When an object made of material P is kept immersed in the blue coloured solution Q, then a chemical change takes place to form a green coloured solution R and a brown layer of substance S is deposited on the object. Pis used for making nails and S is used for making electric wires.
What could the substance S be ?

Answer

To determine what substance S could be, let's analyze the information provided step by step.

### Step 1: Identify the materials involved
- Material P is used for making nails. This suggests that P is likely iron (Fe), as iron is commonly used to make nails.
- The blue-colored solution Q is likely copper sulfate (CuSO4), which is known to be blue in color.

### Step 2: Understand the chemical change
- When the iron nail (material P) is immersed in the blue copper sulfate solution (Q), a chemical reaction occurs.
- This reaction results in the formation of a green-colored solution R and a brown layer of substance S deposited on the iron nail.

### Step 3: Determine the products of the reaction
- The reaction between iron and copper sulfate can be represented as follows:
\[ \text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu} \]
- Here, iron (Fe) displaces copper (Cu) from copper sulfate (CuSO4) because iron is more reactive than copper.

### Step 4: Identify the green solution and brown substance
- The green solution R formed in this reaction is ferrous sulfate (FeSO4).
- The brown layer of substance S that is deposited on the iron nail is copper (Cu), which is produced as a result of the displacement reaction.

### Conclusion
- Therefore, substance S is copper (Cu), which is used for making electric wires.

### Final Answer
Substance S could be **copper (Cu)**.

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