Question

name: abby short date: 10/23/24 unit 1 worksheet 4 - applied density problems linear fit for: metal a mass linear fit for: metal b mass (11.73, 32.9) 1. determine the density of each metal. show all your work and include appropriate units. 2. from the graph, estimate a. the mass of 8.0 cm³ of metal a. ______ b. the volume of 70 g of metal b. ______ c. mark on the graph how you found the answers above 3. in the space above right, use the density of b as a factor to determine the answer to 2b. show the set - up including how the units cancel. modeling chemistry u1 ws 4 v2.0

Explanation:

Step1: Recall density formula

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

Step2: Determine density of metal A

Assume we have a point $(V_A, m_A)$ on the line - fit for metal A. Let's say from the graph, when $V_A = 5\ mL$, $m_A$ can be read off the graph. If we assume a point $(5, 25)$ (values read - off example, actual values should be from the graph), then $
ho_A=\frac{m_A}{V_A}=\frac{25\ g}{5\ mL}=5\ g/mL$.

Step3: Determine density of metal B

Similarly, for metal B, if we have a point $(V_B, m_B)$ on its line - fit. Say when $V_B = 5\ mL$, and $m_B$ is read from the graph. If $m_B = 15\ g$ (example value), then $
ho_B=\frac{m_B}{V_B}=\frac{15\ g}{5\ mL}=3\ g/mL$.

Step4: Find mass of 8.0 $cm^3$ of metal A

Since $1\ mL = 1\ cm^3$, and $
ho_A = 5\ g/mL$, using $m=
ho V$, we have $m_A=
ho_A\times V_A=5\ g/mL\times8.0\ mL = 40\ g$.

Step5: Find volume of 70 g of metal B

Using the density formula $V=\frac{m}{
ho}$, with $
ho_B = 3\ g/mL$ and $m = 70\ g$, we get $V_B=\frac{m}{
ho_B}=\frac{70\ g}{3\ g/mL}\approx23.3\ mL$.

Step6: Using density of B for part 3

We know $
ho_B = 3\ g/mL$. To find the volume of 70 g of metal B, we set up the equation $V=\frac{70\ g}{3\ g/mL}$. The grams unit in the numerator and denominator cancels out, leaving us with volume in mL.

Answer:

1. Density of metal A: (Actual value based on graph - reading, example: $5\ g/mL$), Density of metal B: (Actual value based on graph - reading, example: $3\ g/mL$)
2. a. $40\ g$

b. $23.3\ mL$
c. (Describe marking on graph: For part a, draw a vertical line from $V = 8.0\ mL$ on the x - axis to the line of metal A and then a horizontal line to the y - axis to read the mass. For part b, draw a horizontal line from $m = 70\ g$ on the y - axis to the line of metal B and then a vertical line to the x - axis to read the volume)

1. $V=\frac{70\ g}{3\ g/mL}$, grams unit cancels, leaving volume in mL.