Question

a) you have built a simple two - pan balance shown above to compare the masses of substances a and b. what would happen to the balance if you put equal masses of a and b in the two pans? equal volumes of a and b in the two pans? explain your reasoning.
b) find the slope of the line for both a and b using correct units. state the physical meaning of the slope for each substance.
c) if you put 10.0 ml of a in one balance pan, what mass of b would you need in the other pan to make it balance? explain your reasoning.
d) if you put 35.0 ml of b in one balance pan, what volume of a would you need in the other pan to make it balance? explain your reasoning.
e) water has a density of 1.00 g/ml. sketch the line representing water on the graph in figure 4.

Explanation:

Step1: Analyze balance for equal - mass case

If equal masses of A and B are put in the two pans, the balance will stay level because the gravitational force acting on both pans is the same. Mathematically, if \(m_A=m_B\), the torques on both sides of the balance (assuming equal - arm length) are equal (\(\tau = rF\), where \(r\) is the distance from the pivot and \(F = mg\)).

Step2: Analyze balance for equal - volume case

From the mass - volume graph, the slope of the mass - volume graph gives density (\(
ho=\frac{m}{V}\)). Substance A has a steeper slope, so \(
ho_A>
ho_B\). If equal volumes \(V\) of A and B are put in the pans, \(m_A=
ho_A V\) and \(m_B=
ho_B V\). Since \(
ho_A>
ho_B\), \(m_A > m_B\), and the pan with substance A will go down.

Step3: Calculate slope for substance A

For substance A, choose two points on the line, say \((0,0)\) and \((20,40)\). The slope \(m_A=\frac{\Delta m}{\Delta V}=\frac{40 - 0}{20 - 0}=2\ g/mL\). The physical meaning of the slope is the density of substance A.

Step4: Calculate slope for substance B

For substance B, choose two points on the line, say \((0,0)\) and \((40,20)\). The slope \(m_B=\frac{\Delta m}{\Delta V}=\frac{20 - 0}{40 - 0}=0.5\ g/mL\). The physical meaning of the slope is the density of substance B.

Step5: Solve for mass of B to balance 10.0 mL of A

The mass of 10.0 mL of A is \(m_A=
ho_A V_A\), where \(
ho_A = 2\ g/mL\) and \(V_A = 10.0\ mL\), so \(m_A=2\times10.0 = 20\ g\). To balance it with substance B, we set \(m_B=m_A\). Since \(
ho_B = 0.5\ g/mL\), and \(m_B=
ho_B V_B\), then \(V_B=\frac{m_B}{
ho_B}\). So \(m_B = 20\ g\).

Step6: Solve for volume of A to balance 35.0 mL of B

The mass of 35.0 mL of B is \(m_B=
ho_B V_B\), where \(
ho_B = 0.5\ g/mL\) and \(V_B = 35.0\ mL\), so \(m_B=0.5\times35.0 = 17.5\ g\). To balance it with substance A, since \(
ho_A = 2\ g/mL\) and \(m_A=m_B\), then \(V_A=\frac{m_A}{
ho_A}=\frac{17.5}{2}=8.75\ mL\).

Step7: Sketch line for water

The density of water is \(
ho = 1.00\ g/mL\). The mass - volume relationship is \(m=
ho V\), so when \(V = 10\ mL\), \(m = 10\ g\); when \(V = 20\ mL\), \(m = 20\ g\). Plot the points \((0,0)\), \((10,10)\), \((20,20)\) and draw a straight line through them.

Answer:

a) Equal masses: balance stays level. Equal volumes: pan with A goes down.
b) Slope of A: \(2\ g/mL\) (density of A), Slope of B: \(0.5\ g/mL\) (density of B)
c) \(20\ g\) of B
d) \(8.75\ mL\) of A
e) Plot points \((0,0)\), \((10,10)\), \((20,20)\) and draw a straight - line for water.