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kykrilka [37]
3 years ago
9

Ionic crystals have high boiling points because:

Chemistry
1 answer:
juin [17]3 years ago
8 0

The answer would be B.

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\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag.

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Electrons are conserved in a chemical equation.

The superscript of \rm Ag^{1+} indicates that each of these ions carries a charge of +1. That corresponds to the shortage of one electron for each \rm Ag^{+} ion.

Similarly, the superscript +3 on each \rm Al^{3+} ion indicates a shortage of three electrons per such ion.

Assume that the coefficient of \rm Ag^{+} (among the reactants) is x, and that the coefficient of \rm Al^{3+} (among the reactants) is y.

\rm \mathnormal{x}\; Ag^{1+} + ?\; Al \to \mathnormal{y}\; Al^{3+} + ?\; Ag.

There would thus be x silver (\rm Ag) atoms and y aluminum (\rm Al) atoms on either side of the equation. Hence, the coefficient for \rm Al\! and \rm Ag\! would be y\! and x\!, respectively.

\rm \mathnormal{x}\; Ag^{1+} + \mathnormal{y}\; Al \to \mathnormal{y}\; Al^{3+} + \mathnormal{x}\; Ag.

The x \rm Ag^{1+} ions on the left-hand side of the equation would correspond to the shortage of x electrons. On the other hand, the y Al^{3+} ions on the right-hand side of this equation would correspond to the shortage of 3\, y electrons.

Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of x electrons, the right-hand side should also be x\! electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of 3\, y electrons. These two expressions should have the same value. Therefore, x = 3\, y.

The smallest integer x and y that could satisfy this relation are x = 3 and y = 1. The equation becomes:

\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag.

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