solutionmLH2O2mLKIrate (mL O2 s) 15 010 0 210 010 0 35 020 0 Rate L...

solutionÂ Â Â Â Â Â Â Â Â Â Â Â Â Â mL H2O2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â mL KIÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â rate (mL O2/s) 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ? 2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ? 3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 20.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ? Rate Law: since doubling either [H2O2] or [I-] doubles the reaction rate, what is...

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solutionÂ Â Â Â Â Â Â Â Â Â Â Â Â Â mLH2O2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â mLKIÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â rate (mL O2/s)
1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ?
2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ?
3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 20.0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ?
Rate Law: since doubling either [H2O2] or [I-] doubles thereaction rate, what is the rate law?

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The reaction is first order wrt H2O2 so doubling the H2O2 will double the rate Rate Constant x H2O2 Thus Rate is directly proportional to the volume of H2O2 Hence doubling the volume of hard water would double the rate Detailed calculations 3 H2O2 is about 09 Molar they called for 010Molar KI to be used Molarity of stock solution of appx H2O2 0924 M Molarity of stock solution of KI 010 Molar exp mls H2O2KI H2O 1 50ml 100ml 150 ml 2 100ml 100ml 100 ml 3 50ml 200ml 50 ml Molarities of H2O2 exp See Answer

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