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:
4- A material that transfers heat energy more easily than another material will experience a greater rate of thermal energy loss than an object that does not transfer heat energy easily.
Explanation:
Thermal energy loss has to do with loss of heat energy by a body to another body or its environment. The aim of the process is usually the attainment of thermal equilibrium between the body and its environment.
On a cold day, a material that transfers thermal energy more easily will loose thermal energy faster than an object that does not transfer thermal energy. The rate of heat transfer of a body determines its rate of loss of thermal energy.
Answer:
The answer is
<h2>91.9 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume of copper = 10.3 mL
density = 8.92 g/mL
The mass is
mass = 8.92 × 10.3 = 91.876
We have the final answer as
<h3>91.9 g</h3>
Hope this helps you
Answer:
Demo Mole Quantities
58.5g NaCl(mol/58.5g)(6.02 x 1023/mol) = 6.02 x 1023 Na
+
Cl21 pre-1982 pennies (after 1982 pennies are mostly zinc with copper coating)
63.5g Cu( mol/ 63.5g)(6.02 x 1023/mol) = 6.02 x 1023 Cu
19.0g Al (mol/27.0g)(6.02 x 1023/mol) = 4.24 x 1023 Al
Explanation:
