PH = -log [H3O+]
4.15 = -log [H3O+]
[H3O+] = 10^(-4.15)
[H3O+]= 7.08 × 10^-5
Answer:
Explanation:
MW of NaOH = 40 g/mol
MW of KCl = 74.55 g/mp;
250 mL = .25 L
100g NaOH / 40 g = 25 mol
100g KCl/ 74.55g = 1.34 mol
Molarity of NaOH: 25/.25 = 100M
Molarity of KCl: 1.34/.25 = 5.36 M
The concentration of diluted solution is 0.756M.
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 18.9 mL
Molarity of stock solution (M1) = 10 M
Volume of diluted solution (V2) = 250 mL
Molarity of diluted solution (M2) =?
We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:
M1V1 = M2V2
10 × 18.9 = M2 ×250
189 = M2 × 250
Divide both side by 100
M2 = 189 / 250
M2 = 0.756 M
Therefore, the molarity of the diluted solution is 0.756 M.
Thus the concluded that concentration of the dilute acid is 0.756 M.
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Answer:
Rate expression has been given below
Explanation:
According to the given equation, 1 molecule of A reacts with 1 molecule of B and produces 2 molecules of B at a time.
So, rate of disappearance of both A and B are one half of rate of appearance of B
Hence rate expression can be represented as:
![Rate=\frac{-\Delta [A]}{\Delta t}=\frac{-\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=Rate%3D%5Cfrac%7B-%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B-%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
where
is rate of disappearance of A,
is rate of disappearance of B and
rate of appearance of C