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(a) The body temperature of animal A is constant and does not depend on the environmental temperature. Animal A is homoithermic. The body temperature of animal B increases as the environment. (a) temperature increases. Animal B is poikilothermic (b) Animal B can hibernate or aestivate (c) ()) Man, birds, cat (iü) Snake, lizard toad,

(a) Specimen A is a bean seed(s) Specimen B is tomato fruit (b) Th2 organisms responsible for the damage Observed on specimen A = weevil or beetle. Specimen B= fungus or mould or bacteria (c) ()) Specimen A (bean seed) can be properly stored through (i) Storage in air- tight polythene bags (ii) Storage in silos () Storage in cool dry place (iv) Storage in deep freezer or refrigerator (v) Seed dressing with chemicals or fumigants (vi) Seed processing (vii) Canning (i) Specimen B (tomato fruit) can be properly stored through () drying () canning or processing (iii) freezing (d) (i) Specimen C = Tick Specimen D = Termite (soldier) (i) Specimen C (tick) attacks goats, sheep, dog, man Specimen D (termite) attacks maize, palm tree, rubber, coconut and rice (ii) How Specimnens C and D affect the crop or animal they infest: Specimen C sucks blood from the animal and causes loss or blood (or anaemia) to the animal. It also transmits disease causing organisms to the animal. It causes itching and irritation to the organism. It predisposes animals to diseases Specimen D eats up the roots or stems or woody parts of plants. It then destroys or damages them. It also predisposes plants to diseases (e) Ways by which a farmer can control specimen C (tick) are byenimals graze. Ways by which a farmer can control specimen D (termite) are (i) by the use of insecticide (ii) through ploughing (digging) to remove the queen termite.

(a) Specimen E = earthworm, Specimen F = cockroach Specimen G = twig of flowering plant. Specimen H=rat (b) The excretory organs found in specimens E to H are: Specimen E = nephridium or body surface. Specimen F= malphighian tubule. Specimen G= leaf or stem, Specimen H= lungs kidney and skin (c) The habitats of Specimen E and H are ()) Specimen E (earthworm) soil or damp soil or marshy area. Specimen H (rat) burrows or holes or bush or food stores or house or cupboard. (ii) Ways by which specimen H(rat) is adapted to its habitat are (i) Limbs end in claws for digging the burrow (ii) Forelimbs are short and muscular for absorbing shock on landing (iii) Hind limbs are long and muscular for running fast

- Earlier, molasses act as a raw material which yield ethanol from which ETHENE (gotten is used to) produce a range of poly(ethene) plastics. - Nowadays, starting new raw materials are obtained by the fractional distillation of crude oil or petroleum. - During the thermal cracking of petroleum; long chain alkanes are converted into smaller chain alkanes, ALKENES (with Ethene being the predominate), etc. - The Alkenes such as Ethene and Propene can be made into Poly(ethene) / plastic, nylon, etc.

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option B.

Area of a sector \(\frac{\theta}{360^\circ} * \pir^2\) = \(\frac{135}{360} * 3.142 * 4 * 4\) = \(18\frac{6}{7}cm^2\)

This histogram is transferred into this frequency table \(\begin{array}{c|c} Marks & 20 & 40 & 60 & 80 & 100 \\ \hline students & 9 & 7 & 6 & 6 & 2\end{array}\) Students who scored more than 40 = 6 + 6 + 2 = 14 i.e. \(\frac{14}{30}\) x 100% = 46\(\frac{3}{4}\)%

The heat of fusion is the quantity of heat necessary to change of a solid to a liquid with no temperature change

Total resistance = resistance in series + resistance in parallel = 4 + (\(\frac{4 \times 4}{4 + 4}\)) = 6\(\Omega\) Total power drawn = \(\frac{V^2}{\text{R}}\) = \(\frac{12^2}{6}\) = 24W

When the object is placed inside the focal point (u < f), the image is: → Virtual (formed on the same side as the object) → Erect (upright) → Magnified (larger than the object)

\(\frac{F_1}{F_2}\) = \(\sqrt{\frac{T_1}{T_2}}\) \(T_2 = 2T_1\) \(\frac{F_1}{F_2}\) = \(\sqrt{\frac{T_1}{2T_1}}\) \(\frac{F_1}{F_2} = \frac{1}{ \sqrt2}\) F\(_1\) : F\(_2\) = 1 : \(\sqrt{2}\) F\(_2\) : F\(_1\) = \(\sqrt{2}\) : 1

Al has valency of 3+: 3F will deposit 27g xF will deposit 9g cross multiply → \(\frac{9g * 3F}{27g}\) : 1F = 9g Cu has valency of 2+: 2F will deposit 64g 1F will deposit xg cross multiply → \(\frac{64g * 1F}{2F}\) : 1F = 32g

No expanded working has been attached for this question yet. The verified answer is Option C.

It is vital you rearrange/redraw the structure if confusing this will make naming the compound easier. the name of the compound above is 3 - chloro-2- methyl prop - 1 - ene.