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60ml of mixture of equal volumes of Cl(2...

`60ml` of mixture of equal volumes of `Cl_(2)` and an oxide of chlorine, i.e., `Cl_(2)O_(n)` was heated and then cooled back to the original temperature. The resulting gas mixture was found to have volume of `75ml`. On treatment with `KOH` solution, the volume contracted to `15ml`. Assume that all measurements are made at the same temperature and pressure. Deduce the value of `n` in `Cl_(2)O_(n)`. The oxide of `Cl_(2)` n heating decomposes quantiatively to `O_(2)` and `Cl_(2)`.

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AI Generated Solution

To solve the problem, we will follow these steps: ### Step 1: Understand the Initial Conditions We have a mixture of equal volumes of \( Cl_2 \) and \( Cl_2O_n \). Since the total volume of the mixture is \( 60 \, \text{ml} \), each component contributes \( 30 \, \text{ml} \). ### Step 2: Write the Reaction When \( Cl_2O_n \) is heated, it decomposes into \( Cl_2 \) and \( O_2 \): \[ ...
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