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Determine which substance will be the limiting reagent. Then repeat the same process for the next two trials. Please show sample calculations
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- 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 CdC2O4Chemistry Find the molarity of NaOH, net volume, and average molarity using the data given: KHC8H4O4 + NaOH --> NaKC8H4O4 + H20 trial 1 mass of sample = 0.300 initial volume = 0.0 ml final volume = 8.5 ml trial 2 mass of sample = 0.2959 initial volume = 10.0 ml final volume = 17.4 ml0.600 mole of Ca(NO3) in 700.0 mL of solution MCa2+ = ____M MNO3- = ____M
- Volume of HCL sample = 15ml Molarity standardized NaOh solution = .09929 Initial buret reading = 17.85 mL NaOH final buret reading = 36.16 mL NaOH volume of NaOH used = 17.31 Equation HCl = NaOH=NaCl + H2O What is the molarity of the HCl sample?The molarity for trial 1 will be 0.4035 and trial 2 molarity will be 0.3587 I need some assistance on question 2 if possibleIf 0.9871 g of the primary standard KIO3 is used to prepare 500.00-mL of the standard solution, how much is the KIO3 contained in a 30.00 mL solution? CHOICES 0.05923 g 16.45 g None of the choices 0.09871 g
- H2. How would you prepare 100 mL of a 0.05 M aqueous solution of iron from a 0.7 M solution? It includes a flow diagram of all the steps, the volumetric materials to be used, it includes the disposal of waste (after having used the solution for a colorimetric analysis). Step 1. Take aliquot of: ____ ml Express your result with a decimal and without units.K2CO3 (aq)+ CaCl2 (aq)→ CaCO3 (s) + 2KCl (aq) Data Sheet Table 1: Data and Observations Material Mass CaCl2 2.0g K2CO3 2.5g Filter Paper 1.6g Watch Glass 35.8g Filter Paper + Watch Glass + Precipitate 38.9 Precipitate 1.5g Table 2: Mass of CaCl2 after 24 Hours Initial Observations 24 hour Observation Weigh Boat Mass of Weigh Boat 0.5g Mass of Weigh Boat 0.5g CaCl2 2.0g Mass of CaCl2 2.4g Mass of CaCl2 Calculate the theoretical yield of the solid precipitate. Then use that to calculate the percent yield of the solid precipitate.fill in the blanks solution volume is 50/1000
- Trial 1: [4] Volume of NaOH, dispensed (mL): 0.01608 Trial 1: [5] Molar concentration of NaOH (mol/L): 0.102 Trial 2: [4] Volume of NaOH, dispensed (mL): 0.01423 Trial 2: [5] Molar concentration of NaOH (mol/L): 0.106 Trial 3: [4] Volume of NaOH, dispensed (mL): 0.01721 Trial 3: [5] Molar concentration of NaOH (mol/L): 0.109 Average molar concentration of NaOH (mol/L): 0.1057Volume of HCL sample = 15ml Molarity standardized NaOh solution = .09929 Initial buret reading = 17.85 mL NaOH final buret reading = 36.16 mL NaOH What is the volume of NaOH used?A 15% w/v (NH4)2SO4 standard is needed. How much solute would be needed to make 50 ml?