Answer: The equilibrium concentration of hydrogen gas is 0.0269 M
Explanation:
The chemical reaction follows the equation:

At t = 0 0.044M 0.044M 0.177M
At
(0.044-x)M (0.044-x)M (0.177+x)M
The expression for
for the given reaction follows:
![K_c=\frac{[HI]^2}{[H_2]\times [I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5Ctimes%20%5BI_2%5D%7D)
We are given:

Putting values in above equation, we get:


Hence, the equilibrium concentration of hydrogen gas is (0.044-x) M =(0.044-0.0171) M= 0.0269 M
Answer:
No se pues carnal preguntale a alguien mas.
Explanation:
lo siento:(
Answer:
34.3 g
Explanation:
Step 1: Write the balanced equation
2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O
Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH
The molar mass of CH₃CH₂OH is 46.07 g/mol.
50.0 g × 1 mol/46.07 g = 1.09 mol
Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced
The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.
Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced
The percent yield of the reaction is 85%.
0.545 mol × 85% = 0.463 mol
Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃
The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.
0.463 mol × 74.12 g/mol = 34.3 g
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