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14 Question ( point) Q See page 382 Adding nitric acid to water is quite exothermic. of HNO3 is-33.3 kJ mol, then how much he

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Answer #1

Given:

H soln = -33.3 KJ/mol

number of mol = 0.180 mol

use:

Q = H soln * number of mol

= (-33.3 KJ/mol)*0.180 mol

= -5.99 KJ

= -5990 J

So, heat released = 5990 J

Answer: 5990 J

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14 Question ( point) Q See page 382 Adding nitric acid to water is quite exothermic. of HNO3 is-33.3 kJ mol, then how much heat is evolved by dissolving 0 180 mol HND If the ΔΗΡ οι in 1000m of water?...
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