(2p)2 = 4p2
10
The GCF is 2p.
2p+10 = 8p-14 2p-8p = -14-10 -6p = -24 p = 4
Nitrogen has 2s^3 2p^3 valence electrons so the answer would be 3
2p
no. you would solve and get -80=-2p or 80=2p. thus resulting in p=40
They are the designation for the orbitals present. s and p orbitals in the 2nd energy level.
3p = 2p + 12 subtract 2p from both sides 3p - 2p = 2p - 2p + 12 1p = 12 p = 12 this is how you solve this problem.
20p, 2p, 2p, 2p, 1p.
2s and 2p 2p can be further divided into 2p(x), 2p(y), and 2p(z), depending on which axis you look at.
There are 31 ways:1p × 211p × 19 + 2p × 11p × 17 + 2p × 21p × 16 + 5p × 11p × 15 + 2p × 31p × 14 + 2p × 1 + 5p × 11p × 13 + 2p × 41p × 12 + 2p × 2 + 5p × 11p × 11 + 2p × 51p × 11 + 5p × 21p × 10 + 2p × 3 + 5p × 11p × 9 + 2p × 61p × 9 + 2p × 1 + 5p × 21p × 8 + 2p × 4 + 5p1p × 7 + 2p × 71p × 7 + 2p × 2 + 5p × 21p × 6 + 2p × 5 + 5p1p × 6 + 5p × 31p × 5 + 2p × 81p × 5 + 2p × 3 + 5p × 21p × 4 + 2p × 6 + 5p × 11p × 4 + 2p × 1 + 5p × 31p × 3 + 2p × 91p × 3 + 2p × 4 + 5p × 21p × 2 + 2p × 7 + 5p × 11p × 2 + 2p × 2 + 5p × 31p × 1 + 2p × 101p × 1 + 2p × 5 + 5p × 21p × 1 + 5p × 42p × 8 + 5p × 12p × 3 + 5p × 3
You would have 7p