Answer: Option (C) is the correct answer.
Explanation:
Chemical formula of a secondary amide is R'-CONH-R, where R and R' can be same of different alkyl or aryl groups. Here, the hydrogen atom of amide is attached to more electronegative oxygen atom of the C=O group.
Therefore, the hydrogen atom will be more strongly held by the electronegative oxygen atom. As a result, there will be strongly hydrogen bonded in the liquid phase of secondary amide.
Whereas chemical formula of nitriles is RCN, ester is RCOOR' and acid chlorides are RCOCl. As no hydrogen bonding occurs in any of these compounds because hydrogen atom is not being attached to an electronegative atom.
Thus, we can conclude that secondary amides are strongly hydrogen bonded in the liquid phase.
Answer:
D
Explanation:
Anions are attracted to the positive end of a dipole, while the cations are attracted to the negative end. As the size of the dipole moment or the ionic charge increases, the vastness of the attraction also increases. This type of attraction is important for solutions of ionic substances in polar liquids.
Answer:
Covalent
Explanation:
A molecule of C₂H₅OH has C-C, C-H, C-O, and O-H bonds.
A bond between A and B will be ionic if the difference between their electronegativities (ΔEN) is greater than 1.6.
No bond has a large enough ΔEN to be ionic.
C₂H₅OH is a covalent molecule.
Answer:
Covalent bond.
Explanation:
There are two kinds of chemical bonds: covalent bonds and ionic bonds.
- A covalent bond is formed when two atoms share a pair of electrons (two electrons for each bond.)
- Ions are formed when one atom transfers an electron to another. Ionic bonds refer to the attraction between ions of opposite electric charges.
In this example, since the atoms are sharing atoms, the chemical bond between them would be a covalent bond.
Answer:
The pH of the solution is 5.31.
Explanation:
Let "
is the dissociation of weak acid - HCN.
The dissociation reaction of HCN is as follows.
Initial C 0 0
Equilibrium c(1- ) c c
Dissociation constant = 
In this case weak acids is very small so, (1- ) is taken as 1.
From the given the concentration = 0.050 M
Substitute the given value.
![[H_{3}O^{+}]=c\alpha](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3Dc%5Calpha)
![[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D0.05%5Ctimes%209.8%5Ctimes%2010%5E%7B-4%7D%3D%204.9%5Ctimes10%5E%7B-6%7D)
![pH= -log[H_{3}O^{+}]](https://tex.z-dn.net/?f=pH%3D%20-log%5BH_%7B3%7DO%5E%7B%2B%7D%5D)
![=-log[4.9\times10^{-6}]](https://tex.z-dn.net/?f=%3D-log%5B4.9%5Ctimes10%5E%7B-6%7D%5D)
Therefore, The pH of the solution is 5.31.
