The ionic eqn is as follow:
1 Al(OH)3(s) + 3 H+(aq) + 3 NO3(-1) --> 1 Al(3+)(aq) + 3 NO3(-)(aq) + 3 H2O(l)
3moles of No3- ion on both sides cancels out to give the net ionic eqn:
1 Al(OH)3(s) + 3 H+(aq) --> 1 Al(3+)(aq) + 3 H2O(l)
Answer:
1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵
Explanation:
1) Ca:
Its atomic number is 20. So it has 20 protons and 20 electrons.
Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.
So, you have two more electrons (20 - 18 = 2) to distribute.
Those two electrons go the the orbital 4s.
Finally, the electron configuration is [Ar] 4s².
2) Pm
The atomic number of Pm is 61, so it has 61 protons and 61 electrons.
Pm is in the row (period) 6. So, the noble gas before Pm is Xe.
The atomic number of Xe is 54.
Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.
The resultant distribution for Pm is: [Xe]6s² 4f⁵.
