Answer:
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.
To find the ratio of the the combination for the ion, write the charge of the cation as the subscript for the anion, and the charge of the anion as the subscript of the cation. This will make the charges effectively cancel and you will be left with a neutral ionic compound. Remember, that an ionic compound is made up of a metal and a nonmetal.
For Ca2+ and Cl-, you will get the neutral compound to be CaCl₂.
To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:
Clearing the specific heat of the unknown substance:
Answer:
Explanation: There are 2 moles of C6H12O6 in 300 g C6H12O6 , rounded to one significant figure.
Answer;
-Two chlorine atoms
Explanation;
A barium atom attains a stable electron configuration when it bonds with two chlorine atoms.
-Barium is an alkaline earth metal, in group two of the periodic table. Like other alkaline earth metal it has a valency of two which means it reacts by loosing two electrons.
-Chlorine on the other hand is a halogen (group seven element) it reacts by gaining an electron, thus two chlorine atoms will require two electrons. Therefore, Barium would attain a stable configuration by loosing two electrons to two chlorine atoms.
