Answer:
Explanation:
Hello,
In this case, for the given reaction at equilibrium:
We can write the law of mass action as:
![Keq=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
That in terms of the change 
due to the reaction extent we can write:
![Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7Bx%7D%7B%28%5BCO%5D_0-x%29%28%5BH_2%5D_0-2x%29%5E2%7D)
Nevertheless, for the carbon monoxide, we can directly compute as shown below:
![[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\](https://tex.z-dn.net/?f=%5BCO%5D_0%3D%5Cfrac%7B0.45mol%7D%7B1.00L%7D%3D0.45M%5C%5C)
![[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\](https://tex.z-dn.net/?f=%5BH_2%5D_0%3D%5Cfrac%7B0.57mol%7D%7B1.00L%7D%3D0.57M%5C%5C)
![[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\](https://tex.z-dn.net/?f=%5BCO%5D_%7Beq%7D%3D%5Cfrac%7B0.28mol%7D%7B1.00L%7D%3D0.28M%5C%5C)
![x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M](https://tex.z-dn.net/?f=x%3D%5BCO%5D_0-%5BCO%5D_%7Beq%7D%3D0.45M-0.28M%3D0.17M)
Finally, we can compute the equilibrium constant:
Best regards.
Answer:
I think stirring is the right answer
Expanding in powers 4FCsixteen.
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Answer:
Solids
Explanation:
At standard room temperature and pressure, most of the elements are solids.
Just a few of the elements are liquids and gases at this temperature.
The periodic table is made up of metals, metalloids and non-metals. Most of these substances are actually solid.
Some non-metals are gaseous at standard room temperature.
The net ionic equation : Ba²⁺ + SO₄²⁻ ⇒ BaSO₄ (s)
<h3>Further explanation </h3>
The electrolyte in the solution produces ions.
The equation of a chemical reaction can be expressed in the equation of the ions
In the ion equation, there is a spectator ion that is the ion which does not react because it is present before and after the reaction
When these ions are removed, the ionic equation is called the net ionic equation
For gases and solids including water (H₂O) can be written as an ionized molecule
So only the dissolved compound is ionized ((expressed in symbol aq)
Barium sulfate can be formed from the reaction:
Ba(NO₃)₂(aq) + Na₂SO₄(aq)⇒BaSO₄(s)+2NaNO₃(aq)
For full ionic equation :
Ba²⁺ + 2NO₃⁻ + 2 Na⁺ + SO₄²⁻ ⇒ BaSO₄ (s) + 2 Na⁺ + 2 NO³⁻
by removing spectator ions (2NO₃⁻ and 2 Na⁺), the net ionic equation :
<em>Ba²⁺ + SO₄²⁻ ⇒ BaSO₄ (s) </em>
