Need help with calculating the blank spots thanks mass of Na2CO3 (grams)moles of Na2CO3V final, volume dilute HCl atendpoint (mL)V initial, volume dilute HCl atstart (mL)V total, volume dilute HCl toreach the endpoint (mL)molarity of dilute HCl in buret(mol/L)average molarity of dilute HCI(mol/L)percent difference of trials (%)Average molarity ofconcentrated unknown solution(mol/L)Trial 10.0780g5.52mLa9.77ml14.85ml16g/molTrial 20.0810g11m²17.94mL6.91m4.872 mol/LJacl caq?04.001g/mol 58.44 g/malTrial 30.0930g.10.3mL17.13mL6.83 m

Molar mass of Na2CO3 = 105.99 g/mol Required conversion factor is: 1 mol Na2CO3 = 105.99 g Na2CO3…

mass of Na2CO3 (grams) moles of Na2CO3 V final, volume dilute HCl at endpoint (mL) V initial, volume dilute HCl at start (mL) V total, volume dilute HCl to reach the endpoint (mL) molarity of dilute HCI in buret (mol/L) average molarity of dilute HCI (mol/L) percent difference of trials (%) Average molarity of concentrated unknown solution (mol/L) Trial 1 0.0780g 5.52mL 9.77m2 14..85ml Trial 2 0.0810g 11m² 17,91m4 6.91ml 4.872 mol/L (aq) 84.001 g/mol 58.44 g/m Trial 3 0.0930g. 10.3mL 17.13mL 6.83 me

mass of Na2CO3 (grams) moles of Na2CO3 V final, volume dilute HCl at endpoint (mL) V initial, volume dilute HCl at start (mL) V total, volume dilute HCl to reach the endpoint (mL) molarity of dilute HCI in buret (mol/L) average molarity of dilute HCI (mol/L) percent difference of trials (%) Average molarity of concentrated unknown solution (mol/L) Trial 1 0.0780g 5.52mL 9.77m2 14..85ml Trial 2 0.0810g 11m² 17,91m4 6.91ml 4.872 mol/L (aq) 84.001 g/mol 58.44 g/m Trial 3 0.0930g. 10.3mL 17.13mL 6.83 me

Chemistry for Today: General, Organic, and Biochemistry

9th Edition

ISBN:9781305960060

Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Publisher:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Chapter7: Sollutions And Colloids

Section: Chapter Questions

Problem 7.1E: Many solutions are found in the home. Some are listed below, with the composition as printed on the...

See similar textbooks

Similar questions

Many solutions are found in the home. Some are listed below, with the composition as printed on the label. When no percentage is indicated, components are usually given in order of decreasing amount. When water is present, it is often not mentioned on the label or it is included in the inert ingredients. Identify the solvent and solutes of the following solutions: a.Antiseptic mouthwash: alcohol 25, thymol, eucalyptol, methyl salicylate, menthol, benzoic acid, boric acid b.Paregoric: alcohol 45, opium 0.4 c.Baby oil: mineral oil, lanolin there happens to be no water in this solution-why? d.Distilled vinegar: acetic acid 5
Answers should be in 4th decimal places An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (SG=1.114; MW = 62.07 g/mole) and water at 20oC. The density of the mixture is 1.070 g/mL. Express the concentration of ethylene glycol asa) Volume percentb) Mass percentc) Molarityd) Molalitye) Mole fraction
A 25mL volumetric pipet is used to deliver a sample of the stock solution marked “0.6000M X2SO4” into a 100mL volumetric flask. Distilled water is added to the flask until it is about 3/4thfull, the solution is mixed well, then more water is added to fill it up to the calibration mark, then it is mixed again. Calculate the concentration of X+ion in the dilute solution made above.
Based from the Molarity of the palmitic acid solution that you were able to solve from the previous proble, you are to prepare 250 mL of the said molar concentration. If you have an available 2M stock solution of Palmitic acid, How many mL of the 2M palmitic acid stock solution will you need? Final answer must be rounded off to 2 decimal places, and shall NOT have any units.
What is the final molarity of sodium thiosulfate pentahydrate for each of the runs? Given: the initial volume of each component was diluted to 100mL mass of sodium thiosulfate pentahydrate used: 0.115g 0.0025 M sodium thiosulfate pentahydrate run #1: 10mL run #2: 10mL run #3: 10 mL
Run 1 Run 2 Molarity of KMnO4 solution (M) from bottle 0.00101 0.00101 Initial reading of buret KMnO4 (mL) 0.62 11.35 Final reading of buret KMnO4 (mL) 11.33 21.70 Run 1 Run 2 Volume of KMnO4 solution (mL) Moles of MnO4- used for titration (mol) Moles of C2O42- in 100.0 mL of solution (mol) Molarity of C2O42- (M) Molarity of Cd2+ (M) Ksp of CdC2O4 Average Ksp of CdC2O4 Calculations: Moles of MnO4- used for titration of saturated solution CdC2O4 Moles of C2O42- in 100.0 mL of saturated solution of CdC2O4 Molarity of C2O42- in saturated solution of CdC2O4 Molarity of Cd2+ in saturated solution of CdC2O4 Solubility product, Ksp, of CdC2O4
Given the following data for Mass of test tube and stearic acid = 14.17 gMass of test tube = 11.40 gFreezing point of strearic acid = 69.59o CMass of weighing paper + naphthalene =1.230 gMass of weighing paper = 0.920 gFreezing point solution = 64.00o CKf = 4.5o C/m Determine the following1. mass of stearic acid in g (2 decimal places); _____2. mass of naphthalene in g (2 decimal places); _____3. freezing point depression (2 decimal places); _____4. molality of solution (3 significant figures); _____5. moles of naphthalene (3 significant figures); _____6. molar mass of naphthalene, experimentally (3 significant figures); _____7. % error if theoretical molar mass of naphthalene is 128.17 g/ mole, USE ABSOLUTE VALUE (3 significant figure); _____
and Employees CH STATE GE 90 ENDULERSE IS FOREVER "ON-Tame Name Jodi Rayal Molarity = wt. x1000 3. Calculate the massof Potassium hydrogen phthalate (KHP) to prepare a 250.0 mL solution of 0.1000 M KHP solution. Mol ut. x Voluse the 2500×1000 204.22250 mass Initial Buret Reading Final Buret Reading Vol NaOH Added Moles of NaOH Moles of HCI Volume of unknown HCI solution Concentration of HCI Average concentration of HCI of KHP = 5.11g 4. Complete the table below for the following neutralization reaction of an unknown concentration of hydrochloric acid with 0.100 M sodium hydroxide: Mu spate solution Trial 1 0 11.15 го data and calculate the concentration 15 Trial 2 11.15 22.15 Trial 3 22.15 33.25 abyh 067 LANDINUKASHU TROWTHCENTISSATE ARAD Ma That par މއކނި REMURNAR MISAMARALISERDRY Tile
Volume of an unknown used was 30 mL, Initial Buret volume was 0 and the Final Buret volume was 8.5 mL. What is the molarity of the unknown solution if the Net volume of NaOH being used was 8.5 mL and Millimoles (mmoles) NaOH reacted was 0.791? Then, what is the Mass (g) of Acetic Acid in unknown sample and thr average percent (%) Acetic Acid? (assume density = 1g/mL)
Calculate the percent composition (in m/v %) of 335 mL of solution containing 22 grams of glucose. Report answer with correct sigfigs and units.
Bleach is a 6.0% (mass/volume) solution of NaClO. That means that there are 6 grams of NaClO(MM = 74.44 g/mol) per 100 mL of bleach. Calculate the molarity (mol/L) of NaClO in bleach tothree significant figures (one extra sig fig).
Using the Solubility Data for Report below answer the question: Compound Cold water Solubility Hot Water Solubility Benzoic Acid 0.42 g per 100mL of water 5.79 g per 100mL of water Impurity 0.61 g per 100mL of water 4.62 g per 100mL of water An impure sample contains 1.14 g of impurities and 5.14 g of benzoic acid. The sample is dissloved in 100mL of water and heated. The solution is then cooled. How many grams of impurity will crystalize when the solution has cooled?