If these answers are not the one, don't go off on me because I'm trying my best :
1. A
2. A
3. C
Answer:
2Mg^+ +O2 right arrow 2MgO
Explanation:
Cl03 is a polyatomic ion with a charge of -1. So 2 of those brings that charge to -2. Ca=+2. So Ca=+2 and (Cl03)2= -2. This would balance out the equation. The final result would be Ca(ClO3)2
Answer:-
[Kr] 5s1
Explanation:-
Atomic number of rubidium = 37
Atomic number of Kr = 36
So only 1 electron needs to be assigned an orbital.
Kr is also the last element of the fourth period
So the shell number for the 1 electron will be 5 since it is after Kr
Since s subshell is filled first the electron will go into 5s.
Hence the electron configuration for rubidium (Rb) in noble-gas notation is
[Kr] 5s1