Fractional crystalization of Hydrogen Peroxide would involve lowering the temperature of the solution to just above the melting point. This will depend on the concentration and it is NOT linear. For instance, a 3% solution as is found in most drugstores will freeze at about -2C, 30% at about -25C, 35% at -33C, 50% at -52C, 70% at -40C and finally 90% at about -11C.
Too many unknowns in your question. Is this 3% by mass or by volume? Does the quantity of final solution matter? IE do you need 100 ml, 1 liter or 5000ml. What is the density of the hydrogen peroxide? (needed for a volume % problem) Assuming you mean 3% by mass, then that means 3 g of hydrogen peroxide in 100 g of solution. 300 micromolar = 3 x 10-4 molar. Assuming you want to make one liter then you need 3 x 10-4 moles of peroxide. The molar mass of peroxide is 34 g/mole. 34 g/mole x 3 x 10-4 moles = 1.02 x 10-2 grams of peroxide 1.02 x 10-2 grams / .03 = 0.34 grams of the original solution. Weigh (mass) accurately 0.34 g of the original solution in a 1 liter volumetric flask. Add distilled water until the total volume is 1 liter.
Create a control group by placing a measured amount of catalase solution in a test tube. Prepare test tubes with catalase solution at varying temperatures (e.g., using water baths at different temperatures). Add a known amount of hydrogen peroxide to each test tube and measure the rate of oxygen gas production as the catalase breaks down the hydrogen peroxide. Record and compare the rate of reaction at different temperatures to determine the effect of temperature on catalase activity.
To prepare a 0.1M standard solution of potassium hydrogen phthalate, you would first weigh out the appropriate amount of potassium hydrogen phthalate (KHPh) based on its molar mass, which is 204.22 g/mol. Dissolve this amount in distilled water and make up the solution to the desired volume, typically in a 1 liter volumetric flask. Finally, mix the solution thoroughly to ensure complete dissolution.
To prepare a 5% potassium iodide solution, weigh 5 grams of potassium iodide and dissolve it in 100 mL of water. Stir until the potassium iodide is completely dissolved to achieve a 5% solution.
To prepare a 40.0% lithium nitrate solution, 60mL of the solution would be solvent (water) and 90mL would be solute (lithium nitrate). Therefore, you would need 90mL of lithium nitrate to prepare 150mL of the 40.0% solution.
1 part of hydrogen peroxide out of 30% hydrogenperoxide bottle add 10 parts of distilled water
To prepare 3% hydrogen peroxide from 30% hydrogen peroxide, you need to dilute the 30% solution with water. The formula to calculate the dilution is C1V1 = C2V2, where C1 is the initial concentration (30%), V1 is the volume of the initial solution, C2 is the final concentration (3%), and V2 is the final volume of the solution you want to prepare. By plugging in the values, you can determine the volume of 30% hydrogen peroxide needed and the volume of water needed to achieve a 3% solution.
Oh, dude, preparing 0.2 M hydrogen peroxide is like making lemonade, but with a bit more chemistry. You just take the concentrated hydrogen peroxide solution and dilute it with water until you reach a concentration of 0.2 M. It's not rocket science, just mix and measure, like following a recipe for a boring science experiment.
Barium peroxide is commonly used to prepare hydrogen peroxide because it readily decomposes in the presence of water, releasing oxygen gas which then reacts with hydrogen to produce hydrogen peroxide. This method provides a simple and efficient way to produce hydrogen peroxide on a small scale.
by urea hydrogen peroxide reaction with tri fluro acetic anhydride
See the two Related Questions to the left for the answer.The first is how to prepare a solution starting with a solid substance (and dissolving it). The second question is how to prepare a solution by diluting another solution.
pharmacist
Too many unknowns in your question. Is this 3% by mass or by volume? Does the quantity of final solution matter? IE do you need 100 ml, 1 liter or 5000ml. What is the density of the hydrogen peroxide? (needed for a volume % problem) Assuming you mean 3% by mass, then that means 3 g of hydrogen peroxide in 100 g of solution. 300 micromolar = 3 x 10-4 molar. Assuming you want to make one liter then you need 3 x 10-4 moles of peroxide. The molar mass of peroxide is 34 g/mole. 34 g/mole x 3 x 10-4 moles = 1.02 x 10-2 grams of peroxide 1.02 x 10-2 grams / .03 = 0.34 grams of the original solution. Weigh (mass) accurately 0.34 g of the original solution in a 1 liter volumetric flask. Add distilled water until the total volume is 1 liter.
Given that hydrogen peroxide (H2O2) decomposes easily into water and oxygen, H2O9 would probably not be possible to synthesize in real life.
Create a control group by placing a measured amount of catalase solution in a test tube. Prepare test tubes with catalase solution at varying temperatures (e.g., using water baths at different temperatures). Add a known amount of hydrogen peroxide to each test tube and measure the rate of oxygen gas production as the catalase breaks down the hydrogen peroxide. Record and compare the rate of reaction at different temperatures to determine the effect of temperature on catalase activity.
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The principal uses of the hydrogen peroxide are: - bleaching agent in the paper industry - bleaching agent for textiles - disinfectant /antiseptic - neutralization of hydrogen sulphide - neutralization of nitrogen oxides - treatment of waste waters - treatment of industrial water to delete organic contaminants - starting material to prepare some useful organic and inorganic chemicals - etc.