A) Identify the process being demonstrated.

Here are the answers to the questions about the experiment:

a) Identify the process being demonstrated.
The process being demonstrated is diffusion.

b) Explain the result in set up (ii).
In set up (ii), the inside tubing turned black, and the outside tubing remained brown.

1. The Visking tubing acts as a partially permeable membrane, allowing small molecules to pass through but blocking large molecules.
2. Iodine molecules are small enough to diffuse through the Visking tubing from the beaker (where their concentration is higher) into the tubing (where their concentration is lower).
3. Starch molecules are large and cannot pass through the Visking tubing.
4. Once inside the tubing, the iodine reacts with the starch solution, causing the characteristic blue-black color change, hence the black color inside.
5. The iodine solution outside remains brown because starch cannot diffuse out to react with it.

c) What results could be expected if the experiment was repeated using boiled starch in dilute hydrochloric acid for 5 minutes?
If the experiment was repeated using boiled starch in dilute hydrochloric acid for 5 minutes, the expected results would be:

• Inside tubing: The solution would remain brown.
• Outside tubing: The solution would remain brown.

This is because:

1. Boiling starch in dilute hydrochloric acid causes the hydrolysis of starch, breaking it down into smaller sugar molecules (e.g., glucose).
2. These smaller sugar molecules do not react with iodine to produce a blue-black color.
3. Therefore, even though iodine molecules would still diffuse into the tubing, there would be no starch present to react with, so the solution inside would remain the brown color of iodine.
4. The hydrolyzed starch (sugar molecules) are also small enough to diffuse out of the Visking tubing into the beaker, but they do not react with iodine, so the solution outside would also remain brown.