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a. For a chemistry lab final exam, a high school chemistry student was given a 1-mole sample of CaCl2 and a 1-mole sample of MgCl2 but was not told which sample was which. He was to identify the powders.

He looked up the enthalpies of formation for both of the chemicals and calculated the ΔHreaction for dissolving each powder: CaCl2 (s) Ca 2+ (aq) + 2Cl – (aq), and MgCl2 (s) Mg 2 (aq) + 2Cl – (aq). He then put each powder in a coffee-cup calorimeter and added water.

When sample A dissolved, the temperature increased by 0.74°C. When sample B dissolved, the temperature increased by 0.39°C. Which chemical was A, and which was B? Use the table of enthalpies of formation to help you. Explain your reasoning.

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Ca^2+ ion  is smaller than Mg^2+ ion hence it has a greater heat of formation or lattice energy.

When a crystal lattice is formed, the energy that is released if the component ions of the compound are brought together from infinity is called the lattice energy. It is the energy released when a crystal lattice is formed from its component ions.

The question lets us know that the heat released by the compounds depends on the energy released upon formation of the compound. Hence, the higher the energy released upon formation, the higher the magnitude of heat released upon dissolution of the compound.

Recall that lattice energy depends on the size of the ions. Thus, the smaller the ions, the higher the lattice energy.

Ca^2+ is smaller than Mg^2+ hence more energy is given off when CaCl2  is formed than when MgCl2  is formed.

As stated above, the greater the lattice energy, the greater the heat released when the lattice dissolves and the higher the rise in temperature.

Putting all these together, Sample A must be CaCl2 while sample B must be MgCl2.

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