name:
date
class period
3. is object e or object f more dense? assume the particles are uniformly distributed throughout each object, and particle with a larger size have a larger mass explain your reasoning.
figure 3.
4. in figure 4 below, a graph shows the relationship between mass and volume for two substances, a and b. use the graph to answer questions about these two substances.
two pan balance
fig 4
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 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.
f) determine whether substance a and b will sink or float when placed in a bucket of water:
a: sink float
b: sink float
(circle correct response)
defend your answer using the m - v graph, and your outstanding understanding of density.
©modeling instruction - amta,ans 2019
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Question

Accepted Answer

### Part b)
# Explanation:
## Step 1: Recall the formula for slope
The slope $ m $ of a line on a mass - volume ($ m - V $) graph is given by the formula $ m=\frac{\Delta m}{\Delta V} $, where $ \Delta m $ is the change in mass and $ \Delta V $ is the change in volume.

## Step 2: Calculate the slope for Substance A
For Substance A, we can take two points on the line. Let's take the points $(V_1, m_1)=(0,0)$ and $(V_2, m_2)=(80\space mL, 80\space g)$ (from the graph). Then the slope $ m_A=\frac{\Delta m}{\Delta V}=\frac{80\space g - 0\space g}{80\space mL-0\space mL}=\frac{80}{80} = 1\space g/mL$. The physical meaning of the slope for Substance A is that the density of Substance A is $ 1\space g/mL $ (since density $
ho=\frac{m}{V} $, and the slope of the $ m - V $ graph is equal to the density).

## Step 3: Calculate the slope for Substance B
For Substance B, let's take two points $(V_1, m_1)=(0,0)$ and $(V_2, m_2)=(80\space mL, 40\space g)$ (from the graph). Then the slope $ m_B=\frac{\Delta m}{\Delta V}=\frac{40\space g - 0\space g}{80\space mL - 0\space mL}=\frac{40}{80}=0.5\space g/mL $. The physical meaning of the slope for Substance B is that the density of Substance B is $ 0.5\space g/mL $ (since density $
ho = \frac{m}{V} $, and the slope of the $ m - V $ graph is equal to the density).

# Answer:
The slope of Substance A is $ 1\space g/mL $, which represents its density (mass per unit volume). The slope of Substance B is $ 0.5\space g/mL $, which also represents its density.

### Part c)
# Explanation:
## Step 1: Determine the density of Substance A
From part (b), the density of Substance A, $
ho_A = 1\space g/mL $.

## Step 2: Use the density formula to find the mass
The density formula is $
ho=\frac{m}{V} $, where $
ho $ is density, $ m $ is mass, and $ V $ is volume. We know $ V = 10.0\space mL $ and $
ho_A=1\space g/mL $. Rearranging the formula for mass, we get $ m=
ho\times V $.

## Step 3: Calculate the mass of Substance A
Substitute $
ho = 1\space g/mL $ and $ V = 10.0\space mL $ into the formula: $ m=1\space g/mL\times10.0\space mL = 10.0\space g $. To balance the two - pan balance, the mass of Substance B should be equal to the mass of Substance A. So we need to find the mass of Substance B that is equal to $ 10.0\space g $.

## Step 4: Determine the density of Substance B
From part (b), the density of Substance B, $
ho_B = 0.5\space g/mL $.

## Step 5: Use the density formula to find the volume (or mass) of Substance B
We know that for the balance, $ m_A=m_B $. We can also use the density of B. Let's use the density formula for B: $
ho_B=\frac{m_B}{V_B} $, and since $ m_B = m_A=10.0\space g $, we can find $ V_B $ or we can think in terms of the slope. Since the density of B is $ 0.5\space g/mL $, for a mass of $ 10.0\space g $, using $ m =
ho V $, we can also note that from the graph, when the mass of A is $ 10.0\space g $, the volume of A is $ 10.0\space mL $ (since $
ho_A = 1\space g/mL $). For B, since $
ho_B=0.5\space g/mL $, if we want $ m_B = 10.0\space g $, then $ V_B=\frac{m_B}{
ho_B}=\frac{10.0\space g}{0.5\space g/mL}=20\space mL $. But we can also use the graph: the line for B has a slope of $ 0.5\space g/mL $, so when $ m = 10.0\space g $, $ V = 20\space mL $. But we need the mass of B. Wait, actually, we know that the mass of A is $ m_A=
ho_A\times V_A=1\space g/mL\times10\space mL = 10\space g $. For the balance to be balanced, $ m_B=m_A = 10\space g $. We can also check from the graph of B: when $ m = 10\space g $, what's the volume? But actually, since the density of B is $ 0.5\space g/mL $, if we use the fact that for the balance, the mass of B should be equal to the mass of A ($ 10\space g $).

# Answer:
The mass of Substance B needed is $ 10.0\space g $. (We can also think in terms of the graph: the line for A has a slope of $ 1\space g/mL $, so $ 10\space mL $ of A has a mass of $ 10\space g $. The line for B has a slope of $ 0.5\space g/mL $, so to get a mass of $ 10\space g $, we can see from the density - mass - volume relationship that the mass of B should be $ 10\space g $ to balance the $ 10\space g $ mass of A.)

### Part d)
# Explanation:
## Step 1: Determine the density of Substance B
From part (b), the density of Substance B, $
ho_B=0.5\space g/mL $.

## Step 2: Calculate the mass of $ 35.0\space mL $ of Substance B
Using the density formula $ m=
ho V $, where $
ho = 0.5\space g/mL $ and $ V = 35.0\space mL $, we get $ m_B=
ho_B\times V_B=0.5\space g/mL\times35.0\space mL = 17.5\space g $.

## Step 3: Determine the density of Substance A
From part (b), the density of Substance A, $
ho_A = 1\space g/mL $.

## Step 4: Use the density formula to find the volume of Substance A with the same mass as Substance B
We know that for the balance, $ m_A=m_B $. So $ m_A = 17.5\space g $. Using the density formula for A, $
ho_A=\frac{m_A}{V_A} $, we can rearrange it to $ V_A=\frac{m_A}{
ho_A} $.

## Step 5: Calculate the volume of Substance A
Substitute $ m_A = 17.5\space g $ and $
ho_A = 1\space g/mL $ into the formula: $ V_A=\frac{17.5\space g}{1\space g/mL}=17.5\space mL $.

# Answer:
The volume of Substance A needed is $ 17.5\space mL $.

### Part e)
# Explanation:
## Step 1: Recall the density of water
The density of water is $
ho_{water}=1.00\space g/mL $.

## Step 2: Determine the density of Substance A
From part (b), the density of Substance A, $
ho_A = 1\space g/mL $, which is equal to the density of water.

## Step 3: Determine the density of Substance B
From part (b), the density of Substance B, $
ho_B = 0.5\space g/mL $, which is less than the density of water.

## Step 4: Apply the principle of flotation
An object (or substance) will sink in a fluid if its density is greater than the fluid's density, float if its density is less than the fluid's density, and be neutrally buoyant (neither sink nor float) if its density is equal to the fluid's density. Since $
ho_A=
ho_{water}=1.00\space g/mL $, Substance A will be neutrally buoyant (but for the purpose of sink or float, sometimes neutral buoyancy is considered as neither, but in the context of the question, maybe we can say it will not sink or float in the traditional sense, but if we consider the options, since $
ho_A =
ho_{water} $, and $
ho_B<
ho_{water} $). Wait, the question says "sink or float". For Substance A: density of A is $ 1\space g/mL $, density of water is $ 1.00\space g/mL $, so it will be neutrally buoyant (neither sink nor float, but if we have to choose between sink or float, maybe the question considers that if density is equal, it can be considered as not sinking (floating in a way, but actually neutral). For Substance B: density of B is $ 0.5\space g/mL $, which is less than the density of water ($ 1.00\space g/mL $), so Substance B will float.

# Answer:
To sketch the line for water: The density of water is $ 1.00\space g/mL $, so the equation of the line for water on the $ m - V $ graph is $ m = 1.00\times V $. So it will be a straight line passing through the origin with a slope of $ 1\space g/mL $, same as the line for Substance A (since $
ho_A = 1\space g/mL $).

### Part f)
# Explanation:
## Step 1: Recall the density of water
The density of water, $
ho_{water}=1.00\space g/mL $.

## Step 2: Compare the density of Substance A with water
From part (b), the density of Substance A, $
ho_A = 1\space g/mL $, which is equal to the density of water ($
ho_{water}=1.00\space g/mL $). An object (substance) with a density equal to the fluid's density will be neutrally buoyant (neither sink nor float, but in the context of the question's options, if we have to choose, since $
ho_A=
ho_{water} $, it will not sink (we can consider it as floating in a neutral way).

## Step 3: Compare the density of Substance B with water
From part (b), the density of Substance B, $
ho_B = 0.5\space g/mL $, which is less than the density of water ($
ho_{water}=1.00\space g/mL $). An object (substance) with a density less than the fluid's density will float.

# Answer:
For Substance A: Since $
ho_A=
ho_{water} $, it will be neutrally buoyant (if we consider the options, the answer for A is "float" in a neutral sense, but actually, when density is equal, it neither sinks nor floats. But looking at the options: A: sink float; B: sink float. For Substance A, density is equal to water, so it will not sink (so the correct option for A is "float" (in the sense of not sinking), and for Substance B, since $
ho_B<
ho_{water} $, it will float. So the correct responses are: Substance A: float (because $
ho_A=
ho_{water} $, it is neutrally buoyant, which is a type of floating in place); Substance B: float (because $
ho_B<
ho_{water} $). So we circle the "float" option for both A and B? Wait, no. Wait, the options are: A: sink float (circle correct response); B: sink float (circle correct response). For Substance A, density is equal to water, so it will not sink, so we circle "float" for A. For Substance B, density is less than water, so we circle "float" for B.

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QUESTION IMAGE

Question

Explanation:

Part b)

Step 1: Recall the formula for slope

Step 2: Calculate the slope for Substance A

Step 3: Calculate the slope for Substance B

Step 1: Determine the density of Substance A

Step 2: Use the density formula to find the mass

Step 3: Calculate the mass of Substance A

Step 4: Determine the density of Substance B

Step 5: Use the density formula to find the volume (or mass) of Substance B

Step 1: Determine the density of Substance B

Step 2: Calculate the mass of \( 35.0\space mL \) of Substance B

Step 3: Determine the density of Substance A

Step 4: Use the density formula to find the volume of Substance A with the same mass as Substance B

Step 5: Calculate the volume of Substance A

Answer:

Part c)