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Substance K can be said to be Carbohydrate because starch (carbohydrate) when reacted with iodine, gives a blue-black colour while Protein (substance L) reacts with a Million's reagent to give a deep-red ppt. Hence, the correct answer is A

Take an unknown substance and add a few drops of it to the silver nitrate solution. If the silver nitrate solution turns white due to the formation of silver chloride, the acid is hydrochloric acid. If the solution remains clear, the acid is nitric acid. T is Hydrochloric Acid(an odourless, colourless solution with a pungent smell)

Sodium hydroxide is a strong base that can react with certain salts, especially those that contain amphoteric metals. Zinc is an amphoteric metal, meaning it can react with both acids and bases. Like the hydroxides of other metals, such as lead, aluminium, beryllium, tin and chromium, zinc hydroxide (and zinc oxide), is amphoteric. Thus it will dissolve readily in a dilute solution of a strong acid, such as HCl, and also in a solution of an alkali such as sodium hydroxide. The correct answer in option C

Acid rain is caused by the release of sulphur(iv)oxide and nitrogen oxides (NOx) into the atmosphere, where they react with water, oxygen, and other chemicals to become sulfuric and nitric acid respectively. Sulphur appears in option A, thus it is the correct answer.

From C1V1 = C2V2 C1 = 0.50 mol dm-3; C2 = ?; V1 = 80cm3; V2 = 80 + 20 = 100cm3 :. C2 = C1V2/V2 = 0.50 x 80/100 = 0.40 mol dm-3 - Option B

Biotechnology is the use of biological systems to create products or processes for a specific purpose. It involves manipulating living cells or cell components to produce useful substances. Among the given options, it is only the manufacture of soap that doesn't require manipulating the use of living cells (Bio) to create chemical products. Option D

There are basically two categories of plastics: Thermoplastic can be deformed or bent when heated, such as PVC, nylon, and polythene. Thermoplastics can be melted and re-melted into new forms, making them often recyclable. On the other hand, Thermosetting Plastics cannot be softened again by heating once they are formed, such as Bakelite. The answer is Option A

An allotrope is one or more forms of a chemical element that can exist in the same physical state. Allotropes have different physical properties but same chemical properties since they all emanated from the same element. Therefore, option A is the correct answer.

Mass number = Proton + Neutron = 10 + 12 = 22 - Option A

Adding up the electrons( numbers in the superscript) from the configuration, we have 2+2+6+2+4 = 16. The element with atomic number 16 is Sulphur and it's in group VI. Alternatively, since the atomic number is 16, it can be deduced that we have 2, 8, 6 when arranged in shells or orbits. Because it the last digit is 6, thus the element belongs to group VI. Option A

Electron affinity increases across the period and decreases down the group. This is the same with electronegativity. Consequently, it is easier for elements down the group to lose electrons than the ones at the top. Therefore, metal P will loose electron least. Option A is the correct answer.

(a) Volume of pipette 25.00 cm\(^3\) Indicator used - Methyl orange, colour changed at end point - from yellow to orange/pink Burette readings Rough 1st 2nd 3rd Final reading in cm\(^3\) 16.50 16.00 15.95 26.05 Initial reading in cm\(^3\) 0.00 0.00 0.00 10.00 Difference in cm\(^3\) 16.50 16.00 15.95 16.05 Average volume of acid used = \(\frac{16.00 + 15.95 + 16.05}{3}\) = 16.00 cm\(^3\) Equation for the reaction K\(_2\)CO\(_3\) + 2HNO\(_3\) → 2KNO\(_3\) + CO\(_2\) + H\(_2\)O Molar mass of K\(_2\)CO\(_3\) = 2 x 39 + 2 + 16 x 3 = 138g No. of moles of B = \(\frac{6.90}{138}\) = 0.0500 mol/dm\(^3\) (ii) 1 mole of B contains 6.02 x 10\(^{23}\) mol\(^{-1}\) 0.0500 moles = 6.02 x 10\(^{23}\) x 0.05 = 3.01 x 10\(^{22}\) The number of potassium ions per dm\(^3\) = 3.01 x 10\(^{22}\) (iii) \(\frac{\text{mole concentration of A x volume of A}}{\text{mole concentration of B x volume of B}}\) = mole ratio of \(\frac{A}{B}\) \(\frac{MA \times 16.00}{0.05 \times 25.00} = \frac{2}{1}\) MA = \(\frac{0.05 \times 25.00 \times 2}{16.00}\) = 0.1563 mol/dm\(^3\) (4d.p)

(a)(i) Electronegatively, ionisation energy, electron affinity. (ii) Chemical reaction Nuclear reaction - Involves only the electron - Involves the nucleus - Not usually accompanied by great heat change - Usually accompanied by a lot of heat change - New elements are not formed - New elements are formed - External factors such as pressure, temperature and catalyst affect the rate of reaction. - Not affected by external factors (b) Atom of elements P Q R S T Mass number 16 40 35 18 20 Atom number 8 20 17 8 10 (I) P and S are isotopes of the same elements (ii) R contains 18 neutrons (iii) T is chemically unreactive (iv) Q readily forms an ion with two positive charges (v) R attains an octet structure by accepting one electron (vi) Q forms an ionic bond with R (vii) Q belongs to the s - block in the periodic table (c)(i) To convert CuCO\(_3\) to Cu, heat strongly ; CuCO\(_3\) -> CuO + CO\(_2\) . Pass hydrogen over heated residue of CuO. H\(_2\) + CuO ---> Cu + H\(_2\)O or pass NH\(_3\) over heated copper II oxide. 2CuO + 3NH\(_3\) —› 3Cu + 3H\(_2\)O + N\(_2\) OR add dilute mineral acid to CuCO\(_3\) to form soluble salt: CuCO\(_3\) + 2HCI -> CuCl\(_2\) + H\(_2\) O + CO\(_2\) . Electrolysis using platinum electrodes or any inert electrode. Equation at the cathode Cu\(^{2+}\) + 2e\(^-\) \(\to\) Cu OR add mineral acid to CuCO\(_3\) to form soluble salt CuCO + 2HCI -> CuCl\(_2\) + H\(_2\)O+ CO\(_2\) then add reactive metal e.g. Zinc or Aluminium CuCl\(_2\) + Zn -> Cu + ZnCl\(_2\) (ii) To convert MgO to Mg\(_2\)SO\(_4\) , acid dilute H\(_2\)SO\(_4\) to excess MgO, filter off excess MgO. The filtrate is MgSO\(_4\) , MgO + H\(_2\)SO\(_4\) -> MgSO\(_4\) + H\(_2\)O. Evaporate / heat to dryness to obtain MgSO\(_4\).

(a)(i) Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its density. Rate a \(\frac{1}{\sqrt{d}}\) or \(\frac{1}{\sqrt{m}}\) where m is the molecular mass and d is the density. (ii) 1 molecule = X\(_3\). Relative molecular mass of X = 16 x 3 = 48 Vapour density = 1/2 x Relative molecular mass. Vapour density of x = \(\frac{48}{2}\) = 24 (iii) If equal volumes of Y and Z are at the same temperature and pressure and the mass of Y is twice that of 7, molecules of Z will have the greater average velocity. Since the temperature is the same, molecules of Y and Z have the same average kinetic energy. Hence the smaller molecule will have the greater velocity. (b) Given tha rate curve for the following reaction: MgCO\(_3\) + 2HCI -> MgCl\(_2\) + CO\(_2\) + H\(_2\)O. (i) The reaction occurred for 12 minutes (ii) There was a loss in mass because one of the products was a gay (CO\(_2\)) which escaped to the atmosphere. (iii) The rate was fastest at the beginning of the reaction. The slope of the curve was highest or steepest at the point. (iv) The slope of the curve can be affected by change in the concentration of HCI, change in temperature, change in particle size/surface area of the MgCO\(_3\) (c) PCl\(_5\) -> PCl\(_3\) + Cl\(_2\) \(\Delta\)H = +95 KJmol\(^{-1}\) K\(_{eq}\) = \(\frac{[PCl_3][Cl_2]}{[PCl_5]}\) or K\(_p\) = \(\frac{P^{PCl}_3P^{Cl}_2}{P^{Cl}_5}\) (i) Increased pressure will shift the equilibrium position to the left (ii) increased temperature will shift the equilibrium position to the right (iii) Removal of chlorine will shift the equilibrium position to the right.

(a)(i) Variable oxidation states (oxidation states of +2 or +3) formation of coloured ions (green or brown) Catalytic ability e.g (in the Haber process) formation of complex ions e.g K\(_4\)Fe (CN)\(_6\). Ability to be magnetised strongly (Ferromagnetism / paramagnetism). (ii) C + O\(_2\) --> CO\(_2\) CO + C \(\to\) CO Fe\(_2\)O\(_3\) + 3CO \(\to\) 2Fe + 3CO\(_2\) or Fe\(_2\)O\(_3\) + 3C \(\to\) 2Fe + 3CO CaCO\(_3\) \(\to\) CaO + CO\(_2\) CaO + SiO\(_2\) \(\to\) CaCO\(_3\) (iii) The reaction of iron to produce rust (Fe\(_3\)O\(_2\) x H\(_2\)O) can be prevented by galvanising, painting, electroplating, greasing or oiling, cathodic protection or anodic sacrificing. (iv) The oxidation number of iron in iron rust is +3. (b)(i) Au, Sn, Zn, Mg, Ca, Na in order of reactivity. Mg, Ca, and Na are extracted by electrolysis (ii) Zinc is said to be amphoteric because it can react with acids and with bases to form salts. (c)(i) Oxidation is loss of electrons (ii) If or 96500C is equivalent to 1 mole of electrons 4825 C = \(\frac{4825}{96500}\) moles = 0.05 mole (iii) 1 mole of Cu\(^{2+}\) is discharged by 2F. 1 mole of Cu\(^{2+}\) contains 6.02 x 10\(^{23}\) ions 6.02 x 10\(^{23}\) copper ions is 6.02 x 10\(^{23}\) discharged by 2F. Hence ; 0.25 ions will be discharged by 0.25 F = 7.53 x 10\(^{22}\)