Model Test – Solved ഉത്തരങ്ങൾ
Every question of the model paper, in the same order, with the full answer, the marking scheme, an animated step-by-step solution for the longer questions, and a link to the page where the idea is taught. Try each one again before you open the answer, and mark your own paper honestly.
In equations a coefficient that has just been changed is red; the atom tally turns green when an atom is equal on both sides. Particle pictures show the four reaction types with coloured balls (A + B → AB, AB → A + B, A + BC → AC + B, AB + CD → AD + CB). In energy bars each bar is one particle; the dashed red line is the threshold energy and the bars that reach it turn green. On a volume–time graph the steeper the curve, the faster the reaction.
The chain used in every explanation: factor → more collisions, or more particles with at least the threshold energy → more effective collisions every second → higher rate. A catalyst increases the rate and is left chemically unchanged at the end.
💡 In which one does a new substance form?
When milk sets into curd a new substance with a new taste and smell forms – a chemical change (രാസമാറ്റം). Melting, dissolving and boiling change only the state or the form; no new substance forms – physical changes (ഭൗതികമാറ്റം).
Marking: 1 mark for (c).
📖 Review this on Page 1 · Physical & chemical changes · Page 3 · Signs of a chemical change
💡 Look at the arrow ↓ after Cu(OH)₂.
The blue solid, copper hydroxide Cu(OH)2, does not dissolve in water and sinks – it is a precipitate. The ↓ after its formula shows this. Forming a precipitate is one of the signs of a chemical change.
Marking: 1 mark.
📖 Review this on Page 3 · Signs of a chemical change · Page 5 · Double decomposition & displacement
💡 How many reactants are there on the left?
Carbon and oxygen join to give one product – combination. In 2HgO → 2Hg + O2 one compound, mercury(II) oxide, breaks into two substances, mercury and oxygen – decomposition (here on heating, a thermal decomposition).
Marking: 1 mark.
📖 Review this on Page 4 · Combination & decomposition
💡 Which element is pushed out of the compound?
Magnesium, an element, takes the place of hydrogen in hydrochloric acid and hydrogen is set free – pattern A + BC → AC + B: a displacement reaction (ആദേശ രാസപ്രവർത്തനം).
Marking: 1 mark for (c).
📖 Review this on Page 5 · Double decomposition & displacement
💡 It is the “bar” every collision has to cross.
Only collisions between particles that have at least the threshold energy give products – these are the effective collisions.
Marking: 1 mark.
📖 Review this on Page 8 · Temperature & catalysts · Page 6 · Rate & collision theory
💡 What happens to the number of acid particles in each mL?
Adding water makes the acid dilute (നേർപ്പിച്ച): fewer acid particles per unit volume → fewer collisions → the reaction becomes slower. Warming, powdering and a higher concentration all increase the rate.
Marking: 1 mark for (c).
📖 Review this on Page 7 · Nature, concentration & surface area
💡 Which one is over before you can blink?
An explosion (സ്ഫോടനം) is over in a fraction of a second. Ripening takes days, rusting (തുരുമ്പിക്കൽ) days to months and weathering of rocks many years.
Marking: 1 mark for (c).
📖 Review this on Page 6 · Rate & collision theory
💡 Chew rice for a long time – it tastes sweet.
Amylase – an enzyme (എൻസൈമുകൾ), that is, a biological catalyst. That is why chewed rice begins to taste sweet.
Marking: 1 mark.
📖 Review this on Page 8 · Temperature & catalysts
(b) Which sign of a chemical change is seen here? What is such a reaction called?
💡 Count the products. What do you feel if you touch the container?
(a) Two substances, calcium oxide and water, join to form one new compound, calcium hydroxide – a combination reaction (സംയോജന രാസപ്രവർത്തനം). Pattern A + B → AB.
(b) The mixture becomes hot: heat is given out (an energy change). A reaction that gives out heat is an exothermic reaction (താപമോചക പ്രവർത്തനം).
Marking: (a) type ½, reason ½; (b) sign ½, name ½.
📖 Review this on Page 4 · Combination & decomposition · Page 3 · Signs of a chemical change
(b) Which ions exchange partners in this reaction?
💡 Write the ions of each compound under its formula.
(a) Two compounds react to form two new compounds – a double decomposition reaction (ദ്വിവിഘടന രാസപ്രവർത്തനം) of the precipitation kind (അവക്ഷിപ്തരൂപീകരണ രാസപ്രവർത്തനങ്ങൾ): a white precipitate of barium sulphate, BaSO4, forms.
(b) Before: Ba2+ with Cl−, and Na+ with SO42−. After: Ba2+ with SO42−, and Na+ with Cl−. The barium ion and the sodium ions exchange partners (the chloride and sulphate ions change places) – AB + CD → AD + CB.
Marking: (a) type ½, sign ½; (b) 1 mark.
📖 Review this on Page 5 · Double decomposition & displacement
(b) Why do the gases not react without the spark?
💡 Is every collision an effective collision?
(a) Collision theory (കൊളിഷൻ സിദ്ധാന്തം): a reaction takes place when reactant particles collide; only the collisions in which the particles have at least the threshold energy form products – these are effective collisions (ഫലവത്തായ കൂട്ടിമുട്ടലുകൾ).
(b) At room temperature almost none of the molecules have the threshold energy, so their collisions are not effective. The spark gives some molecules enough energy; their effective collisions start the reaction (the energy it releases then keeps it going – very fast, an explosion).
Marking: (a) 1 mark (collision + threshold energy); (b) 1 mark.
📖 Review this on Page 6 · Rate & collision theory · Page 8 · Temperature & catalysts
(b) Explain, using kinetic energy and threshold energy, how this keeps the fish fresh for longer.
💡 What does cooling do to the speed of particles?
(a) Temperature (താപനില).
(b) Spoiling is a set of chemical reactions. At a low temperature the particles have less kinetic energy (ഗതികോർജം), so fewer particles reach the threshold energy → fewer effective collisions → the reactions of spoiling become slower. (A refrigerator works in the same way.)
Marking: (a) ½ mark; (b) 1½ marks (less kinetic energy ½, fewer reach the threshold energy ½, slower rate ½).
📖 Review this on Page 8 · Temperature & catalysts
(b) What does the ash left behind tell you about it?
💡 A substance that speeds up a reaction but is left at the end is called …?
(a) The ash (its metal compounds) increases the rate of the burning of sugar – it acts as a catalyst (ഉൽപ്രേരകം). (Burning of sugar: C12H22O11 + 12O2 → 12CO2 + 11H2O.)
(b) The ash remains after the reaction – it has not undergone a permanent chemical change. This is the feature of a catalyst: it increases the rate without being used up.
Marking: (a) 1 mark; (b) 1 mark.
📖 Review this on Page 8 · Temperature & catalysts
| What is done and seen | Equation | |
|---|---|---|
| (a) | A zinc strip is dipped in blue copper sulphate solution. The blue colour fades and a reddish-brown layer forms on the zinc. | Zn + CuSO4 → ZnSO4 + Cu |
| (b) | Potassium hydroxide solution is mixed with dilute nitric acid. The test tube becomes warm. | KOH + HNO3 → KNO3 + H2O |
| (c) | Dilute hydrochloric acid is poured on washing soda (sodium carbonate). It fizzes. | Na2CO3 + 2HCl → 2NaCl + H2O + CO2↑ |
💡 Look at how many reactants are elements.
(a) Colour change (നിറവ്യത്യാസം) – the blue colour fades and red-brown copper appears. Zinc takes the place of copper in copper sulphate: displacement.
(b) Heat is given out – the tube becomes warm. Two compounds exchange ions and water forms (a compound that does not split into ions): double decomposition – a neutralisation (നിർവീരീകരണം).
(c) A gas is given off (വാതകം) – effervescence of CO2. Two compounds exchange ions: double decomposition of the gas-forming kind (വാതകരൂപീകരണ രാസപ്രവർത്തനങ്ങൾ).
Marking: ½ for each sign, ½ for each type.
📖 Review this on Page 3 · Signs of a chemical change · Page 5 · Double decomposition & displacement
(b) Fe + Cl2 → FeCl3
(c) NaHCO3 → Na2CO3 + H2O + CO2 (baking soda heated in a cake)
💡 Balance the atom that appears in only one substance on each side first.
(a) Al: 2 on the right → 2Al; Fe: 2 on the left → 2Fe; O 3 = 3: 2Al + Fe2O3 → Al2O3 + 2Fe – aluminium takes the place of iron: displacement (തെർമൈറ്റ്).
| Atom | Reactants | Products |
|---|---|---|
| Al | 2 | 2 |
| Fe | 2 | 2 |
| O | 3 | 3 |
(b) Cl: 2 on the left, 3 on the right – the smallest common number is 6: 3Cl2 and 2FeCl3, then 2Fe: 2Fe + 3Cl2 → 2FeCl3 – two substances give one product: combination.
| Atom | Reactants | Products |
|---|---|---|
| Fe | 2 | 2 |
| Cl | 6 | 6 |
(c) Na: 2 on the right → 2NaHCO3; then H 2 = 2, C 2 = 2, O 6 = 6: 2NaHCO3 → Na2CO3 + H2O + CO2↑ – one compound breaks into three: decomposition (thermal). The CO2 makes the cake rise.
| Atom | Reactants | Products |
|---|---|---|
| Na | 2 | 2 |
| H | 2 | 2 |
| C | 2 | 2 |
| O | 6 | 6 |
Marking: ½ for each balanced equation, ½ for each type.
📖 Review this on Page 4 · Combination & decomposition · Page 5 · Double decomposition & displacement · Page 1 · Physical & chemical changes
| Time (minutes) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Volume of CO2 (mL) | 0 | 36 | 60 | 76 | 86 | 90 | 90 |
(b) Why does the volume stop changing after 5 minutes?
(c) The experiment is repeated with the same mass of marble powder. How will the graph change? Give the reason.
💡 The steepness (slope) of the graph shows the rate.
(a) 1st minute: 36 − 0 = 36 mL. 4th minute: 86 − 76 = 10 mL. The reaction is fastest in the 1st minute – the steepest part of the graph. At the start the reactants are present in the largest amount (highest concentration), so there are the most effective collisions every second; as they are used up the rate falls.
(b) The marble (calcium carbonate) is completely used up (the acid is in excess), so the reaction stops and no more CO2 forms – the graph becomes flat at 90 mL.
(c) The curve is steeper at the start and becomes flat sooner, but the final volume is the same (90 mL) because the mass of calcium carbonate is the same. Reason: the powder has a much larger surface area (പ്രതലപരപ്പളവ്) → more collisions with acid particles every second → a faster rate.
Marking: (a) the two volumes ½, fastest minute with the reason ½; (b) 1 mark; (c) shape ½, reason ½.
📖 Review this on Page 10 · Extended activities · Page 7 · Nature, concentration & surface area
(b) A glowing incense stick bursts into flame when it is put into a jar of pure oxygen.
(c) Potassium reacts violently with cold water, but water can be stored in a copper vessel for years.
💡 Powder – more air – different metals: which factor is each?
(a) Surface area (പ്രതലപരപ്പളവ്): the powder has a far larger surface → many more collisions with oxygen at the same time → it burns very fast with a brilliant flash.
(b) Concentration (ഗാഢത): pure oxygen has about five times as many oxygen molecules per unit volume as air (about 21 % oxygen) → more effective collisions → the stick burns brightly.
(c) Nature of reactants (അഭികാരകങ്ങളുടെ സ്വഭാവം): potassium gives away its single outermost electron very easily, so it is extremely reactive; copper is much less reactive and does not react with water.
Marking: 1 mark each (factor ½, reason ½).
📖 Review this on Page 7 · Nature, concentration & surface area
(b) What is the role of V2O5? What will you find about its amount and its chemical nature at the end?
(c) What does the sign ⇌ tell you about this reaction?
(d) How do the pressure (2 atm) and the temperature (720 K) help the reaction to go faster?
(e) Write the balanced equation for the Haber process and name its catalyst. Where in Kerala is this reaction used on a large scale?
💡 Underline the catalyst and the conditions in the passage first.
(a) Two substances, sulphur dioxide and oxygen, join to form one product, sulphur trioxide – a combination reaction.
(b) V2O5 is a catalyst (ഉൽപ്രേരകം) – it increases the rate. At the end its amount is the same and it is chemically unchanged: a catalyst does not undergo a permanent chemical change.
(c) ⇌ shows a reversible reaction (ഉഭയദിശാ പ്രവർത്തനങ്ങൾ): SO2 and O2 form SO3, and at the same time some SO3 breaks back into SO2 and O2.
(d) Pressure pushes more gas molecules into each unit volume (യൂണിറ്റ് വ്യാപ്തം) – a higher concentration → more collisions → more effective collisions. High temperature gives the molecules more kinetic energy → more of them reach the threshold energy → more effective collisions. Both make the reaction faster.
(e) Haber process (ഹേബർ പ്രക്രിയ): N2(g) + 3H2(g) Fe⇌ 2NH3(g) (200 atm, 700 K); the catalyst is iron (Fe). Ammonia made this way is used to make fertilisers in factories such as FACT at Udyogamandal.
Marking: 1 mark for each part ((d) ½ for pressure, ½ for temperature; (e) ½ equation, ½ catalyst and place).
📖 Review this on Page 4 · Combination & decomposition · Page 8 · Temperature & catalysts · Page 11 · Applications
(b) Name the factor that makes the batter rise faster in April. Where should grandmother keep the batter on a cool July evening so that it is ready by morning?
(c) What are enzymes? Name one enzyme in your body and the change it speeds up.
(d) Why does a split log catch fire faster than a whole log? Explain using collisions.
(e) Why do the flames become bright when air is blown into the fire? Name the factor.
💡 Find the factor in each sentence of the passage before you answer.
(a) Any two: heat and light are given out; smoke and gases are given off; the wood turns into black charcoal and grey ash – new substances that cannot be turned back into wood.
(b) Temperature: in April the particles have more kinetic energy, so more of them reach the threshold energy and the enzyme-catalysed reactions of fermentation go faster. In July she should keep the covered batter in a warm place in the kitchen (for example near the hearth after cooking), not in the refrigerator.
(c) Enzymes (എൻസൈമുകൾ) are biological catalysts – substances made by living things that speed up reactions in them. Example: amylase in saliva speeds up the change of starch into maltose.
(d) Surface area (പ്രതലപരപ്പളവ്): splitting the log exposes much more wood surface to the air, so many more oxygen molecules collide with the wood at the same time → more effective collisions → it catches fire and burns faster.
(e) Concentration (ഗാഢത) of oxygen: blowing pushes more air, so more oxygen molecules reach the burning wood every second → more effective collisions → brighter, faster burning.
Marking: 1 mark for each part (½ + ½ where there are two answers).
📖 Review this on Page 3 · Signs of a chemical change · Page 8 · Temperature & catalysts · Page 7 · Nature, concentration & surface area · Page 11 · Applications
Marks summary
| Section | Questions | Marks each | Total | Topics |
|---|---|---|---|---|
| A | 1–8 | 1 | 8 | chemical change (curd), precipitate, combination : decomposition, displacement (Mg + HCl), threshold energy, diluting an acid, the fastest reaction, amylase |
| B | 9–13 | 2 | 10 | quicklime and water (combination, exothermic), BaCl₂ + Na₂SO₄ (ion exchange, precipitate), collision theory and the H₂–O₂ balloon, fish on ice (temperature), sugar cube and ash (catalyst) |
| C | 14–17 | 3 | 12 | signs and types of three reactions, balance and classify (thermite, FeCl₃, baking soda), volume–time graph of marble and acid, three factors with reasons |
| D | 18–19 | 5 | 10 | contact and Haber processes (catalyst, ⇌, pressure and temperature), Kerala case study: batter, firewood hearth and blowpipe |
| Total | 40 | |||
- Cool-off (15 min) – read the whole paper. Mark the equations to classify (Q3, Q4, Q9, Q10, Q14, Q15, Q18) and the “explain” questions that need collision theory (Q11, Q12, Q16, Q17, Q18, Q19).
- Section A – about 8 minutes. One word or letter each.
- Section B – about 20 minutes. Name the factor or type first, then give the reason in one or two sentences.
- Section C – about 27 minutes. In Q15 check every atom after balancing; in Q16 subtract the readings carefully.
- Section D – about 25 minutes. Underline the substances, the catalyst and the conditions in the passage; answer each part separately.
- Checking – keep 10 minutes: every equation balanced? Every type named in full? Every explanation ends with “more effective collisions → faster”?
Count your marks: 32 or more – excellent; 24–31 – good, revise the pages linked under the questions you missed; below 24 – go back through pages 3 to 8 with their labs and do the scored quiz again. Then return to the chapter home and tick off what you have learned.