St Andrew’s Junior College Revision Questions – Redox 1. Balance the following equations: (a) V2+ + MnO4- → V3+ + Mn2+ (in acidic medium) (b) SO32- + I2 → SO42- + 2I - (in acidic medium) (c) C2O42- + Cr2O72- → CO2 + Cr3+ (in acidic medium) (d) HClO + Fe2+ → Cl2 + Fe3+ (in acidic medium) (e) CH3OH + MnO4- → CO2 + MnO2 (in alkaline medium) (f) Pb2+ + MnO4- → Pb4+ + MnO2 (in alkaline medium) 2. A solution of Y5+ can be oxidized to YO22+ by acidified Cr2O72-. (a) Write the equation for the reaction. (b) Calculate the volume of 0.2M UO3- needed to react with 20.00 cm3 of 0.1M Cr2O72-.
Gravimetric Analysis of a Metal Carbonate Post Lab TABLE Mass of Crucible + M2CO3 14.8145g Mass of Crucible + M2CO3 (1st weighing) 14.7915g Mass of Crucible 12.9405g Mass of M2CO3 1.851g Mass of filter paper + CaCO3 3.8025g Mass of filter paper 1.8120g Mass of CaCO3 1.9905g Moles of CaCO3 0.019887 mol Molar mass of M2CO3 93.08 g/mol Identity of M2CO3 Na2CO3 Percent Error 12.18% error ANALYSIS QUESTIONS 1. Calculate the moles of precipitated calcium carbonate, CaCO3. Enter this value in the Data Table. 1.9905g CaCO3 x 1 mol CaCO3 = 0.019887 mol CaCO3 1 100.09g 2. Calculate the molar mass of the unknown carbonate.
Liquid-liquid extraction Informal report By: Afi Ahmed 100959296 Presented to: Devien Durbano CHM2203 May 26th 2014 Results and calculations Table 1: Mass of mixture containing benzoic acid, 2-naphthol, and hydroquinone dimethyl ether and recrystallization. Mass of mixture: 3.03g Compound | Original mass | Recrystallized mass | Benzoic Acid | 1.01 | 0.14 | 2-naphtol | 1.01 | 0.43 | Hydro. Dimethyl ether | 1.01 | 0.16 | With these values we are able to calculate the percent recovery for Benzoic acid, 2-napthol and the hydroquinone dimethyl ether. 1) Percent recovery for isolation of benzoic acid % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.14/1.01) x100% = 13.86% That concludes that the percent recovery is 13.86%. 2) Percent recovery for isolation of benzoic acid % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.43/1.01) x100% = 42.57% That concludes that the percent recovery is 42,57%.
Repeat the titration until there are two titres within 0.1cm3 of each other. Record results in a suitable table. Results: Rough Titre: 7.653 First Run: 6.553 Second Run: 6.453 Third Run: 6.553 Calculations: During the titration, iron(II) ions are oxidised to iron(III) ions and manganate(VII) ions are reduced to manganese(II) ions. The equation is as follows: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) ? 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) The above equation shows that one mole of manganate(VII) ions reacts with 5 moles of iron(II) ions in acid solution.
Then I added 0.382 to the mass of magnesium ribbon, which is 0.589 to get 0.971. 50.943- 50.561 = 0.382 0.382 + 0.589 = 0.971 2. Determine the percentage of magnesium in the compound. Percentage of magnesium =Mass of magnesium ribbon x100 Mass of magnesium oxide compound Percentage of magnesium = 0.589 x100 0.971 Percentage of magnesium = 0.60659 x100 Percentage of magnesium = 60.66 The percentage of magnesium is 60.66%. 3.
Name: Anh Lan Do Course: CHE-131-92L Date: Feb 24th, 2010 General Chemistry 1 Experiment 4 Percent Composition of a Compound Professor Robert M. Cady Questions: 1. According to the law of definite composition, elements form compounds in the same ratios by mass each time the compound is produced. Using the theoretical percent by mass of magnesium in MgO, calculate the mass of oxygen that will combine with 12.25g of Mg and the mass of MgO that will be formed. 2. Which compound has the greatest percentage by mass of Magnesium?
5.51: Which Reagent is limiting and How Much Precipitate is formed? SCH-3UI-03 David Yu Mrs. Hatton Due Date: May 5, 2012 Cut-Off Date: May 12, 2012 Purpose: To experience and use what you have learned in class about gravimetric stoichiometry by predicting and determining the mass of precipitate of two reactants and then comparing what you experience and what you calculated. Background: Avogadro’s constant is 6.02 x 1023 to find the number of entities. A mole is a useable amount of chemicals that is practical to use. The molar mass of a compound or atom is the mass of 1 mole of anything; this is relative to the atomic mass from the periodic table.
IB HL Chemistry Energetics - End of Topic Test 35 min 37 marks Answer all the questions in the spaces provided 1. What energy changes occur when chemical bonds are formed and broken? A. Energy is absorbed when bonds are formed and when they are broken. B.
c) 0.08 g of oxygen reacted with the magnesium (0.22 g – 0.14 g = 0.08 g). d) Refer to “Determining the Empirical Formula of Magnesium Oxide Lab Solutions” sheet e) Element | % | m (g) | M (g/mol) | n (m÷M) | ÷ by | Ratio | Magnesium | 63.63 | 63.63 | 24.31 | 2.617441382 | 2.273125 | 1 | Oxygen | 36.37 | 36.37 | 16.00 | 2.273125 | 2.273125 | 1 | Since the ratio is 1:1, the empirical formula is MgO. 2. Refer to “Determining the Empirical Formula of Magnesium Oxide Lab Solutions” sheet 3. You need to round the empirical formula to a whole number ratio because you cannot have decimals in the subscripts, which means that you cannot have a fractional amount of molecules in a substance.
In this titration, three major steps involved were first measure out an amount of known solution, add indicator and three, to titrate using unknown solution. At the first colour change the titration would stop. Abstract The purpose to this lab is to standardize an approximately 1 mol•L-1 solution of hydrochloric acid. There were many steps taken in this titration but the main steps were measuring out the amount of known solution, adding the indicator and finally, titrating the solution until the first colour change. In each trial, the initial reading, final reading and the volume of HCl used was recorded down as quantitative results.