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3 years ago

If you needed to make 100 mL of a 0.2 M fruit drink solution from the 1.0 M fruit drink

Chemistry

1 answer:

telo118 [61]3 years ago

8 0

Answer:

We take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to make 100 mL of a 0.2 M fruit drink solution.

Explanation:

Using the rule that: the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution. (MV) before dilution = (MV) after dilution. M before dilution = 1.0 M, V before dilution = ??? mL. M after dilution = 0.2 M, V after dilution = 100 mL. ∴ V before dilution = (MV) after dilution / M before dilution = (0.2 M)(100 mL) / (1.0 M) = 20.0 mL. So, we take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to make 100 mL of a 0.2 M fruit drink solution

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Answer:

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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 $y$ .

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

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

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