# Question d64a6

Aug 27, 2017

${\text{1.2 g mL}}^{- 1}$

#### Explanation:

The idea here is that you can determine the volume of the object by looking at the difference between the water levels.

When you add the object to the water, a volume of water that is equal to the volume of the object will be displaced $\to$ the water level will rise in the cylinder.

In your case, the volume of the water displaced by the object will be equal to

$\text{29.4 mL " - " 25.1 mL = 4.3 mL}$

This means that the volume of the object is equal to $\text{4.3 mL}$.

Now, in order to find its density, you must figure out the mass of exactly $1$ unit of volume, i.e. of $\text{1m L}$, occupied by the object.

Since you know that $\text{4.3 mL}$ have a mass of $\text{5.1 g}$, you can say that $\text{1 mL}$ will have a mass of

1 color(red)(cancel(color(black)("mL"))) * "5.1 g"/(4.3 color(red)(cancel(color(black)("mL")))) = "1.186 g"#

This means that the density of the object, i.e. its mass per unit of volume, will be equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{density = 1.2 g mL}}^{- 1}}}}$

The answer is rounded to two sig figs.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}