Answer: The percent ionization is higher for 0.010 M solution of ![HC_2H_3O_2](https://tex.z-dn.net/?f=HC_2H_3O_2)
Explanation:
cM 0M 0M
So dissociation constant will be:
a) Given c= 0.10 M and
= ?
Putting in the values we get:
![\%(\alpha)=0.013\times 100=1.3\%](https://tex.z-dn.net/?f=%5C%25%28%5Calpha%29%3D0.013%5Ctimes%20100%3D1.3%5C%25)
b) Given c= 0.010 M and
= ?
Putting in the values we get:
![\%(\alpha)=0.041\times 100=4.1\%](https://tex.z-dn.net/?f=%5C%25%28%5Calpha%29%3D0.041%5Ctimes%20100%3D4.1%5C%25)
The percent ionization is higher for 0.010 M solution of ![HC_2H_3O_2](https://tex.z-dn.net/?f=HC_2H_3O_2)
<span>[He] 2s2 2p4 is the EC</span>
Answer:
![S_2F_{10}](https://tex.z-dn.net/?f=S_2F_%7B10%7D)
Explanation:
Hello!
In this case, since the empirical formula of a chemical compound is determined by assuming the by-mass percentages are masses, we can firstly compute the moles of sulfur and fluorine based on their atomic masses:
![n_S=25.24gS*\frac{1molS}{32.01gS}=0.789molS\\\\n_F=74.76gF*\frac{1molF}{19.00gF}=3.93molF](https://tex.z-dn.net/?f=n_S%3D25.24gS%2A%5Cfrac%7B1molS%7D%7B32.01gS%7D%3D0.789molS%5C%5C%5C%5Cn_F%3D74.76gF%2A%5Cfrac%7B1molF%7D%7B19.00gF%7D%3D3.93molF)
Next, we compute the subscript of each element in the formula by dividing each moles by the fewest moles:
![S=\frac{0.789}{0.789}=1\\\\F=\frac{3.93}{0.789} =5](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B0.789%7D%7B0.789%7D%3D1%5C%5C%5C%5CF%3D%5Cfrac%7B3.93%7D%7B0.789%7D%20%3D5)
Whose molar mass is 127.01 g/mol. Now, we can compute the ratio between the molecular and empirical formulas as follows:
![ratio=\frac{254.14}{127.01} =2](https://tex.z-dn.net/?f=ratio%3D%5Cfrac%7B254.14%7D%7B127.01%7D%20%3D2)
Thus, the molecular formula is:
![S_2F_{10}](https://tex.z-dn.net/?f=S_2F_%7B10%7D)
Regards!
Answer:
The smallest particle of an element is called an atom.