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vlada-n [284]
3 years ago
10

What are the oxidation numbers of the elements in the nitrite ion, no2−?

Chemistry
1 answer:
sdas [7]3 years ago
8 0
N = 3
O = -2
1(3) +2(-2)= -1

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Answer:

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\Delta n_g = (12 - 15)/2 = -3/2

We have;

\Delta H = \Delta U + \Delta n_g\cdot R\cdot T

Where R and T are constant, and ΔU is given we can write the relationship as follows;

H = U + \Delta n_g\cdot R\cdot T

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Therefore, we get;

H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ

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