Answer:
Ke = 34570.707
Explanation:
- H2(g) + Br2(g) → 2 HBr(g)
equilibrium constant (Ke):
⇒ Ke = [HBr]² / [Br2] [H2]
∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L
∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L
∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L
⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)
⇒ Ke = 34570.707
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Answer:
can you help mine please
How many molecules of chlorine are needed to react with 56.Og of iron to form Iron (III) chloride (FeCl3)?
Answer:
A
Explanation:
A difference in thermal energy
