(c) Aluminium is reacted with manganese dioxide
Answer: Manganese is obtained along with aluminium oxide. This reaction is utilized during extraction of manganese
3MnO2 + 4Al → 3Mn + 2Al2O3
(d) Ferric oxide is reduced with aluminium
Answer: Iron is obtained. This reaction is known as Thermite reaction and is used for welding of railway tracks.
Fe2O3 + 2Al → 2Fe + Al2O3 + heat
(e) Zinc carbonate undergoes calcinations
Answer: Calcination involves heating carbonate ore in limited supply of air to obtain oxide of metal. Following is the equation:
ZnCO3 → ZnO + CO2
(a) By the transfer of electrons, illustrate the formation of bond in magnesium chloride and identify the ions present in this compound.
(b) Ionic compounds are solids. Give reasons.
Answer: Ionic bond has greater force of attraction because of which ions attract each other strongly. This makes ionic compounds solid.
(c) With the help of a labeled diagram show the experimental set up of action of steam on a metal.
Question 26: (a) Carry out following conversions:
(i) Ethanol to ethene
Answer: Ethanol gives ethene and water when it is heated with concentrated sulphuric acid.
CH3CH2OH + Conc. H2SO4 ⇨ CH2=CH2 + H2O
(ii) Ethanol to ethanoic acid
Answer: Ethanol gives ethanoic acid on oxidation.
CH3CH2OH + (Alkaline KMnO4/Acidified K2Cr2O7) ⇨ CH3COOH
(b) Differentiate between addition reaction and substitution reaction. Give one example of each.
Answer: Formation of larger molecules by addition of more radicals is known as addition reaction. For example; ethene is converted into ethane when heated with the catalyst nickel.
CH2=CH2 + H2 + (Nickel catalyst) ⇨ CH3−CH3
Replacement of a functional group or any atom by another atom or functional group is known as substitution reaction. Substitution reactions are single displacement reactions.
When methane reacts with chlorine gas in the presence of sunlight, it gives chloromethane and hydrogen chloride.
CH4 + Cl2 + Sunlight ⇨ CH3Cl + HCl
Question 27: (a) How do leaves of plants help in excretion? Explain briefly.
Answer: Some useless items are expelled through leaves during transpiration. But major part of excretion through leaves happens when old leaves are shed by plant.
(b) Describe the structure and function of a nephron.
Answer: Nephron: It is composed of a tangled mess of tubes and a filtering part; called glomerulus. Glomerulus is a network of blood capillaries to which renal artery is attached. The artery which takes blood to the glomerulus is called afferent arteriole and the one receiving blood from the glomerulus is called efferent arteriole. Glomerulus is enclosed in a capsule like portion; called Bowman’s capsule. The Bowman’s capsule extends into a fine tube which is highly coiled. Tubes from various nephrons converge into collecting duct; which finally goes to the ureter.
Question 28: (a) Draw a diagram showing germination of pollen on stigma of a flower and mark on it the following organs/parts:
(b) State the significance of pollen tube
Answer: Pollen tube provides a passage to pollen nuclei to reach ovary
(c) Name the parts of flower that develop after fertilization into: (i) Seed (ii) Fruit
Answer: Ovules develop into seeds while ovary develops into fruit
(a) “Use of a condom is beneficial for both the sexes involved in sexual act”. Justify this statement giving two reasons.
Answer: Use of condom helps in preventing STDs to both partners, it also helps in preventing unwanted pregnancy.
(b) How do oral contraceptives help in avoiding pregnancies?
Answer: Oral contraceptives contain hormones which prevent ovulation in females. Thus, they help in preventing pregnancy.
(c) What is sex selective abortion? How does it affect a healthy society? (State any two consequences)
Answer: Abortion of female foetus to obey the social custom of preference to a male child can be termed as sex selective abortion. This results in eschewed gender ratio which is against women. It also results in mental and physical torture of the girl child and her mother.
Question 29: (a) Define power and state its SI unit.
Answer: Rate of doing work or transferring energy per unit time is called power. SI unit of power is Watt which is equal to 1 Joule per second.
(b) A torch bulb is rated 5 V and 500 mA. Calculate its
Answer: P = V × A
`= 5 V × 0.5 A=2.5` W
Answer: R = `V/I`
(iii) Energy consumed when it is lighted for `21/2` hours.
Answer: power consumed in 1 sec = 2.5 W
So, power consumed in2.5 hours
`=2.5× 60×60×2.5=225` Wh
= 0.225 kWh
Question 30: (a) A security mirror used in a big showroom has radius of curvature 5 m. If a customer is standing at a distance of 20 m from the cash counter, find the position, nature and size of the image formed in the security mirror.
Answer: Given, u = -20m, f = 2.5m (Assuming that mirror is near cash counter)
Image distance can be calculated as follows:
Or, `v=(100)/(35)=2.85` m
Positive sign shows that image is formed behind the mirror, is virtual and erect.
Magnification can be calculated as follows:
Image is about one sixth the size of object.
(b) Neha visited a dentist in his clinic. She observed that the dentist was holding an instrument fitted with a mirror. State the nature of this mirror and reason for its use in the instrument used by dentist.
Answer: Dentists use concave mirror because when object is placed between F and P of concave mirror then a virtual, erect and enlarged image is formed behind the mirror. It helps the doctor see a larger image of tooth.
Rishi went to a palmist to show his palm. The palmist used a special lens for this purpose
(i) State the nature of the lens and reason for its use.
Answer: Palmists use convex lens because it shows enlarged, virtual and erect image when object is between F and O of convex lens.
(ii) Where should the palmist place/hold the lens so as to have a real and magnified image of an object?
Answer: If the palmist wants real and magnified image, he should put object between F1 and F2 or on F. But in that case he will have to use a screen to see the image. So, for convenience, palmists use the option as stated in previous answer.
(iii) If the focal length of this lens is 10 cm and the lens is held at a distance of 5 cm form the palm, use lens formula to find the position and size of the image.
Answer: Given, f = 10 cm, u = -5 cm
Using lens formula:
Or, `v=-10` cm
Image is formed at 10 cm on the same side of lens (as object). It is erect and virtual.
Image size `=v/u=(10)/5=2`
Image is twice as big as object.
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