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kondor19780726 [428]
3 years ago
13

The question is below

Chemistry
1 answer:
rodikova [14]3 years ago
3 0
The anode is the electrode where the oxidation occurs.

Cathode is the electrode where the reducction occurs.

Equations:

  Mn(2+) + 2e-  --->  Mn(s)                      Eo = - 1.18 V
2Fe(3+) + 2e-  ----> 2 Fe(2+)                2Eo = + 1.54 V

The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.

Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72

Answer: 2.72 V

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Explanation:

The equation for above decomposition is

2N_2O \rightarrow 2N_2 + O_2

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2N_2O \rightarrow 2N_2 + O_2

initial           4.70                         0             0

change        -2x                          +2x           +x

final             4.70 -2x                     2x           x

pressure ofO_2 after first half life  = 2.35 = 4.70 - 2x

                                                          x = 1.175

pressure of N_2 after first half life  =  2x = 2(1.175) = 2.35 ATM

Total pressure  = 2.35 + 2.35 + 1.175

                          = 5.875 atm

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