Answer: ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
of 0.056 M HF solution is 
Explanation:
cM 0 0
So dissociation constant will be:
Give c= 0.056 M and 
= ?
Putting in the values we get:
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=0.056\times 0.0016=8.96\times 10^{-5}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.056%5Ctimes%200.0016%3D8.96%5Ctimes%2010%5E%7B-5%7D)
Thus of 0.056 M HF solution is
Answer:
Type of hybridisation
Explanation:
For example , saturated hydrocarbons have sp3 hybridisation while unsaturated hydrocarbons have either sp2 or sp hybridisation.
Example: Ethane( C2H6) has sp3 hybridisation and ethene(C2H4) has sp2 hybridisation and Ethyne(C2H2) has sp hybridization.
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