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Tambaya 1 Rahoto
(a) In an experiment, 25.0 cm\(^{3}\) of H\(_{2}\)SO\(_{4}\) completely neutralized 24.0 cm\(^{3}\) of a 0.1 50 mol dm\(^{-3}\) aqueous KOH using a suitable indicator.
(i) Write a balanced chemical equation for the reaction.
(ii) Calculate the concentration of the acid solution.
(b)i. A burning magnesium ribbon was placed in a gas jar containing Carbon (IV) oxide.
- Write an equation for the reaction.
- Explain briefly why the magnesium ribbon burns in carbon(lV) oxide although the gas does not support combustion.
- Calculate the percentage mass of nitrogen in magnesium trioxonitrate (V). |N = 140. O = 16.0. Mg = 24.01]
(c) Consider the following organic compound: CH\(_{3}\)CH\(_{2}\)CH = CHCOOH.
(i) State two chemical reactions which could he sed te identify the compound.
(ii) What would be observed in each of the reactions stated in c(i)
(d) Describe briefly how soap is manufactured using pellets of sodium hydroxide and vegetable oil
(e) Define the term electronegativity
(a)(i) Equation for the neutralization
\[H_2SO_{4(aq)} + 2KOH_{(aq)} \to K_2SO_{4(aq)} + 2H_2O_{(l)}\](a)(ii) Concentration of the acid
Moles of KOH \(= \dfrac{24.0}{1000} \times 0.150 = 3.6 \times 10^{-3}\,mol\).
From the equation, 2 mol KOH react with 1 mol \(H_2SO_4\), so moles of acid \(= \dfrac{3.6 \times 10^{-3}}{2} = 1.8 \times 10^{-3}\,mol\).
\[[H_2SO_4] = \frac{1.8 \times 10^{-3}}{25.0/1000} = 0.072\,mol\,dm^{-3}\](b) Magnesium burning in carbon(IV) oxide
Equation:
\[2Mg_{(s)} + CO_{2(g)} \to 2MgO_{(s)} + C_{(s)}\]Explanation: magnesium is a very reactive metal and a powerful reducing agent. The heat of the burning ribbon is enough to decompose carbon(IV) oxide; magnesium removes the oxygen from the \(CO_2\), reducing it to carbon (seen as black specks) while itself being oxidized to white magnesium oxide. So the ribbon continues to burn even though \(CO_2\) does not normally support combustion.
Percentage mass of nitrogen in magnesium trioxonitrate(V), \(Mg(NO_3)_2\):
\[M = 24 + 2(14 + 48) = 148\]\[\%N = \frac{2 \times 14}{148} \times 100 = 18.92\%\](c) The compound \(CH_3CH_2CH=CHCOOH\)
This compound has both a carbon-carbon double bond (unsaturation) and a carboxylic acid (\(-COOH\)) group.
(d) Manufacture of soap
Vegetable oil is boiled with concentrated sodium hydroxide (caustic soda) solution. The oil is hydrolyzed (saponified) to give soap (the sodium salt of the fatty acid) and glycerol. Common salt (sodium chloride) is then added to the mixture to salt out (precipitate) the soap, which floats to the top; it is filtered off, pressed and moulded.
(e) Electronegativity
Electronegativity is the relative tendency (power) of an atom to attract the shared pair of electrons in a covalent bond towards itself.
Bayanin Amsa
(a)(i) Equation for the neutralization
\[H_2SO_{4(aq)} + 2KOH_{(aq)} \to K_2SO_{4(aq)} + 2H_2O_{(l)}\](a)(ii) Concentration of the acid
Moles of KOH \(= \dfrac{24.0}{1000} \times 0.150 = 3.6 \times 10^{-3}\,mol\).
From the equation, 2 mol KOH react with 1 mol \(H_2SO_4\), so moles of acid \(= \dfrac{3.6 \times 10^{-3}}{2} = 1.8 \times 10^{-3}\,mol\).
\[[H_2SO_4] = \frac{1.8 \times 10^{-3}}{25.0/1000} = 0.072\,mol\,dm^{-3}\](b) Magnesium burning in carbon(IV) oxide
Equation:
\[2Mg_{(s)} + CO_{2(g)} \to 2MgO_{(s)} + C_{(s)}\]Explanation: magnesium is a very reactive metal and a powerful reducing agent. The heat of the burning ribbon is enough to decompose carbon(IV) oxide; magnesium removes the oxygen from the \(CO_2\), reducing it to carbon (seen as black specks) while itself being oxidized to white magnesium oxide. So the ribbon continues to burn even though \(CO_2\) does not normally support combustion.
Percentage mass of nitrogen in magnesium trioxonitrate(V), \(Mg(NO_3)_2\):
\[M = 24 + 2(14 + 48) = 148\]\[\%N = \frac{2 \times 14}{148} \times 100 = 18.92\%\](c) The compound \(CH_3CH_2CH=CHCOOH\)
This compound has both a carbon-carbon double bond (unsaturation) and a carboxylic acid (\(-COOH\)) group.
(d) Manufacture of soap
Vegetable oil is boiled with concentrated sodium hydroxide (caustic soda) solution. The oil is hydrolyzed (saponified) to give soap (the sodium salt of the fatty acid) and glycerol. Common salt (sodium chloride) is then added to the mixture to salt out (precipitate) the soap, which floats to the top; it is filtered off, pressed and moulded.
(e) Electronegativity
Electronegativity is the relative tendency (power) of an atom to attract the shared pair of electrons in a covalent bond towards itself.
Tambaya 2 Rahoto
(a)i. With the aid of an equation, explain briefly why aluminum metal is not affected by air.
(ii) In the extraction of aluminum from bauxite, state the:
- substance used for purifying the ore;
- composition of the mixture electrolyzed.
(b) ZnO is an amphoteric oxide. Write equations to illustrate this statement.
(c)i) List three uses of sodium trioxocarbonate(IV).
(ii) Explain briefly why a solution of trioxonitrate(V) acid turns yellowish on storage for some time.
(ii) Describe briefly how trioxonitrate(V) ions could be tested for in the laboratory.
(d) Write balanced chemical equations for the preparation of hydrogen chloride.
(i) using concentrated H\(_{2}\)SO\(_{4}\):
(ii) by direct combination of its constituent elements.
(iii) State one use of hydrogen chloride.
(a)(i) Why aluminium is not affected by air
When exposed to air, aluminium quickly forms a thin, tough, coherent and impervious layer of aluminium oxide on its surface. This oxide film sticks firmly to the metal and prevents air and moisture from reaching the metal beneath, so the metal is protected from further attack.
\[4Al_{(s)} + 3O_{2(g)} \to 2Al_2O_{3(s)}\](a)(ii) Extraction of aluminium from bauxite
(b) ZnO as an amphoteric oxide
Reacting as a base (with acid):
\[ZnO_{(s)} + 2HCl_{(aq)} \to ZnCl_{2(aq)} + H_2O_{(l)}\]Reacting as an acid (with alkali):
\[ZnO_{(s)} + 2NaOH_{(aq)} \to Na_2ZnO_{2(aq)} + H_2O_{(l)}\](c)(i) Three uses of sodium trioxocarbonate(IV)
Manufacture of glass; manufacture of soap and detergents; softening of hard water (also used in paper manufacture).
(c)(ii) Why trioxonitrate(V) acid turns yellow on storage
On standing, especially in light and warmth, the acid partly decomposes, releasing brown nitrogen(IV) oxide which dissolves back in the acid and gives it a yellow colour:
\[4HNO_{3} \to 4NO_{2} + 2H_2O + O_{2}\](c)(iii) Test for trioxonitrate(V) ions
Brown-ring test: to the solution add freshly prepared iron(II) tetraoxosulphate(VI) solution, then carefully pour concentrated tetraoxosulphate(VI) acid down the side of the tilted test tube so it forms a layer below. A brown ring forms at the junction of the two liquids, confirming the presence of nitrate ions.
(d) Preparation of hydrogen chloride
\[\text{(i) } NaCl_{(s)} + H_2SO_{4(l)} \to NaHSO_{4(s)} + HCl_{(g)}\]\[\text{(ii) } H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}\](iii) One use of hydrogen chloride: it is used to prepare hydrochloric acid and metal chlorides, and industrially in the manufacture of chlorine and in cleaning (pickling) metals.
Bayanin Amsa
(a)(i) Why aluminium is not affected by air
When exposed to air, aluminium quickly forms a thin, tough, coherent and impervious layer of aluminium oxide on its surface. This oxide film sticks firmly to the metal and prevents air and moisture from reaching the metal beneath, so the metal is protected from further attack.
\[4Al_{(s)} + 3O_{2(g)} \to 2Al_2O_{3(s)}\](a)(ii) Extraction of aluminium from bauxite
(b) ZnO as an amphoteric oxide
Reacting as a base (with acid):
\[ZnO_{(s)} + 2HCl_{(aq)} \to ZnCl_{2(aq)} + H_2O_{(l)}\]Reacting as an acid (with alkali):
\[ZnO_{(s)} + 2NaOH_{(aq)} \to Na_2ZnO_{2(aq)} + H_2O_{(l)}\](c)(i) Three uses of sodium trioxocarbonate(IV)
Manufacture of glass; manufacture of soap and detergents; softening of hard water (also used in paper manufacture).
(c)(ii) Why trioxonitrate(V) acid turns yellow on storage
On standing, especially in light and warmth, the acid partly decomposes, releasing brown nitrogen(IV) oxide which dissolves back in the acid and gives it a yellow colour:
\[4HNO_{3} \to 4NO_{2} + 2H_2O + O_{2}\](c)(iii) Test for trioxonitrate(V) ions
Brown-ring test: to the solution add freshly prepared iron(II) tetraoxosulphate(VI) solution, then carefully pour concentrated tetraoxosulphate(VI) acid down the side of the tilted test tube so it forms a layer below. A brown ring forms at the junction of the two liquids, confirming the presence of nitrate ions.
(d) Preparation of hydrogen chloride
\[\text{(i) } NaCl_{(s)} + H_2SO_{4(l)} \to NaHSO_{4(s)} + HCl_{(g)}\]\[\text{(ii) } H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}\](iii) One use of hydrogen chloride: it is used to prepare hydrochloric acid and metal chlorides, and industrially in the manufacture of chlorine and in cleaning (pickling) metals.
Tambaya 3 Rahoto
(a) Distinguish between molecular formula and structural formula
(b) List three factors that determine the ionization energy of an atom.
(c) State the two conditions necessary for the establishment of a chemical equilibrium
(d) Consider the following table
| Element | A | B | C |
| Ionization energy KJ mol\(^{-1}\) | 619 | 518 | 594 |
(d)(i) State which of the elements is the strongest reducing agent.
(ii) Give a reason for the answer stated in (d)(i)
(e) State Graham's law of diffusion
(f) Consider the following salts: \(\mathrm{Mg(NO_3)_2}\), \(\mathrm{CaCO_3}\), \(\mathrm{Na_2SO_4}\). State which of the salts is/are:
(i) readily soluble in water:
(ii) insoluble in water.
(g) Classily each of the following products as addition polymer or condensation polymer:
(i) protein:
(ii) perspex:
(iii) nylon.
(h) Define atomic radius.
(i) Explain briefly why ethanol has a higher boiling point than propane even though they both have comparable molar masses.
(j) State three significance of the pH value in everyday life.
(a) Molecular formula and structural formula
A molecular formula shows the actual number of atoms of each element present in one molecule of a compound (for example ethanol \( C_2H_6O \)). A structural formula shows how those atoms are arranged and joined together by bonds (for example \( CH_3CH_2OH \)).
(b) Three factors that determine ionization energy
(c) Two conditions for chemical equilibrium
(d) Ionization energies: A = 619, B = 518, C = 594 \( \text{kJ mol}^{-1} \).
| Element | A | B | C |
|---|---|---|---|
| Ionization energy / kJ mol\(^{-1}\) | 619 | 518 | 594 |
(i) The strongest reducing agent is B.
(ii) B has the lowest ionization energy, so it loses (donates) its outermost electron most easily; a good reducing agent is one that readily gives up electrons.
(e) Graham's law of diffusion
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or relative molecular mass): \( \text{rate} \propto \dfrac{1}{\sqrt{M}} \).
(f) (i) Readily soluble in water: \( Mg(NO_3)_2 \) and \( Na_2SO_4 \) (all nitrates and all sodium salts are soluble). (ii) Insoluble in water: \( CaCO_3 \).
(g) (i) Protein: condensation polymer. (ii) Perspex: addition polymer. (iii) Nylon: condensation polymer.
(h) Atomic radius
Atomic radius is half the distance between the nuclei of two identical atoms joined by a single covalent bond (half the internuclear distance).
(i) Ethanol versus propane
Ethanol molecules contain an \( -OH \) group and form hydrogen bonds between neighbouring molecules. These hydrogen bonds are much stronger than the weak van der Waals (dispersion) forces between propane molecules, so more energy is needed to separate ethanol molecules, giving it the higher boiling point even though the molar masses are comparable.
(j) Three significances of pH in everyday life
Bayanin Amsa
(a) Molecular formula and structural formula
A molecular formula shows the actual number of atoms of each element present in one molecule of a compound (for example ethanol \( C_2H_6O \)). A structural formula shows how those atoms are arranged and joined together by bonds (for example \( CH_3CH_2OH \)).
(b) Three factors that determine ionization energy
(c) Two conditions for chemical equilibrium
(d) Ionization energies: A = 619, B = 518, C = 594 \( \text{kJ mol}^{-1} \).
| Element | A | B | C |
|---|---|---|---|
| Ionization energy / kJ mol\(^{-1}\) | 619 | 518 | 594 |
(i) The strongest reducing agent is B.
(ii) B has the lowest ionization energy, so it loses (donates) its outermost electron most easily; a good reducing agent is one that readily gives up electrons.
(e) Graham's law of diffusion
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or relative molecular mass): \( \text{rate} \propto \dfrac{1}{\sqrt{M}} \).
(f) (i) Readily soluble in water: \( Mg(NO_3)_2 \) and \( Na_2SO_4 \) (all nitrates and all sodium salts are soluble). (ii) Insoluble in water: \( CaCO_3 \).
(g) (i) Protein: condensation polymer. (ii) Perspex: addition polymer. (iii) Nylon: condensation polymer.
(h) Atomic radius
Atomic radius is half the distance between the nuclei of two identical atoms joined by a single covalent bond (half the internuclear distance).
(i) Ethanol versus propane
Ethanol molecules contain an \( -OH \) group and form hydrogen bonds between neighbouring molecules. These hydrogen bonds are much stronger than the weak van der Waals (dispersion) forces between propane molecules, so more energy is needed to separate ethanol molecules, giving it the higher boiling point even though the molar masses are comparable.
(j) Three significances of pH in everyday life
Tambaya 4 Rahoto
(a) What is the structure of:
(i) graphite:
(ii) diamond
(ii) Explain briefly why diamond is hard and a non-conductor of electricity while graphite is soft and an electrical conductor,
(b)i. State what is achieved at each of the following stages in the purification of town water supply:
- aeration;
- screening
- sedimentation.
(ii) Name two substances responsible for hardness in water.
(iii) State two methods for the removal of hardness in water.
(iv) Give one disadvantage of hard water
(c)i). Describe briefly the extraction of tin from its ore.
(ii) Write a balanced chemical equation for the reaction.
(iii) Write an equation for the reaction of tin with:
- oxygen;
- chlorine.
(a)(i) Structures
(a)(iii) Hardness and conductivity
Diamond is hard because its carbon atoms are held in a rigid three-dimensional network of strong covalent bonds, and it is a non-conductor because all four valence electrons of each carbon are used in bonding, leaving no free electrons to carry charge. Graphite is soft because its layers are held by weak forces and can slide over one another, and it conducts electricity because each carbon uses only three electrons in bonding, leaving one delocalized electron per atom free to move and carry charge.
(b)(i) Water purification stages
(b)(ii) Substances causing hardness
Dissolved calcium and magnesium salts, that is, the hydrogencarbonates and sulphates of calcium and magnesium (\(Ca^{2+}\) and \(Mg^{2+}\) ions).
(b)(iii) Methods of removing hardness
Boiling (removes temporary hardness), adding washing soda (\(Na_2CO_3\)), ion exchange, or distillation.
(b)(iv) One disadvantage of hard water
It wastes soap by forming an insoluble scum, and it deposits scale (fur) in kettles, boilers and pipes.
(c)(i) Extraction of tin
Tin is obtained from its ore cassiterite, \(SnO_2\). The ore is concentrated by washing and froth flotation to remove earthy impurities, then roasted in air to remove sulphur and arsenic. The concentrated oxide is then reduced with carbon (coke) in a reverberatory or blast furnace, and the molten tin is run off and refined.
(c)(ii) Equation for the reduction
\[SnO_{2(s)} + 2C_{(s)} \to Sn_{(l)} + 2CO_{(g)}\](c)(iii) Reactions of tin
\[\text{With oxygen: } Sn_{(s)} + O_{2(g)} \to SnO_{2(s)}\]\[\text{With chlorine: } Sn_{(s)} + 2Cl_{2(g)} \to SnCl_{4(l)}\]Bayanin Amsa
(a)(i) Structures
(a)(iii) Hardness and conductivity
Diamond is hard because its carbon atoms are held in a rigid three-dimensional network of strong covalent bonds, and it is a non-conductor because all four valence electrons of each carbon are used in bonding, leaving no free electrons to carry charge. Graphite is soft because its layers are held by weak forces and can slide over one another, and it conducts electricity because each carbon uses only three electrons in bonding, leaving one delocalized electron per atom free to move and carry charge.
(b)(i) Water purification stages
(b)(ii) Substances causing hardness
Dissolved calcium and magnesium salts, that is, the hydrogencarbonates and sulphates of calcium and magnesium (\(Ca^{2+}\) and \(Mg^{2+}\) ions).
(b)(iii) Methods of removing hardness
Boiling (removes temporary hardness), adding washing soda (\(Na_2CO_3\)), ion exchange, or distillation.
(b)(iv) One disadvantage of hard water
It wastes soap by forming an insoluble scum, and it deposits scale (fur) in kettles, boilers and pipes.
(c)(i) Extraction of tin
Tin is obtained from its ore cassiterite, \(SnO_2\). The ore is concentrated by washing and froth flotation to remove earthy impurities, then roasted in air to remove sulphur and arsenic. The concentrated oxide is then reduced with carbon (coke) in a reverberatory or blast furnace, and the molten tin is run off and refined.
(c)(ii) Equation for the reduction
\[SnO_{2(s)} + 2C_{(s)} \to Sn_{(l)} + 2CO_{(g)}\](c)(iii) Reactions of tin
\[\text{With oxygen: } Sn_{(s)} + O_{2(g)} \to SnO_{2(s)}\]\[\text{With chlorine: } Sn_{(s)} + 2Cl_{2(g)} \to SnCl_{4(l)}\]Tambaya 5 Rahoto
(a)i. State two characteristics of a homologous series.
(ii) Explain briefly why there are differences in the reaction of ethane and ethene.
(b) When crystals of sodium chloride were warmed with concentrated tetraoxosulphate( VI) acid, a gas was evolved.
(i) Name the gas.
(ii) State two physical properties of the gas.
(iii) Write a balanced chemical equation for the reaction.
(c)i. What are hydrocarbons?
(ii) State two natural sources of hydrocarbons.
(iii) A hydrocarbon contains 83% of carbon by mass. Calculate its empirical formula. [H=1.0, C=12.0]
(d) Draw and label a diagram of a set-up that could be used to electroplate a copper ornament with silver
(a)(i) Characteristics of a homologous series
(a)(ii) Ethane is a saturated hydrocarbon containing only carbon-carbon single bonds. It undergoes mainly substitution reactions. Ethene is unsaturated and contains a carbon-carbon double bond, \(C=C\), which is reactive; it therefore readily undergoes addition reactions.
(b)(i) The gas evolved is hydrogen chloride, \(HCl\).
(b)(ii) Hydrogen chloride is a colourless gas with a pungent, choking smell. It is denser than air and is very soluble in water.
(b)(iii)
(c)(i) Hydrocarbons are organic compounds that contain carbon and hydrogen only.
(c)(ii) Natural sources of hydrocarbons include crude oil (petroleum) and natural gas.
(c)(iii) Empirical formula
Assume 100 g of the hydrocarbon.
| Element | Mass (g) | Moles | Mole ratio |
|---|---|---|---|
| C | 83 | \(\frac{83}{12}=6.92\) | \(\frac{6.92}{6.92}=1\) |
| H | \(100-83=17\) | \(\frac{17}{1}=17\) | \(\frac{17}{6.92}=2.46\approx2.5\) |
The simplest ratio is \(\mathrm{C:H}=1:2.5\). Multiplying throughout by 2 gives \(2:5\).
(d) Electroplating a copper ornament with silver
The silver electrode is connected to the positive terminal and is the anode. The copper ornament is connected to the negative terminal and is the cathode. Aqueous silver trioxonitrate(V), \(\mathrm{AgNO_3(aq)}\), is used as the electrolyte. Silver dissolves from the anode and is deposited on the copper ornament.
Bayanin Amsa
(a)(i) Characteristics of a homologous series
(a)(ii) Ethane is a saturated hydrocarbon containing only carbon-carbon single bonds. It undergoes mainly substitution reactions. Ethene is unsaturated and contains a carbon-carbon double bond, \(C=C\), which is reactive; it therefore readily undergoes addition reactions.
(b)(i) The gas evolved is hydrogen chloride, \(HCl\).
(b)(ii) Hydrogen chloride is a colourless gas with a pungent, choking smell. It is denser than air and is very soluble in water.
(b)(iii)
(c)(i) Hydrocarbons are organic compounds that contain carbon and hydrogen only.
(c)(ii) Natural sources of hydrocarbons include crude oil (petroleum) and natural gas.
(c)(iii) Empirical formula
Assume 100 g of the hydrocarbon.
| Element | Mass (g) | Moles | Mole ratio |
|---|---|---|---|
| C | 83 | \(\frac{83}{12}=6.92\) | \(\frac{6.92}{6.92}=1\) |
| H | \(100-83=17\) | \(\frac{17}{1}=17\) | \(\frac{17}{6.92}=2.46\approx2.5\) |
The simplest ratio is \(\mathrm{C:H}=1:2.5\). Multiplying throughout by 2 gives \(2:5\).
(d) Electroplating a copper ornament with silver
The silver electrode is connected to the positive terminal and is the anode. The copper ornament is connected to the negative terminal and is the cathode. Aqueous silver trioxonitrate(V), \(\mathrm{AgNO_3(aq)}\), is used as the electrolyte. Silver dissolves from the anode and is deposited on the copper ornament.
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