It could be: 33-(3+3) = 21
7
4=4(3s) 4=12s s= 1/3
There are: 21/3 = 7
3 * 3 / ( 3 * 3 ) = 1 but that uses only four 3s, so 33 / ( 3 * 3 * 3 ) = 1 uses five 3s
The are two electrons in the 3s orbital of magnesium (Mg.)
33 + (3/3)
There are four 3s in 12. This can be determined by dividing 12 by 3, which equals 4. Each division represents one occurrence of the number 3 within 12.
Is done by (3/3) plus (3/3) = 2
3 + 3 + 3/3 = 7
Acc to modern olympic rules, each game is played for 21 points, best of 3s
You can't unless something is missing in your question.
The element that has four electrons in energy level n=3 is arsenic (As), which has the atomic number 33. In its electron configuration, arsenic has the distribution of electrons as follows: 1s² 2s² 2p⁶ 3s² 3p³, where the 3s and 3p orbitals together account for the four electrons in the n=3 energy level.