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Anon25 [30]
2 years ago
7

Determine the poh of a 0.227 m c5h5n solution at 25°c. The kb of c5h5n is 1.7 × 10-9.

Chemistry
1 answer:
Juliette [100K]2 years ago
3 0

4.71

hope this helps!

(:

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A geologist is studying an area where stream erosion and deposition are the dominant surface processes. He notices that all of t
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PLEASE HELP <br> What is the element dot structure for Li+1? Really confused about the +1 part.
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Read 2 more answers
A galvanic (voltaic) cell consists of an electrode composed of chromium in a 1.0 M chromium(III) ion solution and another electr
svetlana [45]

Answer: The potential of the following electrochemical cell is 1.08 V.

Explanation:

E^0_(Cr^{3+}/Cr)=-0.74V[/tex]

E^0_(Cu^{2+}/Cu)=0.34V[/tex]

The element  with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

Here Cr undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.

2Cr+3Cu^{2+}\rightarrow 2Cr^{3+}+3Cu

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials, when concentration is 1M.

E^0=E^0_{[Cu^{2+}/Ni]}- E^0_{[Cr^{3+}/Cr]}

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Thus the potential of the following electrochemical cell is 1.08 V.

6 0
3 years ago
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