Answer : The value of equilibrium constant (K) is, 424.3
Explanation : Given,
Concentration of
at equilibrium = 0.067 mol
Concentration of
at equilibrium = 0.021 mol
Concentration of
at equilibrium = 0.040 mol
The given chemical reaction is:

The expression for equilibrium constant is:
![K_c=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
Now put all the given values in this expression, we get:


Thus, the value of equilibrium constant (K) is, 424.3
Answer: The entire water/ice solution is at the melting/freezing point, 32°F (0°C). Adding rock salt — or any substance that dissolves in water — disrupts this equilibrium.
Explanation: Hope this helps! Have a great day :)
There are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.
<h3>HOW TO CALCULATE NUMBER OF MOLES?</h3>
The number of moles of a substance can be calculated by multiplying the molarity by the volume.
No. of moles = Molarity × volume
According to this question, 3L of a KBr solution are contained in a 0.4M.
no. of moles = 3L × 0.4M = 1.2moles
Therefore, there are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.
Learn more about no. of moles at: brainly.com/question/14919968
Answer:
The direct channels by which extraction can affect the economy are of particular interest. The development of mines or oil and gas fields can result in significant shocks to a regional economy, generating jobs and drawing in capital from other regions and countries.
Explanation:
<u>Answer:</u> The volume when the pressure and temperature has changed is 
<u>Explanation:</u>
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.
The equation follows:

where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
Let us assume:
![P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K](https://tex.z-dn.net/?f=P_1%3D1.20atm%5C%5CV_1%3D795mL%5C%5CT_1%3D116%5EoC%3D%5B116%2B273%5DK%3D389K%5C%5CP_2%3D0.55atm%5C%5CV_2%3D%3FmL%5C%5CT_2%3D75%5EoC%3D%5B75%2B273%5DK%3D348K)
Putting values in above equation, we get:

Hence, the volume when the pressure and temperature has changed is 