Answer:
283.725 kJ ⋅ mol − 1
Explanation:
C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1
Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1
C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1
4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1
eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1
so,
average bond enthalpy is 
= 283.725 kJ ⋅ mol − 1
Answer:
Molarity of NaOH = 0.025 M
Explanation:
Given data:
Molarity of HCl = C₁ = 0.05 M
Volume of HCl = V₁= 50 mL
Molarity of NaOH = C₂=?
Volume of NaOH =V₂= 100 mL
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Molarity of HCl
V₁ = Volume of HCl
C₂ = Molarity of NaOH
V₂ = Volume of NaOH
Now we will put the values:
C₁V₁ = C₂V₂
0.05 M × 50 mL = C₂ × 100 mL
2.5 M.mL =C₂ × 100 mL
C₂ = 2.5 M.mL /100 mL
C₂ = 0.025 M
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