Question

below are animal cells placed in beakers of various concentrations. follow the steps in order. 1. go through each image and add the missing percentages. make sure the percentages inside the cell add up to 100%. then make sure the percentages outside the cell add up to 100%. 2. circle the highest solute percentage in each image 3. add your arrow showing which way water will move. remember, water moves from high to low concentration. 4. circle the type of solution for each image (hypo, iso, hyper). if water goes into the cell it is hypotonic. if water goes out of the cell it is hypertonic. images of beakers with cells and percentages (some missing) follow, e.g., first row: cell with 90% h2o, 10% solute; beaker with 40% h2o, 60% solute, etc.

Answer:

Step 1: Calculate Missing Percentages (Cell and Beaker)

For a cell or beaker, \( \text{Water}\% + \text{Solute}\% = 100\% \), so \( \text{Missing}\% = 100\% - \text{Known}\% \).

• 2nd Row, 1st Beaker (Solute):

Cell: \( 55\% \, \text{H}_2\text{O} + 45\% \, \text{Solute} = 100\% \) (good).
Beaker: \( 55\% \, \text{H}_2\text{O} + x = 100\% \) → \( x = 100 - 55 = 45\% \, \text{Solute} \).

• 2nd Row, 2nd Beaker (Solute):

Cell: \( 15\% \, \text{H}_2\text{O} + 85\% \, \text{Solute} = 100\% \) (good).
Beaker: \( 35\% \, \text{H}_2\text{O} + x = 100\% \) → \( x = 100 - 35 = 65\% \, \text{Solute} \).

• 2nd Row, 3rd Beaker (Water):

Cell: \( x + 95\% \, \text{Solute} = 100\% \) → \( x = 100 - 95 = 5\% \, \text{H}_2\text{O} \).
Beaker: \( 25\% \, \text{H}_2\text{O} + 75\% \, \text{Solute} = 100\% \) (good).

• 3rd Row, 1st Beaker (Water):

Cell: \( x + 55\% \, \text{Solute} = 100\% \) → \( x = 100 - 55 = 45\% \, \text{H}_2\text{O} \).
Beaker: \( x + 25\% \, \text{Solute} = 100\% \) → \( x = 100 - 25 = 75\% \, \text{H}_2\text{O} \).

• 3rd Row, 2nd Beaker (Solute/Beaker Water):

Cell: \( 80\% \, \text{H}_2\text{O} + x = 100\% \) → \( x = 20\% \, \text{Solute} \).
Beaker: \( x + 90\% \, \text{Solute} = 100\% \) → \( x = 10\% \, \text{H}_2\text{O} \).

• 3rd Row, 3rd Beaker (Solute):

Cell: \( 5\% \, \text{H}_2\text{O} + x = 100\% \) → \( x = 95\% \, \text{Solute} \).
Beaker: \( 65\% \, \text{H}_2\text{O} + x = 100\% \) → \( x = 35\% \, \text{Solute} \).

Step 2: Circle Highest Solute Percentage (Cell vs. Beaker)

• 1st Row, 1st Beaker:

Cell Solute: \( 10\% \), Beaker Solute: \( 60\% \) → Circle \( 60\% \).

• 1st Row, 2nd Beaker:

Cell Solute: \( 60\% \), Beaker Solute: \( 30\% \) → Circle \( 60\% \).

• 1st Row, 3rd Beaker:

Cell Solute: \( 80\% \), Beaker Solute: \( 80\% \) → Circle \( 80\% \) (both).

• 2nd Row, 1st Beaker:

Cell Solute: \( 45\% \), Beaker Solute: \( 45\% \) → Circle \( 45\% \) (both).

• 2nd Row, 2nd Beaker:

Cell Solute: \( 85\% \), Beaker Solute: \( 65\% \) → Circle \( 85\% \).

• 2nd Row, 3rd Beaker:

Cell Solute: \( 95\% \), Beaker Solute: \( 75\% \) → Circle \( 95\% \).

• 3rd Row, 1st Beaker:

Cell Solute: \( 55\% \), Beaker Solute: \( 25\% \) → Circle \( 55\% \).

• 3rd Row, 2nd Beaker:

Cell Solute: \( 20\% \), Beaker Solute: \( 90\% \) → Circle \( 90\% \).

• 3rd Row, 3rd Beaker:

Cell Solute: \( 95\% \), Beaker Solute: \( 35\% \) → Circle \( 95\% \).

Step 3: Water Movement (High \( \text{H}_2\text{O} \) → Low \( \text{H}_2\text{O} \))

• 1st Row, 1st Beaker:

Cell \( \text{H}_2\text{O} = 90\% \), Beaker \( \text{H}_2\text{O} = 40\% \) → Arrow out of cell (water moves to beaker, lower \( \text{H}_2\text{O} \)).

• 1st Row, 2nd Beaker:

Cell \( \text{H}_2\text{O} = 40\% \), Beaker \( \text{H}_2\text{O} = 70\% \) → Arrow into cell (water moves to cell, lower \( \text{H}_2\text{O} \) in cell).

• 1st Row, 3rd Beaker:

Cell \( \text{H}_2\text{O} = 20\% \), Beaker \( \text{H}_2\text{O} = 20\% \) → No net movement (iso).

• 2nd Row, 1st Beaker:

Cell \( \text{H}_2\text{O} = 55\% \), Beaker \( \text{H}_2\text{O} = 55\% \) → No net movement (iso).

• 2nd Row, 2nd Beaker:

Cell \( \text{H}_2\text{O} = 15\% \), Beaker \( \text{H}_2\text{O} = 35\% \) → Arrow into cell (water moves to cell, lower \( \text{H}_2\text{O} \) in cell).

• 2nd Row, 3rd Beaker:

Cell \( \text{H}_2\text{O} = 5\% \), Beaker \( \text{H}_2\text{O} = 25\% \) → Arrow into cell (water moves to cell, lower \( \text{H}_2\text{O} \) in cell).

• 3rd Row, 1st Beaker:

Cell \( \text{H}_2\text{O} = 45\% \), Beaker \( \text{H}_2\text{O} = 75\% \) → Arrow into cell (water moves to cell, lower \( \text{H}_2\text{O} \) in cell).

• 3rd Row, 2nd Beaker:

Cell \( \text{H}_2\text{O} = 80\% \), Beaker \( \text{H}_2\text{O} = 10\% \) → Arrow out of cell (water moves to beaker, lower \( \text{H}_2\text{O} \) in beaker).

• 3rd Row, 3rd Beaker:

Cell \( \text{H}_2\text{O} = 5\% \), Beaker \( \text{H}_2\text{O} = 65\% \) → Arrow into cell (water moves to cell, lower \( \text{H}_2\text{O} \) in cell).

Step 4: Classify Solution (Hypo, Iso, Hyper)

• Hypotonic: Water enters cell (beaker \( \text{H}_2\text{O} > \) cell \( \text{H}_2\text{O} \)).
• Isotonic: Equal \( \text{H}_2\text{O} \) (no net movement).
• Hypertonic: Water exits cell (beaker \( \text{H}_2\text{O} < \) cell \( \text{H}_2\text{O} \)).

Beaker #	Cell \( \text{H}_2\text{O} \)	Beaker \( \text{H}_2\text{O} \)	Solution Type
1st Row 2	\( 40\% \)	\( 70\% \)	Hypotonic
1st Row 3	\( 20\% \)	\( 20\% \)	Isotonic
2nd Row 1	\( 55\% \)	\( 55\% \)	Isotonic
2nd Row 2	\( 15\% \)	\( 35\% \)	Hypotonic
2nd Row 3	\( 5\% \)	\( 25\% \)	Hypotonic
3rd Row 1	\( 45\% \)	\( 75\% \)	Hypotonic
3rd Row 2	\( 80\% \)	\( 10\% \)	Hypertonic
3rd Row 3	\( 5\% \)	\( 65\% \)	Hypotonic

Final Answers (Missing Percentages)

• 2nd Row, 1st Beaker (Solute): \( \boldsymbol{45\%} \)
• 2nd Row, 2nd Beaker (Solute): \( \boldsymbol{65\%} \)
• 2nd Row, 3rd Beaker (Water): \( \boldsymbol{5\%} \)
• 3rd Row, 1st Beaker (Cell Water): \( \boldsymbol{45\%} \); (Beaker Water): \( \boldsymbol{75\%} \)
• 3rd Row, 2nd Beaker (Cell Solute): \( \boldsymbol{20\%} \); (Beaker Water): \( \boldsymbol{10\%} \)
• 3rd Row, 3rd Beaker (Cell Solute): \( \boldsymbol{95\%} \); (Beaker Solute): \( \boldsymbol{35\%} \)

(For solution type and arrows, follow the explanations above.)