# Question #51727

##### 1 Answer

#### Answer:

The answer must be rounded to three significant figures.

#### Explanation:

It looks to me like you're trying to determine the mass of a given volume of a substance by using a **density** of the substance.

On that note, I assume that your calculation looks like this

#7.379 xx 10^(-4) color(red)(cancel(color(black)("m"^3))) xx (100^3 color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("m"^3)))) xx ("19.3 g")/(1 color(red)(cancel(color(black)("cm"^3)))) = ?#

The first thing to note here is that the original measurement has **significant figures**, the number of sig figs you have in its *mantissa*.

#7.379 * 10^(-4)" " -> " 4 sig figs: " {7, 3, 7, 9}#

Now, you're using a conversion factor to take you from *cubic meters* to *cubic centimeters*

#"1 m"^3 = 100^3 quad "cm"^3#

Because **defined** as being equivalent to **as many significant figures** as you want for this conversion factor.

In other words, *infinite number* of significant figures because they are **defined** that way.

Finally, you have the **density** of the substance, which is equal to **for every** **significant figures** in the mass of the substance present in

#"19.3 g " -> " 3 sig figs: " {1, 9, 3}#

You can use the fact that density is defined as the mass of **exactly** one unit of volume to use an infinite number of sig figs for

So you can say that in terms of the number of sig figs, your calculation comes down to

#"4 sig figs" xx "as many sig figs as you need"/"as many sig figs as you need" xx "3 sig figs"/"as many sig figs as you need"#

Since you're dealing with a **multiplication**, the answer must be rounded to the number of sig figs of your **least precise** measurement.

In this case, you have

#"3 sig figs " ("19.3 g") " " < " " "4 sig figs " (7.379 * 10^(-4))#

so the result must be rounded to **significant figures**. This means that

#7.379 xx 10^(-4) color(red)(cancel(color(black)("m"^3))) xx (100^3 color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("m"^3)))) xx ("19.3 g")/(1 color(red)(cancel(color(black)("cm"^3)))) = "14,241.47 g"#

will be rounded to *scientific notation*

#7.379 xx 10^(-4) color(red)(cancel(color(black)("m"^3))) xx (100^3 color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("m"^3)))) xx ("19.3 g")/(1 color(red)(cancel(color(black)("cm"^3)))) = color(darkgreen)(ul(color(black)(1.42 xx 10^(4) quad "g")))#