Question

d = m/v
d = density in (g/cm³ or g cm⁻³) (g/ml or g ml⁻¹) (kg/m³ or kg m⁻³)
m = mass in (g)
v = volume in (cm³ or ml) note that 1 ml = 1 cm³
as the mass of an object increases, the density of that object
as the volume of that object increases, the density of that object
pure water has a density of 1.00 g cm⁻³. anything larger than this number will while any object with a density lower than this number would
we measure the mass of an object with the use of a
the volume of an object that is a liquid can be measured with the use of a
the volume of an object that are solids and have a square or rectangular shape can be measured by measuring the sides and multiplying them (v = l x w x h) (be sure to measure in cm).
level 1 practice problems: (set up the problem, give answer and units):

1. if an object had a mass of 345 g and took up a space of 100 ml, then what would be the density?
2. if an object had a mass of 70 g and took up a space of 1500 cm³, then what would be the density?
3. if an object took up 500 cm³ of space and had a mass of 15 g, then what would be the density?
4. would any of these objects float on pure water? explain your answer.

Explanation:

Step1: Recall density formula

The density formula is $D=\frac{M}{V}$, where $D$ is density, $M$ is mass and $V$ is volume.

Step2: Solve problem 1

Given $M = 345\ g$ and $V=100\ mL$. Since $1\ mL = 1\ cm^{3}$, then $D=\frac{345\ g}{100\ cm^{3}} = 3.45\ g/cm^{3}$.

Step3: Solve problem 2

Given $M = 70\ g$ and $V = 1500\ cm^{3}$. Then $D=\frac{70\ g}{1500\ cm^{3}}\approx0.047\ g/cm^{3}$.

Step4: Solve problem 3

Given $M = 15\ g$ and $V = 500\ cm^{3}$. Then $D=\frac{15\ g}{500\ cm^{3}}=0.03\ g/cm^{3}$.

Step5: Analyze floating - ability

Pure water has a density of $1.00\ g/cm^{3}$. Objects with density less than $1.00\ g/cm^{3}$ will float and objects with density greater than $1.00\ g/cm^{3}$ will sink. The objects in problem 2 ($0.047\ g/cm^{3}$) and problem 3 ($0.03\ g/cm^{3}$) will float because their densities are less than that of pure water. The object in problem 1 ($3.45\ g/cm^{3}$) will sink because its density is greater than that of pure water.

Answer:

1. $D = 3.45\ g/cm^{3}$
2. $D\approx0.047\ g/cm^{3}$
3. $D = 0.03\ g/cm^{3}$
4. The objects in problem 2 and 3 will float as their densities ($0.047\ g/cm^{3}$ and $0.03\ g/cm^{3}$) are less than the density of pure water ($1.00\ g/cm^{3}$). The object in problem 1 will sink as its density ($3.45\ g/cm^{3}$) is greater than the density of pure water.