Answer:
![[H^+]=0.00332M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.00332M)
Explanation:
Hello,
In this case, considering the dissociation of valeric acid as:
Its corresponding law of mass action is:
![Ka=\frac{[H^+][C_5H_9O_2^-]}{[HC_5H_9O_2]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BC_5H_9O_2%5E-%5D%7D%7B%5BHC_5H_9O_2%5D%7D)
Now, by means of the change 
due to dissociation, it becomes:
Solving for we obtain:
Thus, since the concentration of hydronium equals , the answer is:
![[H^+]=x=0.00332M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.00332M)
Best regards.
1) concentration or partial pressure of species
involved. 2) temperature • 3) presence of catalyst
4) nature of reactants.
Answer:
1.73 M
Explanation:
We must first obtain the concentration of the concentrated acid from the formula;
Co= 10pd/M
Where
Co= concentration of concentrated acid = (the unknown)
p= percentage concentration of concentrated acid= 37.3%
d= density of concentrated acid = 1.19 g/ml
M= Molar mass of the anhydrous acid
Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1
Substituting values;
Co= 10 × 37.3 × 1.19/36.5
Co= 443.87/36.6
Co= 12.16 M
We can now use the dilution formula
CoVo= CdVd
Where;
Co= concentration of concentrated acid= 12.16 M
Vo= volume of concentrated acid = 35.5 ml
Cd= concentration of dilute acid =(the unknown)
Vd= volume of dilute acid = 250ml
Substituting values and making Cd the subject of the formula;
Cd= CoVo/Vd
Cd= 12.16 × 35.5/250
Cd= 1.73 M
Hihi!
During Rutherford's experiment<span> of shooting particles through a sheet of </span>gold foil <span>he had discovered that some were deflected! This had proved that an </span>atom<span> is actually a small dense nucleus surrounded by orbiting electrons!
I hope I helped!
-Loliarual</span>
