(a) A compound X reacts with excess HNO\(_{3(aq)}\) to give carbon (IV) oxide and another compound Y. A solution of Y reacts with NaOH\(_{(aq)}\) to form a white precipitate which is insoluble in excess NaOH\(_{(aq)}\). Identify X and Y.
(b) (i) Write a balanced equation to illustrate the reducing property of ammonia in its reaction with CuO.
(ii) Explain why it is not advisable to heat ammonium dioxonitrate (III) directly.
(iii) Give two uses nitrogen.
(c) Give the reason why (i) dilute H\(_{2}\)SO\(_{4}\) is not suitable for the preparation of CO\(_{2(g)}\) from CaCO\(_{3(s)}\) (ii) concentrated H\(_{2}\)SO\(_{4}\) cannot be used to dry ammonia gas.
(d) State two: (i) physical properties; (ii) chemical properties of metals.
(e) What is the oxidation number of: (i) chlorine in I. Cl\(_{2}\). II. ClO\(_{-(3)}\)
(ii) vanadium in V\(_{2}\)O\(_{5}\)
(f)(i) Explain the term half-life. (ii) Two radioactive elements, P and Q have half-life of 1200 seconds and 3600 seconds respectively.
I. Which of the elements is more stable? II. Give a reason for your answer.
(a) X gives carbon(IV) oxide with acid, so X is a trioxocarbonate(IV); Y forms a white hydroxide precipitate insoluble in excess NaOH. Taking the metal as calcium: X is calcium trioxocarbonate(IV), CaCO3, and Y is calcium trioxonitrate(V), Ca(NO3)2. \(CaCO_3 + 2HNO_3 \rightarrow Ca(NO_3)_2 + H_2O + CO_2\); \(Ca(NO_3)_2 + 2NaOH \rightarrow Ca(OH)_2 + 2NaNO_3\).
(b)(i) \(2NH_3 + 3CuO \rightarrow 3Cu + N_2 + 3H_2O\) (ammonia reduces copper(II) oxide to copper).
(ii) Ammonium dioxonitrate(III), NH4NO2, decomposes explosively when heated directly, so direct heating is dangerous. \(NH_4NO_2 \rightarrow N_2 + 2H_2O\)
(iii) Two uses of nitrogen: manufacture of ammonia; providing an inert atmosphere (also food packaging/preservation, or liquid nitrogen as a refrigerant).
(c)(i) Dilute H2SO4 is unsuitable because insoluble calcium tetraoxosulphate(VI) forms and coats the marble (CaCO3), stopping further reaction.
(ii) Concentrated H2SO4 cannot dry ammonia because it reacts with the (basic) ammonia to form ammonium tetraoxosulphate(VI). \(2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4\)
(d)(i) Two physical properties of metals: they are good conductors of heat and electricity; they are malleable and ductile (also lustrous, sonorous, high density).
(ii) Two chemical properties: they form basic oxides; they act as reducing agents (are electropositive) and displace hydrogen from dilute acids.
(e)(i) Chlorine: in Cl2 it is 0; in ClO3- it is +5.
(ii) Vanadium in V2O5: \(2V + 5(-2) = 0 \Rightarrow V = +5\).
(f)(i) Half-life is the time taken for half of the atoms (nuclei) of a radioactive element to decay.
(ii) I. Q is more stable. II. Because Q has the longer half-life (3600 s), it decays more slowly, so it is more stable.
(a) X gives carbon(IV) oxide with acid, so X is a trioxocarbonate(IV); Y forms a white hydroxide precipitate insoluble in excess NaOH. Taking the metal as calcium: X is calcium trioxocarbonate(IV), CaCO3, and Y is calcium trioxonitrate(V), Ca(NO3)2. \(CaCO_3 + 2HNO_3 \rightarrow Ca(NO_3)_2 + H_2O + CO_2\); \(Ca(NO_3)_2 + 2NaOH \rightarrow Ca(OH)_2 + 2NaNO_3\).
(b)(i) \(2NH_3 + 3CuO \rightarrow 3Cu + N_2 + 3H_2O\) (ammonia reduces copper(II) oxide to copper).
(ii) Ammonium dioxonitrate(III), NH4NO2, decomposes explosively when heated directly, so direct heating is dangerous. \(NH_4NO_2 \rightarrow N_2 + 2H_2O\)
(iii) Two uses of nitrogen: manufacture of ammonia; providing an inert atmosphere (also food packaging/preservation, or liquid nitrogen as a refrigerant).
(c)(i) Dilute H2SO4 is unsuitable because insoluble calcium tetraoxosulphate(VI) forms and coats the marble (CaCO3), stopping further reaction.
(ii) Concentrated H2SO4 cannot dry ammonia because it reacts with the (basic) ammonia to form ammonium tetraoxosulphate(VI). \(2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4\)
(d)(i) Two physical properties of metals: they are good conductors of heat and electricity; they are malleable and ductile (also lustrous, sonorous, high density).
(ii) Two chemical properties: they form basic oxides; they act as reducing agents (are electropositive) and displace hydrogen from dilute acids.
(e)(i) Chlorine: in Cl2 it is 0; in ClO3- it is +5.
(ii) Vanadium in V2O5: \(2V + 5(-2) = 0 \Rightarrow V = +5\).
(f)(i) Half-life is the time taken for half of the atoms (nuclei) of a radioactive element to decay.
(ii) I. Q is more stable. II. Because Q has the longer half-life (3600 s), it decays more slowly, so it is more stable.
