Question

calculations questions:
density = \frac{mass}{volume}
volume = length × width × height

1. an object has a mass of 28 g and a volume of 9 cm³. what is its density? circle your answer and show your work for full credit. (2 points)
2. an object has a mass of 30 grams. it is placed in a graduated cylinder that contains 20 ml of water. the water level rises to 30 ml. calculate the density of the object. circle your answer and show your work for full credit. (2 points)
3. what is the mass of a cube of copper if the density is 9.0 g/ml and the volume of the cube is 64 cm³? circle your answer and show your work for full credit. (2 points)
4. a rectangular solid made of an unknown metal is 12.0 cm long, 3.0 cm wide and 1.39 cm tall. its mass is 135.0 g. use the cart below to determine the metal the solid is made from. circle your answer and show your work for full credit. (3 points)

name of metal	density (g/cm³)
zinc	7.1
copper	9.0
silver	10.5
lead	11.4
gold	19.3

density = ________
name of the metal is: ________

Explanation:

Step1: Recall density formula

The formula for density is $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass and $V$ is volume.

Step2: Solve problem 19

Given $m = 28\ g$ and $V=9\ cm^{3}$, we substitute into the density - formula: $
ho=\frac{28\ g}{9\ cm^{3}}\approx3.11\ g/cm^{3}$.

Step3: Solve problem 20

First, find the volume of the object. The volume of the object is equal to the volume of water displaced. The initial volume of water is $V_1 = 20\ mL$ and the final volume is $V_2 = 30\ mL$. So, $V=V_2 - V_1=30\ mL - 20\ mL = 10\ mL=10\ cm^{3}$ (since $1\ mL = 1\ cm^{3}$). Given $m = 30\ g$, using the density formula $
ho=\frac{m}{V}=\frac{30\ g}{10\ cm^{3}} = 3\ g/cm^{3}$.

Step4: Solve problem 21

We know that $
ho=\frac{m}{V}$, and we want to find $m$. Rearranging the formula gives $m=
ho V$. Given $
ho = 9.0\ g/mL$ and $V = 64\ cm^{3}=64\ mL$ (since $1\ cm^{3}=1\ mL$), then $m=9.0\ g/mL\times64\ mL = 576\ g$.

Step5: Solve problem 22

First, find the volume of the rectangular solid. Using $V=l\times w\times h$, where $l = 12.0\ cm$, $w = 3.0\ cm$ and $h = 1.39\ cm$. So, $V=12.0\ cm\times3.0\ cm\times1.39\ cm=49.84\ cm^{3}\approx49.8\ cm^{3}$. Given $m = 135.0\ g$, using the density formula $
ho=\frac{m}{V}=\frac{135.0\ g}{49.8\ cm^{3}}\approx2.7\ g/cm^{3}$. From the table, the metal is aluminum.

Answer:

1. $\approx3.11\ g/cm^{3}$
2. $3\ g/cm^{3}$
3. $576\ g$
4. Density $\approx2.7\ g/cm^{3}$, Name of the metal is: Aluminum