The subscripts indicate the number of atoms of the element immediately before it.
Answer : The pH of the solution is, 2.67
Explanation :
The equilibrium chemical reaction is:
![Al^{3+}+H_2O\rightarrow Al(OH)^{2+}+H^+](https://tex.z-dn.net/?f=Al%5E%7B3%2B%7D%2BH_2O%5Crightarrow%20Al%28OH%29%5E%7B2%2B%7D%2BH%5E%2B)
Initial conc. 0.450 0 0
At eqm. (0.450-x) x x
As we are given:
![K_a=1.00\times 10^{-5}](https://tex.z-dn.net/?f=K_a%3D1.00%5Ctimes%2010%5E%7B-5%7D)
The expression for equilibrium constant is:
![K_a=\frac{(x)\times (x)}{(0.450-x)}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%280.450-x%29%7D)
Now put all the given values in this expression, we get:
![1.00\times 10^{-5}=\frac{(x)\times (x)}{(0.450-x)}](https://tex.z-dn.net/?f=1.00%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%280.450-x%29%7D)
![x=0.00212M](https://tex.z-dn.net/?f=x%3D0.00212M)
The concentration of
= x = 0.00212 M
Now we have to calculate the pH of solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![pH=-\log (0.00212)](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%280.00212%29)
![pH=2.67](https://tex.z-dn.net/?f=pH%3D2.67)
Therefore, the pH of the solution is, 2.67
Given :
A compound has a molar mass of 129 g/mol .
Empirical formula of compound is C₂H₅N .
To Find :
The molecular formula of the compound.
Solution :
Empirical mass of compound :
![M_e = ( 2 \times 12 ) + ( 5 \times 1 ) + ( 1 \times 14 )\\\\M_e = 43\ gram/mol](https://tex.z-dn.net/?f=M_e%20%3D%20%28%202%20%5Ctimes%2012%20%29%20%2B%20%28%205%20%5Ctimes%201%20%29%20%2B%20%28%20%201%20%20%5Ctimes%2014%20%29%5C%5C%5C%5CM_e%20%3D%2043%5C%20gram%2Fmol)
Now, n-factor is :
![n = \dfrac{M}{M_e}\\\\n = \dfrac{129}{43}\\\\n = 3](https://tex.z-dn.net/?f=n%20%3D%20%5Cdfrac%7BM%7D%7BM_e%7D%5C%5C%5C%5Cn%20%3D%20%5Cdfrac%7B129%7D%7B43%7D%5C%5C%5C%5Cn%20%3D%203)
Multiplying each atom in the formula by 3 , we get :
Molecular Formula, C₆H₁₅N₃
Gas to liquid solids are going to break the liquid