0620_s14_qp_32
A paper of Chemistry, 0620
Questions:
7
Year:
2014
Paper:
3
Variant:
2

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Propanoic acid is a carboxylic acid. Its formula is CH3–CH2–COOH. Propanoic acid is the third member of the homologous series of carboxylic acids. Give the name and structural formula of the fourth member of this series. name formula Members of a homologous series have very similar chemical properties. State three other characteristics of a homologous series. Carboxylic acids can be made by the oxidation of alcohols. Draw the structural formula of the alcohol which can be oxidised to propanoic acid. Show all atoms and bonds. Name a reagent, other than oxygen, which can oxidise alcohols to carboxylic acids. Complete the following equations for some of the reactions of propanoic acid. The salts of this acid are called propanoates. zinc + propanoic acid → + hydrogen calcium + propanoic → + oxide acid LiOH + CH3CH2COOH → + A piece of magnesium was added to 100 cm3 of an aqueous acid. The time taken for the metal to react completely was measured. This experiment was repeated using different aqueous acids. The same volume of acid was used in each experiment and the pieces of magnesium used were identical. In one experiment the reaction was carried out at a different temperature. experiment acid concentration in mol / dm3 temperature / °C time / minutes A propanoic 1.0 B propanoic 1.0 C propanoic 0.5 D hydrochloric 1.0 Explain the following in terms of collision rate between reacting particles. Why is the rate in experiment C slower than the rate in experiment A? Why is the rate in experiment B faster than the rate in experiment A? Why is the rate in experiment D faster than the rate in experiment A?
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The soluble salt hydrated lithium sulfate is made by titration from the soluble base lithium hydroxide. burette filled with sulfuric acid aqueous lithium hydroxide and a few drops of a pH indicator conical flask The sulfuric acid is added slowly from the burette until the indicator just changes colour. The volume of sulfuric acid needed to just neutralise the lithium hydroxide is noted. Describe how you would continue the experiment to obtain pure dry crystals of hydrated lithium sulfate. Using 25.0 cm3 of aqueous lithium hydroxide, concentration 2.48 mol / dm3, 2.20 g of hydrated lithium sulfate was obtained. Calculate the percentage yield, giving your answer to one decimal place. 2LiOH + H2SO4 → Li2SO4 + 2H2O Li2SO4 + H2O → Li2SO4.H2O Number of moles of LiOH used = Number of moles of Li2SO4.H2O which could be formed = Mass of one mole of Li2SO4.H2O = 128 g Maximum yield of Li2SO4.H2O = g Percentage yield = % An experiment was carried out to show that the formula of the hydrated salt is Li2SO4.H2O. A sample of the hydrated salt was weighed and its mass recorded. It was then heated and the anhydrous salt was weighed. This procedure was repeated until two consecutive masses were the same. This procedure is called ‘heating to constant mass’. What is the reason for heating to constant mass? The mass of the hydrated salt is m1 and the mass of the anhydrous salt is m2. Explain how you could show that the hydrated salt has one mole of water of crystallisation per mole of the anhydrous salt.