H₂ + ½O₂ → H₂O
2g → 18g
So, 18g water needs 2g H₂
So, 1g water needs 2/18g H₂
So, 180g water needs 2/18 x 180g H₂
→ 20 litres of H₂
the answer is B. Atmosphere
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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
They certainly can. However, they have other groups that are used to classify a compound.