Question

purpose: in this activity you will practice reading a graduated cylinder and calculate density based on mass and volume.
part one:
read each graduated cylinder and write down the amount. be sure to include ml in your answer.

1. ______ 2. ____ 3. ____ 4. ______

part three:
density is the mass of a substance contained in a given volume. if the mass and volume of an object are known, then its density can be found. the formula for finding density is: density = mass/volume
the mass of an eraser is 50 grams. its volume is 25 cm³. what is its density? 50g/25cm³ = 2 g/cm³
find the density of the objects with the following values for mass and volume. show your work.
mass: 100g volume: 50cm³ 1. ______ mass: 150g volume: 75cm³ 2. ____ mass: 230g volume: 100cm³ 3. ____ mass: 376g volume: 98cm³ 4. ____ mass: 6750g volume: 237cm³ 5. ______
show work here.
mass: 950g volume: 375cm³ 6. ______ mass: 500g volume: 500cm³ 7. ____ mass: 1,876g volume: 475cm³ 8. ____ mass: 872g volume: 546cm³ 9. ____ mass: 886g volume: 947cm³ 10. ______
making sense
look at the drawing of a balance. the pan on the left holds a piece of lead. the pan on the right holds a box full of feathers.

1. which has the greater mass? (look at the weight)
2. which has the greater volume?
3. which has the greater density?

Explanation:

Step1: Recall density formula

Density $
ho=\frac{m}{V}$, where $m$ is mass and $V$ is volume.

Step2: Calculate densities for given mass - volume pairs

For the first set of values (Mass = 100g, Volume = 50$cm^{3}$)

$
ho_1=\frac{100g}{50cm^{3}} = 2g/cm^{3}$

For the second set of values (Mass = 150g, Volume = 75$cm^{3}$)

$
ho_2=\frac{150g}{75cm^{3}} = 2g/cm^{3}$

For the third set of values (Mass = 230g, Volume = 100$cm^{3}$)

$
ho_3=\frac{230g}{100cm^{3}} = 2.3g/cm^{3}$

For the fourth set of values (Mass = 376g, Volume = 98$cm^{3}$)

$
ho_4=\frac{376g}{98cm^{3}}\approx3.84g/cm^{3}$

For the fifth set of values (Mass = 6750g, Volume = 237$cm^{3}$)

$
ho_5=\frac{6750g}{237cm^{3}}\approx28.48g/cm^{3}$

For the sixth set of values (Mass = 950g, Volume = 375$cm^{3}$)

$
ho_6=\frac{950g}{375cm^{3}}\approx2.53g/cm^{3}$

For the seventh set of values (Mass = 500g, Volume = 500$cm^{3}$)

$
ho_7=\frac{500g}{500cm^{3}} = 1g/cm^{3}$

For the eighth set of values (Mass = 1876g, Volume = 475$cm^{3}$)

$
ho_8=\frac{1876g}{475cm^{3}}\approx3.95g/cm^{3}$

For the ninth set of values (Mass = 872g, Volume = 546$cm^{3}$)

$
ho_9=\frac{872g}{546cm^{3}}\approx1.60g/cm^{3}$

For the tenth set of values (Mass = 886g, Volume = 947$cm^{3}$)

$
ho_{10}=\frac{886g}{947cm^{3}}\approx0.94g/cm^{3}$

For the balance - related questions:

1. Since the balance is in equilibrium, the mass of the lead and the mass of the feathers are equal.
2. Feathers are less dense and will occupy a much larger volume than the lead for the same mass. So, the feathers have a greater volume.
3. Lead is a dense material compared to feathers. So, the lead has a greater density.

Answer:

1. $2g/cm^{3}$
2. $2g/cm^{3}$
3. $2.3g/cm^{3}$
4. $\approx3.84g/cm^{3}$
5. $\approx28.48g/cm^{3}$
6. $\approx2.53g/cm^{3}$
7. $1g/cm^{3}$
8. $\approx3.95g/cm^{3}$
9. $\approx1.60g/cm^{3}$
10. $\approx0.94g/cm^{3}$

For balance questions:

1. They have equal mass.
2. Feathers have greater volume.
3. Lead has greater density.