Answer:
The equilibrium value of [CO] is 1.04 M
Explanation:
Chemical equilibrium is the state to which a spontaneously evolving chemical system, in which a reversible chemical reaction takes place. When this situation is reached, it is observed that the concentrations of substances, both reagents and reaction products, they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.
Reagent concentrations and products in equilibrium are related by the equilibrium constant Kc. Being:
aA + bB ⇔ cC + dD
![Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
In this case:
![Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_%7B3%7DOH%20%5D%7D%7B%5BCO%5D%2A%5BH_%7B2%7D%20%5D%5E%7B2%7D%20%7D)
You know:
- Kc= 14.5
- [H₂]= 0.322 M
- [CH₃OH] =1.56 M
Replacing:
![14.5=\frac{1.56}{[CO]*0.322^{2} }](https://tex.z-dn.net/?f=14.5%3D%5Cfrac%7B1.56%7D%7B%5BCO%5D%2A0.322%5E%7B2%7D%20%7D)
Solving:
![[CO]=\frac{1.56}{14.5*0.322^{2} }](https://tex.z-dn.net/?f=%5BCO%5D%3D%5Cfrac%7B1.56%7D%7B14.5%2A0.322%5E%7B2%7D%20%7D)
[CO]= 1.04 M
The equilibrium value of [CO] is 1.04 M
Answer:
c. carbon monoxide
Explanation:
no one bothers to read why anyways so look at the picture
Explanation:
Below is an attachment containing the solution.
Answer: Esta tendencia es tan regular que el poder de combinación, o valencia, de un elemento se definió una vez como el número de átomos de hidrógeno unidos al elemento en su hidruro. El hidrógeno es el único elemento que forma compuestos en los que los electrones de valencia están en la capa n = 1.
Explanation:
¡Espero que esto ayude!
Answer:
0.500 mole of Xe (g) occupies 11.2 L at STP.
General Formulas and Concepts:
<u>Gas Laws</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Stoichiometry</u>
- Mole ratio
- Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify.</em>
0.500 mole Xe (g)
<u>Step 2: Convert</u>
- [DA] Set up:

- [DA] Evaluate:

Topic: AP Chemistry
Unit: Stoichiometry