WAEC SSCE - Chemistry - 2011 (Essay)
Question 1 Report
(a)Define the term solubility.
(b) The table below gives the solubility of salt Z at various mperatures
| Temperature (°C) | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
| Solubility (mol dm\(^{-3}\)) | 0.13 | 0.21 | 0.31 | 0.45 | 0.63 | 0.85 | 1.10 |
(i) Plot a graph of solubility against temperature.
(ii) From the graph determine the solubility of salt Z at 35°C.
(iii) If 100cm\(^{-3}\) of the saturated solution is cooled from 55°C to 35°C, calculate the mass of salt Z that would crystallize out. [Molar mass of salt Z = 100 g]
(c)(i) Write a balanced equation to illustrate the reaction of AI\(_2\)O\(_3\) with dilute I. HCI; II. NaOH.
(ii) What is the name given to an oxide that exhibits both acidic and basic properties?
(iii) Give one metallic oxide which exhibite these properties.
(d)(i) Determine the oxidation number of: I. Al in [Al (H\(_2\)O)\(_6\)]\(^{3+}\); II. H in NaH.
(ii) Give the IUPAC name of each of the following substances; I. CuSO\(_4\).5H\(_2\)O; II. CaCO\(_3\); Ill. KMnO\(_4\).
(a) Solubility is the maximum amount of a solute that dissolves in 1 dm3 of solution at a specified temperature to form a saturated solution.
(b)(i) The solubility curve of salt Z is plotted below. Temperature is on the horizontal axis and solubility is on the vertical axis.
(ii) From the graph, at 35°C:
Solubility of Z = 0.54 mol dm−3.
(iii) From the graph:
- Solubility at 55°C = 0.97 mol dm−3
- Solubility at 35°C = 0.54 mol dm−3
Decrease in solubility:
\[0.97-0.54=0.43\ \text{mol dm}^{-3}\]
Volume of saturated solution = \(100\text{ cm}^3=0.100\text{ dm}^3\).
Moles of salt crystallized:
\[0.43\times0.100=0.043\ \text{mol}\]
\[\text{Mass crystallized}=0.043\times100=\boxed{4.3\ \text{g}}\]
(c)(i)
I. Reaction with dilute hydrochloric acid:
\[\mathrm{Al_2O_3(s)+6HCl(aq)\rightarrow2AlCl_3(aq)+3H_2O(l)}\]
II. Reaction with sodium hydroxide solution:
\[\mathrm{Al_2O_3(s)+2NaOH(aq)+3H_2O(l)\rightarrow2NaAl(OH)_4(aq)}\]
(c)(ii) Such an oxide is called an amphoteric oxide.
(c)(iii) Zinc oxide, \(\mathrm{ZnO}\), is an amphoteric metallic oxide.
(d)(i)
I. In \(\mathrm{[Al(H_2O)_6]^{3+}}\), water is neutral. Therefore, oxidation number of Al = +3.
II. In \(\mathrm{NaH}\), Na is \(+1\). Therefore, oxidation number of H = −1.
(d)(ii)
| Formula | IUPAC name |
|---|---|
| \(\mathrm{CuSO_4\cdot5H_2O}\) | Copper(II) tetraoxosulfate(VI) pentahydrate |
| \(\mathrm{CaCO_3}\) | Calcium trioxocarbonate(IV) |
| \(\mathrm{KMnO_4}\) | Potassium tetraoxomanganate(VII) |
Answer Details
(a) Solubility is the maximum amount of a solute that dissolves in 1 dm3 of solution at a specified temperature to form a saturated solution.
(b)(i) The solubility curve of salt Z is plotted below. Temperature is on the horizontal axis and solubility is on the vertical axis.
(ii) From the graph, at 35°C:
Solubility of Z = 0.54 mol dm−3.
(iii) From the graph:
- Solubility at 55°C = 0.97 mol dm−3
- Solubility at 35°C = 0.54 mol dm−3
Decrease in solubility:
\[0.97-0.54=0.43\ \text{mol dm}^{-3}\]
Volume of saturated solution = \(100\text{ cm}^3=0.100\text{ dm}^3\).
Moles of salt crystallized:
\[0.43\times0.100=0.043\ \text{mol}\]
\[\text{Mass crystallized}=0.043\times100=\boxed{4.3\ \text{g}}\]
(c)(i)
I. Reaction with dilute hydrochloric acid:
\[\mathrm{Al_2O_3(s)+6HCl(aq)\rightarrow2AlCl_3(aq)+3H_2O(l)}\]
II. Reaction with sodium hydroxide solution:
\[\mathrm{Al_2O_3(s)+2NaOH(aq)+3H_2O(l)\rightarrow2NaAl(OH)_4(aq)}\]
(c)(ii) Such an oxide is called an amphoteric oxide.
(c)(iii) Zinc oxide, \(\mathrm{ZnO}\), is an amphoteric metallic oxide.
(d)(i)
I. In \(\mathrm{[Al(H_2O)_6]^{3+}}\), water is neutral. Therefore, oxidation number of Al = +3.
II. In \(\mathrm{NaH}\), Na is \(+1\). Therefore, oxidation number of H = −1.
(d)(ii)
| Formula | IUPAC name |
|---|---|
| \(\mathrm{CuSO_4\cdot5H_2O}\) | Copper(II) tetraoxosulfate(VI) pentahydrate |
| \(\mathrm{CaCO_3}\) | Calcium trioxocarbonate(IV) |
| \(\mathrm{KMnO_4}\) | Potassium tetraoxomanganate(VII) |
Question 2 Report
(a)(i) What is a functional group?
(ii) State the functional group in each of the following compounds: I. CH\(_3\)CH\(_2\)CH(CH\(_3\))OH; II. CH\(_3\)CH\(_2\)CH\(_2\)COOH.
(b)(i) Complete the following equations:
(ii) Draw the structure(s) of the product(s) formed in (b)(i) above.
(c)(i).Write an equation for the prepzration of butan-2-ol from butene.
(d) (i) Give the reagents required for the following conversions to take place: I. CH\(_2\) = CH\(_2\) to CH\(_3\)CH\(_2\)OH; II. CH\(_3\)CH\(_2\)OH to CH\(_3\)COOH; Ill. CH\(_3\)COOH to CH\(_3\)COOCH\(_2\)CH\(_3\).
(e) Consider the following organic structure; CH\(_3\)OHCHCH = CHCOOH
(i) State what would be observed when the organic compound is treated with each of the following reagents: I. cold NaHCO\(_{3(aq)}\); II. hot solution of I\(_2\) in NaOH\(_{(aq)}\); Ill. bromine water.
(a)(i) A functional group is an atom or group of atoms in an organic compound that determines its characteristic chemical reactions.
(a)(ii)
- In CH3CH2CH(CH3)OH, the functional group is the hydroxyl group, OH.
- In CH3CH2CH2COOH, the functional group is the carboxyl group, COOH.
(b)(i) The completed equations are:
- \[\mathrm{CH_3COOCH_2CH_2CH_3 + H_2O \xrightarrow[\text{heat}]{H^+} CH_3COOH + CH_3CH_2CH_2OH}\]
- \[\mathrm{CH_3CH_2CH_2CH_2COOCH_3 + NaOH \xrightarrow{\text{heat}} CH_3CH_2CH_2CH_2COONa + CH_3OH}\]
(b)(ii) The displayed structures of the products are shown below.
(c)(i) Butan-2-ol is prepared from butene by acid-catalysed hydration:
\[\mathrm{CH_2{=}CHCH_2CH_3 + H_2O \xrightarrow{H^+\,/\,H_2SO_4} CH_3CH(OH)CH_2CH_3}\](d)(i)
- CH2=CH2 CH3CH2OH: steam and concentrated H2SO4 or H3PO4 catalyst.
- CH3CH2OH CH3COOH: acidified potassium dichromate(VI), K2Cr2O7/H2SO4, heat under reflux. Acidified KMnO4 is also suitable.
- CH3COOH CH3COOCH2CH3: ethanol and concentrated H2SO4, warmed.
(e)(i) For CH3CH(OH)CH=CHCOOH:
- With cold aqueous NaHCO3, brisk effervescence occurs as carbon dioxide is evolved.
\[\mathrm{RCOOH + NaHCO_3 \rightarrow RCOONa + H_2O + CO_2}\] - With hot iodine solution in aqueous sodium hydroxide, a yellow precipitate of iodoform, CHI3, is formed.
- Bromine water is decolourised, showing the presence of the carbon-carbon double bond.
Answer Details
(a)(i) A functional group is an atom or group of atoms in an organic compound that determines its characteristic chemical reactions.
(a)(ii)
- In CH3CH2CH(CH3)OH, the functional group is the hydroxyl group, OH.
- In CH3CH2CH2COOH, the functional group is the carboxyl group, COOH.
(b)(i) The completed equations are:
- \[\mathrm{CH_3COOCH_2CH_2CH_3 + H_2O \xrightarrow[\text{heat}]{H^+} CH_3COOH + CH_3CH_2CH_2OH}\]
- \[\mathrm{CH_3CH_2CH_2CH_2COOCH_3 + NaOH \xrightarrow{\text{heat}} CH_3CH_2CH_2CH_2COONa + CH_3OH}\]
(b)(ii) The displayed structures of the products are shown below.
(c)(i) Butan-2-ol is prepared from butene by acid-catalysed hydration:
\[\mathrm{CH_2{=}CHCH_2CH_3 + H_2O \xrightarrow{H^+\,/\,H_2SO_4} CH_3CH(OH)CH_2CH_3}\](d)(i)
- CH2=CH2 CH3CH2OH: steam and concentrated H2SO4 or H3PO4 catalyst.
- CH3CH2OH CH3COOH: acidified potassium dichromate(VI), K2Cr2O7/H2SO4, heat under reflux. Acidified KMnO4 is also suitable.
- CH3COOH CH3COOCH2CH3: ethanol and concentrated H2SO4, warmed.
(e)(i) For CH3CH(OH)CH=CHCOOH:
- With cold aqueous NaHCO3, brisk effervescence occurs as carbon dioxide is evolved.
\[\mathrm{RCOOH + NaHCO_3 \rightarrow RCOONa + H_2O + CO_2}\] - With hot iodine solution in aqueous sodium hydroxide, a yellow precipitate of iodoform, CHI3, is formed.
- Bromine water is decolourised, showing the presence of the carbon-carbon double bond.
Question 3 Report
(a)The table below gives the volume/pressure data for a particular sample of a gas-at a given temperature.
| Volumedm\(^3\) (V) | 4.00 | 2.00 | 1.00 |
| Pressure/atm (P) | 1.00 | 2.00 | 4.00 |
(i) Deduce a mathematical relationship between volume (V) and pressure (P).
(ii) Name law that can be deduced from the data.
(iii) Calculate the pressure of the gas when the volume is 3.20 dm\(^3\)
(b)(i) What is the role of a salt bridge in an electrochemical cell?
(ii) What type of ions must flow into the cathode? Give a reason for your answer.
(iii)A standard galvanic cell constructed with \( \mathrm{Ag^+_{(aq)}} \) \( \mathrm{Ag_{(s)}} \) and \( \mathrm{Zn^{2+}_{(aq)}} \) couple is discharged until 3.3 g of Ag forms.
I. Write the overall cell reaction and standard cell potential?
II. How many moles of electrons flowed through the circuit during the discharge?
III. How many coulombs of charges flowed through the circuit?
\( \mathrm{Ag^+_{(aq)} + e^- \rightleftharpoons Ag_{(s)}} \); E° = + 0.80V
\( \mathrm{Zn^{2+}_{(aq)} + 2^- \rightleftharpoons Zn_{(s)}} \), E° = – 0.76 V [Ag = 108]
(c)(i) Define each of the following terms: I. Activation energy; II. Exothermic reaction.
(ii) Give one example of an endothermic process.
(iii) What is the significance of activated complex in a chemical reaction?
Question 4 Report
(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.
Answer Details
(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.
Question 5 Report
The following table gives the atomic numbers of elements V, W, X, Y and Z.
| Element | V | W | X | Y | Z |
| Atomic number | 11 | 16 | 18 | 19 | 24 |
(a)Which of the elements: (i) belong(s) to group 1?
(ii) is/are riotle gas(es)?
(iii) form(s) coloured compound(s)?
(iv) form(s) an anion?
(v) react(s) with water to liberate hydrogen?
(vi) react(s) with water to form alkaline solution?
(b) What is the:
(i) charge on the ion formed in
(a)(iv) above?
(ii) group of the element(s) in (a)(i) above?
(c)(i) Write the formula of the compound formed between element V and element W.
(ii) Sate the type of bond formed in (c)(i) above. Explain your answer.
(d)(i) What is a covalent compound?
(ii) Give two factors that influence covalent bonding.
(iii) State the type of bond that exists in each of the following substances: MgO, \( \mathrm{NH_3} \) and Fe.
(iv) What are intermolecular forces?
First identify each element from its atomic number, then place it in the Periodic Table.
| Element | Atomic number | Identity | Group / block |
|---|---|---|---|
| V | 11 | Sodium, Na | Group 1 |
| W | 16 | Sulfur, S | Group 6 |
| X | 18 | Argon, Ar | Group 8 (noble gas) |
| Y | 19 | Potassium, K | Group 1 |
| Z | 24 | Chromium, Cr | Transition (d-block) metal |
(a)
- (i) Group 1: V and Y (sodium and potassium). Each has a single electron in its outermost shell.
- (ii) Noble gas: X (argon). Its electron configuration 2,8,8 is a complete, stable octet.
- (iii) Coloured compound(s): Z (chromium). Transition metals have partly filled d-orbitals, so d-d electron transitions absorb visible light and give coloured compounds.
- (iv) Forms an anion: W (sulfur). With 6 outer electrons it gains 2 electrons to complete its octet, forming S2-.
- (v) Reacts with water to liberate hydrogen: V and Y (sodium and potassium).
- (vi) Reacts with water to form an alkaline solution: V and Y. They form NaOH and KOH respectively, which are alkalis.
The reaction of a Group 1 metal with water is illustrated by:
\[2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)\](b)
- (i) The ion in (a)(iv) is the sulfide ion S2-; its charge is 2- (a double negative charge).
- (ii) The elements in (a)(i) belong to Group 1.
(c)
- (i) V is Na (forms Na+) and W is S (forms S2-). Balancing the charges gives the formula Na2S.
- (ii) The bond is ionic (electrovalent). Sodium, a Group 1 metal, loses one electron to become Na+; sulfur, a Group 6 non-metal, gains two electrons to become S2-. The complete transfer of electrons from the metal to the non-metal, and the strong electrostatic attraction between the oppositely charged ions, is the mark of ionic bonding.
(d)
- (i) A covalent compound is a compound in which the atoms are held together by the sharing of one or more pairs of electrons, usually between non-metallic atoms.
- (ii) Two factors that influence covalent bonding:
- High ionisation energy of the combining atoms (they hold their electrons too tightly to transfer them, so they share instead).
- A small difference in electronegativity between the atoms (electrons are shared rather than fully transferred).
- (iii) Type of bond in each substance:
Substance Bond type MgO Ionic (electrovalent) NH3 Covalent Fe Metallic - (iv) Intermolecular forces are the relatively weak forces of attraction that exist between separate molecules of a substance, as distinct from the strong bonds that hold atoms together within a molecule.
Answer Details
First identify each element from its atomic number, then place it in the Periodic Table.
| Element | Atomic number | Identity | Group / block |
|---|---|---|---|
| V | 11 | Sodium, Na | Group 1 |
| W | 16 | Sulfur, S | Group 6 |
| X | 18 | Argon, Ar | Group 8 (noble gas) |
| Y | 19 | Potassium, K | Group 1 |
| Z | 24 | Chromium, Cr | Transition (d-block) metal |
(a)
- (i) Group 1: V and Y (sodium and potassium). Each has a single electron in its outermost shell.
- (ii) Noble gas: X (argon). Its electron configuration 2,8,8 is a complete, stable octet.
- (iii) Coloured compound(s): Z (chromium). Transition metals have partly filled d-orbitals, so d-d electron transitions absorb visible light and give coloured compounds.
- (iv) Forms an anion: W (sulfur). With 6 outer electrons it gains 2 electrons to complete its octet, forming S2-.
- (v) Reacts with water to liberate hydrogen: V and Y (sodium and potassium).
- (vi) Reacts with water to form an alkaline solution: V and Y. They form NaOH and KOH respectively, which are alkalis.
The reaction of a Group 1 metal with water is illustrated by:
\[2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)\](b)
- (i) The ion in (a)(iv) is the sulfide ion S2-; its charge is 2- (a double negative charge).
- (ii) The elements in (a)(i) belong to Group 1.
(c)
- (i) V is Na (forms Na+) and W is S (forms S2-). Balancing the charges gives the formula Na2S.
- (ii) The bond is ionic (electrovalent). Sodium, a Group 1 metal, loses one electron to become Na+; sulfur, a Group 6 non-metal, gains two electrons to become S2-. The complete transfer of electrons from the metal to the non-metal, and the strong electrostatic attraction between the oppositely charged ions, is the mark of ionic bonding.
(d)
- (i) A covalent compound is a compound in which the atoms are held together by the sharing of one or more pairs of electrons, usually between non-metallic atoms.
- (ii) Two factors that influence covalent bonding:
- High ionisation energy of the combining atoms (they hold their electrons too tightly to transfer them, so they share instead).
- A small difference in electronegativity between the atoms (electrons are shared rather than fully transferred).
- (iii) Type of bond in each substance:
Substance Bond type MgO Ionic (electrovalent) NH3 Covalent Fe Metallic - (iv) Intermolecular forces are the relatively weak forces of attraction that exist between separate molecules of a substance, as distinct from the strong bonds that hold atoms together within a molecule.
Question 6 Report
(a)(i) Draw the energy profile diagram for the reaction; H\(_{2(g)}\) + I\(_{2(g)}\) \(\to\) 2Hl\(_{(g)}\); \(\Delta\)H = –13 KJ mol\(^{-1}\)
(ii) If the concentration of HI\(_{(g)}\) increases from 0.000 to 0.002 mol dm\(^{-3}\) in 80 seconds, what is the rate of t reaction?
(b)(i)Give one use of each of the following compounds: I. NaHCO\(_{3}\); II. CaSO\(_{4}\); III. CaCO\(_{3}\).
(ii) State a drying agent that can be used for each of the following gases: I. SO\(_{2}\); II. HCI; Ill. NH\(_{3}\)
(c)(i) Write an equation for the complete combustion of carbon in oxygen.
(ii) Calculate the number moles of carbon (IV) oxide produced from the complete combustion of 2.5 g of carbon. [ C = 12.0, O = 16]
(iii) Mention one use of I. carbon (II) oxide; II. carbon (IV) oxide.
(d) An industrial raw material has the following composition by mass:
Iron = 28.1%; Chlorine = 35.7%; Water cf crystallization = 36.2%.
Calculate the formula for the material. [H = 1.00, O = 16.0, CI = 35.5, Fe = 56.0]
(e) Give one example of a (i) metal that is liquid at room temperature,
(ii) non-metal that is liquid room temperature.
(a)(i) The reaction is exothermic because \(\Delta H=-13\ \text{kJ mol}^{-1}\). The products are therefore at a lower energy level than the reactants.
(ii)
\[\text{Rate of reaction}=\frac{\text{change in concentration of HI}}{\text{time}}\]
\[=\frac{0.002-0.000}{80}=2.5\times10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}\]
Rate of reaction = \(2.5\times10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}\).
(b)(i)
- \(\mathrm{NaHCO_3}\): manufacture of baking powder.
- \(\mathrm{CaSO_4}\): manufacture of plaster of Paris.
- \(\mathrm{CaCO_3}\): manufacture of cement.
(ii)
- \(\mathrm{SO_2}\): concentrated \(\mathrm{H_2SO_4}\).
- \(\mathrm{HCl}\): concentrated \(\mathrm{H_2SO_4}\).
- \(\mathrm{NH_3}\): calcium oxide, \(\mathrm{CaO}\).
(c)(i)
\[\mathrm{C_{(s)}+O_{2(g)}\rightarrow CO_{2(g)}}\]
(ii)
From the equation, 1 mol of carbon produces 1 mol of \(\mathrm{CO_2}\).
\[n(\mathrm C)=\frac{\text{mass}}{\text{molar mass}}=\frac{2.5}{12.0}=0.2083\ \text{mol}\]
\[n(\mathrm{CO_2})=0.2083\ \text{mol}\approx\boxed{0.208\ \text{mol}}\]
(iii)
- Carbon(II) oxide, \(\mathrm{CO}\): reducing agent in the extraction of metals.
- Carbon(IV) oxide, \(\mathrm{CO_2}\): used in fire extinguishers.
(d)
Using 100 g of the material:
| Constituent | Mass (g) | Number of moles | Mole ratio |
|---|---|---|---|
| Fe | 28.1 | \(\frac{28.1}{56.0}=0.502\) | \(\frac{0.502}{0.502}=1\) |
| Cl | 35.7 | \(\frac{35.7}{35.5}=1.006\) | \(\frac{1.006}{0.502}=2\) |
| \(\mathrm{H_2O}\) | 36.2 | \(\frac{36.2}{18.0}=2.011\) | \(\frac{2.011}{0.502}=4\) |
Therefore, the formula of the material is \(\boxed{\mathrm{FeCl_2\cdot4H_2O}}\).
(e)
(i) Mercury, \(\mathrm{Hg}\).
(ii) Bromine, \(\mathrm{Br_2}\).
Answer Details
(a)(i) The reaction is exothermic because \(\Delta H=-13\ \text{kJ mol}^{-1}\). The products are therefore at a lower energy level than the reactants.
(ii)
\[\text{Rate of reaction}=\frac{\text{change in concentration of HI}}{\text{time}}\]
\[=\frac{0.002-0.000}{80}=2.5\times10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}\]
Rate of reaction = \(2.5\times10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}\).
(b)(i)
- \(\mathrm{NaHCO_3}\): manufacture of baking powder.
- \(\mathrm{CaSO_4}\): manufacture of plaster of Paris.
- \(\mathrm{CaCO_3}\): manufacture of cement.
(ii)
- \(\mathrm{SO_2}\): concentrated \(\mathrm{H_2SO_4}\).
- \(\mathrm{HCl}\): concentrated \(\mathrm{H_2SO_4}\).
- \(\mathrm{NH_3}\): calcium oxide, \(\mathrm{CaO}\).
(c)(i)
\[\mathrm{C_{(s)}+O_{2(g)}\rightarrow CO_{2(g)}}\]
(ii)
From the equation, 1 mol of carbon produces 1 mol of \(\mathrm{CO_2}\).
\[n(\mathrm C)=\frac{\text{mass}}{\text{molar mass}}=\frac{2.5}{12.0}=0.2083\ \text{mol}\]
\[n(\mathrm{CO_2})=0.2083\ \text{mol}\approx\boxed{0.208\ \text{mol}}\]
(iii)
- Carbon(II) oxide, \(\mathrm{CO}\): reducing agent in the extraction of metals.
- Carbon(IV) oxide, \(\mathrm{CO_2}\): used in fire extinguishers.
(d)
Using 100 g of the material:
| Constituent | Mass (g) | Number of moles | Mole ratio |
|---|---|---|---|
| Fe | 28.1 | \(\frac{28.1}{56.0}=0.502\) | \(\frac{0.502}{0.502}=1\) |
| Cl | 35.7 | \(\frac{35.7}{35.5}=1.006\) | \(\frac{1.006}{0.502}=2\) |
| \(\mathrm{H_2O}\) | 36.2 | \(\frac{36.2}{18.0}=2.011\) | \(\frac{2.011}{0.502}=4\) |
Therefore, the formula of the material is \(\boxed{\mathrm{FeCl_2\cdot4H_2O}}\).
(e)
(i) Mercury, \(\mathrm{Hg}\).
(ii) Bromine, \(\mathrm{Br_2}\).
