N2 + 3H2 ---> 2NH3
nitrogen + hydrogen ---> ammonia
Answer:
A small still is separating propane and butane at 135 °C, and initially contains 10 kg moles of a mixture whose composition is x = 0.3 (x = mole fraction butane). Additional mixture (x = 0.3) is fed at the rate of 5 kg mole/hr. The total volume of the liquid in the still is constant, and the concentration of the vapor from the still (xp) is related to x, as follows: Xp = How long will it take for X, to change from 0.3 to 0.35.
Answer:
No one is correct. The correct expression is:
Keq = [H₂]² . [O₂]² / [H₂O]²
Explanation:
To build the Keq expression in a chemical equilibrium you must consider the molar concentrations of reactants / products, and they must be elevated to the stoichiometric coefficient.
The balance reaction is:
<u>2</u> H₂O (g) ⇄ <u>2</u> H₂ (g) + O₂ (g)
Keq = [H₂]² . [O₂] / [H₂O]²
In opposite side: <u>2</u> H₂ (g) + O₂ (g) ⇄ <u>2</u> H₂O (g)
Keq = [H₂O]² / [H₂]² . [O₂]
ACIDIC BEHAVIOR OF SOLUTION
Answer:
The percent yield of this reaction is 84.8 % (option A is correct)
Explanation:
Step 1: Data given
The student isolated 15.6 grams of the product = the actual yield
She calculated the reaction should have produced 18.4 grams of product = the theoretical yield = 18.4 grams
Step 2: Calculate the percent yield
Percent yield = (actual yield / theoretical yield ) * 100 %
Percent yield = (15.6 grams / 18.4 grams ) * 100 %
Percent yield = 84.8 %
The percent yield of this reaction is 84.8 % (option A is correct)
