It depends if you are taking the class in high school or college. For college doing the experiments is fun but the calculations is not. Also if you are taking it, bring friends.
Answer:
more people may make more changes to it.
there is the answer to it!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
[H2]eq = 0.0129 M
[F2]eq = 1.0129 M
[HF]eq = 0.9871 M
Explanation:
∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2
experiment:
∴ n H2 = 3.00 mol
∴ n F2 = 6.00 mol
∴ V sln = 3.00 L
⇒ [H2]i = 3.00 mol / 3.00 L = 1 M
⇒ [F2]i = 6.00 mol / 3.00 L = 2 M
[ ]i change [ ]eq
H2 1 1 - x 1 - x
F2 2 2 - x 2 - x
HF - x x
⇒ K = (x)² / (1 - x)*(2 - x) = 1.15 E2
⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115
⇒ x² = (2 - 3x + x²)(115)
⇒ x² = 230 - 345x + 115x²
⇒ 0 = 230 - 345x + 114x²
⇒ x = 0.9871
equilibrium:
⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M
⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M
⇒ [HF] = x = 0.9871 M
Answer:
specialized if you add one more I cause I'm pretty sure there is supposed to be one more i.
