I believe it would be a compound.
Answer: 
Explanation:

cM 0 0
So dissociation constant will be:

Given: c = 0.15 M
pH = 1.86
= ?
Putting in the values we get:
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![1.86=-log[H^+]](https://tex.z-dn.net/?f=1.86%3D-log%5BH%5E%2B%5D)
![[H^+]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01)
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)


As ![[H^+]=[ClCH_2COO^-]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BClCH_2COO%5E-%5D%3D0.01)

![K_a=1.67\times 10^{-3]](https://tex.z-dn.net/?f=K_a%3D1.67%5Ctimes%2010%5E%7B-3%5D)
Thus the vale of
for the acid is 
Answers:
Density = 0.8 g/mol.
Given data:
v = 25 ml
m = 20 g
δ = ?
Solution:
Formula for calculating density is given as,
Density = Mass / Volume
putting values
Density = 20.0 g / 25 ml
Density = 0.8 g/mol.
The question ask for the percentage of the abundance of galium-69 where there is two isotopes of galium: the 69Ga and the 71Ga. The average atomic mass of gallium is 69.723 amu. So the formula would be <span>69.723amu=(%x)∗(68.926amu)+(1−%x)∗(70.025amu) and the answer to this is 1.58%</span>