Answer:
London Dispersion Force<span> is the strongest interparticle force in a sample of Kr.
Explanation:
Krypton belongs to Noble Gases. They exists in monoatomic form as they are inert in nature and are very less reactive. So, as there is no polarity in Krypton so it will fail to create either Dipole-Dipole or Hydrogen Bond Interactions between its atoms. While, London Dispersion Forces or Van Der Waals forces can exist between Kr atoms. When Kr atoms approaches one other they create Instantaneous dipole. This</span> Instantaneous dipole induces dipole in second Kr atom and the process starts propagating. Hence, interactions are generated between Kr atoms.
Answer:
c
Explanation:
i think if not tell me ok
a) Alkali metals
=> group 1
=> Li: 1s2 2s => 1s
Na: [Ne] 3s => 3s
K: [Ar] 4s => 4s
Rb: [Kr] 5s => 5s
Cs: [Xe] 6s => 6s
Fr: [Rn] 7s => 7s
=> outer electron configuration is ns, where n is the main energy level: 1, 2, 3, 4, 5, 6,7.
b) Alkaline earth metals
=> group 2 => you have to add 1 electron to the alkaly metal of the same row.
=> Be: [He] 2s2 => 2s2
Mg: [Ne] 3s2 => 3s2
Ca: [Ar] 4s2 => 4s2
Sr: [Kr] 5s2 => 5s2
Ba: [Xe] 6s2 => 6s2
Ra: [Rn[ 7s2 => 7s2
=>outer electron configuration is n s2, where n is the main energy level: 1, 2, 3, 4, 5, 6, 7
c) halogens
=> group 7
=> F: [He] 2s2 2p5 => 2s2 2p5
Cl: [Ne] 3s2 3p5 => 3s2 3p5
Br: [Ar] 3d10 4s2 4p5 => 4s2 4p5
I: [Kr] 4d10 5s2 5p6 => 5s2 5p5
At: [Xe] 4f14 5d10 6s2 6p5
=> outer electron configuration is ns2 np5, where n is the main energy level 1, 2, 3, 4, 5, 6, 7
d) Noble gases
=> group 8
I will show only the outer shell which is what is requested
=> He: 1s2
Ne: ... 2s2 2p6
Ar: ... 3s2 3p6
Kr: ... 4s2 4p6
Xe: ... 5s2 5p6
Rn: ... 6s2 6p6
=> the outer electron configuration is ns2 np6, except for He for which it is 1s2
answer should be 1.4 cm³
1 L = 10 and so
dL = 100 and then
cL = 1,000
mL = 0.001 m³
1 m³ = 1,000
dm³ = 1,000,000
cm³ = 1,000,000,000
mm³ = 1,000 L
So, 1 mL = 1 cm³ = 0.001 L = 0.1 cL
1,400 mL = 1,400 cm³ = 1.4 L = 140 cL