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:
The u (amu is the old unit name) is 1/12 of the weight of an 12C atom. The way the u is chosen ensures that all core and atom masses are multiples of 1(±0.1) u.
Explanation:
Further explanation if needed...
Carbon 12 was chosen because the chemical atomic weights based on C12 are almost identical to the chemical atomic weights based on the natural mix of oxygen. Simply because the atomic mass is defined as 1/12 of the mass of 12C. Others isotopes of carbon (13C mostly, with an abundance of 1.1% approximately) account for an average atomic mass slightly above 12.
<span>Begin by classifying which energy level, and indirectly principal quantum number, n, resembles to the N shell.
no. of orbitals =n2
In your case, the fourth energy level will contain
n=4⇒no. of orbitals= 4^2=16
The number of subshells is given by the principal quantum number.
no. of subshells=n
In your case, the fourth energy level will have
no. of subshells = 4 this is the answer
to check:
the fourth energy shell will can hold a thoroughgoing of no. of electrons=2⋅42=32 e−</span>
