Which separation of matter technique is this a picture of? type your answer,
2 answers:
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In Methyl Amine the Nitrogen atom is attached to two Hydrogen Atoms and one Methyl (CH₃-) group. From Lewis structure it is shown that the Nitrogen atom also contains a lone pair of electron.
So,
According to Valence Shell Electron Pair Repulsion (VSEPR) Theory, those central atoms which have four electron pairs and all of them are bonding electron pairs gives Tetrahedral Geometry.
Methylamine also contains four electron pairs, but out of four only three are bonding pair electrons and one lone pair electron. Now, due to presence of lone pair of electron (which has greater repulsion effect than bonding electron pair) the bond angle between H-N-H will decrease from 109° to 107° and results in the formation of <span>Trigonal Pyramidal geometry as shown below,</span>
So,
According to Valence Shell Electron Pair Repulsion (VSEPR) Theory, those central atoms which have four electron pairs and all of them are bonding electron pairs gives Tetrahedral Geometry.
Methylamine also contains four electron pairs, but out of four only three are bonding pair electrons and one lone pair electron. Now, due to presence of lone pair of electron (which has greater repulsion effect than bonding electron pair) the bond angle between H-N-H will decrease from 109° to 107° and results in the formation of <span>Trigonal Pyramidal geometry as shown below,</span>
Answer:
E cell = +1.95 V
Explanation:
At Anode : Oxidation reaction takes place
At Cathode : Reduction reaction takes place
The reaction with lower value of reduction potential will undergo Oxidation
E = -1.18 V
This equation undergo oxidation reaction and become:
Anode(Oxidation-Half) :
E = +1.18 V
Cathode(Reduction-Half) :
E =+0.77 V
To balance the reaction multiply reduction-Half with 2.We get :
Note that E is intensive property , do not multiply E of oxidation-half with 2
Ecell = 0.77 -(-1.18)
E = +1.95 V