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Murrr4er [49]
3 years ago
7

To a 0.0001 m solution of mg(no3)2, naoh was added to a final concentration of 0.001m did a precipitate form?

Chemistry
1 answer:
Natalija [7]3 years ago
8 0

I looked on a solubility chart to answer this question, and hydroxides are generally insoluble (with some exceptions of course). However, it says to consider Mg(OH)_{2} as an insoluble substance, though it may be moderately soluble.


The answer that you are most likely looking for is: Yes, a precipitate does form - this is due to the double placement reaction:


Mg(NO_{3})_{2}_{(aq)} + 2NaOH_{(aq)} → Mg(OH)_{2} {(s)} + 2NaNO_{3}_{(aq)}

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Answer:

ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of  NH₄NO₃(s) given) (exothermic)

Explanation:

           3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g)  +  17H₂O(g)   +  10CO₂(g)

ΔH⁰(f):  3(-365.6)Kj     1(-301)Kj    14(0)Kj       3(0)Kj    17(-241.8)Kj    10(-393.5)Kj

            = -1096.8Kj     = -301Kj     = 0Kj         = 0Kj      = -4110.6Kj    = -3930.5Kj

ΔHₙ°(rxn) = ∑ (ΔH˚(f)products) - ∑(ΔH˚(f)reactants)

= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]

= [-(8041.1) - (-1397.8)]Kj

= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)  

∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)

ΔH°(rxn for 14.11g  NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of  NH₄NO₃(s) given)

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