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(a)(i) A dielectric is an insulating material (solid, liquid or gas) separating the two plates of a capacitor. (ii) Using C = \(\frac{\xi_1.\xi_oA}{d} = \frac{900 \times 8.85 \times 10^{-12} \times 9.6 \times 10^{-3}}{2.25 \times 10^{-4}}\) = 3.4 x 10\(^{-7}\) (b)(i) A voltmeter has a higher resistance. Reason: A voltmeter is connected inparallel in a circuit to allow passage of minimal current through it; hence the circuit current is not significantly affected. (ii) \(\frac{1}{R_1} = \frac{1}{400} + \frac{1}{400}\) \(R_1 = 200\Omega\) \(\frac{1}{R_2} = \frac{1}{800} + \frac{1}{800}\) \(R_1 = 400\Omega\) RT= 200 + 400 = 600\(\Omega\) I = \(\frac{V_T}{R_T} = \frac{6}{600}\) = 0.01A Voltmeter reading = Pd across R\(_2\) = 0.01A Voltmeter reading = Pd across R\(_2\) = 0.01 x 400 = 4v (c) \(\frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = \frac{1 + 1}{2} = \frac{2}{2}\) Note: This catadation is for the two resistors in parallel. R\(_T = 2 + 1 + 2 = 5\Omega\)

(a)(i) Nuclear Fission is the splitting up of a heavy nucleus into lighter nuclei with release of a huge amount of energy and neutrons. (ii) (\(\alpha\)) graphite; (\(\beta\)) Boron rods; (\(\gamma\) )Liquid sodium Material Function Graphite Slows down high energetic neutrons in the reaction chamber. Boron rods To absorb excess neutrons so as to control fission rate Liquid sodium Extraction / conduction of excess heat from the reaction chamber. (b)(i) E\(_n\), n = \(\alpha\) ---------------------------------- 0.eV E\(_4\), n = 5 ---------------------------------------------------- - 0.54eV E\(_3\), n = 4 ---------------------------------------------------- - 0.85eV E\(_2\), n = 3 ---------------------------------------------------- 1.51eV E\(_1\), n = 2 ------------------------------------------------------ 3.39eV (ii) Using \(\lambda\) = \(\frac{hc}{\Delta E}\) = \(\frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{(-1.51 - ( - 13.60)) \times 1.6 \times 10^{-19}}\) = 1.02 x 10\(^{-7}\)m (c) \(\lambda\) = \(\frac{0.693}{ T^{\frac{1}{2}}}\) = \(\frac{0.693}{5572}\) = 0.0001244 years\(^{-1}\) N = Noe\(^{- \lambda t}\) e\(^{0.0001244t}\) = 1.2632 0.0001244t = 0.234 t = 1878 years

S/N I/A V/V LogV/V LogI/A 1.0 0.10 0.30 -0.552 -1.000 2.0 0.15 0.50 -0.301 -0.823 3.0 0.20 0.70 -0.154 -0.698 4.0 0.25 0.80 -0.096 0.602 5.0 0.30 0.90 -0.045 0.522[ Slopes S = \(\frac{\bigtriangleup \text{logI}}{\bigtriangleup \text{logV}}\) = \(\frac{\text{logl}_{2}} {\text{logV}_{2}}\) - \(\frac{\text{logl}_{1}}{\text{logV}_{2}}\) S = \(\frac{-1.00 - (-0.70)}{-0.15 - (0.50)}\) = \(\frac{0.30}{0.10}\) S= 3 PRECUATIONS 1. avoided parallax error when reading ammeter and voltmeter 2. opened the key when readings are not being taken to ensure that the cells do not run down. 3. ensured that the jockey is not dragged along the potentiometer wire. (b)i. As x increases, the brightness of the bulb increases. As x increases, voltage and current through the bulb also increases, hence brightness increases ii. Electrical devices whose action does not obey ohm's law. They are: Diode, Capacitor, Transistor, Thermistor and Thyristor. SEE GRAPH ABOVE

S/N L/CM t/s T=t/o/S LogT/S LogL/cm 1 80.0 17.90 1.79 0.252 1.903 2 70.0 17.30 1.73 0.238 1.845 3 60.0 15.34 1.53 0.185 1.775 4 50.0 13.90 1.39 0.145 1.698 5 40.0 13.06 1.31 0.155 1.602 Slope (s) = \(\frac{\bigtriangleup \text{logT}_{2}}{\bigtriangleup\text{logL}_{2}}\) - \(\frac{\bigtriangleup \text{logT}_{1}}{\bigtriangleup\text{logL}_{1}}\) = \(\frac{0.15 - 0.05}{0.164 - 0.148}\) = \(\frac{0.100}{0.016}\) = 6.25 PRECAUTIONS 1. avoided zero error in the reciding of the stop watch 2. avoided Parallax in reading the metre rule and the stop watch 3. ensured that the support of the bifilar pendulum is rigid 4. ensured smooth and regular oscillations in a horizontal plane. b. i. Simple harmonic motion is defined as the motion of an object whose acceleration is proportional to the distance from a fixed point and is always directed towards that point see graph above

bo = 102^\circc S/N M/g bi/ ^\circc \(\theta_{1}\)=(bi-bo)/^\circ 1.0 10.0 104.00 104 - 102 = 2.00 2.0 20.0 106.00 106 - 102 = 4.00 3.0 30.0 108.00 108 - 102 = 6.00 4.0 40.0 110.00 110 - 102 = 8.00 5.0 50.0 112.00 112 - 102 = 10.00 Slope, s = \(\frac{\bigtriangleup M_{2} - \bigtriangleup M_{1}}{\bigtriangleup\theta_{2} - \bigtriangleup\theta_{1}}\) = \(\frac{40}{8} - \frac{10}{2} = \frac{30}{6}\) S = 5 PRECAUTIONS 1. stirred the water gently to ensure uniform temperature. 2. avoided splashing of the water 3. avoided error due to parallax in reading the thermometer and the stop clock. (b) i. Boiling point is defined as the point when liquid changes to vapour when the saturation vapour pressure is equal to the atmospheric pressure. ii. Impurities affect the boiling of a liquid because their presence inicreases the boiling rboint of the liquidelcom- pared to the boiling point of a pure solvent.

S/N I/A V/V LogV/V LogI/A 1.0 0.10 0.30 -0.552 -1.000 2.0 0.15 0.50 -0.301 -0.823 3.0 0.20 0.70 -0.154 -0.698 4.0 0.25 0.80 -0.096 0.602 5.0 0.30 0.90 -0.045 0.522[ Slopes S = \(\frac{\bigtriangleup \text{logI}}{\bigtriangleup \text{logV}}\) = \(\frac{\text{logl}_{2}} {\text{logV}_{2}}\) - \(\frac{\text{logl}_{1}}{\text{logV}_{2}}\) S = \(\frac{-1.00 - (-0.70)}{-0.15 - (0.50)}\) = \(\frac{0.30}{0.10}\) S= 3 PRECUATIONS 1. avoided parallax error when reading ammeter and voltmeter 2. opened the key when readings are not being taken to ensure that the cells do not run down. 3. ensured that the jockey is not dragged along the potentiometer wire. (b)i. As x increases, the brightness of the bulb increases. As x increases, voltage and current through the bulb also increases, hence brightness increases ii. Electrical devices whose action does not obey ohm's law. They are: Diode, Capacitor, Transistor, Thermistor and Thyristor. SEE GRAPH ABOVE

An element's chemical properties depend on its valence electrons.The chemical properties of elements depend on an element's electron configuration.

Test tubes are convenient containers for heating small amounts of liquids or solids with a Bunsen burner or alcohol burner.

Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers.

Scandium has electronic configuration \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{1}4s^{2}\) hence, \(Sc^{3+}\) has configuration \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\). Hence, no electron in the 3d-orbital.

The various gases vaporize as their boiling points are reached, then they are pumped away and separately re-liquefied for convenient storage.

The purity of a solid can be determined by measuring the melting point. A pure solid will have a single, sharp melting point.

This is same as asking for the molar mass of the compound = \((2\times 14.0)+ (8\times 1.0) + 12.0+(16.0\times 3)\) = 28.0+8.0+12.0+48.0 = 96.0g

The orbital is a path or region around the nucleus. The electrons move in the orbitals and not through it.

The ionic radius decreases across a period but increases along a group.