1. Tardigrade
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  3. Chemistry
  4. The ionic product of water at 25° C is 10-14. Its ionic product at 90° C will be,

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Q. The ionic product of water at $25^{\circ} C$ is $10^{-14}$. Its ionic product at $90^{\circ} C$ will be,

Solution:

as temperature increases Ionic product increases.