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Molar Mass / Molecular Weight of KOH: Potassium hydroxide
WebDetermine the volume of KOH required to reach a pH of 4.20 for the titration of 50.00 mL of 0.100 M benzoic acid (Ka = 6.3 x 10-5) with 0.0500 M KOH. Determine the volume of KOH required to reach a pH of 11.50 for the titration of 50.00 mL of 0.100 M benzoic acid (Ka = 6.3 x 10-5) with 0.0500 M KOH. WebProcedure of KOH Wet Mount Preparation Emulsify the specimen in a drop of 20% potassium hydroxide on a glass slide with the help of inoculating loop. To assist clearing, hairs should not be more than 5 mm long, and skin scales, crusts, and nail snips should not be more than 2 mm across. tiffany toomer
A 1.87 L aqueous solution of KOH contains 155 g of KOH . The …
Web10 nov. 2016 · 1 answer KOH + KHP ==> H2O + K2P. mols KHP = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols KHP to mols KOH. That's a 1:1 ratio; therefore, ? mols KHP = the same mols KOH. Then M KOH = mols KOH/L KOH. You know mols and L, solve for M answered by DrBob222 November 10, 2016 Answer this … Web3/15/2024, 9:56:54 PM To prepare 1000 mL of a 0.1 mol/L solution of Potassium hydroxide we have to dissolve 5.6105 g of KOH (100 % purity) in deionized or distilled water. After the solid is completely dissolved, dilute the solution to … Web22 mei 2016 · 1 Answer David G. May 22, 2016 250 cm3 = 0.25 dm3 (= 0.25 L) For a solution, C = n V (concentration = number of moles/volume). Rearranging, n = CV = 0.50× 0.25 = 0.125 mol We need 0.125 mol of KOH and the molar mass of KOH is 56.1 g mol−1, so 56.1 ×0.125 = 7.0 g. We need 7.0 g of KOH. Answer link tiffany tomkinson