NSMQ 2023 SEMI-FINALS CHEMISTRY PAST QUESTIONS AND ANSWERS.

#01

1. What type of reaction takes place between alkenes and bromine molecules? Give the general type of product formed?

Ans:

Type of reaction: Addition reactions.

Product: (1,2-) Dibromoalkane

2. Name the type of reaction and the product formed when sodium hydroxide solution reacts with a primary alkyl halide.

Ans:

Type of reaction: It is a substitution reaction.

Product: An alkanol

3. Give the reaction mechanism or the chemical steps followed when hydrogen bromide gas reacts with an alkene.

Ans:      

In Step 1, the pi-cloud of the alkene attacks/attracts the proton of the HBr to form a carbocation.

In Step 2, the bromide ion reacts with the carbocation to give a bromoalkane.

[In both steps, it is the electron pair that attacks]

#02

Preamble to all schools:

Ammonium dichromate (NH₄)₂Cr₂O₇ decomposes on heating according to the following balanced equation:

(NH₄)₂Cr₂O₇ →   N₂         +              4H₂O     +              Cr₂O₃

Cr = 52.0;            O = 16.0;             N = 14.0;              H = 1.00

1. Calculate the percentage loss in mass when the salt decomposes to give Cr₂O₃.

Ans:

(NH₄)₂Cr₂O₇ →   N₂  +       4H₂O     +  Cr₂O₃

252                                        152

Loss in mass  =   252 – 152 = 100

%Loss in mass    = (100/252) x 100 = 39.7

2. Calculate the mass of the oxide that can be obtained from 378g of the ammonium compound if the yield is 90.0%.

Ans:

(NH₄)₂Cr₂O₇ →   N₂ + 4H₂O + Cr₂O₃.

252                                        152

Mass of the oxide from 378g of the salt  = (378/252) x 152    =      228g

Actual yield         = 0.900 x 228     = 205g

3. Calculate the volume of nitrogen gas at STP that can be generated from 2.016kg of the ammonium compound. The molar volume of an ideal gas at STP is 22.4dm³.

Ans:      

(NH₄)₂Cr₂O₇ →   N₂ + 4H₂O + Cr₂O₃ .

252                                        22.4dm³

Volume of nitrogen at STP = (2016/252) x 22.4    = 1.79 x 10²dm³.

#03

1. Give the units of the rate constant of a second-order reaction for which the concentrations of reactants are given in ppm or parts per million.

Ans:

The general units are conc^-1 s^-1, hence (ppm)^-1s^-1  or million per part second.

2. If it is the intention to prepare metallic sodium by electrolysis, which electrolyte is better and why: concentrated sodium chloride solution or fused sodium chloride?

Ans:

Fused sodium chloride is preferred.

Reason: In an aqueous solution, Na⁺ and H⁺ are available to be reduced at the cathode; H⁺ has the higher electrode/reduction potential, hence H₂ will be formed instead of metallic sodium.

3. 20.0dm³ of nitrogen gas kept at a temperature of 87.0^oC and a pressure of 95.0kPa is allowed to cool to -3.00^oC at a pressure of 100kPa. Calculate the new volume of the gas.

Ans:

Use the Ideal Gas Equation:

P₁V₁/T₁ = P₂V₂/T₂;            V₂ =      V₁(P₁/P₂)(T₂/T₁)

P₁  =  95.0kPa;      V₁ =  20.0dm³; T₁  =  360K; P₂  =  100kPa;    T₂  =  270;   V₂  =  ?

V₂ = (20.0 x 95.0 x 270)/(360 x 100) = 14.25 = 14.3dm³

#04

1. How do the atomic number and mass number change when a radioactive element decays by electron emission?

Ans:

Atomic number increases by 1, mass number remains unchanged.

2. How do the atomic number and mass number change when a radioactive element decays by electron capture?

Ans:

Atomic number decreases by 1, mass number remains unchanged.

3. How do the atomic number and the mass number change when a radioactive nuclide emits a neutron and a gamma ray at the same time?

Ans:

Atomic number remains unchanged, and mass number decreases by 1.

#05

What mass of magnesium Tetraoxosulphate(VI) heptahydrate is needed to prepare a 400 cm³ solution of 0.150 moldm^-3 concentration?

Mg = 24.0;          S = 32.0;              O = 16.0;             H = 1.00

Ans:      

MgSO₄•7H₂O     =  56.0 + 64.0 + 7*18.0 = 120 + 126 = 246

Mass required to prepare 1 dm³ solution of 0.150moldm^-3  =  246 x 0.150 =  36.9g

Therefore, mass needed for 400 cm³ solution of 0.150moldm^-3  =  36.9 x 0.400     =  14.8g

2. Calculate the mass of CuSO₄•5H₂O needed to prepare a 250 cm³ solution which is 0.450 moldm^-3 with respect to Cu²⁺ ions.

Cu = 64.0;           S = 32.0;              O = 16.0;             H = 1.00

Ans:

CuSO₄•5H₂O =   96.0 + 64.0 + 5 x 18.0 =  250

Mass of the compound needed to prepare 1dm³ of 0.450 moldm^-3 Cu²⁺

= 0.450 x 250

= 112.5g

Therefore,  mass needed for 250 cm³  =  112.5/4 = 28.1g

3. Calculate the mass of sodium trioxocarbonate(IV) decahydrate needed to prepare a 500cm³ solution, which is 0.240moldm^-3 with respect to Na⁺ ions.

Na = 23.0;           O = 16.0;             C = 12.0;              H = 1.00

Ans       

Na₂CO₃•10H₂O = 46.0   + 60.0   + 180    = 286.

1.00 mol Na₂CO₃•10H₂O = 2mol Na⁺ ions.

Mass of the compound needed for 1dm³ of 0.120moldm^-3 solution

= 286*0.120

=34.32g

Mass for 500cm³ of 0.240moldm^-3 solution of Na⁺ ions

= 34.32/2

= 17.2g

#06

1. For a substance to be regarded as a suitable catalyst for a reaction, that substance must satisfy some criteria. Give two such criteria.

Ans:       Any two of the following:

• The catalyst must increase the rate of the reaction.
• The catalyst is not consumed by the reaction.
• A small quantity of the catalyst should be able to affect the rate of the reaction
• The catalyst does not change the equilibrium constant for the reaction

2. An element X exists as tetra-atomic molecules X₄. If X has two natural isotopes, how many peaks will be observed in its Mass Spectrum? The spectrum is recorded such that there is no fragmentation.

Ans: 5 peaks.

Assume isotopes X₁, X₂

Molecules that are possible are:

X₁-X₁-X₁-X₁;  X₁-X₁-X₁-X₂; X₁-X₁-X₂-X₂;   X₁-X₂-X₂-X₂;  X₂-X₂-X₂-X₂

3. What is the study of metallurgy about?

Ans:

It is the scientific study of the production of metals from their ores (and the making of alloys).

#07

Preamble to all schools:

Use the Kinetic Theory of Gases to explain the following experimental observations:

1. For a given gas at constant temperature and volume, the pressure increases when the molar quantity of the gas increases.

Ans:

The Theory predicts that the pressure of a gas results from collisions between the gas particles and the walls of the container. When the number of particles increases, the number of collisions per unit area increases even at constant temperature.

2. For a given amount of gas at a constant volume, the pressure of the gas increases with temperature.

Ans:

A postulate of the Theory states that the average kinetic energy of a gas particle depends only on the temperature of the gas; hence, the average kinetic energy increases as the gas gets warmer. The higher average kinetic energy means gaseous particles move faster and collide with the wall of their container with greater force, and more frequently, hence an increase in pressure.

3. For a given quantity of a gas, at constant temperature, the pressure is inversely proportional to the volume.

Ans:      

As the volume decreases at constant temperature, the distances the gaseous molecules have to travel before colliding with the walls of the container decrease, hence the frequency of collisions increases, leading to an increase in pressure.

#08

1. If phosphorus is the fifth member of the third Period of the Periodic Table, what is its atomic number?

Ans: 15 [2 + 8 + 5]

2. If calcium, the third member of Group II of the Periodic Table, has the atomic number 20, what is the atomic number of barium, the fifth member of the same Group?

Ans: 56 [20 + 18 + 18]

3. If krypton, the last member of Period 4, has the atomic number 36, what is the atomic number of zinc, a member of the same Period?

Ans: 30 [36 – 6; Ga, Ge, As, Se, Br, Kr]

#09

Preamble to all schools.

Gaseous butanone will burn in oxygen to give carbon(IV) oxide and water vapour.

Consider the following bond energies, all in kJmol^-1.

C – C 347; C – H 413; C – O 358;  C = O 805; O – H 464; O₂ 494

1. Give a balanced equation for the combustion of one mole of 2-butanone and give the type and respective number of bonds to be broken.

Ans:      

CH₃COCH₂CH_3(g) + 11/2O_2(g) →     4CO_2(g)  + 4H₂O_(g).

Bonds to be broken  = 3C – C  + 8 C -H  + 1 C = O + 5.5 O₂.

2. Calculate the enthalpy change for the formation of the necessary bonds in the combustion of gaseous 2-butanone.

Ans:      

Bonds to be formed =     8 C = O  + 8 O -H

Enthalpy change =  -8 x 805  + (-8 x 464)

= -10,152kJ or -10,200 or -1.02*10⁴kJ

3. Calculate the energy required to break all the necessary bonds in the combustion of 2-butanone.

Ans:

Bonds to be broken = 3 C – C +    8 C -H + 1 C = O + 5.50 O₂.

Energy required =  3 x 347  +  8 x 413  +  805  +  5.50 x 494

= +7867kJ  or  +7,870 or  +7.87 x 10³ Kj

#10

1. A 2.00dm³ flask is filled with argon gas at 27.0 °C until the pressure is 70.0 kPa. Calculate the total pressure when 6.40 g of O₂ gas at 27.0 °C is added to the flask. Ideal Gas constant R is 8.31 JK^-1mol^-1. O = 16.0

Ans:

Moles of O₂ in 6.40g       = 6.40/32.0 =     0.200

Pressure of 0.200 mol of O₂;

P_O2 =      (nRT)/V

P_O2 = (0.200 *8.31 *300)/2.0       = 8.31*30.0        = 249.3 kPa

Total pressure    = (70.0  +  249.3) kPa      = 319 kPa

2. State the hybridization of the carbon orbitals in the ion HCO₃^–.

Ans:  sp².

3. Which of the following elements has the lowest first ionisation energy: chlorine, phosphorus, and sulphur?

Ans: Sulphur (P   1012;  S   1000;  Cl   1251kJmol-1)

#11

Preamble to all schools

Values of the first ionisation energies for the first 20 elements (hydrogen to calcium) show several clear patterns. State any one of them and give reasons for that pattern.

Ans:

• The first ionisation energy (IE) generally increases appreciably as we go across a Period.

Reason:  Across a Period, the effective nuclear charge increases, and the atomic radius decreases. Hence, it becomes more difficult to remove an electron from the valence shell.

• There is a gradual decrease in the first ionisation energy as we go down a Group.

Reason:  Down a Group, the atomic radius increases, hence the attractive force of the nucleus on the outermost electrons decreases. It becomes easier to remove an electron from the valence shell.

• There are minor exceptions in a Period. e.g. Be, B or  N, O  or Mg, Al

Reason: The Relative Stability of the electron configuration becomes significant in some instances. Removing an electron to give a half- or fully-filled sub-shell, like B or O, respectively, is more favourable despite a higher effective nuclear charge and smaller atomic radius than removing an electron from a half- or fully-filled sub-shell-like N or Be.

• There is a big drop in the first ionisation (IE) as we go from the end of one period to the beginning of the next period

Reason:  The end of a Period is occupied by an inert gas. Its atoms have filled shells and hence possess a very stable electronic configuration and high ionisation potential. The element next after an inert gas is an element that begins a new Period and will have a valence shell of ns¹. Losing that electron will be thermodynamically very favourable.

#12

1. Give the systematic names of the compounds that will be obtained by acid hydrolysis of N-methylpropanamide.

Ans:

• Propanoic acid and
• Methylamine or methanamine

2. Name the ester that will be formed when pentanoic acid is made to react with 1-butanol.

Ans:

• Butyl pentanoate

3. Name the major product formed when 1-pentene undergoes an addition reaction under appropriate conditions with a molecule of water.

Ans:

• 2-Pentanol

#13

1. Calculate the pH of a 0.100 moldm³ solution of chloroethanoic acid whose pK_a is 2.86.

Ans:

pH =       ½ pK_a     –              ½ logC_a

= 2.86/2 + 0.500 = 1.43 + 0.500 = 1.93

2. Calculate the pH of a 0.100 moldm³ solution of ethanolamine, whose pK_b is 4.50.

Ans:      

pOH =    ½ pK_b     –   ½ logC_b

= 4.50/2 + 0.5    = 2.25   + 0.500 = 2.75

pH = 14.0            – pOH    = 14.0   – 2.75    = 11.3.

3. Oxoiodate (I) acid, HOI, is a weak inorganic acid. If it’s a 0.100 moldm^-3 solution has a pH of 5.82, find the pK_a of the acid.

Ans:

pH =       ½ pK_a         –          ½ logC_a

Hence   5.82       = ½ pK_a    +          0.500

pKa =     2*5.32  = 10.6

#14

1. Name the elements in Period 3 that form pure ionic hydrides.

Ans:      

Sodium and magnesium.

2. Give the hybridisation of the bonding orbitals of beryllium in its hydride.

Ans: sp

3. Calcium ethanedioate has a solubility product of 4.00 *10^-10 at about 30 ⁰C. Calculate the solubility of this salt in water and 0.100 moldm^-3 Ca²⁺ solution. Account for the difference, if any.

Ans:

In water Ksp of CaC₂O₄ = 4.00*10^-10;        Solubility = √(4.00*10^-10)

Hence    [Ca²⁺]     or [C₂O₄^2-] = 2.00 *10^-5 moldm^-3

In 0.100 moldm^-3 Ca²⁺:

Ksp =      [0.100][C₂O₄^2-]  = 4.00*10^-10

[C₂O₄^2-] = 4.00 *10^-9 moldm^-3

Solubility in Ca²⁺ solution is lower, due to the common ion effect.

#15

1. Extraction of gold involves two general processes. What are these?

Ans:      

1. i) Concentration and ii) Purification.
2. To concentrate gold from its ore, the ore in some cases is roasted in kilns. Explain how this leads to the concentration of gold.

Ans:      

For ores obtained from deep mines (in Obuasi, Ghana), the ores contain arsenic sulphide, and roasting expels the arsenic and sulphur as gaseous oxides.

2. Explain which process in the extraction of aluminium from bauxite may be regarded as the concentration of the metal.

Ans:

The raw bauxite is treated with concentrated NaOH solution to remove sand and other impurities.

#16

1. Give the products of decomposition when ammonium trioxonitrate(V) is heated.

Ans:      

Nitrogen (I) oxide and water/steam

[NH₄NO₃   →       N₂O        +     2H₂O]

2. Give the products of decomposition when potassium trioxonitrate(V) is heated.

Ans:      

Potassium dioxonitrate (III) and oxygen gas

[2KNO₃                 →           2KNO₂                  +              O₂]

3. Give the products of decomposition when barium trioxonitrate(V) is heated.

Ans:      

Barium oxide, nitrogen(IV) oxide, and oxygen gas.

[2Ba(NO₃)₂         →           2BaO     + 4NO₂ + O₂]

#17

1. Radium–226, atomic number 88, undergoes alpha emission to give a new nuclide. Give a balanced equation for this decay.

Ans:

₈₈²²⁶Ra  →           ₈₆²²²X     +              ₂⁴α

2. Radium–226, atomic number 88, can undergo an alternative decay instead of alpha emission, yielding the same daughter nuclide but different emissions. State the emissions and give a balanced equation for this decay.

Ans:      

Other decay: emission of two protons and two neutrons

₈₈²²⁶Ra                   →           ₈₆²²²X     +              2₁¹p       +              2₀¹n

3. Plutonium–239, atomic number 94, undergoes radioactive decay accompanied by a type of emission that can be used as a source of power for a heart pacemaker. If the new nuclide is uranium–235, atomic number 92, give a balanced equation for the decay and state the type of emission which can serve as the source of energy.

Ans :

₉₄²³⁹Pu  →           ₉₂²³⁵U                    +              ₂⁴α

Hence, source of energy                =              α – emission.

#18

1. Chlorofluoromethanes were used as refrigerants and spray-can propellants, but they have been banned. Why?

Ans:      

Chlorofluoromethanes decompose to give chlorine atoms, which catalyse the decomposition of ozone in the stratosphere.

2. Hydrogen peroxide concentration of 6.00% w/v is a good antiseptic. If you buy a bottle of hydrogen peroxide labelled 10.0% w/v, how would you prepare 500 cm³ of 6.00% w/v H₂O₂ from that?

Ans:      

Dilution:   10.0% w/v to 6.00% w/v;         Dilution factor    3 in 5

Measure 300 cm³ of the 10.0% w/v and make it up to 500 cm³ with clean water.

3. In a solution of butanal in tetrachloromethane, what will be the attractive forces between butanal and tetrachloromethane molecules?

Ans:

Dipole – Induced dipole forces

#19

1. The Group I elements are soft, metallic solids with low melting points. What accounts for this physical nature?

Ans:

They are made up of large atoms, which result in weak metal bonds.

2. The Group I elements are the most reactive metallic elements. What could be the reason for this?

Ans:

They have low first ionisation energies, thus losing the ns¹ electron readily to form +1 cations.

3. Use the reactions with water to illustrate the reactivity of Group I metals down the Group.

Ans:      

Reactivity increases down the Group. They all react with water. Lithium reacts with water gently but readily. Sodium and Potassium react vigorously. Reactions with Rubidium and Caesium are explosive/violent.

#20

Preamble to all schools:

Each school will be presented with two named organic compounds. i) Give the molecular formula of each and hence state if they are isomers.  ii) If they are isomers, determine the type of isomers that they are.

1. Cyclopentane and 2-pentene.

Ans:

They both have the same formula, C₅H_10, hence they are isomers. They are structural isomers.

2. Cyclohexene and 1-hexyne

Ans:

They both have the formula C₆H_10, hence they are isomers. They are functional Group isomers.

3. d-2-Butanol and l-2-butanol

Ans:      

They both have the formula C₄H₁₀O, hence they are isomers. They are stereoisomers or enantiomers.

#21

Preamble to all schools:

The IO₃^– ion reacts with the iodide ion in an acidic medium according to the following equation:

IO₃^– +      5I^– + 6H⁺ → 3I₂   + 3H₂O  (1)

The iodine liberated can be titrated against Na₂S₂O₃ solution using starch as an indicator. The equation for that reaction is as follows:

I₂ + 2S₂O₃^2-          →           2I^– +        S₄O₆^2-     (2)

1. To 20.0 cm³ of KIO₃ solution, excess KI solution and dilute H₂SO₄ are added, and the liberated iodine is titrated against 0.400 moldm^-3 solution of Na₂S₂O₃. If the titre is 24.0 cm³, calculate the concentration in moldm^-3 of the KIO₃ solution.

Ans:

From equations (1) and (2);  n (IO₃^–)/n (S₂O₃^2-)     = 1/6

(20.0 *M)/(0.400 *24.0) = 1/6;

M =        (0.400*24.0)/(6*20.0)

Concentration of KIO₃ solution, M =          0.0800moldm^-3

2. 0.00064 mol of KIO₃ is dissolved in enough water, excess dilute H₂SO₄ and KI solution are added, and the iodine liberated is titrated against 0.200 moldm^-3 solution of Na₂S₂O₃. Calculate the expected titre.

Ans:      

Let Vcm³ be the titre;      n (KIO₃)/n (Na₂S₂O₃) =    1/6

Millimoles of Na₂S₂O₃     = (0.200*V);       Millimoles of KIO₃ =         0.640

0.640/(0.200*V) =           1/6;       V =          (6 *0.640)/0.200

Titre V   =              19.2 cm³

3. Calculate the volume of 0.500 moldm^-3 H₂SO₄ that has to be added to a titration in which 0.000450 mol of KIO₃ is used. Note that in practice, the amount of acid has to be ten times in excess.

Ans:      

Let the exact volume of H⁺ be Vcm³.

From the H₂SO₄, the H⁺ concentration =  1.00 moldm^-3

mmol(IO₃^–)/mmol H⁺       = 1/6;    (0.450)/(V *1.00) =          1/6

Volume V of H⁺ =              (6 *0.450)/1      =              2.70 cm³

The exact volume of H₂SO₄, =      1.35 cm³;

Ten times in excess =       13.5 cm³

#22

1. State Avogadro’s Law

Ans:      

Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules.

2. NO₂ dimerises in a reversible process to give N₂O₄ according to the following equation:

2NO₂                     ⇌                            N₂O₄

In one such reaction, a 1.00dm³ flask was charged with 2 moles of NO₂ and heated to 150 ^oC. At equilibrium, it was found that 0.400 mol of N₂O₄ was formed. Find the equilibrium constant.

Ans:

2NO₂                     →                           N₂O₄

Initial:    2.00 mol                              0

Equil:     2 – 0.800                             0.400

K_C =        (0.400/(1.20)²)  = 0.400/1.44 =  0.278 or 2.78*10^-1.

3. Give the oxidation state of chromium in the chromium complex cation [Cr(NH₃)₅NO₂]²⁺.

Ans:       +3

#23

1. Give the relative positions of the slag and the molten metallic iron in the Blast Furnace during the extraction of iron, and explain why those relative positions.

Ans:      

The slag floats on top of the iron at the bottom of the furnace.

Explanation: The slag is less dense than the molten iron.

2. Biotechnology may be used to extract gold from its ore. Explain how this is done?

Ans:

Some microbes are introduced into a suspension of the powdered ore in water.

The microbes feed on the impurities and free the gold.

3. Bauxite usually has sand and iron(III) oxide as impurities. Explain how aluminium oxide is separated from these impurities.

Ans:      

The ore is treated with a concentrated solution of NaOH. Aluminium oxide and silica (or silicon dioxide) dissolve in the concentrated solution of NaOH, while iron (III) oxide remains undissolved.  Aluminium hydroxide Al(OH)₃ precipitates when the solution is seeded.

#24

1. Name the reagent that can be used to convert ethylbenzene into benzoic acid.

Ans:

Hot, acidified KMnO₄ solution.

2. What reagents are needed for the iodoform test?

Ans:

KOH solution and iodine

3. Name the product formed when 1-butene is treated with dilute, neutral KMnO₄ solution.

Ans:

1,2-butandiol      or            butan-1,2-diol

#25

Preamble to all schools:

Ammonia gas burns in pure oxygen gas to nitrogen gas and steam. The balanced equation for the reaction is as follows: 4NH_3(g)  +              3O_2(g)     → 2N_2(g)    +         6H₂O_(g).

Also consider the following bond energies, all in kJmol^-1: N – H 386; O – H 459; N ≡ N 942;                         N = O   607; O₂                494.

1. Calculate the energy required to break all the necessary bonds in the combustion of ammonia gas in pure oxygen

Ans:      

4NH_3(g)  + 3O_2(g) →           2N_2(g)     +              6H₂O_(g).

Bonds broken     =              12 N – H  +          3O₂

Energy required = 12*386  +  3*494

= 6,114kJ or 6.11*10³kJ

2. If the energy required to break all the bonds in the combustion is 6,114kJ, calculate the enthalpy change for the reaction.

Ans:      

4NH_3(g)  + 3O_2(g) →           2N_2(g)     +              6H₂O_(g).

Bonds to be formed =     2N ≡ N  + 12O – H.

Energy given out = -2*942            +   -12*459         = -7,392kJ

Enthalpy of reaction =     -7,392 + 6,114kJ

= -1,278kJ or -1.28*10³kJ

3. In the presence of a platinum catalyst, ammonia burns in oxygen to give nitrogen(II) oxide and steam and the balanced equation is as follows: 4NH_3(g) + 5O_2(g)     →           4NO_(g)   +              6H₂O_(g). If the energy required to break all the necessary bonds is 7,102kJ and assuming that the NO molecule contains a nitrogen–oxygen double bond, calculate the enthalpy of reaction.

Ans:      

4NH_3(g)  + 5O_2(g) →           4NO_(g)   +              6H₂O_(g)

Bonds to be formed =     4N = O  +              12O – H.

Energy given out  = -4*607 + -12*459     = -7,936kJ

Enthalpy of reaction        = 7,102 – 7,936

= -834kJ

#26

1. A metal alloy made from aluminium (Al), magnesium (Mg), copper (Cu) and steel has the composition:

Al            18.0%;  Mg         12.0%;  Cu           25.0%. The rest is steel, which is 98.5% iron and 1.50% carbon. Calculate the mass in grams of carbon in a 2.00 kg alloy.

Ans:

% steel  = 100 – (18.0 + 12.0 + 25.0)        = 45.0

Therefore, in 2.00 kg alloy, the mass of steel         = 450*2                = 900g

Mass of carbon = (1.5/100)*900                               = 13.5 g

2. Both CaO and anhydrous CaCl₂ can be used to dry wet gases. Which solid would be suitable for CO₂ gas?

Ans:

CaCl₂

3. Whereas the water molecules in the salt ZnSO₄•7H₂O are not taken into consideration when determining the oxidation number of zinc, the water molecules in the complex ion [V(H₂O)₆]³⁺ have to be considered when determining the oxidation state of vanadium. Why?

Ans:      

The water molecules in the salt are not directly bonded to zinc, whereas those in the vanadium complex ion are bonded to vanadium.

#27

1. State the Aufbau Principle (or Building up Principle).

Ans:      

Electrons always occupy the lowest empty energy level.

2. State Pauli’s Exclusion Principle.

Ans:      

No two electrons in an atom can have exactly the same energy.

3. State Hund’s Rule

Ans:      

When electrons fill a subshell, every orbital in the subshell is occupied by a single electron before any orbital is doubly occupied (and all electrons in singly occupied orbitals have their spins in the same direction).

#28

1. Extraction of metals goes through three stages or processes. Give the three processes.

Ans:

Concentration of the ore/metal/metal compound.

Chemical reduction of the ore/metal compound.

Purification of the metal.

2. Pick the pairs whose solutions will act as a buffer.

H₃PO₄,   NaHCO₃,   HCl,   NH₃,   HPO₄^2-,   H₂PO₄^–,   NaOH,   NH₄Cl ,

Ans:

1. i) H₃PO₄/ H₂PO₄^–
2. ii) H₂PO₄^– / HPO₄^2-

iii) NH₃/ NH₄Cl.

3. Explain why SO₂ gas is not the anhydride of H₂SO₄ acid.

Ans:      

An anhydride of a substance reacts with water without going through any redox reaction. The sulphur in SO₂ and H₂SO₄ is in different oxidation states, so conversion of the gas to the acid will involve a redox reaction.

#29

1. Calculate the concentration of an Na₂CO₃ solution if 20.0 cm³ of it requires a titre of 24.0 cm³ of 0.0950 moldm^-3 of an HCl solution in a titration using methyl orange as indicator.

Ans:      

2HCl       +  Na₂CO₃            →           2NaCl        +         CO₂        +    H₂O

24.0, 0.0950      20.0, M

(24.0*0.0950)/(20.0*M)               = 2/1

Concentration of Na₂CO₃ solution, M =  (24.0*0.0950)/(20.0*2)  = 0.0570 moldm^-3.

2. Iron (II) reacts with an acidified solution of MnO₄^– in the ratio 5:1. Calculate the concentration of a solution of Fe²⁺ ions if 20.0cm³ of it requires 15.0 cm³ of 0.120 moldm^-3 of acidified solution of MnO₄^– for complete reaction.

Ans:      

MnO₄^–   + 5 Fe²⁺ + 8 H⁺     →           products

15.0, 0.120                         20.0, M

(15.0*0.120)/(20.0*M)  =  1/5

Concentration of Fe²⁺ , M = (5*15.0*0.120)/20.0                = 0.450 moldm^-3

3. Iodine reacts with the S₂O₃^2- ion in a 1:2 ratio. If 20.0 cm³ of an iodine solution of unknown concentration reacted completely with 18.0 cm³ of 0.0640 moldm^-3 of S₂O₃^2- solution, then what is the concentration of the iodine solution?

Ans:

I₂ + 2 S₂O₃^2-         →           product

20.0, M                18.0, 0.0640

(18.0*0.064)/(20.0*M)                  = 2/1

Concentration of iodine, M = (18.0*0.0640)/(2*20.0)       = 0.0288 moldm^-3

#30

1. Two half–cells, M⁺/M and Q⁺/Q, where M and Q are metals, have the electrode potentials 1.20 and -0.850 volts, respectively. Which of the two metals can react with dilute mineral acid to release hydrogen gas? Give your reason.

Ans:

2. Reason: The potential for the reaction Q →   Q⁺  +  e    is positive, whereas the potential for a similar reaction of M is negative. The one with the positive oxidation potential can oxidise H⁺ to H₂.
3. Give the main difference in the definitions of an acid according to the Arrhenius Theory and according to the Bronsted–Lowry Theory.

Ans:

The Arrhenius concept of acid is limited to a substance that increases H⁺ concentration in water but the Bronsted–Lowry concept makes any proton donor an acid irrespective of the medium.

3. Butane isomerises to 2-methylpropane or isobutane in an equilibrium process. If a 1.00dm³ flask is charged with 2.00 moles of butane at 30 °C and the gas allowed to come to equilibrium, calculate the equilibrium concentration of butane and isobutane at 30 °C given that the equilibrium constant, K_c is 2.50.

Ans:

Butane                  ⇌            isobutane            K_c            =              2.50

Initial:                    2.00 moldm^-3                                     0.00 moldm^-3

At eq.:                   2.00 – x moldm^-3                              x moldm^-3

K_c            =              x/(2.00 – x) = 2.50;           x =5.00 – 2.50x

3.50x     = 5.00;                                                                  x              =              1.43.

Hence concentration of butane, 2.00-x   = 0.570 moldm^-3

Concentration of isobutane, x      = 1.43moldm^-3.

#31

Preamble to all schools.

Each school will be presented with an incomplete statement. You are to complete the statement.

1. In the Periodic Table, sodium is to phosphorus as potassium is to

Ans:       Arsenic

2. In the Periodic Table, titanium is to chromium as iron is to

Ans:       Nickel

3. In the Periodic Table, lithium is to magnesium as boron is to

Ans:       Silicon.

#32

1. What is the percentage by mass of oxygen in the Earth’s crust?

Ans:       47.0%   (Accept ±1)

2. What is the percentage by volume of nitrogen in the Earth’s atmosphere?

Ans:       78.1%   (Accept ± 1)

3. What is the percentage by volume of argon in the Earth’s atmosphere?

Ans:       0.93%   (Accept ± 0.05)

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