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  1. Sep 23, 2022 · The first conversion factor comes from the balanced chemical equation, the second conversion factor is the molar mass of H 2 O 2, and the third conversion factor comes from the definition of percentage concentration by mass.

  2. Balance the following unbalanced equation and determine how many moles of \(\ce{H2O}\) are produced when 1.65 mol of NH 3 react: \[\ce{NH3 + O2 → N2 + H2O} \nonumber \] Answer a 0.391 mol H 2 O

  3. The first conversion factor comes from the balanced chemical equation, the second conversion factor is the molar mass of H 2 O 2, and the third conversion factor comes from the definition of percentage concentration by mass.

  4. The first conversion factor comes from the balanced chemical equation, the second conversion factor is the molar mass of H 2 O 2, and the third conversion factor comes from the definition of percentage concentration by mass.

  5. Prepare a concept map and use the proper conversion factor. The conversion factors are 1 mol C2H6O over 46.07 g C2H6O, 1 mol O over 1 mol C2H6O, and 16.00 g O over 1 mole O. Cancel units and calculate.

  6. Mole ratios are used as conversion factors between products and reactants in stoichiometry calculations. For example, in the reaction 2H₂(g) + O₂(g) → 2H₂O(g)

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  8. Apr 4, 2022 · Assuming it is 3.0% (mass/volume), this means you have 3.0 g H 2 O 2 in 100 mls of solution. Molar mass H 2 O 2 = 34.0 g / mole. 3.0 g H 2 O 2 / 100 mls x 1 mol H 2 O 2 / 34.0 g x 1000 mls / L = 0.88 mol / L = 0.88 M H 2 O 2

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