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Veseljchak [2.6K]
3 years ago
6

The mixing of which pair of reactants will result in a precipitation reaction? the mixing of which pair of reactants will result

in a precipitation reaction? nh4br(aq) + nh4i(aq) naclo4(aq) + (nh4)2s(aq) ni(no3)2(aq) + na2co3(aq) k2so4(aq) + cu(no3)2(aq)
Chemistry
2 answers:
Ahat [919]3 years ago
6 0

<u>Answer:</u> The pair undergoing a precipitation reaction is Ni(NO_3)_2(aq.)+Na_2CO_3(aq.)

<u>Explanation:</u>

Precipitation reaction is defined as the reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

For the given options:

<u>Option 1:</u>  NH_4Br(aq.)+NH_4I(aq.)\rightarrow \text{No reaction}

As, the cation is same in both of the reactants. Thus, the reactants will remain as such. Thus, no precipitation reaction occurs.

<u>Option 2:</u>  2NaClO_4(aq.)+(NH_4)_2S(aq.)\rightarrow Na_2S(aq.)+2NH_4ClO_4(aq.)

As, no solid product is getting formed. Thus, no precipitation reaction occurs.

<u>Option 3:</u>  Ni(NO_3)_2(aq.)+Na_2CO_3(aq.)\rightarrow NiCO_3(s)+2Na(NO_3)(aq.)

As, nickel (II) carbonate is getting formed as a solid. It is considered as a precipitate. Thus, it is a precipitation reaction

<u>Option 4:</u>  K_2SO_4(aq.)+Cu(NO_3)_2(aq.)\rightarrow CuSO_4(aq.)+2KNO_3(aq.)

As, no solid product is getting formed. Thus, no precipitation reaction occurs.

Hence, the pair undergoing a precipitation reaction is Ni(NO_3)_2(aq.)+Na_2CO_3(aq.)

IceJOKER [234]3 years ago
5 0
The mixing of Ni(NO3)2(aq) and Na2CO3(aq) will result in a precipitation reaction.

Ni(NO3)2(aq) + Na2CO3(aq) =====> NiCO3(s) + 2NaNO3(aq)

:-) ;-)
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Also, we know that the relation between enthalpy and temperature change is as follows.

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Now, calculate entropy change for block 1 as follows.

     \Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

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            = 10000 g \times 0.385 J/K g \times -0.143

            = -554.12 J/K

Now, entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

            \Delta S_{total} = \Delta S_{1} + \Delta S_{2}

                       = -554.12 J/K + 647.49 J/K

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