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3 years ago

How many grams of solid K2CO3 are needed to prepare 150. mL of a 0.50% by volume K2CO3 solution ?

Chemistry

1 answer:

serg [7]3 years ago

3 0

Answer:

0.75 g of K2CO3

Explanation:

To prepare 150. mL of a 0.50% by volume K2CO3 solution;
It means 0.5% w/v in 150 mL.
Therefore, there would be 0.5 of K2CO3 in 100 mL of the solution.

Taking the density of the solution to be equivalent to the density of water, that is, 1 g/ml.

The mass of the solution will be; 150 mL × 1 g/mL = 150 g

Therefore; the mass of K2CO3 will be;

= 0.5/100 × 150 g

= 0.75 g

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Monica [59]

Answer:

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

Explanation:

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 $y$ .

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

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

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, $x = 3\, y$ .