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!
Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol.
<span>[mW (m) - mW (g)]/mW (m) x 100% </span>
<span>(they want % error so, if it is negative, just get rid of the sign) </span>
The pressure will not affect the rate of solution.
You need to find the whole molar mass of the compound using the periodic table to add the values.
Na2CO3= (2 x 23.0) + 12.0 + (3 x 16.0)= 106 g/mol
H2O= 10 x [ (2 x 1.01 ) + (16.0) ]= 180.2 g/mol
the total molar mass is 106 + 180.2 = 286.2 g/mol
the percentage of water you can find by doing "parts over the whole"
H2O%= 180.2 / 286.2 X 100= 63.0%
