We have that for the Question it can be said that the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)
From the question we are told
how should the ph of a 0.1 m solution of <em>nac2h3o2</em> compare with that of a 0.1 m <u>solution </u>of kc2h3o2?
Generally
with the ph of a 0.1 m solution of <em>nac2h3o2</em> compared with that of a 0.1 m <u>solution </u>of kc2h3o2 ,we see that the salt produce is a weak acid and strong akali salt
We see that the salt produced in water gives a base from the derived weak acid
The salt produce is CH_3COONa
Therefore
the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)
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Answer:
The bond angles between the axial bonding groups are slightly less than 180°.
The bond angles between the equatorial bonding groups are slightly less than 120°.
Explanation:
Accordign to VSEPR theory, a molecule with four bonding groups and one lone pair on the central atom has a trigonal bipyramidal electronic geometry.
The position of the lone pair can be located in the equatorial position or axial position.
When the lone pair is found in equatorial position, it has two axial groups that repel it and the angle of the lone pair between each axial group is 90°.
When the lone pair is in axial position it has 3 equatorial groups that repel it and the angle of the lone pair between each equatorial group is 90°.
Since the molecule has a lone pair, the most stable geometric structure is when the lone pair is in the equatorial position, because it has fewer repulsions than in the axial position.
The molecular geometry is "seesaw"
The bond angles between the axial bonding groups are slightly less than 180°.
The bond angles between the equatorial bonding groups are slightly less than 120°.
Answer:
D) N-Cl.
Explanation:
The electronegativity of the elements in Group 5 (N. P As and Bi) decreases as we go down the Group so N is the most electronegative and chlorine is less able to draw electrons away from N than from the other elements in Group 5.
