Question

using density as a conversion factor
the following table displays prefixes used for si units

factor	name	symbol
$10^{3}$	kilo	k
$10^{2}$	hecto	h
$10^{1}$	deca	da
$10^{-1}$	deci	d
$10^{-2}$	centi	c
$10^{-3}$	milli	m
$10^{-6}$	micro	$mu$

you can use these relationships to create equalities and conversion factors. for example, the prefix deci is denoted with a d and is associated with a factor of $10^{-1}$. if you were to apply this prefix to liter, you would obtain the equality $1 dl = 10^{-1} l$. this could be used to create the conversion - factor $\frac{1 dl}{10^{-1} l}$ and $\frac{10^{-1} l}{1 dl}$.
the conversion factor used will depend on whether the final unit should be in l or dl, and whether the initial value has l or dl in the numerator or denominator. make sure the units cancel, leaving you with the units you need.
part c
iron has a density of $7.86 g/cm^{3}$ ($1 cm^{3}=1 ml$). calculate the volume (in dl) of a piece of iron having a mass of $3.17 kg$. note that the density is provided in different units of volume and mass than the desired units of volume (dl) and the given units of mass (kg). first convert the density to $kg/dl$ ($1 cm^{3}=1 ml$) before calculating the volume for the piece of iron.
express your answer to three significant figures.

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Explanation:

Step1: Convert mass to grams

Given mass $m = 3.7\ kg$. Since $1\ kg=1000\ g$, then $m = 3.7\times1000\ g=3700\ g$.

Step2: Use density formula

The density formula is $
ho=\frac{m}{V}$, where $
ho = 7.86\ g/cm^{3}$, and $m = 3700\ g$. We can solve for volume $V$, so $V=\frac{m}{
ho}=\frac{3700\ g}{7.86\ g/cm^{3}}\approx470.74\ cm^{3}$.

Step3: Convert $cm^{3}$ to $mL$

Since $1\ cm^{3}=1\ mL$, then $V = 470.74\ mL$.

Step4: Convert $mL$ to $dL$

We know that $1\ dL = 100\ mL$, so $V=\frac{470.74\ mL}{100}=4.71\ dL$ (rounded to three - significant figures).

Answer:

$4.71$