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(a) Conditions necessary for: Constructive interference of waves i. lt must involve at least two waves. ii. Waves must vibrate in phase. iii. The distance between the sources of the wave must be of the order of the wavelength. Total internal reflection i. The waves must travel from a denser medium to a less dense medium. ii. The angle of incidence (in the denser medium) must be greater than the critical angle. iii. It must involve two media of different optical densities. Production of beats 1. The amplitude of the two interfering waves should be the same. ii. The difference between the frequencies of interfering waves should be small. iii. The two sources of waves must be sounded together. (b) Calculation of velocity of sound in air V = f\(\lambda\) V = 2f(\(l_2\) - \(l_1\)) = 2 x 256(100 - 35) = 33280 cm s\(^{-1}\) = 332.80 m s\(^{-1}\) (c)(i) Classification of musical instruments i. Wind ii. Stringed iii. Percussion (ii) Examples of classified instruments saxophone I. Wind: flute, trumpet, pipe, organ, iii. Stringed: piano, guitar, violin iii. Percussion: piano, drum, bell, gong. xylophone (d) Calculation of critical angle \(\eta\) = \(\frac{1}{\text {Sin C}}\) 1.5 = \(\frac{1}{\text {Sin C}}\) Sin C = 0.667 C = 41.80º (e) Calculation of temperature \(\frac{V_{1}}{\sqrt T_{1}} = \frac{V_{2}}{\sqrt T_{2}}\) \(\frac{240} {\sqrt 375} = \frac {310} {\sqrt T_{2}}\) 240 \(\sqrt T_{2}\) = 310 x 19.36 \(\sqrt T_{2}\) = 625 K = 352 ºC

(a) Factors affecting the rate i. Surface area of the liquid (exposed surface area) ii. Temperature of the liquid iii. Wind/ drought. iv. Humidity/ amount of water vapour v. Nature of the liquid vi. Air pressure (b) Explanation of latent heat; The energy needed to change the state of a substance without a change in temperature. (c)(i) Water in a clay pot cooler than in a rubber container; A clay pot has pores which allows evaporation to occur while a plastic container has no pores. (ii) Pressure cooker cooks faster than ordinary pot; Pressure cooker raises the pressure within it, which raises the temperature to/above the boiling point thus cooks faster. (d)(i) Calculation of latent heat of fusion of the body Ivt = mL mL = 12 x 40 x 1400 = 6.72 x 10\^{5}\)j (ii) Calculation of specific heat capacity of the body C = \(\frac{Ivt}{m \bigtriangleup \theta}\) = \(\frac{12 \times 40 \times 72}{1.5 \times 60}\) = 384 jkg\(^{-1}\) K\(^{-1}\) (e) difference between evaporation and boiling Evaporation Boiling It takes place at the surface of the liquid It occurs throughout the entire volume of the liquid It takes place across all temperatures It takes place at specific temperatures It produces a cooling effect It produces no cooling effect It is wind aided It is not wind aided

(a)() Meaning of artificial radioactivity; The process by which a stable nucleus is bombarded with a neutron to make it unstable and so disintegrates/decays with the emission of particles/radiation and. energy. Emission Nature Charge Ionizing Beta (\(\beta\)) High speed electron Negative Moderately ionizing Gamma (\(\gamma\)) Electro-magnetic radiation Neutral Negligible ionizing ability Alpha particles Helium nucleus Positive Highly ionizing (b) Calculation of kinetic energy K.E = eV = 1.6 x 10\(^{-19}\) x 30 x 10\(^{3}\) = 4.8 \times 10\(^{-15}\)J (c)(i) graph of K.E versus f i. Both axes correctly distinguished ii. Reasonable scale on both axes (ii) (I) Planck's constant. h = slope of the graph Slope i. \(\bigtriangleup\)K.E and \(\bigtriangleup\)f correctly determined ii. \(\frac{\bigtriangleup K.E}{\bigtriangleup f}\) corectly determined and recorded within range. (II) Threshold frequency, \(f_0\) . The intercept on the horizontal or frequency axis. (III) Threshold frequency, \(W_o\) Negative intercept on K.E axis.

(a) OBSERVATIONS (i) Value of \(M_0\) correctly determined and recorded to at least 1 d.p in grammes (ii) Value of m correctly determined and recorded to at least 1 d.p in grammes (iii) Value of d correctly read and recorded to at least 1 d.p in cm (iv) Five values of M correctly recorded in grammes (v) Five values of Z = AK correctly read and recorded to at least1 d.p in cm and in trend. Trend: As M increases Z decreases. (vi) Five values of y = (Z - 5) correctly evaluated and recorded. (vii) Five values of x = (d - Z) correctly evaluated and recorded (vii) Five values of v = \(\frac{x}{y}\) correctly evaluated and recorded to at least 3 d.p (ix) Composite table showing at least M, Z,y, x and v. (b)(i) Condition for a body to be in stable - The algebraic sum of all external forces on the object is zero OR - The algebraic sum of all external torques from the external forces acting on the object is zero. OR - The sum of clockwise moments about a point in the object is equal to the sum of anticlockwise moment about the same point. OR - The algebraic sum of all moments about a point on the object is zero. (ii) If the distance is y, 100y = 50(6 - y) 100y = 300 - 50y 150y = 300 y = 2m

(a) OBSERVATIONS (i) Value of D correctly measure and recorded to at least 2 d.p in cm (ii) Value of A = 0.25\(\pi\)D\(^{2}\) correctly evaluated (iii) Value of \(y_0\) correctly read and recorded to at least 1 d.p in cm. (iv) Five values of m correctly recorded in grammes. (v) Five values of y correctly read and recorded to at least 1 d.p in cm and in trend. (vi) Five values of h = (y - yo) correctly evaluated. (vii) Five values of log h correctly evaluated to at least 3 d.p (viii) Five values of log m correctly evaluated to 3d. (ix) Composite table showing at least m, y, h, log h and log m. (b)(i) The Law of Flotation; A floating body displaces its own weight of fluid in which it floats. (ii) Weight of rod = 0.63 x 0.1 x 720 x 10 = 453.6 Let y be the depth of immersion \(\therefore\) Weight of water displaced = (0.1y) x 1.26 x 1000 x 10 1260y = 453.6 y=453.6=9.36 = 1260y = 453.6 y = \(\frac{453.6}{1260}\) = 9.36

(a) OBSERVATIONS (i) Value of E correctly read and recorded to at least 1 d.p in volts. (ii) Value of i correctly read and recorded to at least 1 d.pin amperes. (iii) Five values of d correctly determined and recorded to at least 1 d.p in cm. (iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend. Trend; As d increases, I decreases (v) Five values of I\(^{1}\) correctly evaluated to at least 3 s.f. (vi) Composite table showing at least d, I, and I\(^{-1}\). (ii) P = IV Power P = \(\frac{240 \times 0.75}{100}\) 0.18kw Energy consumed in 10hr = 0.18 x 10 = 1.8kWh Cost of energy = 1.8 x 0.5 = $0.9 OR Cost of energy = \(\frac{\text {P x cost x time}}{100}\) \(\frac{0.7 \times 240 \times 10}{1000}\) = $0.9

(a) OBSERVATIONS (i) Value of E correctly read and recorded to at least 1 d.p in volts. (ii) Value of i correctly read and recorded to at least 1 d.pin amperes. (iii) Five values of d correctly determined and recorded to at least 1 d.p in cm. (iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend. Trend; As d increases, I decreases (v) Five values of I\(^{1}\) correctly evaluated to at least 3 s.f. (vi) Composite table showing at least d, I, and I\(^{-1}\). (ii) P = IV Power P = \(\frac{240 \times 0.75}{100}\) 0.18kw Energy consumed in 10hr = 0.18 x 10 = 1.8kWh Cost of energy = 1.8 x 0.5 = $0.9 OR Cost of energy = \(\frac{\text {P x cost x time}}{100}\) \(\frac{0.7 \times 240 \times 10}{1000}\) = $0.9

(a)(i) Is the energy required to remove one mole of an electron from a gaseous atom. OR Is the minimum energy required to convert one mole of a gas of atom into one mole of a gaseous plus 1 ions (ii) \(_{4}\)Be: Is\(^{2}\) 2s\(^{2}\) \(_{5}\)B :Is\(^{2}\) 2s\(^{2}\) 2p\(^{1}\) The electron to be removed from Be is in the 2s orbital which is closer to the nucleus, nuclear attraction is greater hence first ionization energy is greater but the electron to be removed from B is in the 2p orbital which is farther away from the nucleus, hence nuclear attraction is weaker hence first ionization energy is smaller. OR 2p has less energy than 2s So it is easier to remove an electron from 2p as the nuclear charge is weaker in 2p because 2p is further away from the nucleus. Hence, first ionization energy is small. (b) Q - It Q = 0.12 x 500 = 60 C 96500 = 1F = 60 C liberates 0.015g \(\therefore\) 96500 will liberate \(\frac{96500}{60}\) x 0.015 = 24 g n((\(T_1\)) = \(\frac{m}{M}\) = \(\frac{24}{48}\) = 0.5mol 1F = 0.5mol 2F = 1mol hence charge = +2 Alternative A Q = lt = 0.12 x 500 = 60 C 0.015 g of Ti is deposited by 60 C . \(\therefore\) 48 g will be deposited by = \(\frac{60}{0.015}\) x 45 = 1920000 96500 C\(\equiv\) i mole of electron \(\therefore\) 192000 C \(\frac{1}{96500}\) x 192 = 1.989 \(\approx\) 2.0 moles of electron \(\therefore\) charge on Ti ion is +2 Alternate B \(\frac{m}{M} = \frac{It}{nF}\) n = \(\frac{Mlt}{mF}\) = \(\frac{48 \times 0.12 \times 500}{0.015 \times 96500}\) = 1.989 \(\equiv\) 2 \(\therefore\) Change on Ti ion is +2 (c)(i) Alumina (A\(l_2\)\(_{3}\)) mixed with molten cryolite (N\(a_3\)AI\(F_6\)) (i) - abundant deposits of bauxite - cheap source of electricity (d)(i) Is the ability of a substance to be attracted strongly into a magnetic field. (ii) I. \(_{24}\)Cr\(^{2+}\) - Is\(^{2}\) 2P\(^{6}\) 3S\(^{2}\) 3P\(^{6}\) 3d\(^{4}\) unpaired electrons = 4 \(_{24}\)Cr\(^{6}\) - Is\(^{2}\) 2s\(^{2}\) 2p\(^{6}\) 3s\(^{2}\) 3d\(^{6}\) unpaired electrons = 0 (II) \(_{24}\)Cr\(^{2+}\) III. The greater the number of unpaired electrons in the 3d - orbital, the| greater-the paramagnetism.

(a)(i) This can be defined as the number of atoms or molecules in one mole of a substance which is equal to 6.02 x 10\(^{23}\) (ii) Mass/m.m = no. of molecules / Avogadro's number (2.30 / m.m) x (3.01x10\(^{22}\) / 6.02 x 10\(^{23}\)) m.m = (2.30 x 6.02 x 10\(^{23}\)) / 3.01 x 10\(^{22}\) m.m = 46g / mol \(\therefore\) 14+('16 x 2) 14 + 32 = 46g /mol \(\therefore\) N\(o_2\) (b)(i) I. Two layers are formed / no reaction / immiscible II. there is a reaction / hydrolysis with steamy fumes /white fumes (ii) The reaction occurs by interaction of the tones pair of electrons of water binding to the central atom (silicon or carbon) / attaching to the central atom In silicon, the 3p - orbital is available to accept the pairs of electrons but in carbon, there is no vacant orbitals for bonding. OR Water is polar hence the reaction with SiC\(I_4\). But CC\(I_4\) is non-polar hence there is no reaction (c) (i) - Purification of metals/ copper - Electroplating

(a)i. An acid is any compound which in a chemical reaction is able to attach itself to an unshared pair of electrons in another molecule-a base, to form a new compound ii. - AlCl3 - BF3 - BeF2 - H+ - K+ (b). It is the addition of NaCl to the reaction mixture in soap production to facilitate the precipitation of soap from the mixture (c) Reagent -- AgNO3 in NH3(aq) / Ammonical silver trioxonitrate condition-- heat (d) - It is the fraction of a given isotope in amixture of isotopes of the same element - The % of an isotope with a specific atomic mass found in a naturally occuring sample of an element. (e)i. Because the element is - Unreactive - Stable - A noble gas - Inert - Completely filled outermost shell ii. Because of weak forces between the molecules hence not much heat is needed to break the forces (f) - (Conc) HNO3 produces brown fumes (of NO2) when copper tunings are added to it/copper (II) salt is added but H2SO4 will not react with the copper tuning/Copper (II) salt - When Conc. H2SO4 is added to sugar it chars the sugar but Conc. HNO3 will not. - Add BaCl2 solution to each acid white precipitate indicates H2SO4 no precipitate indicates HNO3 (g) Electrochemical cell Electrolytic cell anode is negative/cathode is positive anode is positive/cathode is negative reaction is spontaneous non-spontaneous chemical energy is converted to electrical energy electrical energy is converted to chemical energy current is generated from within the cell current is generated from external source/battery the two half cells are in separate containers and connected by a salt bridge the two half cells are in the same container (h) Ionization energy decreases down group 1 elements, the lower the ionization energy, the easier to loose an electron hence the higher the reactive of the elements (i) Molecular formula shows the actual number of atoms of each element in a molecule (j) NH3 will deviate more because it has a larger volume and also it has a stronger intermolecular forces

(a)i. Temperature: 350-5000C/high temperature Pressure: 150-1000 atm/high pressure Catalyst: (Finely divided) Iron ii. Rate of the reaction increases as particles are closer together/more particles per unit volume. There are more collisions per seconds/collisions are more often (b)i. 4NO2 + O2 + 2H2O -----> 4HNO3 ii. 2NO2 + 2H2O -----> HNO3 + HNO2 A misture of two acids will be produced iii. 4HNO3(aq) → 4NO2 + O2 + 2H2O i. - Oxygen is neutral while NO2 is acidic ii. - NO2 is soluble while O2 is sparingly soluble in water iii. - O2 reacts with CO to form CO2 while NO2 reacts with CO to form a mixture of N2 and CO2 (c)i. Sulphur(IV) oxide Sulphur (VI) oxide ii. SO2(g) + H2O(l) ---->H2SO3(aq) SO3(g) + H2O(l) ---->H2SO4(aq) (d)i. - calcium oxide - quick lime b. - Limestone - Quick lime - Calcium trioxocarbonat(IV) - Calcium oxide c. - Calcium chloride - Calcium tetraoxosulphate (VI) - Calcium hydrogen trioxocarbonate (IV) d. Calcium tetraoxosulphate (VI) (dihydrate)

(a)i. Its colour darkens from pale yellow to amber colour while its crystals become needle-like and at 119oC it begins to melt. ii. The solution turns green, bubbles of colourless gas rise from the bottom of the mixture, but brown fumes of NO2 are produced and the solutions later turns blue. (b)i - SO2 - CO2 - CO - NO2 - H2S ii. Methane iii. - Contamination of soil - Soil erosion - Contamination of ground water - Loss of biodiversity - Formation of sink holes iv. Because of the large surface area and the higher the area the faster the rate of reaction. v. Coke (c)i. The sample is added to anhydrous copper (II) tetraoxosulphate (VI) when it turns from white to blue confirms the presence of water/sample is added to anhydrous cobalt chloride/cobalt chloride paper when it turns from blue to pink indicates the presence of water. ii(a) It softens temporary hard water (b) To remove permanent hardness/soften hard water iii. - Ca(HCO3)2 ---heat--->CaCO3 + CO2 + H2O + H2O or Mg(HCO3)2 ---heat--->MgCO3 + CO2 + H2O + H2O

Ethanol fermentation, also called alcoholic fermentation, is a biological process which converts sugars such as glucose, fructose, and sucrose into cellular energy, producing ethanol and carbon dioxide as by-products

Maltose is produced by the enzymatic hydrolysis of starch (a homopolysaccharide) catalyzed by the enzyme amylase.

Soaps cannot be used in acidic condition because they lose their cleansing effect due to the formation of long chain fatty acids