Answer:
C
Explanation:
I just aswered this question and i got it right
Answer : The maximum amount of nickel(II) cyanide is ![5.84\times 10^{-12}M](https://tex.z-dn.net/?f=5.84%5Ctimes%2010%5E%7B-12%7DM)
Explanation :
The solubility equilibrium reaction will be:
![Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-](https://tex.z-dn.net/?f=Ni%28CN%29_2%5Crightleftharpoons%20Ni%5E%7B2%2B%7D%2B2CN%5E-)
Initial conc. 0.220 0
At eqm. (0.220+s) 2s
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ni^{2+}][CN^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BNi%5E%7B2%2B%7D%5D%5BCN%5E-%5D%5E2)
Now put all the given values in this expression, we get:
![3.0\times 10^{-23}=(0.220+s)\times (2s)^2](https://tex.z-dn.net/?f=3.0%5Ctimes%2010%5E%7B-23%7D%3D%280.220%2Bs%29%5Ctimes%20%282s%29%5E2)
![s=5.84\times 10^{-12}M](https://tex.z-dn.net/?f=s%3D5.84%5Ctimes%2010%5E%7B-12%7DM)
Therefore, the maximum amount of nickel(II) cyanide is ![5.84\times 10^{-12}M](https://tex.z-dn.net/?f=5.84%5Ctimes%2010%5E%7B-12%7DM)
Some examples of malleable materials are gold, silver, iron, aluminum, copper and tin.
Answer:
= 0.28M
Explanation:
data:
volume = 0.250 L
= 0.250dm^3 ( 1litre = 1dm^3)
moles = 0.70 moles
Solution:
molarity = ![\frac{no. of moles}{volume in dm^3}](https://tex.z-dn.net/?f=%5Cfrac%7Bno.%20of%20moles%7D%7Bvolume%20in%20dm%5E3%7D)
= 0.70 / 0.250
molarity = 0.28 M