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
The moon, man has not been to mars yet
Answer:
12.99
Explanation:
<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of KOH: 716. mg (0.716 g)
- Volume of the solution: 130. mL (0.130 L)
Step 2: Calculate the moles corresponding to 0.716 g of KOH
The molar mass of KOH is 56.11 g/mol.
0.716 g × 1 mol/56.11 g = 0.0128 mol
Step 3: Calculate the molar concentration of KOH
[KOH] = 0.0128 mol/0.130 L = 0.0985 M
Step 4: Write the ionization reaction of KOH
KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)
The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M
Step 5: Calculate the pOH
We will use the following expression.
pOH = -log [OH⁻] = -log 0.0985 = 1.01
Step 6: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - pOH = 14 -1.01 = 12.99