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Sorry I need 20 characters to submit this Answer that’s why I’m adding more words
The balanced equation would be 
<h3>Electrochemical equations</h3>
Zn reacts with Cu solution according to the following equation:

In the reaction,
is reduced according to the following: 
While Zn is oxidized according to the following: 
Thus, giving the overall equation of; 
More oxidation-reduction equations can be found here: brainly.com/question/13699873
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Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't
Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to
solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per
kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
Answer:
i think that the answer is solid liquid and gas.
Explanation: