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  • Does H2O2 break down into H2O - Chemistry Stack Exchange
    $\ce{H2O2}$ -> $\ce{H2O +1 2O2}$ This might make it easier to understand $\ce{H2O2}$ will homolytically cleave for form two $\ce{ OH}$ radicals Radicals are very reactive and will start a chain reaction, but ultimately you will end up with water and oxygen products from $\ce{H2O2}$
  • redox - Half equations for H2O2 for its reducing and oxidising nature . . .
    Is there a complete list of all the half equations for $\ce{H2O2}$ - both oxidation and reduction, in acidic and alkaline conditions? I've looked on the internet but can't seem to find a list with all of them These are my first attempts: $\ce{H2O2 + 2e^- -> 2OH^-}$ (in alkaline conditions) $\ce{H2O2 + 2H^+ + 2e^- -> 2H2O}$ (in acidic conditions)
  • How do I make 100 milliliters of 100mM H2O2 using 3% H2O2?
    Assume density 1 g ml for 3% solution which gives 0 03 g $\ce{H2O2}$ per ml, or 30 g $\ce{H2O2}$ per liter The molecular weight of $\ce{H2O2}$ is about 34 g mol so: $\dfrac{30}{34} \approx 1$ molar You want 100 mMolar which is 0 100 molar Thus you need a 10 fold dilution So add 10 ml of the 3% to 90 ml water to get 100 ml of solution
  • Adding H2O2 in water does what to the pH of the water?
    The link above says if the water is acidic, $\ce{H2O2}$ will dissociate into hydronium + hydroperoxide ($\ce{H3O+ + HOO-}$) I imagine the production of hydronium will increase the acidity For example, if the water is slightly acidic, say $\mathrm{pH}$ $6 5$, will the addition of $\ce{H2O2}$ make the solution even more acidic?
  • How does KI catalyze the reaction with H2O2?
    Since H2O2 decomposition happens incredibly slowly, the latter seems very unlikely On the other hand, I can't seem to visualize why an anion would abstract the uncharged oxygen either Yes, O has a lone pair, but that should give it a slight negative dipole moment and make it even harder to abstract, in my mind
  • inorganic chemistry - What is the n factor of H2O2 undergoing . . .
    implying that n-factor of $\ce{H2O2}$ is 2 but I am calculating n-factor of $\ce{H2O2}$ as 1 because the definition of n-factor is "the no of electron gained or lost per mole" As we can see 2 electrons are participating in this reaction so electron exchanged per mole is 1 which is equal to n-factor
  • Why is oxygenated water not H2O2? - Chemistry Stack Exchange
    More precisely, the presence of H2O2 and other reactive oxygen species (ROS) have been cited in the literature as occurring in illuminated natural waters (see, for example, ‘Reactive Oxygen Species in Natural Waters ‘, by Neil V Blough and Richard G Zepp) To quote from an online two-page preview :
  • inorganic chemistry - Reaction intermediates of MnO2 catalyzed H2O2 . . .
    The kinetic experiments on the decomposition of $\ce{H2O2}$ together with the experiments on $\ce{HO•}$ detection show the existence of an adsorption step prior to decomposition This type of process is also predicted with the DFT calculations The decomposition of $\ce{H2O2}$ follows a similar mechanism for the three metal oxides studied
  • acid base - When does hydrogen peroxide act as a reducing and when as . . .
    Let us say that $\ce{H2O2}$ reacts with some ion $\ce{Y+}$ If $+1$ is yttrium's highest oxidation state then it will have no other option but to to get reduced to one of its lower oxidation states, and hydrogen peroxide acts as reducing agent





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