Question #cddba

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Question #cddba

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Answer 1 · 2017-08-31T20:07:53

Using conversion factors, we get $23 m L$ $23mL$ .

Explanation:

We can answer this question using conversion factors.

In short, since we know that $1.48$ $1.48$ grams of the substance is equal to $1 c m^{3}$ $1cm^3$ of the substance, we know that $\frac{1 c m^{3}}{1.48 g} = 1$ $(1cm^3)/(1.48g)=1$

Also, since $1 m L$ $1mL$ is equal to $1 c m^{3}$ $1cm^3$ , we can say that $\frac{1 m L}{1 c m^{3}} = 1$ $(1mL)/(1cm^3)=1$

Then, all we need to do is multiply our value ( $34 g$ $34g$ ) by the conversion factors.

$34 g \cdot \frac{1 c m^{3}}{1.48 g} \cdot \frac{1 m L}{1 c m^{3}}$ $34color(green)g*color(red)(1cm^3)/(1.48color(green)g)*(1mL)/color(red)(1cm^3)$

The same-colored units cancel out, and we're left with:

$\frac{34 m L}{1.48}$ $(34mL)/1.48$

$34 m L$ $34mL$ has two significant figures, so we give our answer to two significant figures:

$23 m L$ $23mL$

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Question #cddba

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