Answer:
1s2 2s2 2p6 3s2 3p6 3d6 4s2
Explanation:
Answer:
The equilibrium constant for the reversible reaction = 0.0164
Explanation:
At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.
The reaction is given as
A ⇌ B
Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]
The rate of forward reaction = |r₁| = k₁ [A]
The rate of backward reaction = |r₂| = k₂ [B]
(Taking only the magnitudes)
where k₁ and k₂ are the forward and backward rate constants respectively.
k₁ = 0.010 s⁻¹
k₂ = 0.0610 s⁻¹
|r₁| = 0.010 [A]
|r₂| = 0.016 [B]
At equilibrium, the rate of forward and backward reactions are equal
|r₁| = |r₂|
k₁ [A] = k₂ [B] (eqn 1)
Note that equilibrium constant, K, is given as
K = [B]/[A]
So, from eqn 1
k₁ [A] = k₂ [B]
[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164
K = [B]/[A] = (k₁/k₂) = 0.0164
Hope this Helps!!!
Answer:
B
Explanation:
the fluorine has an high tendency to gain electrons from other elements with lower electronegativities
Answer:
C) 712 KJ/mol
Explanation:
- ΔH°r = Σ Eb broken - Σ Eb formed
- 1/2Br2(g) + 3/2F2(g) → BrF3(g)
∴ ΔH°r = - 384 KJ/mol
∴ Br2 Eb = 193 KJ/mol
∴ F2 Eb = 154 KJ/mol
⇒ Σ Eb broken = (1/2)(Br-Br) + (3/2)(F-F)
⇒ Σ Eb broken = (1/2)(193 KJ/mol) + (3/2)(154 KJ/mol) = 327.5 KJ/mol
∴ Eb formed: Br-F
⇒ Σ Eb formed (Br-F) = Σ Eb broken - ΔH°r
⇒ Eb (Br-F) = 327.5 KJ/mol - ( - 384 KJ/mol )
⇒ Eb Br-F = 327.5 KJ/mol + 384 KJ/mol = 711.5 KJ/mol ≅ 712 KJ/mol
