Trial 3 Mass of oxalic acid 0.2467g dihydrate (g) Initial NaOH 0.2ml burette reading (ml) Final NaOH burette 37.80ml reading (ml) Observations Light pink (i.e. light pink, dark pink, overshot) Volume of NaOH added (ml) Moles of oxalic acid dihydrate (mol) Mole ratio of NaOH to oxalic acid Moles of sodium hydroxide (mol) Molarity of NaOH, M

Trial 3 Mass of oxalic acid 0.2467g dihydrate (g) Initial NaOH 0.2ml burette reading (ml) Final NaOH burette 37.80ml reading (ml) Observations Light pink (i.e. light pink, dark pink, overshot) Volume of NaOH added (ml) Moles of oxalic acid dihydrate (mol) Mole ratio of NaOH to oxalic acid Moles of sodium hydroxide (mol) Molarity of NaOH, M

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter7: Statistical Data Treatment And Evaluation

Section: Chapter Questions

Problem 7.25QAP

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trials. Firstly, the student used potassium hydrogen phthalate (KHP, 204.22 g/mol) as standard to determine the exact concentration of the NaOHtitrant. The results of the student are summarized in the table below:Note: The stoichiometric relationship of KHP to NaOH is 1:1Standardization Trial 1 Trial 2 Trial 3 Trial 4KHP Mass, g 0.5033 0.5066 0.6989 0.6843Volume NaOH used, mL 24.32 25.61 24.67 24.56After the standardization, the student weighed different amounts of the unknown acid. In a similar fashion, the student performed acid-base titrationusing phenolphthalein as indicator. The results of the molecular weight determination are summarized below:Note: The stoichiometric relationship of Unknown acid to NaOH is 1:2Molecular WeightDeterminationTrial 1 Trial 2 Trial 3 Trial 4Unknown Mass, g 0.1234 0.1034 0.1178 0.1322Volume NaOH used, mL 21.75 20.56 24.39 25.08The potential unknown given by the professor to the student are as follows:Diprotic acid Fumaric acid Oxalic acid Succinic acid…
Deionized water is added to a 138.0 mL sample of 1.103 M oxalic acid to dilute it to 0.436 M. What volume of water, in mL, was added to dilute the sample? Round answer to correct sig figs.
What is the concentration of citrix acid in lime juice mm of citrix acid 192.14mol given : 2.07 mole of citrix acid lime juice= 0.0296L pls show work answrr needs to be moles/ L pls show rounded significant figures
EX 2.Given-> Volume of Na+ = 500 ml Molarity of Na+= 0.0100M Molar mass of Na2CO3 = 105.99 gm/mole Millimole of Na+ = molarity × volume Number of millimole = 0.0100 × 500 = 5 millimole Na2CO3 ---> 2Na+ + CO32- Millimole of Na2CO3 = millimole of Na+/2 Millimole of Na2CO3 =5/2 = 2.5 millimole Mole of Na2CO3 = 2.5 × 10-3mole (1 mole = 10^3 millimole) Weight of Na2CO3 required = mole × molar mass = 2.5 × 10-3 × 105.99 =0.26 gm Hence, 0.26 gm Na2CO3 must dissolve in 500 ml of water.
(a) During the analysis of water sample by argentometric titration, results obtained are as follows: Experiment1 2 3 4 Volume 15.5, 15.2 ,15.1 ,15.4 A)Calculate average deviation and relative average deviation for the given data. (b)Calculate the molecular weight of an unknown acid if 8.5 g of it is dissolved in 200.0 ml of water and requires 50.0 ml of 1.5 M sodium hydroxide for complete neutralization .
Trial one Trial two KIO3 solution taken (mL) 1.70 16.5 Moles of KIO3 1.06 x 10^-3 1.06 x 10^-3 Final buret reading (mL) 17.0 33.5 Initial buret reading 0 17.0 Volume of Na2S2O3 solution (mL) 17.0 16.5 Molarity of Na2S2O3 solution (MOL/L) 0.374 0.389 MEAN molarity (mol/L) 0.382 Filtrate Trail one Trail two Final buret reading (mL) 24.0 47.9 Initial buret reading (mL) 0 24.0 Volume of Na2S2O3 solution (mL) 24.0 23.9 Moles of Na2S2O3 used 0.00917 0.00913 Moles if IO3^- present initially 7.05 x 10^-4 7.02 x 10^-4 Molarity of IO3^- (mol/L) 3.53 x 10^-2 3.51 x 10^-2 Average 1.7 x 10^-4 1a. Calculate the solubility of Ca(IO3)2 from your results with the first and second filtrates 1c. Calculate Ksp for Ca(IO3)2 using the mean solubilitity
Give typed full explanation After heating 1.1929g CaCO3 and MgCO3 for some minutes the mass of the mixture decreased to 0.8015. The %comp of CaCO3 turned out to be 54.29% and MgCO3 %comp is 45.71%. Calculate the percent error/percent yield.
A certain mineral contains the substance Cu(NO3)2 .Cu2SO4. Cu3N2. The copper present in a 1.5000g sample of this mineral is analyzed by adding aqueous ammonia to precipitate the copper as Cu(OH)2 and then incinerated and weighed as CuO. What is the weight percentage of Cu in this mineral if 0.85g of CuO were obtained? Please answer fast I give you upvote.
If exactly 200.0 mL of an aqueous solution that is 1.037E3 ppb Pb(NO3)2 (331.2 g/mol) is mixed with 100.0 mL of an aqueous solution that is 3.090E9 fM KI (166.00 g/mol), how many micrograms of solid PbI2 (461.01 g/mol, 6.16 g/mL) are formed? Assume complete reaction.
% Recovery of Ferrocene = (0.0459/0.102g) x 100 = 45% % Recovery of Acetylferrocene = (0.054/0.102g) x 100 = 52.9 % Question 1 : According to the data, what does % recovery means? can we say that acetylferrocene has a higher % recovery because the proportion of the amount of acetylferrocene present in the analytical portion of the 50:50 mixture is 7.9% higher than Ferrocene????? Question 2 :Am I calculating my %recovery correctly????
A mixture of Al2O3(s) and CuO(s) weighing 18.371 mg was heated under H2(g) at 1 0008C to give 17.462 mg of Al2O3(s) 1 Cu(s). The other product is H2O(g). Find wt% Al2O3 in the original mixture.
0.070 grams of nickel (II) hydroxide (MM•92.71g/mol) dissolve in 500.00 mL of water at 20.0 Celsius. What is the Ksp for nickel (II) hydroxide? Show all work, give correct sig figs, thank you!