Question #9c740

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Question #9c740

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Answer 1 · 2017-07-08T00:02:32

${55.5 g CaCl}_{2}$ $"55.5 g CaCl"_2$

Explanation:

For starters, notice that your sample contains

$3.01 \cdot 10^{23} = \frac{1}{2} \cdot 6.02 \cdot 10^{23}$ $3.01 * 10^(23) = 1/2 * color(blue)(6.02 * 10^(23))$

formula units of calcium chloride. As you know, in order to have $1$ $1$ mole of calcium chloride, you need to have a sample that contains $6.02 \cdot 10^{23}$ $color(blue)(6.02 * 10^(23))$ formula units of calcium chloride $\to$ $->$ this is known as Avogadro's constant.

In your case, the sample contains exactly half, $\frac{1}{2}$ $1/2$ , of a number equal to Avogadro's constant of formula units of calcium chloride, which means that it contains $0.5$ $0.5$ moles of calcium chloride.

To convert the number of moles to grams, you must use the compound's molar mass.

You will end p with

$0.5 color(red)(cancel(color(black)("moles CaCl"_2))) * "110.98 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = color(darkgreen)(ul(color(black)("55.5 g")))$

The answer is rounded to three sig figs, the number of significant figures you have for the number of formula units present in the sample.

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Question #9c740

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