CAIE
Chemistry
(9701)
Questions & Answers
Radium is a Group 2 element.
Define . Some data relating to radium and sulfur are listed. Select relevant data from this list for use in your answers to parts (b) to (e).
+ 2e +1619 first ionisation energy of sulfur +1000 second ionisation energy of sulfur +2260 first electron affinity of sulfur -200 second electron affinity of sulfur +532 enthalpy change for S
+555 lattice energy of RaS
-2612
Write an equation for the process corresponding to the second electron affinity of sulfur. Include state symbols.
Sulfur exists as S molecules in the solid state. Use the data in this question to calculate the enthalpy change for the reaction S 8S
. enthalpy change = kJ mol
Calculate the standard enthalpy change of formation, , of radium sulfide.
(i) State the two major factors that affect the numerical magnitude of a lattice energy.
For each factor you have identified in (e)(i), state whether it tends to make the lattice energy of radium sulfide more or less exothermic than that of sodium chloride. Explain your answer. , respectively. Identify the dominant factor in determining the relative numerical magnitudes of the lattice energies of radium sulfide and sodium chloride. Explain your answer. [Total: 13] 2 Ethoxyethane, C , can dissolve both in water and in octan-1-ol. The expression and numerical value for the partition coefficient of ethoxyethane between water and octan-1-ol are given. Water and octan-1-ol are immiscible. concentration of C in octan-1-ol concentration of C in water The mixture is analysed immediately and a value of K is calculated. The calculation is performed correctly; the value calculated is 5.625. Explain why the value calculated is less than 6.760.
A second experiment is performed and the value of K is found to be 6.760. The concentration , of ethoxyethane in the aqueous layer. [1] of water. Calculate the maximum amount, in mol, of ethoxyethane that can be extracted into the water.
An aqueous solution of lead( II) nitrate is mixed with an aqueous solution of sodium iodide. A yellow precipitate of lead( II) iodide is formed and is filtered out, leaving solution X. The concentration of Pb in solution X is 5.68 × 10 The concentration of I in solution X is 4.20 × 10
Use these data to calculate a value for the solubility product, K , of lead(II) iodide. State the units of K = units =
Potassium iodide is very soluble in water. Describe and explain what is seen if a few drops of saturated potassium iodide solution are added to a portion of solution X. [Total: 10]
Pure water is a very poor conductor of electricity. However, when hydrogen chloride gas is dissolved in water, ions are formed and a current flows during electrolysis.
The overall change after electrolysis is that hydrogen chloride gas is converted into hydrogen and chlorine.
of chlorine are produced at the anode, measured under room conditions.
Calculate the number of chlorine molecules produced during the electrolysis. number of chlorine molecules =
Calculate the total number of electrons transferred to produce this number of chlorine molecules. total number of electrons =
Calculate the quantity of charge, in coulombs, of the total number of electrons calculated in
. quantity of charge = C
Calculate the time, Y, in minutes, for which the current flows. Y = minutes [1] , of three species are given in the table.HCl
Predict the effect of increasing temperature on the spontaneity of this reaction. Explain your answer. [Total: 9] until completely decomposed to the metal oxide, nitrogen dioxide and oxygen.
State which of these two Group 2 nitrates requires the higher temperature before it begins to decompose. Explain your answer. increasing amounts of water are added to it, with stirring, until no solid remains. Identify the sample to which most water must be added to cause all the solid to dissolve. Explain your answer by reference to the solubilities of the products formed when water is added to the oxides. You should refer to relevant energy terms in your answer. dissolved in water. The oxidising agent H of a solution in which nitric acid, HNO , is the only nitrogen-containing product. , of HNO of solution.
The HNO NaOH(aq). Calculate the minimum volume, in cm , of NaOH(aq) needed. Give your answer to three significant figures. volume = cm [1] [Total: 7]
Transition elements form complexes.
Molybdenum, Mo, forms an octahedral complex consisting of one Mo atom surrounded by carbon monoxide, CO, molecules. CO is a monodentate ligand. Iron forms an octahedral complex consisting of one Fe and a number of cyanide, CN , ions. CN is a monodentate ligand.
Define the term monodentate ligand .
Complete the table by stating the formulae and charges of the complexes described. formula charge molybdenum complex iron( III) complex
Draw a three-dimensional diagram to show the shape of this iron( III) complex. Label one 180° bond angle on your diagram.Fe
An excess of aqueous ammonia is added to dilute copper( II) sulfate solution. A dark blue complex, [Cu(NH , is formed.
Write an ionic equation for this reaction.
Explain the origin of colour in copper( II) complexes.
An excess of concentrated hydrochloric acid is added to the dark blue solution of [Cu(NH . A new complex, Z, is formed. The colour of the solution changes.
Write an equation for the formation of Z from the solution of [Cu(NH . Include the formula and charge of Z.
Name the type of reaction when Z forms from [Cu(NH
State the geometry of Z.
State the colour of a solution of Z.
Explain why the colour of a solution of [Cu(NH is different from the colour of a solution of Z. You should refer to the energies of the orbitals involved in your answer. [Total: 15]
An excess of aqueous ammonia is added to a solution containing the complex ion [Co(H [Co(H + 6NH [Co(NH + 6H
Complete the sentence to describe the colour change that will be seen during this reaction.
Write an expression for the stability constant, K stab , of [Co(NH stab
The numerical value of K stab of [Co(NH is 7.7 × 10 What deduction about the properties of [Co(NH and [Co(H can be made from this stab value?
Oxygen can oxidise [Co(NH to [Co(NH under standard conditions in alkaline solutions. [Co(NH + e [Co(NH
Use this information and the Data Booklet to calculate the value for this oxidation of [Co(NH = V
Write an ionic equation for this oxidation of [Co(NH
Predict, by selecting suitable data from the Data Booklet, whether oxygen can oxidise
in acidic solution, in the absence of ammonia. Explain your answer. [Total: 7]
The structure of benzene-1,3-dicarboxylic acid is shown. benzene-1,3-dicarboxylic acid
State the empirical formula of benzene-1,3-dicarboxylic acid.
Benzene-1,3-dicarboxylic acid is an isomer of benzene-1,4-dicarboxylic acid. These two isomers can be distinguished by carbon-13 ( C) NMR spectroscopy. State the number of peaks in the carbon-13 ( C) NMR spectrum of each compound. benzene-1,3-dicarboxylic acid benzene-1,4-dicarboxylic acid
Benzene-1,3-dicarboxylic acid can be made by the two-step synthesis shown below. benzene-1,3-dicarboxylic acid compound P step 2step 1 compound Q
Name compound P.
Explain why the major product of this two-step synthesis is benzene-1,3-dicarboxylic acid and not benzene-1,4-dicarboxylic acid.
The reagents used for step 1 are CH . These reagents give rise to C=O ions which react with compound P. Name the mechanism of this reaction.
Draw the mechanism of the reaction of CHC=O ions with compound P. Include all relevant curly arrows and charges, the structure of the intermediate and the structure of compound Q.C=O intermediate compound Q [3] acid, in step 2. [Total: 10] 8 Alanine, H NCH(CH)CO H, and glutamic acid, H NCH(CH H)CO H, are two naturally occurring amino acids.
H NCH(CH)CO H exists as two optical isomers. Draw three-dimensional structures of these two optical isomers.
The proton ( H) NMR spectrum of either alanine in D O or glutamic acid in D O is shown. 9 8 7 6 5 4 3 2 1 0 / ppm State whether this is the spectrum of alanine in D O or the spectrum of glutamic acid in D Explain your answer by reference to the number of peaks and splitting patterns.
The mass spectrum of glutamic acid, H NCH(CH H)CO H, is obtained.
State the m/e value of the molecular ion peak in this spectrum.
The spectrum has peaks with m/e values of 88 and 131. Draw the structures of the ions responsible for these peaks. m/e structure of ion
At pH 11 alanine exists as H NCH(CH)CO ions and glutamic acid exists as NCH(CH)CO ions. A mixture of alanine and glutamic acid at pH 11 is subjected to electrophoresis.
State how the mixture can be maintained at pH 11 during electrophoresis.
Draw a fully labelled diagram for the apparatus that would be used to carry out this electrophoresis. Your diagram should include the position of the mixture of alanine and glutamic acid at the start of the electrophoresis experiment.
Identify the electrode that each amino acid travels towards during electrophoresis at pH 11. alanine glutamic acid [1] Alanine travels a shorter distance. Explain the factors that account for the difference in the distances travelled. [Total: 14] 9 Butylamine, CH , can be synthesised from different organic compounds by using suitable reagents. Each reaction involves one step.
Complete the table to describe three different syntheses. group. starting organic compound reagent and conditions
Compare and explain the relative basicities of ammonia, butylamine and phenylamine. most basic least basic [Total: 10]
Complete the table to show the structure of the organic product from each reaction of phenol,OH. reaction reaction mixture structure of organic product 1 phenol + NaOH(aq) 2 phenol + Na
phenol + (aq) an excess of phenol +
Identify two reactions from the table in which ethanol would behave in a similar way to phenol. [Total: 5] Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every UHDVRQDEOHH?RUWKDVEHHQPDGHE\WKHSXEOLVKHU8&/(6WRWUDFHFRS\ULJKWKROGHUVEXWLIDQ\LWHPVUHTXLULQJFOHDUDQFHKDYHXQZLWWLQJO\EHHQLQFOXGHGWKH publisher will be pleased to make amends at the earliest possible opportunity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