9. Sequence the following electron configurations in order from least to greatest atomic number: Hint: Exponents
1. 1s 2s22p%3s23p6452
II.1s22s22p63523p64523d104p3
III. 1s22s22p%3s23p64523d104p6
1.1s22s22p3s23p4523d4
ol|ol|ol|O
CLEAR ALL

The sequence : 1,4,2,3

In an atom there are levels of energy in the shell and subshell  

This energy level is expressed in the form of electron configurations.  

So the electron configuration shows the number of electrons it has

Atomic number = number of electrons in neutral atoms

Charging electrons in the sub shell uses the following sequence:  

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.  

1. 1s² 2s²2p⁶3s²3p⁶4s² : 2+2+6+2+6+2=20

2. 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p³ = 2+2+6+2++6+2+10+3=33

3.  1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶ = 2+2+6+2++6+2+10+6=36

4.  1s²2s²2p⁶3s²3p⁶4s²3d⁴ = 2+2+6+2++6+2+4=24

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied