Explanation:
the correct empirical formula for C6H12O6 is CH2O...
hope it will help....
A very fatal disese that causes you to poop blood
Answer:
See explanation below
Explanation:
The question is incomplete. However, here's the missing part of the question:
<em>"For the following reaction, Kp = 0.455 at 945 °C: </em>
<em>C(s) + 2H2(g) <--> CH4(g). </em>
<em>At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"</em>
With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.
The expression of Kp for this reaction is:
Kp = PpCH4 / (PpH2)²
We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:
PpCH4 = Kp * PpH2²
*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.
Now solving for PpCH4:
PpCH4 = 0.455 * (1.78)²
<u><em>PpCH4 = 1.44 atm</em></u>
Answer:
A. 
B. 
Explanation:
Hello!
In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:
Thus we proceed as follows:
A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:
Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:
B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:
Best regards!
Lithium-Chloride, Carbon-oxide, Barium Bromide, Ferrous Iodide, and Ammonium Chloride. Did this help?
