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bazaltina [42]
3 years ago
14

N2(g) + 2 O2(g) ------- 2 NO2(g) ΔH = +66.4 kJ

Chemistry
1 answer:
SpyIntel [72]3 years ago
5 0

Answer:

The answer to your question is the letter d. -132.8 kJ

Explanation:

To solve this problem, just observe that the ΔH is originally equal to +66.4 kJ.

If we change the order of the reaction, ΔH changes its sign but the value is the same.

Finally, if we duplicate the number of moles, also the ΔH increases twice.

For this problem

                           ΔH = 2(-66.4 kJ)

                          ΔH = - 132.8 kJ

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3 years ago
Is the indicator generally added to the titrant or the analyte in a titration?
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The decomposition of NI3 to form N2 and I2 releases −290.0 kJ of energy. The reaction can be represented as 2NI3(s)→N2(g)+3I2(g)
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Answer:

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

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According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

\frac{-290.0 kJ}{2}\times 0.05063=-7.34 kJ

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3 0
3 years ago
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