What is the mass of sodium oxalate (Na2C2O4, MM = 134,00 g/mol), in mg, should be taken so that in its titration 25,00 mL of a solution of potassium permanganate (KMnO4) 0,032 mol/ L in strong acid medium?   21,4 mg 268 mg 536 mg 107 mg 134 mg

A 0.5504-g sample of KHP (potassium hydrogen phthalate, KHC8H4O4; molar mass = 204.22 g/mol) is completely dissolved in enough deionized water to make 50.00 mL of solution. The resulting solution is titrated with a NaOH solution of unknown concentration. If 21.50 mL of the base solution is needed to reach the end-point, what is the molar concentration of NaOH solution? The reaction occurs as follows:                 NaOH(aq)  +  KHC8H4O4(aq)  -->  NaKC8H4O4(aq)  +  H2O(l)               (A) 0.2457 M           (B) 0.1254 M           (C) 0.05390 M           (D) 0.03823 M

A solution of hydrogen peroxide was titrated with a solution of potassium permanganate (see reaction).5 H 2O 2(aq) + 2 KMnO 4(aq) + 3 H 2SO 4 -> 5 O 2(g) + 2 MnSO 4(aq) + K 2SO 4(aq) + 8 H 2O(l)It required 75.00 mL of 0.1456 M KMnO 4 to titrate 2.623 g of the solution containing hydrogen peroxide. What was the mass percentage of H 2O 2 in the solution?

You have been provided with a 2.08 x 10-2 L  sample of lithium hydroxide (LiOH) of unknown concentration. You perform a titration with 2.19 M nitric acid (HNO3), and find that 24.6 mL are required to reach the equivalence point (as determined using a coloured indicator). What is the concentration of the lithium hydroxide solution in mol L-1?

Oxalic acid (H2C2O4) is present in many plants. If 25.6 mL of 0.0200 M KMnO4 is needed to titrate a 2.75 g sample of plant extract to reach the equivalence point, what is the mass percent of oxalic acid (H2C2O4) in the sample? 2 MnO4- + 16 H+ + 5 C2O42- → 2 Mn2+ + 10 CO2 + 8 H2O

A 500.0-mg sample of impure sodium bicarbonate (NaHCO3; MM = 84.01 g/mol) is analyzed by titrating the sample with standard 0.1288 M HCI solution, requiring 38.27 mL. The reaction is NaCl + H₂O + CO₂ NaHCO3 + HCI Calculate the percent sodium bicarbonate in the sample. 0.9858% 82.82% 4.929% 0.08282% -

A solution of oxalic acid dihydrate (H₂C₂O₄･2H₂O) with a known concentration of 0.400 M H₂C₂O₄･2H₂O is titrated with a 0.333 M NaO solution. How many L NaOH are required to reach the second equivalence point with a starting volume of 65.0 mL H₂C₂O₄･2H₂O, according to the following balanced chemical equation:   H₂C₂O₄･2H₂O + 2NaOH → Na₂C₂O₄ + 4H₂O

Tartaric acid, H2C4H4O6, is a diprotic acid that naturally occurs in the production of apple wine. Emma was tasked to find out the acid content in a 50.0 mL wine sample by titrating it with a standardized solution of NaOH. She was able to establish the working concentration of the NaOH solution by using 17.61 mL of NaOH to titrate a primary standard of KHP that weighed 1.501 g. The grams and %w/w of Tartaric acid present in the 50.0 mL wine sample were then determined by titrating a 12.50 mL aliquot using 21.78 mL of the standardized NaOH solution. H2C4H4O6 (aq) + 2NaOH (aq) ↔  Na2C4H4O6 (aq) + 2H2O (l) Assume that the density of the wine sample is 1.000 g/mL MM of KHP = 204.22 g/mol MM of H2C4H4O6 = 150.087 g/mol                                                                                                                                                     What is the molar concentration of the prepared NaOH solution?…

A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 54. mg of oxalic acid (H₂C₂O4), a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in 250. mL of distilled water. The student then titrates the oxalic acid solution with his sodium hydroxide solution. When the titration reaches the equivalence point, the student finds he has used 92.3 mL of sodium hydroxide solution. Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 2 significant digits. do O Ar M X ?

Calculate the number of grams of AgCl will precipitate out of solution if 125 mL of 0.500 M AgNO3 is combined with 275 mL of 0.235 M NaCl according to the following         AgNO3 (aq)   +    NaCl (aq)  -->  AgCl (s)   +  NaNO3 (aq)