Dumping waste is different from dilution and dispersion. Dumping waste is directly disposing your waste and not regulating its effect to the environment. Dilution is done usually before waste water disposal, meaning adding water to the waste to minimize its concentration.
Answer:
are electrical signals
Explanation:
Feel free to correct me, I'm just trying to help
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:
![[CO]=[Cl_2]=0.01436M](https://tex.z-dn.net/?f=%5BCO%5D%3D%5BCl_2%5D%3D0.01436M)
![[COCl_2]=0.015M-0.01436M=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.015M-0.01436M%3D0.00064M)
Regards!
The rate law for this reaction is [A]².
Balanced chemical reaction used in this experiment: A + B → P
The reaction rate is the speed at which reactants are converted into products.
Comparing first and second experiment, there is no change in initial rate. The concentration of reactant B is increased by double. Initial rate does not depands on concentration of reactant B.
Comparing first and third experiment, initial rate is nine times greater, while concentration of reactant A is three times greater. Conclusion is that concentration of reactant A is squared and the rate is [A]².
More info about rate law: brainly.com/question/16981791
#SPJ4