Double replacement because H and K are both switching
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:
Kc = [Pb²⁺]³.[PO₄³⁻]²
Explanation:
Let's consider the following reaction at equilibrium.
Pb₃(PO₄)₂(s) ⇄ 3 Pb²⁺(aq) + 2 PO₄³⁻(aq)
The concentration equilibrium constant is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [Pb²⁺]³.[PO₄³⁻]²
This equilibrium constant is known as the solubility product of Pb₃(PO₄)₂.
During a collision, the person would most likely experience the same conditions with or without a seatbelt.
Due to the temperature of the surrounding earth
