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

Explanation:

Ksp for Ca(OH)2 has already been stated in the question as 8.0 x 10-8mol2dm-6

The value of the reaction quotient depends heavily on the concentration of the reactants. As the initial concentration of the calcium carbide decreases considerably, the reaction quotient decreases until Q<Ksp hence the Ca(OH)2 will not precipitate from solution.

The reaction equation is:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

From

Ca(OH)2= Ca2+ + 2OH-

Concentration of solution= 0.064×1/64= 1×10-3

Since [Ca2+] = 1×10-3

[OH-]= (2×10-3)^2= 4×10^-6

Hence Q= 4×10^-9

This is less than the Ksp hence the answer.

fire could start, buildings collapsing, and ash falling from the sky.

explanation:

this is because the buildings or plants could catch on fire and the buildings would most likely collapse. lastly ash would fall because all the burning would cause ash to fall from the sky.

hope this .