Answer:
I'm sorry, you didn't post the pictures.
Answer:
Explanation:
Calculate the pH of a hydrochloride acid solution, HCl, whose hydronium ion (H3O)+ concentration is 8.29 X 10-4 M.
Note: answer should have three significant figures
Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
Answer:
Kc = 2.34 mol*L
Explanation:
The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.
A + B ⇄ C + D
Kc = [C] * [D] / [A] * [B]
According to the reaction
Kc = [SO2]^2 * [O2]^2 / [SO3]^2
Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3
0.450 --> 0 + 0 (Beginning of the reaction)
0.260 --> 0.260 + 0.130 (During the reaction)
0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)
Kc = [0.260]^2 + [0.130]^2 / [0.190]^2
Kc = 2.34 mol*L
![pH=-log [H_3O^{+}]](https://tex.z-dn.net/?f=pH%3D-log%20%5BH_3O%5E%7B%2B%7D%5D%20)
![[H_3O^{+}]= 10^{-pH}=10^{-3,24}=0,00058M](https://tex.z-dn.net/?f=%5BH_3O%5E%7B%2B%7D%5D%3D%2010%5E%7B-pH%7D%3D10%5E%7B-3%2C24%7D%3D0%2C00058M%20)