The two factors of 2p are 2 and p.
4 to fill the 2p shell
2. the 2s ordital and 2p
Aluminum is the group 12, period three chemical element. It has an electron configuration of 1s2 2s2 2p6 3s2 3p1. That means that its 2p subshell is full, having 6 electrons.
2s^2 2p^6 8 electrons total
C = 1s2,2s2,2p2 so the outermost (2p) orbital has 2 electrons in Ground State
It has four factors: 1, 2, p, and 2p.
There are six 2p electrons in Iron
There are 6 2p electrons in argon.
p2 - 2p + 2 can be factored as (p - 1)(p - 1)which can be written as ( p - 1)2.
There are 50 2p coins in 1.00.
The GCF is 2p.
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
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Nitrogen has 2s^3 2p^3 valence electrons so the answer would be 3
29 Ways: 20(1p) 18(1p),1(2p) 16(1p),2(2p) 14(1p),3(2p) 12(1p),4(2p) 10(1p),5(2p) 8(1p),6(2p) 6(1p),7(2p) 4(1p),8(2p) 2(1p),9(2p) 10(2p) 4(5p) 3(5p),2(2p),1(1p) 3(5p),1(2p),3(1p) 3(5p),5(1p) 2(5p),5(2p) 2(5p),4(2p),2(1p) 2(5p),3(2p),4(1p) 2(5p),2(2p),6(1p) 2(5p),1(2p),8(1p) 2(5p),10(1p) 1(5p),7(2p),1(1p) 1(5p),6(2p),3(1p) 1(5p),5(2p),5(1p) 1(5p),4(2p),7(1p) 1(5p),3(2p),9(1p) 1(5p),2(2p),11(1p) 1(5p),1(2p),13(1p) 1(5p),15(1p)
3p = 2p + 12 subtract 2p from both sides 3p - 2p = 2p - 2p + 12 1p = 12 p = 12 this is how you solve this problem.
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