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Lab: Analysis of Hydrogen Peroxide SolutionsFill a burette with 50 mL of 0.2M potassium permanganate solution. Add 10 mL of NEW hydrogen peroxide and 2 mL of 6M sulfuric acid to a flask. Titrate the hydrogen peroxide with the KMnO4 solution to the purple endpoint of excess MnO4- ion.
Questions:(a) Volume of potassium permanganate used in each titration (mL):50mL(b) The volume of potassium permanganate required to titrate 10 mL of the new hydrogen peroxide (mL):20mL(c) Calculate the concentration of H2O2 from the volume of KMnO4 used and the stoichiometry of the reaction (MW of H2O2 is 34.01).
In these calculations I got lost.-I found 0.12655 mol KMnO4. Is this the amount in mol of KMnO4 which reacted? So now how I get the concentration of the H2O2?-2(C2x0.02L)KMnO4=5(34.01xV1)H2O2. What is C2 and V1?
Determine the mass of solid sodium formate of MW 68.0 must be combined to 150 mL of 0.66 mol/L
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A zero-order reaction has a constant rate of 3.20×10-4 . If after 60.0 seconds the concentration has dropped to 6.50×10-2 , what was the initial concentration?
Find the theoretical, percent yield and percent error of the reaction - Calculate the percent error of this reaction.
A particular diatomic molecule which contains H exhibits a 16.95 cm^-1 spacing between its rotational lines and bond length of 141.5 picometers. Determine the mass (in amu) of the unknown atom in the molecule.
Explain why dichloroethene has a boiling point of 60 degrees C while trans-dichloroethene as a boiling point of 48 degrees C. Include drawings of the structures with dipoles in your answer.
A 322 mL container holds 0.146 g of Ne and an unknown amount of Ar at 35°C and a total pressure of 635 mmHg. Calculate the moles of Ar present.
Five metal rods (Copper, Iron, Nickel, Tin and Zinc), at room temperature, of equal mass are place in a 100 oC water bath. Which metal will reach 80 oC first.
For a gas sample of N molecules that consists of N1 molecules of mass m1 and N2 molecules of mass m2, follow through a derivation which led to the result PV=1/3mv2 and PV = nRT.
The surface area of 100 g of glucose > than that of 1 g. give explanation why the solubility does not increase with the amount of solid deposited at the bottom of the flask?
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