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raketka [301]
2 years ago
7

Select the compound that is most likely to increase the solubility of ZnSe when added to water.NaCnMgBr2NaClKClO4

Chemistry
1 answer:
vlada-n [284]2 years ago
6 0

Answer:

NaCl.

Explanation:

In the solution, ZnSe ionizes to Zn^2^+ and Se^2^- . Following reaction represents the ionization of ZnSe in solution -

ZnSe ⇄ Zn^2^+ + Se^2^-

As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.

If we add NaCl to this solution, then we have Na^+ and Cl^- in the solution which will be formed by the ionization of NaCl.

Now, Zn^2^+ in the solution will react with two Cl^- ions to form ZnCl_2 as follows -

Zn^2^++2Cl^- ⇄ ZnCl_2

Due to this reaction the concentration of Zn^2^+ will decrease in the solution and more ZnSe can be soluble in the solution.

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Answer:

The enthalpy change in the the reaction is -47.014 kJ/mol.

Explanation:

X(s)+H_2O(l)\rightarrow X(aq)

Volume of water in calorimeter = 22.0 mL

Density of water = 1.00 g/mL

Mass of the water in calorimeter = m

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Heat released during the reaction be Q

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Specific heat of the solution is equal to that of water :

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Q=Mc\times \Delta T

Q=24.50 g\times 4.18 J/g ^oC\times 14.0^oC=1,433.74 J=1.433 kJ

Heat released during the reaction is equal to the heat absorbed by the water or solution.

Heat released during the reaction =-1.433 kJ

Moles of substance X= \frac{2.50 g}{82.0 g/mol}=0.03048 mol

The enthalpy change, ΔH, for this reaction per mole of X:

\Delta H=\frac{-1.433 kJ}{0.03048 mol}=-47.014 kJ/mol

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