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:
351.43mL
Explanation:
To calculate the original volume of hydrogen gas in this question, the Boyle's law equation will be used. Boyle's law equation is:
P1V1 = P2V2
Where; P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume
According to this question, the P1= 1.56atm, V1 = ?, P2 = 0.73atm, V2 = 751mL
Hence;
P1V1 = P2V2
1.56 × V1 = 0.73 × 751
1.56 V1 = 548.23
V1 = 548.23/1.56
V1 = 351.43mL
Therefore, the original volume of hydrogen gas is 351.43 mL.
Answer: Chemical composition modification (or, physical signal would be color).
Answer:
son 12.6 gramos de HF
Explanation:
Tienes que saber qual es el reactor limitante en este caso es fluoruro con los 20 gramos puedes producer .631 mol qual son 12.6 gramos