A 66.7 mL sample of 18.0 M sulfuric acid was diluted with enough water in a volumetric flask to make 5.00 x 102 mL of solution. A 25.0 ml aliquot of this solution was then further diluted to a volume of 1.50 x 102 ml. What is the molarity of the solution after the second dilution?
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Dilution - In dilution of the any sample the number of moles of the sample will not be changed. Change only in volume.
n1=n2
M1V1=M2V2
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- 2. In Avogadro’s Number lab, we used oleic acid to create a monolayer on the surface of water. The oleic acid solution was prepared by dissolving oleic acid in ethanol and it has a concentration of 0.50% by volume. The following parameters of oleic acid will be helpful to solve this question: Molar mass: 282.47 g/mol Density = 0.895 g/mL A) What’s the mass of oleic acid in 1.00 mL of this oleic acid solution? B) How many oleic acid molecules are there in 1.00 mL of this oleic acid solution?By pipet, 5.00 mLmL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask and diluted to the calibration mark. Determine the molarity of the resulting solution.BSA for concentration measurements is often prepared as a 10 mg/mL stock solution. Suppose you wish to prepare a series of 10 mL solutions of BSA at 1.0, 0.8, 0.6, 0.4, and 0.2 mg/mL. Using the dilution equation C1 × V1 = C2 × V2, you plan to do this by transferring a calculated volume (V1) of the 10 mg/mL (C1) into a clean test tube, then adding water in sufficient volume to produce 10 mL (V2) at the desired concentration (C2). Complete the table below, showing the volume of stock 10 mg/mL solution you need to add to each tube, and the volume of water you need to add to each. Also, convert the desired mg/mL concentration values to their equivalent in μM (micromolar; 10-6 M).
- What volume (in mL) of 30.0% hydrogen peroxide solution is required to provide the mass required in question 8 above? Mass=10.2g of a 30% hydrogen peroxide soltuion. Density of 30.0% H2O2 = 1.11 g/mL Answer with 3 sig figs.0.500 cm3 of concentrated H2O2 solution diluted exactly to 50.00 cm3 by water. 10.00 cm3 samples were taking out, and titrated by 0.02003 mol/dm3 KMnO4. The average of the individual result is 18.15 cm3. Give the molarity (mol/dm3) and the mass concentration (g/dm3) of the original concentrated H2O2 sample! Mr = 34.02 g/mol.Mg +2HCL > Mg cl2 +H2 If 1.505×10 exponent 23 particles of Mg react with 300ml 0.9mol.dm3 solutions of HCl Determine the limiting reagent
- In a Spec-20 experiment a student took 4.3178g of an unknown material containing cobalt and dissolved it in 250.00 ml of water. The student determined the concentration of the solution to be 0.075M in cobalt. What was the mass percent cobalt in the original unknown?By pipet, 19.00 mL of a 0.823 M stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask and diluted to the calibration mark. Determine the molarity of the resulting solution.Is their a true chemistry defeinition of what "percent by volume means?" If so, can it be described as this definition: Volume/volume percentage (v/v percent) is a measure of the concentration of a substance in a solution. It is expressed as the ratio of the volume of the solute to the total volume of the solution multiplied by 100.
- Answer showing all steps. The answer should be typewritten with a computer keyboard Calculate the mass of Na2S203.5H20 needed to prepare 250 cm3 of a 0.011 M solution in a volumetric flask. (RMM = 248)using table 1, answer the questiosn below. Mass of Crushed Antacid (g) .500g Concentration of HCl (M) 1.0 M Volume HCl (mL) 5.01ml Concentration of NaOH (M) 1.0M Initial NaOH Volume (mL) 4.10ml Final NaOH Volume (mL) 2.60ml Total Volume of NaOH Used (mL) 1.5ml HCl available for neutralization (g): Moles of NaOH required to reach stoichiometric point (mol): HCl neutralized by antacid (g): HCl neutralized per gram of antacid (g):A 32.5 mL sample of an 5.6 % (m/v) KBr solution is diluted with water so that the final volume is 236.5 mL.