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log([R]₀/[R]) = kt/2.303 = (0.0115×60)/2.303 ≈ 0.301 → [R]₀/[R] ≈ 2 → % remaining = 50%.
Ref: NCERT Class 12 Chemistry > Chapter 3: Chemical Kinetics > Topic: Rate of Reaction - Average and Instantaneous Rate and Rate Law
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