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SEE Model Paper 2

Full marks: 75Time: 3 hours

All questions are compulsory.

Write your answers on paper first. Then open each answer and its marking scheme, and give yourself one mark for each point you got. In multiple choice, your first choice counts.

Section A

10 × 1 = 10

1. Choose the correct option for each of the following questions.

a.

How many bits make one byte?

Pick an answer to see why.

b.

Hydra and the jellyfish are placed in the phylum Coelenterata. Why?

Pick an answer to see why.

c.

In which method of genetic technology is an egg fertilized by a sperm outside the body, in a laboratory dish?

Pick an answer to see why.

d.

Villagers in Nawalparasi run a “vulture restaurant”: they look after old cattle until they die and leave the bodies for the vultures. How does this help the vultures?

Pick an answer to see why.

e.

A statement and two arguments are given below.

Statement: A feather and a coin dropped together in a tube with the air pumped out reach the bottom at the same time.

Argument 1: In a vacuum the Earth pulls the feather and the coin with equal forces.

Argument 2: With no air resistance, both fall with the same acceleration g, which does not depend on the mass of the falling body.

Pick an answer to see why.

f.

Prakash sees his book clearly but can’t read the whiteboard from the back of the class. Where does his eye focus the light from the whiteboard, and what kind of lens corrects it?

Pick an answer to see why.

g.

If the average density of the universe were more than the critical density, what would happen to it?

Pick an answer to see why.

h.

In which group and period of the modern periodic table is the element with atomic number 20?

Pick an answer to see why.

i.

Crushed magnetite (Fe₃O₄) ore is mixed with sand and rock. Which method is best to concentrate it?

Pick an answer to see why.

j.

The tube well at a family’s house in the Terai gives hard water. To wash their clothes well while wasting the least, what should they use, and why?

Pick an answer to see why.

Section B

9 × 1 = 9

2. Write very short answers to the following questions.

a.

How many bytes are there in 3 KB?

Answer

1 KB = 1024 bytes, so 3 KB = 3 × 1024 = 3072 bytes.

Marking scheme
  • 3072 bytes.
b.

The two daughter cells formed by mitosis are identical to the parent cell. Why?

Answer

Before the cell divides, the DNA of every chromosome is copied, and each daughter cell then gets one exact copy of every chromosome. So both have the same number of chromosomes and the same genes as the parent cell.

Marking scheme
  • Each chromosome is copied and each daughter cell gets one copy of every chromosome (same number, same genes).
c.

What is the normal blood pressure of a healthy adult?

Answer

About 120/80 mmHg: a systolic pressure of about 120 mmHg and a diastolic pressure of about 80 mmHg.

Marking scheme
  • About 120/80 mmHg.
d.

A glass bottle filled to the top with water cracks when it is left in a freezer. Why?

Answer

Water expands as it cools below 4 °C, and expands by about 9% more when it freezes. The ice needs more space than the bottle has, and it pushes on the glass hard enough to crack it.

Marking scheme
  • Water expands when it cools below 4 °C and freezes, and the expanding ice cracks the bottle.
e.

It takes 18,000 J of heat to warm a 2 kg block of a metal from 20 °C to 30 °C. Calculate the specific heat capacity of the metal.

Answer

From Q = ms(T₂ − T₁):
s=Qm(T2−T1)=18,0002×(30−20)=900 J/kg °C s = \frac{Q}{m(T_2 - T_1)} = \frac{18{,}000}{2 \times (30 - 20)} = 900\ \text{J/kg °C}

Marking scheme
  • 900 J/kg °C, from Q = ms(T₂ − T₁).
f.

What is the near point of a normal adult eye?

Answer

About 25 cm from the eye: the nearest point at which the eye can see clearly (the least distance of distinct vision).

Marking scheme
  • About 25 cm.
g.

What would happen to the Earth if the Sun’s gravitational pull on it suddenly disappeared?

Answer

Nothing would bend its path any more, so the Earth would stop going round the Sun and move off in a straight line, in the direction it was moving at that moment (along a tangent to its orbit), into space.

Marking scheme
  • It would leave its orbit and move away in a straight line (along a tangent).
h.

Carbon dioxide is used to put out fires. Why?

Answer

Carbon dioxide does not burn and does not help things burn, and it is heavier than air. It settles over the fire like a blanket and keeps the oxygen away, so the fire goes out.

Marking scheme
  • It doesn’t burn or support burning and, being heavier than air, covers the fire and cuts off its oxygen.
i.

Impure copper containing a little silver and gold is purified by electro-refining. Where do the silver and gold end up?

Answer

Silver and gold are less reactive than copper, so they do not go into the solution as ions. As the impure copper anode wears away, they fall to the bottom below the anode as anode mud, from which these valuable metals can be collected.

Marking scheme
  • They fall below the anode as anode mud (being less reactive, they don’t dissolve).

Section C

14 × 2 = 28

Write short answers to the following questions.

3.
[2]

At night, a medium-wave (MW) radio station often comes with a hiss and crackle, but a song played on a mobile phone over the internet sounds as clear as the original.

  • i.Why does noise spoil the MW radio’s analog signal?
  • ii.Why does a little noise do no harm to the digital signal?
Answer
  • i.An analog signal carries the sound as the shape of a wave that can take any value. Noise from lightning, other waves and the atmosphere changes that shape on the way, and the receiver can’t tell the added noise from the sound, so it plays the hiss and crackle too.
  • ii.A digital signal carries only 0s and 1s. The receiver only has to decide “is this a 0 or a 1?”, so a little noise doesn’t change the answer, and the data is recreated exactly.
Marking scheme
  • i. The analog signal can take any value, so the noise added on the way becomes part of it and is heard.
  • ii. The digital signal is only 0s and 1s; the receiver only decides 0 or 1, so a little noise does no harm.
4.
[2]

Write any two features of the body of a worker bee that suit its work.

Answer

Any two:

  • a.A pollen basket on each hind leg, ringed with stiff hairs, to carry pollen home.
  • b.Hairy legs and body that catch pollen.
  • c.Wax glands under the abdomen, to make wax for the comb.
  • d.A long proboscis to suck nectar, and mandibles to shape wax.
  • e.A barbed sting to guard the hive.
Marking scheme
  • One correct feature.
  • A second correct feature.
5.
[2]

In pea plants, tall (T) is dominant over dwarf (t). A farmer wants to know whether his tall pea plant is pure (TT) or a hybrid (Tt). He crosses it with a dwarf plant, and about half of the offspring are dwarf.

  • i.What is the genotype of his tall plant? Give a reason.
  • ii.Show the cross in a Punnett square.
Answer
  • i.Tt. A dwarf offspring (tt) needs a t from each parent. The dwarf parent gives only t, so the tall plant must have given the other t. A TT plant gives only T, and all its offspring would have been tall.
  • ii.Tt × tt: the tall plant’s gametes are T and t; the dwarf plant’s are all t.
T t
t Tt (tall) tt (dwarf)

The offspring are 1 tall (Tt) : 1 dwarf (tt), about half of each, as he found.

Marking scheme
  • i. Tt, because the dwarf offspring got a t from it (a TT plant would give only tall offspring).
  • ii. A correct Punnett square of Tt × tt giving Tt (tall) and tt (dwarf), 1 : 1.
6.
[2]

A rhinoceros horn was seized from smugglers, and the body of a rhino killed by poachers was found in Chitwan National Park. Scientists want to know whether the horn came from that rhino.

  • i.How can a DNA test show whether the horn came from that rhino and not another?
  • ii.The forest guard who found the horn held it in his bare hands for a long time. Why might this spoil the test, and what should he have done?
Answer
  • i.Every individual has its own DNA (except identical twins), and every cell of an animal carries the same DNA. So if the DNA from the horn matches the DNA from the dead rhino’s body, the horn came from that rhino; if it doesn’t match, it came from another.
  • ii.A DNA test is very sensitive: cells from his skin could contaminate the sample and give a wrong answer. He should have handled it with clean gloves, and sealed and labeled it carefully for the laboratory.
Marking scheme
  • i. Each individual’s DNA is different but the same in all its cells, so a match shows it is the same rhino.
  • ii. His cells can contaminate the sample (a wrong result); it should be handled with gloves and sealed and labeled carefully.
7.
[2]

Using the sex chromosomes, explain why it is the father’s sperm, not the mother’s egg, that decides whether a baby is a boy or a girl.

Answer

A woman’s body cells are 44 + XX, so every egg she makes is 22 + X. A man’s body cells are 44 + XY, so he makes two kinds of sperm in equal numbers: 22 + X and 22 + Y.
An X sperm and the egg give 44 + XX, a daughter; a Y sperm and the egg give 44 + XY, a son. Every egg carries an X, so which kind of sperm fertilizes it decides the baby’s sex.

Marking scheme
  • Every egg carries X (the woman is XX), but the sperm carry X or Y in equal numbers (the man is XY).
  • X sperm + egg → XX (girl); Y sperm + egg → XY (boy); so the sperm decides.
8.
[2]

In a blood test, Pramila’s blood clumps with anti-A serum and with anti-D serum, but not with anti-B serum.

  • i.What is her blood group?
  • ii.Can she safely be given red cells from a donor who is AB+ve? Give a reason.
Answer
  • i.Clumping with anti-A shows antigen A, and with anti-D the Rh factor; no clumping with anti-B shows there is no antigen B. Her blood group is A+ve (A positive).
  • ii.No. AB+ve red cells carry antigen B, which her own cells lack. The anti-B antibodies in her plasma would clump and destroy them (a transfusion reaction).
Marking scheme
  • i. A+ve.
  • ii. No: AB red cells carry antigen B, and her anti-B antibodies would clump them.
9.
[2]

Hari, a 50-year-old bus driver, smokes, eats fried food every day and has had high blood pressure for years. One morning he feels a sudden tight pain in the center of his chest that spreads to his left arm, with a cold sweat.

  • i.What is most likely happening to him, and what should be done at once?
  • ii.Write any two things in his way of life that raised his risk, with a change for each.
Answer
  • i.Most likely a heart attack: a coronary artery has been blocked, and part of the heart muscle is getting no blood. An ambulance should be called at once (102 in Nepal), or he should be taken quickly to a hospital, sitting and resting, not walking.
  • ii.Any two: smoking (stop smoking); fried food, rich in fat (eat more vegetables, fruit, whole grains and pulses, and less fried food, salt and sugar); high blood pressure for years (have it checked and treated regularly); sitting all day as a driver (exercise for at least 30 minutes on most days).
Marking scheme
  • i. Heart attack; call an ambulance (102) or take him to hospital at once.
  • ii. Any two risk factors from his life, each with a sensible change.
10.
[2]

Write any two uses of neem.

Answer

Any two:

  • a.Neem juice and paste are used for skin problems.
  • b.Neem twigs are used as toothbrushes (datiun).
  • c.Neem leaves are kept with stored grain and clothes to keep insects away.
  • d.Neem oil is used as a natural pesticide.
  • e.Its leaves, root, bark, flowers and fruit are used in Ayurvedic medicines.
Marking scheme
  • One correct use.
  • A second correct use.
11.
[2]

A thermometer brought from abroad shows a child’s temperature as 104 °F.

  • i.Convert this temperature to degrees Celsius.
  • ii.Convert it to kelvin.
Answer
  • i.C100=F−32180=104−32180=72180 \dfrac{C}{100} = \dfrac{F - 32}{180} = \dfrac{104 - 32}{180} = \dfrac{72}{180} , so C=0.4×100=40 °C C = 0.4 \times 100 = 40\ °\text{C} : a high fever.
  • ii.K=C+273=40+273=313 K K = C + 273 = 40 + 273 = 313\ \text{K}
Marking scheme
  • i. 40 °C.
  • ii. 313 K.
12.
[2]

Sudip says, “At noon on a clear June day after rain, I will see a rainbow if I stand with my back to the Sun.”

  • i.Is he right? Give a reason.
  • ii.Why is red always on the outside of a rainbow and violet on the inside?
Answer
  • i.No. The eye gets red light only from drops at about 42° from the line from the Sun through the head, so the rainbow is a circle centered on the point directly opposite the Sun. At noon in June the Sun is very high, much more than 42° above the horizon, so that circle lies wholly below the horizon. A rainbow is seen only when the Sun is low: in the morning or the late afternoon.
  • ii.Each color leaves the drops at its own angle to the sunlight: red at about 42° and violet, which bends most, at about 40°. So red reaches the eye from drops on a bigger circle, and violet from a slightly smaller circle inside it.
Marking scheme
  • i. No: the rainbow lies 42° from the point opposite the Sun, which at noon is so far below the horizon that the bow is too.
  • ii. Red comes out at about 42° and violet at about 40°, so red is on the bigger (outer) circle.
13.
[2]

Light travels at 3 × 10⁸ m/s in air and at 2.25 × 10⁸ m/s in water.

  • i.Calculate the refractive index of water.
  • ii.The refractive index of a kind of plastic is 1.6. Calculate the speed of light in it.
Answer
  • i.μ=cv=3×1082.25×108=1.33 \mu = \dfrac{c}{v} = \dfrac{3 \times 10^8}{2.25 \times 10^8} = 1.33
  • ii.v=cμ=3×1081.6=1.875×108 m/s v = \dfrac{c}{\mu} = \dfrac{3 \times 10^8}{1.6} = 1.875 \times 10^8\ \text{m/s}
Marking scheme
  • i. μ = c/v = 1.33.
  • ii. v = c/μ = 1.875 × 10⁸ m/s.
14.
[2]

Argon has atomic number 18 and atomic weight 39.9; potassium has atomic number 19 and atomic weight 39.1.

  • i.What would go wrong if these two elements were placed in order of atomic weight?
  • ii.How does the modern periodic law put them right?
Answer
  • i.Potassium (39.1) would come before argon (39.9). Potassium, a very reactive metal, would then fall among the noble gases, and argon, an unreactive gas, among the alkali metals: each would sit in a group whose properties it doesn’t share.
  • ii.The modern periodic law arranges the elements by atomic number, not atomic weight. Argon (18) comes before potassium (19), so argon falls in group 0 with the noble gases and potassium in group IA with the alkali metals, where their properties belong.
Marking scheme
  • i. Potassium would come before argon, so each would fall in a group of elements unlike it.
  • ii. Arranged by atomic number, argon (18) comes first: argon in group 0, potassium in group IA.
15.
[2]

In the ammonia fountain experiment, a flask of dry ammonia is held upside down with a long glass tube dipping into water that contains phenolphthalein. When a few drops of water are squeezed into the flask, water shoots up the tube like a fountain and turns pink.

  • i.Why does the water rush up into the flask?
  • ii.Why does it turn pink?
Answer
  • i.Ammonia is very soluble in water: the first few drops dissolve so much of it that the pressure inside the flask falls far below the air pressure outside. The air pressing on the water in the beaker pushes the water up the tube.
  • ii.The ammonia dissolves to make ammonium hydroxide, an alkali (a base), and phenolphthalein turns pink in a base.
Marking scheme
  • i. Ammonia is very soluble, so the pressure inside falls and the air pressure outside pushes the water up.
  • ii. Its solution (ammonium hydroxide) is a base, which turns phenolphthalein pink.
16.
[2]

Sort the following food preservatives into first-class and second-class preservatives: vinegar, sulphur dioxide, sugar, potassium nitrate.

Answer

First-class (from natural sources, used in the amount needed): vinegar and sugar.
Second-class (chemicals made in factories, used only in small amounts set by law): sulphur dioxide and potassium nitrate.

Marking scheme
  • First-class: vinegar and sugar.
  • Second-class: sulphur dioxide and potassium nitrate.

Section D

7 × 4 = 28

Write long answers to the following questions.

17.
[4]

The work done in lifting a load is W = mgh, where m is the mass of the load, g the acceleration due to gravity and h the height it is lifted.

  • i.Find the fundamental units of work from this formula. (1)
  • ii.Pressure is force per unit area. Write the fundamental units of pressure. (1)
  • iii.Weight = mg, so g can also be given in newtons per kilogram (N/kg). Show that N/kg is the same unit as m s⁻². (1)
  • iv.Asmita writes the formula for density as ρ = m × V, where V is the volume. Show, using units, that it cannot be correct. (1)
Answer
  • i.Unit of W = kg × m s⁻² × m = kg m² s⁻² (the joule).
  • ii.Unit of pressure = unit of force ÷ unit of area = kg m s⁻² ÷ m² = kg m⁻¹ s⁻² (the pascal).
  • iii.1 N = 1 kg m s⁻², so N/kg = kg m s⁻² ÷ kg = m s⁻², the unit of acceleration.
  • iv.Density is measured in kg m⁻³, but m × V would be in kg × m³ = kg m³. The two sides have different units, so the formula is wrong. (The correct one is ρ = m/V.)
Marking scheme
  • i. kg m² s⁻².
  • ii. kg m⁻¹ s⁻².
  • iii. N = kg m s⁻², so N/kg = m s⁻².
  • iv. m × V gives kg m³, not kg m⁻³ (density), so it is wrong.
18.
[4]

Bimala studied four plants, P, Q, R and S, and noted their features in the table below.

Plant Features
P grows on a damp wall; small stem and simple leaves held down by rhizoids; no vascular tissue; a capsule on a stalk at its top
Q large feather-like leaves rise from an underground rhizome with true roots; brown spots (sori) under some leaves; no seeds
R needle-like leaves; seeds lie open on the scales of woody cones
S flowers with parts in threes; parallel veins in the leaves; fibrous roots; seeds inside a fruit
  • i.Write the division or subdivision to which P and Q belong. (1)
  • ii.Write the subdivision to which R and S belong, and the feature of their seeds that separates them. (1)
  • iii.Q belongs to the division Tracheophyta, but P does not. Write a feature from the table that shows this. (1)
  • iv.Is S a monocot or a dicot? Give a reason from the table. (1)
Answer
  • i.P belongs to Bryophyta (it is a moss) and Q to Pteridophyta (it is a fern).
  • ii.R is a gymnosperm and S an angiosperm. A gymnosperm’s seeds are naked (on the cone scales, with no fruit); an angiosperm’s seeds are enclosed in a fruit.
  • iii.Q has true roots, and with them true stems and leaves with vascular tissue (xylem and phloem); P has only rhizoids and no vascular tissue.
  • iv.A monocot: its leaves have parallel veins (it also has fibrous roots, and its flower parts are in threes).
Marking scheme
  • i. P: Bryophyta; Q: Pteridophyta.
  • ii. R: gymnosperm (naked seeds); S: angiosperm (seeds in a fruit).
  • iii. Q has true roots (vascular tissue); P has only rhizoids and no vascular tissue.
  • iv. Monocot, with one reason (parallel veins, fibrous roots, flower parts in threes).
19.
[4]

A stone was dropped from rest and its velocity was measured every second (line A). A rover did the same experiment on Mars (line B). The graph shows the results. Ignore air resistance.

051015202530350123time (s)velocity (m/s)29.4 m/s11.1 m/sAB
Figure 1 The velocity of a falling stone on the Earth (A) and on Mars (B).
  • i.Calculate the acceleration of the stone from line A. (1)
  • ii.Calculate the acceleration due to gravity on Mars from line B. (1)
  • iii.Calculate how far the stone fell on the Earth in these 3 seconds. (1)
  • iv.The mass of the stone is 2 kg. Calculate its weight on Mars. (1)
Answer
  • i.g=v−ut=29.4−03=9.8 m/s2 g = \dfrac{v - u}{t} = \dfrac{29.4 - 0}{3} = 9.8\ \text{m/s}^2
  • ii.gMars=11.1−03=3.7 m/s2 g_{\text{Mars}} = \dfrac{11.1 - 0}{3} = 3.7\ \text{m/s}^2 , a little over a third of the Earth’s.
  • iii.h=ut+12gt2=0+12×9.8×32=44.1 m h = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2} \times 9.8 \times 3^2 = 44.1\ \text{m} . (The area under line A gives the same: 12×3×29.4=44.1 \tfrac{1}{2} \times 3 \times 29.4 = 44.1 m.)
  • iv.W=mg=2×3.7=7.4 N W = mg = 2 \times 3.7 = 7.4\ \text{N} . (Its mass is still 2 kg; only its weight changes.)
Marking scheme
  • i. 9.8 m/s².
  • ii. 3.7 m/s².
  • iii. h = ut + ½gt² = 44.1 m.
  • iv. W = mg = 7.4 N.
20.
[4]

A mechanic in Butwal lifts one side of a truck with a hydraulic jack. Its pumping piston has an area of 4 cm² and its lifting piston an area of 600 cm². The load on the lifting piston is 30,000 N.

  • i.Calculate the force needed on the pumping piston. (1)
  • ii.To raise the truck by 2 cm, how far must the pumping piston be pushed down in all? (1)
  • iii.His helper says, “Let’s fill the jack with air instead of oil, because air is lighter.” Is he right? Give a reason. (1)
  • iv.One day the mechanic pumps many times, but the truck doesn’t rise at all. Write one likely reason. (1)
Answer
  • i.By Pascal’s law, F1A1=F2A2 \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2} , so F1=F2×A1A2=30,000×4600=200 N F_1 = F_2 \times \dfrac{A_1}{A_2} = 30{,}000 \times \dfrac{4}{600} = 200\ \text{N} .
  • ii.The lifting piston has 600 ÷ 4 = 150 times the area, so the oil pushed out by the small piston raises it only 1/150 as far. To raise it 2 cm, the small piston must go down 150 × 2 = 300 cm (3 m) in all, stroke after stroke. The jack multiplies force, not energy.
  • iii.No. Air can be compressed: pushing the piston would only squeeze the air, and very little of the push would reach the lifting piston. Oil can’t be compressed, so it passes on the pressure fully.
  • iv.Any one: the release valve is open, so the oil flows back into the reservoir; the oil has leaked out or the reservoir is empty; air has got into the jack.
Marking scheme
  • i. F₁ = F₂ × A₁/A₂ = 200 N.
  • ii. 300 cm (3 m), because the large piston has 150 times the area.
  • iii. No: air is compressed instead of passing on the pressure (oil can’t be compressed).
  • iv. One likely reason (release valve open, oil leaked or used up, air in the jack).
21.
[4]

Sujata wound insulated copper wire round a soft iron nail and connected it to a cell and a switch, as shown in the figure. When she closed the switch, the nail picked up some pins. The arrows show the direction of the current.

+−cellswitchcurrentpinsABsoft iron nailinsulated copper wire
Figure 2 An electromagnet made with a soft iron nail.
  • i.Which end of the nail, A or B, is the north pole? Name the rule you used. (1)
  • ii.What happens to the pins when the switch is opened? Why? (1)
  • iii.Write any two ways to make the electromagnet hold more pins. (2)
Answer
  • i.End A. By Maxwell’s right-hand grip rule: hold the coil in the right hand with the fingers curling the way the current goes round the turns (up the front of each turn), and the thumb points to the north pole, at A.
  • ii.The pins fall off. The magnetism of soft iron is temporary: when the current stops, the coil makes no field and the nail loses its magnetism.
  • iii.Any two: increase the current (use two or more cells); wind more turns of wire round the nail; use a thicker soft iron core.
Marking scheme
  • i. A, by the right-hand grip rule.
  • ii. They fall off: with no current the soft iron loses its magnetism (temporary magnet).
  • iii. One way: more current (more cells).
  • iii. A second way: more turns of wire (or a thicker soft iron core).
22.
[4]

Sapana dropped equal pieces of magnesium ribbon into equal volumes of hydrochloric acid in three test tubes, and timed how long each piece took to disappear.

Test tube Acid Temperature Time taken
P dilute 20 °C 120 s
Q dilute 30 °C 60 s
R twice as concentrated 20 °C 55 s
  • i.Write the balanced chemical equation of the reaction, and name its type. (1)
  • ii.What do test tubes P and Q show? Explain in terms of the collisions of particles. (1)
  • iii.Why was the reaction faster in R than in P? (1)
  • iv.Predict about how long the ribbon would take in the dilute acid at 40 °C. Give a reason. (1)
Answer
  • i.Mg+2HCl→MgCl2+H2 \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 . It is a single displacement reaction: magnesium, above hydrogen in the reactivity series, displaces hydrogen from the acid.
  • ii.A rise in temperature makes the reaction faster: 10 °C more halved the time. Hotter particles move faster, so they collide more often, and many more of the collisions have enough energy (the activation energy) to react.
  • iii.The acid in R is more concentrated: there are more acid particles in the same volume, so they collide with the magnesium more often.
  • iv.About 30 s. In P and Q, a rise of 10 °C halved the time (doubled the rate), so another 10 °C should halve it again: 60 ÷ 2 = 30 s.
Marking scheme
  • i. Mg + 2HCl → MgCl₂ + H₂; single displacement.
  • ii. Higher temperature, faster reaction: faster particles collide more often and with more energy.
  • iii. Higher concentration: more acid particles, so more collisions.
  • iv. About 30 s, since 10 °C more halves the time.
23.
[4]
  • i.What is a homologous series? (1)
  • ii.Write any two characteristics of the members of a homologous series. (2)
  • iii.Write one use each of methyl alcohol and ethyl alcohol. (1)
Answer
  • i.A homologous series is a series of organic compounds that can be written with the same general formula, in which each member differs from the next by a –CH₂– group (for example, the alkanes CH₄, C₂H₆, C₃H₈…).
  • ii.Any two:
  • a.They have the same general formula (CₙH₂ₙ₊₂ for the alkanes).
  • b.Neighboring members differ by CH₂, and by 14 in molecular mass.
  • c.They have the same functional group, so they have similar chemical properties.
  • d.Their physical properties change gradually: the bigger the molecule, the higher the boiling point.
  • iii.Methyl alcohol: any one, such as making methylated spirit, as a solvent for paints and varnishes, or making formaldehyde. Ethyl alcohol: any one, such as sterilizing skin and wounds (hand sanitizers), preserving biological specimens, in red thermometers, or as a fuel in spirit lamps.
Marking scheme
  • i. Compounds with the same general formula, each member differing from the next by –CH₂–.
  • ii. One correct characteristic.
  • ii. A second correct characteristic.
  • iii. One correct use of methyl alcohol and one of ethyl alcohol.

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