QUESTION IMAGE
Question
fill in the blanks in the following table
| substance | mass (g) | volume (cm³) | density (g/cm³) |
|---|---|---|---|
| y | 50.0 | 10.0 | |
| z | 5.0 | 4.0 |
level 7
- water has a density of 1g/ml. there are four rocks that all have the same volume of 10 cubic centimeters. the mass for each of these rocks is given below. which one of the four rocks will float when placed in water?
a. 30 grams
b. 20 grams
c. 12 grams
d. 8 grams
Step1: Recall density formula
The density formula is $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass and $V$ is volume.
Step2: Calculate density of substance X
For substance X, $m = 10.0$ g and $V=20.0$ cm³. Using the density formula $
ho_X=\frac{m_X}{V_X}=\frac{10.0}{20.0}=0.5$ g/cm³.
Step3: Calculate volume of substance Y
For substance Y, $
ho_Y = 10.0$ g/cm³ and $m_Y = 50.0$ g. Rearranging the density formula to find volume $V_Y=\frac{m_Y}{
ho_Y}=\frac{50.0}{10.0}=5.0$ cm³.
Step4: Calculate mass of substance Z
For substance Z, $
ho_Z = 4.0$ g/cm³ and $V_Z = 5.0$ cm³. Using the density formula $m_Z=
ho_Z\times V_Z=4.0\times5.0 = 20.0$ g.
Step5: Determine floating - rock
The density of water is $
ho_{water}=1$ g/cm³. A rock will float if its density is less than that of water. For each rock with $V = 10$ cm³:
- For option A: $
ho_A=\frac{m_A}{V}=\frac{30}{10}=3$ g/cm³.
- For option B: $
ho_B=\frac{m_B}{V}=\frac{20}{10}=2$ g/cm³.
- For option C: $
ho_C=\frac{m_C}{V}=\frac{12}{10}=1.2$ g/cm³.
- For option D: $
ho_D=\frac{m_D}{V}=\frac{8}{10}=0.8$ g/cm³. Since $
ho_D<
ho_{water}$, the rock with 8 grams will float.
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| Substance | Mass (g) | Volume (cm³) | Density (g/cm³) |
|---|---|---|---|
| Y | 50.0 | 5.0 | 10.0 |
| Z | 20.0 | 5.0 | 4.0 |
D. 8 grams