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Eva8 [605]
3 years ago
10

Ans with solution...

Chemistry
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:

The answer to your question is:

Vol of NO2 = 11.19 L

Vol of O2 = 2.8 L

Explanation:

Data

N2O5 = 56 g

STP     T = 0°C = 273°K

           P = 1 atm

MW N2O5 = 216 g

Gases law = PV = nRT

Process

                   216 g of N2O5 ---------------- 1 mol

                     54 g               -----------------  x

                    x = (54 x 1) / 216

                    x = 0.25 mol of N2O5

                   2 mol of N2O5 -----------------  4 mol of NO2

                   0.25 mol          ------------------    x

                   x = (0.25 x 4) / 2 = 0.5 mol of NO2

                   V = nRT/P

                   V = (0.5)(0.082)(273) / 1 = 11.19 L

                   2 mol of N2O5 ----------------- 1 O2

                   0.25 N2O5 ----------------------  x

                   x = (0.25 x 1) / 2 = 0.125 mol

                   Vol = (0.125)((0.082)(273) / 1 = 2.8 L

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inn [45]

Answer:

3.47 ×10^-10

Explanation:

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E°cathode = -0.41 V

E°anode = -0.13 V

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From

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4 0
3 years ago
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