Homogeneous:
denoting a process involving substances in the same phase (solid, liquid, or gaseous).
Herogeneous:
of or denoting a process involving substances in different phases (solid, liquid, or gaseous).
Answer:
![[CO]=[Cl_2]=0.01436M](https://tex.z-dn.net/?f=%5BCO%5D%3D%5BCl_2%5D%3D0.01436M)
![[COCl_2]=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.00064M)
Explanation:
Hello there!
In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:
![K=\frac{[CO][Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BCO%5D%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Which can be written in terms of x, according to the ICE table:
Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:
![[COCl_2]=0.015M-0.01436M=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.015M-0.01436M%3D0.00064M)
Regards!
Pure magnesium's formula would just be Mg because all elements except for 7 nonmetals are just left alone when they are by themselves in a formula. The 7 diatomic elements( means they have to have two of them without another element attached to it aka. a subscript two after it when it's by itself) are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. An easy way to remember the diatomic seven is that when looking at a periodic table if you trace over them from nitrogen over to fluorine and down to iodine all of those elements are diatomic + hydrogen.
And your unbalanced and balanced equations are correct.
(sorry I went on a tangent with the diatomic rules hopefully it will help you in the future though)
Answer: Volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration.
Explanation:
According to the dilution law,
where,
= molarity of stock solution = 1M
= volume of stock solution = ?
= molarity of diluted solution = 0.075 M (1mM=0.001M)
= volume of diluted solution = 10 ml
Putting in the values we get:
Thus 0.75 ml of 1M EtOH is taken and (10-0.75)ml = 9.25 ml of water is added to make the volume 10ml.
Therefore, volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration
