How many ions of calcium and nitride are present in 0.0592 g Calcium Nitride?
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Answered by
1
Answer:
n
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Correct option is C)
The overall cell reaction is given by:
Cu
2+
+H
2
→Cu+2H
+
E
cell
∘
=E
Cu
2+
/Cu
∘
−E
H
+
/H
2
∘
⟹E
cell
∘
=0.337−0=0.337V [standard hydrogen electrode potential is zero].
Now, E
cell
=0.337−
2
0.059
log[
[Cu
2+
]
[H
+
]
2
]
E
cell
=0.337−
2
0.059
log(
(0.1)
(0.01)
2
)
E
cell
=0.425V
Option C is correct.
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Answered by
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Answer:
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