Answer : The molarity after a reaction time of 5.00 days is, 0.109 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = 5.00 days
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.110 M
Now put all the given values in above equation, we get:
![9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)](https://tex.z-dn.net/?f=9.7%5Ctimes%2010%5E%7B-6%7D%3D%5Cfrac%7B1%7D%7B5.00%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.110%29%7D%5Cright%29)
![[A]=0.109M](https://tex.z-dn.net/?f=%5BA%5D%3D0.109M)
Hence, the molarity after a reaction time of 5.00 days is, 0.109 M
Hi Sydney!
I can't draw in this question, but there is a picture showing this phase for you to follow when you draw it.
Hope This Helps :)
Veces is Spanish word for times.Hope I helped.
B. This is because the Hydrogen and Oxygen need balanced out.
Current-
C-1 | C-1
H-4 | H-2
O-2 | O-3
Adding a coefficient of 2 before oxygen in the reactants and H2O in the products would balance this equation
<span>CH4 + 2O2 → CO2 + 2H2O</span>
C-1 | C-1
H-4 | H-4
O-4 | O-4
C. The mass number is the amount of Protons and electrons there are combined, so 200 - 80 is 120.
