This is called the pedigree chart.
Answer : The maximum amount of nickel(II) cyanide is 
Explanation :
The solubility equilibrium reaction will be:
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:
Therefore, the maximum amount of nickel(II) cyanide is
Answer:
Period
Explanation:
If we ubicate te period one on the alkaline metals, we can see the lithium, so we go to the borans on te group AIII an we see Boron, move to the rigth on the same sense tou the group AVI we see the Oxygen, we can know that also for the electronic configuration Li:1s²2s¹ B:1s²2s²3p
The CaCO3 produced if 47.5 moles of NH3 produced is calculated as follows
CaCN2 +3H2O = CaCO3 + 2NH3
by use of mole ratio between CaCO3 to NH3 which is 1:2 the moles of CaCO3 is therefore = 47.5 /2= 23.75 moles
mass of CaCO3 is therefore = moles x molar mass
= 23.75 moles x 100g/mol= 2375 grams which is approximate 2380 grams(answer 6)
