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These NCERT Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Acids, Bases and Salts NCERT Solutions for Class 10 Science Chapter 2

Class 10 Science Chapter 2 Acids, Bases and Salts InText Questions and Answers

In-text Questions (Page 18)

Question 1.
You have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?
Answer:
We dip the red litmus paper in any one of the three test tubes. If there is no change in the colour of the litmus paper, it means this test tube contains water or acidic solution.

If the colour of the red litmus paper turns blue, it means this test tube contains basic solution.

Now, again we dip this litmus paper in any one of the remaining two test tubes. If we do not observe any change in the colour of the litmus paper it is surely water and if a colour change is noticed, then it will be an acidic solution.

In-text Questions (Page 22)

Question 1.
Why should curd and sour substances not be kept in brass and copper vessels?
Answer:
Brass is an alloy of copper and zinc, Brass and copper vessels reacts with atmospheric oxygen and to form metal oxides. The nature of metal oxides is basic. Curd and sour substances contains acids, e.g., curd contains lactic acid and orange contains citric acid. If we keep curd and sour substances in brass and copper vessels the layer of oxides present on the surface of brass and copper reacts with the acids present in curd and sour substances to form salt and water and taste of sour substances chages from sour to bitter. So curd and sour substance are not be kept in brass and copper vessels.

Question 2.
Which gas is usually liberated when an acid reacts with a metal? Illustrate with an example. How will you test for the presence of this gas?
Answer:
Hydrogen has is usually liberated when an acid reacts with a metal.
e.g.

This gas buns with a pop sound when a burning candle is placed near the evolved gas.

Question 3.
A Metal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
Answer:
Metal compound ‘A’ is CaCO₃

In-text Questions (Page 25)

Question 1.
Why do HCl, HNO₃, etc show acidic character in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?
Answer:
A substance is said to be an acid, when it produces H⁺ ions in its aqueous solution and turns blue litmus paper into red. HCl, HNO₃, etc. ionise in their aqueous solutions and produce H⁺ ions. They turn the colour of blue litmus paper into red. When electric current is passed through the aqueous solutions of acids, the H⁺ ions reach the cathode and form hydrogen. But in their aqueous solutions alcohol and glucose do not produce H⁺ ions and they do turn blue litmus into red. So they do not show acidic character in their aqueous solution.

Question 2.
Why does an aqueous solution of an add conduct electricity?
Answer:
Acids are ionised in their aqueous solution and produce H⁺ (aq) ions. These ions are responsible for the flow of electricity in the aqueous solutions of acids. H⁺ ions migrates towards cathode and gain electron to produce hydrogen gas. So when electricity is passed through the aqueous solution of an acid, hydrogen gas is liberated at cathode.

So an aqueous solution of an acid conducts electricity because it contains H⁺ (aq) ions or hydronium ions.

Question 3.
Why does dry HCl gas not change the colour of the dry litmus paper?
Answer:
The separation of H⁺ ion from HCl molecules cannot occur in the absence of water. In fact H⁺ ion is an atom that has lost an eletron, it is simply isolated proton and is smallest positive ion. It is attracted very strongly towards electrons of anything. So the separation of H⁺ ion from HCl molecules cannot occur and the reaction.
H – Cl(g) H⁺ → + Cl^–

cannot occur in the absence of water, because highly concentrated positive charge of TP would join back with the Cl^– ion to get back electron and reform covalent ‘H – Cl‘ molecules. It means dry HCl does not contain H⁺ ions so it does not change the colour of the dry litmus paper.

Question 4.
While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid ?
Answer:
The process of dissolving an add or a base in water is a highly exothermic process. Be careful while mixing concentrated sulphuric acid with water During dilution of an acid, the acid must always be added slowly to water with constant stirring. It water is added to a concentrated acid the heat generated may cause the mixture lo splash out from the container and cause bums. The glass container may also break due to heal release during dilution. So it is recommended that the acid should be added to water and not water to the acid.

Question 5.
How is the concentration of hydronium ions (H₃O)⁺ affected when a solution of an acid is diluted?
Answer:
An aqueous solution of an acid contains the specific number of H₃O⁺ ions per unit volume. These H₃O⁺ ions are responsible for the acidic character of an acid. When water is added in an acid it results in decrease in inconcentration of H₃O⁺ ions per unit volume of the acid because the specified volume of an acid contains the definite no. of H₃O⁺ ions, when its volume increases by addition of water, the no. of H₃O⁺ ions decreases per unit volume of the acid. So we can say that on dilution the concentration of H₃O⁺ ions decreases per unit volume of the acid.

Question 6.
How is the concentration of hydroxide ions (OH^–) affected when excess base is dissolved in a solution of sodium hydroxide?
Answer:
A solution of sodium hydroxide contain Na⁺ and OH^– ions. When a base is dissolved in a solution of sodium hydroxide, it results in increase of concentration of OH^– ions per unit volume because added base is ionised and produce a large no. of hydroxide ions (OH^–) in the solution of sodium hydroxide.

In-text Questions (Page 28)

Question 1.
You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
Answer:
The hydrogen ion concentration of a solution is calculated by the formula :
[H⁺] = 10^-pH
∴ The conc. of H⁺ ions in solution ‘A’ = 10^-6
and the conc, of H⁺ ions in solution ‘B’ = 10^-8
So solution ‘A’ has more hydrogen ion concentration than ‘B’.
Solution A’ is acidic and solution ‘B’ is basic.

Question 2.
What effect does the concentration of H⁺(aq) ions have on the nature of the solution?
Answer:
On the basis of concentration of H⁺ (aq) ions in a solution we can justify the nature of the solution i.e., whether it is acidic, basic or neutral. A solution which has the concentration of H⁺ (aq) ion equal to 10^-7 mole/litre will be neutral and if the concentration is slightly less than 10^-7 then it will be basic and if the cone, of H⁺ (aq) ions in a solution is slightly higher than 10^-7 then it will be acidic.

Question 3.
Do basic solutions also have H⁺(aq) ions? If yes, then why are these basic?
Answer:
Yes, basic solutions also have H⁺ (aq) ions and similarly acidic solutions have OH^– (aq). because acids or bases shows acidic or basic character in water. Distilled water also has 10^-7 mol/L, H⁺ and 10^-7 mole/L, OH^– ions at 298K.

If an aqueous solution has an equal concentration of hydrogen ions and hydroxide ions in it, it is said to be neutral. Now, if an aqueous solution has more of hydrogen ions and less of hydroxide ions, it will be acidic solution. On the other hand, if an aqueous solution has more of hydroxide ions and less of hydrogen ions it will be basic in nature.

Question 4.
Under what soil condition do you think a fanner would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?
Answer:
If the soil has acidic nature, then famer would treat the soil of his fields with quick lime or slaked lime or chalk to maintain its pH near about neutral.

In-text Questions (Page 33)

Question 1.
What is the common name of the compound CaOCl₂?
Answer:
Bleaching powder.

Question 2.
Name the substance which cm treatment with chlorine yields bleaching powder,
Answer:
Dry slaked lime : Ca(OH)₂
Chlorine : Cl₂

Question 3.
Name the sodium compound which is used for softening hard water.
Answer:
Washing soda : Sodium carbonate (Na₂CO₃)

Question 4.
What will happen if a solution of sodium hydrocarbonate is heated? Give the equation of the reaction involved.
Answer:
On heating sodium hydrocarbonate, it produces sodium carbonate, carbondioxide and water.

Question 5.
Write an equation to show the reaction between Plaster of Paris and water,
Answer:

Class 10 Science Chapter 2 Acids, Bases and Salts Textbook Questions and Answers

Page No. 34

Question 1.
A solution turns red litmus blue, its pH is likely to be:
(a) 1
(b) 4
(c) 5
(d) 10
Answer:
(d) 10

Question 2.
A solution reacts with crushed egg-shells to give a gas that turns lime-water milky. The solution contains.
(a) NaCl
(b) HCl
(c) LiCl
(d) KCl
Answer:
(b) HCl

Question 3.
10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a given solution of HCl. If we take 20 mL of the same solution of NaOH, the amount of HCl solution (the same solution as before) required to neutralise it will be:
(a) 4 mL
(b) 8 mL
(c) 12 mL
(d) 16 mL
Answer:
(d) 16 mL

Question 4.
Which one of the following types of medicines is used for treating indigestion ?
(a) Antibiotic
(b) Analgesic
(c) Antarid
(d) Antiseptic
Answer:
(c) Antacid.

Question 5.
Write word equations and then balanced equations for the reactions taking place when:
(a) dilute sulphuric add reacts with zinc granules.
(b) dilute hydrochloric acid reacts with magnesium ribbon.
(c) dilute sulphuric acid reacts with aluminium powder.
(d) dilute hydrochloric acid reacts with iron fillings.
Answer:
(a) Dilute sulphuric acid reacts with zinc granules to form zinc sulphate and hydrogen.
Sulphuric acid + Zinc → Zinc Sulphate + Hydrogen
H₂SO₄ + Zn → ZnSO₄ + H₂

Step I.

The above equation is a balanced equation because it contains the equal no. of atoms of different elements in reactants to the no. of atoms of different elements in products.

(b) Hydrochloric add + Magnesium → Magnesium chloride + Hydrogen
HCl + Mg → MgCl₂ + H₂
Step I.

Step II. Balance hydrogen and chlorine by multiplying ‘HCl’ with ‘2′ we get
2 HCl + Mg → MgCl₂ = H₂

(c) Sulphuric acid + Aluminium → Aluminium sulphate + Hydrogen
H₂SO₄ + Al → Al₂(SO₄)₃ + H₂
Step I.

Step II. Balance oxygen by multiplying ‘H₂SO₄‘ with ‘3’ we get:
3H₂SO₄ + Al → Al₂(SO₄)₃ + H₂
Step III. Balance aluminium by multiplying ‘Al’ in reactants side with ‘2’ we get;
3H₂SO₄ + 2Al → Al₂(SO₄)₃ + H₂
Step IV. Balance hydrogen by multiplying ‘H₂‘ in product side with ‘3’ we get:
3H₂SO₄ + 2Al → Al₂(SO₄)₃ + 3H₂
This is a balance equation.

(d) Hydrochloric acid + Iron → Iron (III) chloride + Hydrogen
HCl + Fe → FeCl₃ + H₂
Step I.

Step II. Balance chlorine by multiplying ‘FeCl₃‘ with ‘2’ and ‘HCl’ with ‘6’ we get:
6HCl + Fe → 2FeCl₃ + H2
Step III. Balance hydrogen and iron by multiplying ‘Fe’ with ‘2’ and ‘H₂‘ with ‘3’ we get:
6HCl + 2Fe → 2FeCl₃ + 3H₂
This is a balance equation.

Question 6.
Compounds such as alcohols and glucose also contain hydrogen but are not categorised as adds. Describe an activity to prove it.
Answer:
A substance is said to be an acid when it can produce H⁺ ions in its aqueous solution and turns blue litmus paper to red.

HCl, HNO₃ etc. are ionized in their aqueous solution and furnished H⁺ ions. When electricity is passed through aqueous solutions of acids, hydrogen liberates at cathode.

Compounds such as alcohols and glucose contain hydrogen but they do not ionise in their aqueous solution and not produce H⁺ ions, and they do not turn blue litmus to red so they are not categorised as acids.

Activity: Take four test tubes namely, A, B, C and D. Add ethyl alcohol (alcohol) aq. solution of glucose ; hydrochloric acid and nitric acid, 2 mL of each in test tube A, B, C and D respectively. Now add 1-2 drops of blue litmus solution in each test tube. We observe that the colour of test tube ‘A’ and ‘B’ remain unchanged however the colour of test tube and ‘C’ and ‘D’ turns red. If proves that alcohol and glucose are not categorised as acid.

Question 7.
Why does distilled water not conduct electricity, whereas rainwater does ?
Answer:
Distilled water does not contain ions (H⁺ or OH^–) or it contains a very poor concentration of H⁺ or OH^– ions ≈ 1 × 10^-7 moles per litre. So it does not conduct electricity, whereas rainwater has some acidic character due to the presence of acids such as H₂SO₄, HNO₃ etc. So rainwater has a sufficient amount or concentration of H⁺ ions, so it conducts electricity.

Question 8.
Why do acids not show acidic behaviour in the absence of water ?
Answer:
A substance is said to be an acid when it can furnish H⁺ ions in its aqueous solution. The separation of H⁺ ions from the molecules of an acid cannot occur in the absence of water. The molecule of water are polar in nature, so the negative pole of ‘H₂O’ molecule pull the H⁺ ions from acids towards itself and separate them from acids e.g,
HCl + H₂O → H₃O⁺ + Cl^–. Hydrogen ions cannot exist alone, but they exist after combining with water molecules. Thus hydrogen ions must always be shown as H⁺ (aq) of hydronium ion (H₃O⁺).

In absence of water ‘H⁺‘ are does not furnish by an acid so it does not show acidic behaviour.

Question 9.
Five solutions A, B, C, D and E when tested with universal indicater showed pH as 4, 1, 11, 7 and 9, respectively. Which solution is:
(a) neutral ?
(b) strongly alkaline ?
(c) strongly acidic
(d) weakly acidic ?
(e) weakly alkaline ?
Arrange the pH in increasing order of hydrogen ion concentration.
Answer:
(a) Solution ‘D’ is neutral because it has pH 7.
(b) Solution ‘C’ is strongly alkaline because it has pH 11.
(c) Solution ‘B’ is strongly acidic because it has pH 1.
(d) Solution ‘A’ is weakly acidic because it has pH 4.
(e) Solution ‘E’ is weakly alkalin because it has pH 9.

Hydrogen ion concentration is inversely proportional to the pH of the solution. It means, higher the ‘pH’ lower will be the hydrogen ion concentration.

The ‘pH’ of the given solution can be arranged m increasing order of hydrogen-ion concentration as follows :
pH : 11 > 9 > 7 > 4 > 1
Hydrogen ion concentration : C < E < D < A < B

Question 10.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A which acetic acid (CH₃COOH) is added to test tube B. In which test tube will the fizzing occurs more vigorously and why ?
Answer:
Fizzing occurs more vigourosly in test tube A’ because hydrochlonic acid is a strong acid while acetic acid is a weak acid.

Magnesium metal reacts with acids and produce repective salt and hydrogen gas.

Question 11.
Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd ? Explain your answer.
Answer:
The fresh milk has a pH of 6. Milk has slightly sweet taste But when it turns into curd its taste has changed from lightly sweet taste to intense sour taste. It means its pH has changed. The intense sour taste of curd is the presence of the strong acidic character. So when milk changes into curd its pH decreases because acidic character increases.

Question 12.
A milkman adds a very small amount of baking soda to fresh milk.
(a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline ?
(b) Why does this milk take a long time a set as curd ?
Answer:
(a) Milk contains lactose and pH of milk is approximately 6. The micro-organism which converts milk into curd work effectively in slightly acidic medium. Baking soda has slightly alkaline nature. So milkman adds small quantity of baking soda to protect it from curding because in alkaline medium micro-organism do not function properly and thus milk is prevented from rancidity or curding. So milkman makes milk alkaline.

(b) The process of converting milk into curd takes place in slightly acidic medium, Micro-organism converts lactose of milk into lactic acid. But in alkaline medium curding takes place very slowly, so this milk takes long to set as curd.

Question 13.
Plaster of Paris should be stored in moisture-proof container. Explain why ?
Answer:
Plaster of Paris (CaSO₄. \(\frac {1}{2}\)H₂O) has a very remarkable Property of setting into a hard mass on wetting with water or moisture. The setting of Plaster of Paris is due to its hydration to form crystals of gypsum which set to form a hard, solid mass as follows :

So, to protect the hydration of Plaster of Paris it is stored in moisture proof container.

Question 14.
What is a neutralisation reaction ? Given two examples.
Answer:
Neutralisation reaction : A reaction in which an acid reacts with a base and to form a salt and water is called neutralisation reaction.
Examples:
(i)

(ii)

Question 15.
Given two important uses of washing soda and baking soda.
Answer:
Uses of washing soda :

• Washing soda is used in laundary. In other words, washing soda is used as a cleansing agent for domestic purposes. It is a main component of many dry soap powders.
• It is used in the manufacture of many useful sodium compounds like caustic soda, borax, glass and soap.

Uses of baking soda :
(i) Baking soda (sodium hydrogen carbonate) is used as an antacid in medicine to remove acidity of the stomach.

(ii) Baking soda is used in making powder. Baking powder contains baking soda in acid like tartaric acid. When baking powder is mixed in the dough for preparing cake or bread is heated, sodium hydrogen carbonate present it decomposes it give carbon dioxide and sodium carbonate. This carbon dioxide gas bubbles out causing the cake or bread to rise and becomes light and spongy. The tartaric acid present in baking powder reacts with sodium carbonate and neutralises it. If tartaric acid is not present in baking power, then the cake will taste bitter due to the presence of sodium carbonate in it.

Class 10 Science Chapter 2 Acids, Bases and Salts Textbook Activities

Activity 2.1 (Page 18)

• Collect the following samples from the science laboratory-hydrochloric acid (HCl), sulphuric acid (H₂SO₄), nitric acid (HNO₃), acetic add (CH₃COOH), sodium hydroxide (NaOH), calcium hydroxide [Ca(OH)₂], potassium hydroxide (KOH), magnesium hydroxide [Mg(OH)₂], and ammonium hydroxide (NH₄OH).
• Put a drop of each of the above solutions on a watch-glass and test with a drop of the following indicators as shown in Table 2.1.
• What change in colour did you observe with red litmus, blue litmus, phenolphthalein and methyl orange solutions for each of the solutions taken?
• Tabulate your observations in Table 2.1.

Table 2.1

Activity 2.2 (Page 18)

• Take some finely chopped onions in a plastic bag along with some strips of clean cloth. Tie up the bag tightly and leave overnight in the fridge. The cloth strips can now be used to test for acids and bases.
• Take two of these cloth strips and check their odour.
• Keep them on a clean surface and put a few drops of dilute HCl solution on one strip and a few drops of dilute NaOH solution on the other.
• Rinse both cloth strips with water and again check their odour.
• Note your observations.
• Now take some dilute vanilla essence and clove oil and check their odour.
• Take some dilute HCl solution in one test tube and dilute NaOH solution in another. Add a few drops of dilute vanilla essence to both test tubes and shake well. Check the odour once again and record changes in odour, if any.
• Similarly, test the change in the odour of clove oil with dilute HCl and dilute NaOH solutions and record your observations.

Observations: Vanilla indicator.

Activity 2.3 (Page 19)

• Set the apparatus as shown in Fig.
• Take about 5 mL of dilute sulphuric acid in a test tube and add a few pieces of zinc granules to it.
• What do you observe on the surface of zinc granules?
• Pass the gas being evolved through the soap solution.
• Why are bubbles formed in the soap solution?
• Take a burning candle near a gas filled bubble.
• What do you observe?
• Repeat this Activity with some more acids like HCl, HNO₃, and CH₃COOH,
• Are the observations in all the cases the same or differed ?

Observation : Bubbles of hydrogen gas are evolved from the surface of zinc granules. When hydrogen gas is passed through a soap solution it forms bubbles because hydrogen is lighter than soap solution and it filled in the bubbles of soap. Hydrogen gas burns with a pop sound.

Hydrogen evolves vigorously with acids like HCl and HNO₃ but releases slowly with CH₃COOH.

Activity 2.4 (Page 20)

• Place a few pieces of granulated zinc metal in a test tube.
• Add 2 mL of sodium hydroxide solution and warm the contents of the test tube.
• Repeat the rest of the steps as in Activity 2.3 and record your observations.

Observations : Zinc reacts with ‘NaOH’ to forms sodium zincate and hydrogen.

Hydrogen bums with a pop near the candle.

Activity 2.5 (Page 20)

• Take two test tubes, label them as A and B.
• Take about 0.5 g of sodium carbonate (Na₂CO₃) in test tube A and about 0.5 g of sodium hydrogen carbonate (NaHCO₃) in test tube B.
• Add about 2 mL of dilute HCl to both the test tubes.
• What do you observe?
• Pass the gas produced in each case through lime water (calcium hydroxide solution) as shown in Fig. and record your observations.

Observations : Sodium carbonate (Na₂CO₃) and sodium bicarbonate (NaHCO₃) reacts with dilute HCl to form NaCl, H₂O and CO₂, Carbondioxide evolves with effervescence.

Test Tube A:
Na₂CO₃(s) + 2 HCl (aq) → 2NaCl (aq) H₂O(l) + CO₂
Test Tube B:
NaHCO₃ (s) + HCl (aq) → NaCl (aq) + H₂O(l) + CO₂ (↑)

On passing carbondioxide gas through lime water e.g. aq. Ca(OH)₂, it turns milky.

On Passing carbondioxide for long time through lime water, the milky colour of water disapperas due to the formation of soluble Ca(HCO₃)₂ (aq).

Activity 2.6 (Page 21)

• Take about 2 mL of dilute NaOH solution in a test tube and add two drops of phenolphthalein solution.
• What is the colour of the solution?
• Add dilute HCl solution to the above solution drop by drop.
• Is there any colour change for the reaction mixture?
• Why did the colour of phenolphthalein change after the addition of an acid?
• Now add a few drops of NaOH to the above mixture.
• Does the pink colour of phenolphthalein reappear?
• Why do you think this has happened?

Observation : (i) The colour of the solution is pink because phenolphthalein given pink colour in basic medium.
(ii) As dil. HCl is added drop by drop in the above solution it turns pink to colourless.
(iiï) The colour of the phenolphthalein change after the addition of an acid because in neutral or acidic medium phenolphthaLein does not give any colour.
(iv) Yes the pink colour of phenolphthalein reappears because when NaOH is added in the neutral mixture it becomes basic and in basic medium phenolphthalein gives pink colour.

Activity 2.7 (Page 21)

• Take a small amount of copper oxide in a beaker and add dilute hydrochloric acid slowly while stirring.
• Note the colour of the solution. What has happened to the copper oxide?

Observation : We observe that the colour of the solution becomes blue-green and the copper oxide dissolves. The blue-green colour of the solution is due to the formation of copper (II) Chloride.

Activity 2.8 (Page 22)

• Take solutions of glucose, alcohol, hydrochloric acid, sulphuric acid, etc.
• Fix two nails on a cork, and place the cork in a 100 mL beaker.
• Connect the nails to the two terminals of a 6 volt battery through a bulb and a switch, as shown in Fig,

• Now pour some dilute HCl in the beaker and switch on the current,
• Repeat with dilute sulphuric acid.
• What do you observe?
• Repeat the experiment separately with glucose and alcohol solutions. What do you observe now? Does the bulb glow in all cases?
  

The Bulb will start glowing in the case of acids like HCl, H₂SO₄ etc. But in case of glucose and alcohol bulb does not glow because these solutions do not conduct electricity. The electric current in carried through the solution by ions. Since the cation present in acids is H⁺, this suggests that acid produce hydrogen ions in solution, which are responsible for their acidic character.

Activity 2.9 (Page 23)

• Take about 1g solid NaCl in a clean and dry test tube and set up the apparatus as shown in Fig.
• Add some concentrated sulphuric acid to the test tube.
• What do you observe ? Is there a gas coming out of the delivery tube?
• Test the gas evolved successively with dry and wet blue litmus paper.
• In which case does the litmus paper change colour?
• On the basis of the above Activity, what do you infer about the acidic character of: (i)dry HCl gas (ii) HCl solution?

Note : If the climate is very humid, you will have to pass the gas produced through a guard tube (drying tube) containing calcium chloride to dry the gas.
Observation : A pungent smelling gas is evolved from the test tube.
2NaCl + H₂SO₄ → Na₂SO₄ + HCl (g)
The colour of dry blue litmus paper remain unchanged while the colour of wet blue litmus paper turns red.

This activity suggests that .hydrogen ions in HCl are produced in the presence of water. The separation of K⁺ ion from HCl molecules cannot occur in the absence of water.
HCl + H₂O → H₃O⁺ + Cl^–

Hydrogen ions cannot exist alone, but they exist after combining with water molecules. Thus hydrogen ions must always be shown as H⁺(aq) or hydronium ion (H₃O⁺).
H⁺ + H₂O → H₃O⁺

Activity 2.10 (Page 24)

• Take 10 mL water in a beakér.
• Add a few drops of concentrated H₂SO₄ to it and swirl the beaker slowly.
• Touch the base of the beaker.

Question 1.
Is there a change in temperature?
Answer:
Yes, there is a change in temperature. Temperature of the beaker increases.

Question 2.
Is this an exothermic or endothermic process?
Answer:
This is an enothermic reaction

Question 3.
Repeat the above Activity with sodium hydroxide pellets and record your observations.
Answer:
When sodium hydroxide pellets are added in water, the temperature of beaker increases. This is also an exothermic reaction.

Activity 2.11 (Page 26)

• Test the pH values of solutions given in Table.
• Record your observations.
• What is the nature of each substance on the basis of your observations?

Observations :

Activity 2.12 (Page 27)

• Put about 2g soil in a test tube add 5 mL water to it.
• Shake the contents of the test tube.
• Filter the contents and collect the filtrate in a test tube.
• Check the pH of this filtrate with the help of universal indicator paper.
• What can you conclude about the ideal soil pH for the growth of plants in your region?

Observations:
(i) The pH of the sample of soil is about 7.6. It means it is alkaline in nature.
(ii)The pH of good soils should be about 7 but due to pollution the pH increases or decreases. For the normal growth of plants pH should be 6 to 8.

Activity 2.13 (Page 28)

• Write the chemical formulae of the salts given below. Potassium sulphate, sodium sulphate, calcium sulphate, magnesium sulphate, copper sulphate, sodium chloride, sodium nitrate, sodium carbonate and ammonium chloride.
• Identify the acids and bases from which the above salts may be obtained.
• Salts having the same positive or negative radicals are said to belong to a family. For example, NaCl and Na₂SO₄ belong to the family of sodium salts. Similarly, NaCl and KCl belong to the family of chloride salts. How many families can you identify among the salts given in this Activity?

Answer:

(i) Sulphate Family
K₂SO₄, Na₂SO₄, CaSO₄, MgSO₄, Cu₄SO₄
(ii) Chloride Family NaCl, NH₂Cl
(iii) Sodium Family
Na₂SO₄, NaCl, NaNO₃, Na₂CO₃

Activity 2.14 (Page 29)

• Collect the following salt, samples – sodium chloride, potassium nitrate, aluminium chloride, zinc sulphate, copper sulphate, sodium acetate, sodium carbonate and sodium hydrogen carbonate (some other salts available can also be taken).
• Check their solubility in water (use distilled water only).
• Check the action of these solutions on litmus and find the pH using a pH paper.
• Which of the salts are acidic, basic or neutral?
• Identify the acid or base used to form the salt.
• Report your observations in Table 2.4.

Observations:

Activity 2.15 (Page 32)

• Heat a few crystals of copper sulphate in a dry boiling tube.
• What is the colour of the copper sulphate after heating?
• Do you notice water droplets in the boiling tube? Where have these come from?
• Add 2-3 drops of water on the sample of copper sulphate obtained after heating.
• What do you observe? Is the blue colour of copper sulphate restored?

Observation : Copper sulphate crystals have blue colour which seems to be dry but contain water of crystallisation. When we heat the crystals, this water is removed and the salt turns white.

Yes we noticed some water droplets in the boiling tube. Hydrated copper sulphate contains ‘5’ molecules of water of crystallisation per molecule- It has chemical formula CuSO₄, 5H₂O. On heating, this water of molecules is removed and form water droplets in the boiling tube.

When 2-3 water droplets are added in the anhydrous CuSO₄, we find that the blue colour of the crystals reappears.

Class 10 Science Chapter 2 Acids, Bases and Salts Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is litmus solution ?
Answer:
Litmus solution is a purple dye, extracted from lichen.

Question 2.
Define an acid.
Answer:
A substance which is furnished H⁺_caq ions in their aqueous solution, e.g. Hd, H₂SO₄ etc.

Question 3.
What is a base ?
Answer:
Base is a substance which is produced OH^– _caq ions in their aqueous solution e.g. NaOH, KOH etc.

Question 4.
What is an alkali ?
Answer:
Water soluble base is called alkali, e.g. NaOH, KOH etc.

Question 5.
A substance change blue litmus to red. What is the nature of the substance ?
Answer:
Acidic.

Short Answer Type Questions

Question 1.
What are non-electrolytes ? Give some examples.
Answer:
A compound which does not conduct electricity when dissolved in water or in molten state is called a non-electrolyte.

For example, an aqueous solution of sugar does not conduct electricity, so it is a non-electrolyte. Other examples of non-electrolytes are glucose, urea, benzene, acetone, etc.

Question 2.
What are strong acids ? Explain.
Answer:
An acid which completely dissociated in its aqueous solution and produces a large number of hydrogen ions is called a strong acid.
e.g. Hydrochloric acid, sulphuric acid and nitric acid.

Question 3.
What are weak acids ? Explain.
Answer:
An acid which is partially ionised dissociated in its aqueous solution and produces a small number of hydrogen ions is called a weak acid.

Long Answer Type Questions

Question 1.
How washing soda is prepared ? Give its properties and uses.
Answer:
The Solvay proces gives us anhydrous sodium carbonate Na₂CO₃, which is called soda ash, Soda ash is obtained by heating sodium carbonate. Now, washing soda contains 10 molecules of water of crystallization. So, anhydrous sodium carbonate is dissolved in water and recrystallized to get washing soda crystals containing 10 molecules of water crystallization :

Properties:

• It is a transparent crystalline solid.
• It is fairly soluble in water.
• The solution of washing soda in water is alkaline which turns red litmus to blue.
• When washing soda crystals are heated strongly, they lose all the water of crystallization and form soda ash.
  

Uses:

• Sodium carbonate is used in laundry as washing soda.
• It is a component of many dry soap powders.
• It is used for softening of hard water.
• It is used in the manufacture of many useful sodium compounds like caustic soda, borax, glass and soap.
• It is also used in the manufacture of paper.

Multiple Choice Questions

Question 1.
The pH of distilled water is:
(a) 7
(b) 8
(c) 10
(d) 6.8
Answer:
(a) 7

Question 2.
The chemical formula of P.O.P is:
(a) CaSO₄
(b) CaSO₄. \(\frac {1}{2}\)H₂O
(c) CaSO₄.H₂O
(d) CaSO₄.2H₂O
Answer:
(b) CaSO₄. \(\frac {1}{2}\)H₂O

Question 3.
The Plaster of Paris is prepared by heating;
(a) Gypsum
(b) Limestone
(c) Lime
(d) Soda ash
Answer:
(a) Gypsum

Question 4.
The chemical formula of gypsum is :
(a) CaSO₄
(b) CaSO₄.\(\frac {1}{2}\)H₂O
(c) CaSO₄. \(\frac {3}{2}\)H₂O
(d) CaSO₄.2H₂O
Answer:
(d) CaSO₄.2H₂O

Question 5.
Baking powder contains baking soda and:
(a) Oxalic acid
(b) Tartaric acid
(c) Carbonic acid
(d) Methanoic acid
Answer:
(b) Tartaric acid

These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3

Question 1.
State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :


Solution:

Question 2.
In the given figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:

Question 3.
Diagonals AC and BD of a trape∠ium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac { OA }{ OC }\) = \(\frac { OB }{ OD }\)
Solution:

Question 4.
In the adjoining figure, \(\frac { QR }{ QS }\) = \(\frac { QT }{ PR }\) and ∠1 = ∠2 show that ∆PQS ~ ∆TQR.

Solution:

Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:

Question 6.
In the given figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.

Solution:

Question 7.
In the given figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
Solution:

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Solution:
Given: ABCD is a parallelogram, whose side AD is produced to E. BE is joined which intersects CD at F.

TO Prove: ∆ABE ~ ∆CFB
Proof: In ∆ABE and ∆CFB
∠A = ∠C [opposite angles of a parallelogram are equal]
∠AEB = ∠EBC [alternate angle]
Thus, by AA-similarity criterion
∆ABE ~ ∆CFB
Hence Proved.

Question 9.
In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) ∆ABC ~ ∆AMP
(ii) \(\frac { CA }{ PA }\) = \(\frac { BC }{ MP }\)
Solution:

Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that
Solution:

Question 11.
In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.

Solution:

Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see in given figure). Show that ∆ABC ~ ∆bPQR.

Solution:

Question 13.
D is a point on the side BC of a triangle ABC, such that ∠ADC = ∠BAC. Show that CA² = CB.CD.
Solution:

Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
Solution:

Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:

DE is a vertical pole of length = 6 m
Length of its shadow = 4 m
Let height of tower AB = h m
Length of its shadow = 28 m
In ∆ABC and ∆DEC,
∠ABC = ∠DEC [Each 90°]
∠C = ∠C [Common]
∆ABC ~ ∆DEC [AA]
\(\frac { AB }{ DE }\) = \(\frac { 6 }{ x }\) = \(\frac { 4 }{ 28 }\)
⇒ x = 42 cm.

Question 16.
If AD and PM are medians of triangles ABC and PQR respectively, where ∆ABC ~ ∆PQR Prove that  \(\frac { AB }{ PQ }\) = \(\frac { AD }{ PM }\)
Solution:

These NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-Metals Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Metals and Non-Metals NCERT Solutions for Class 10 Science Chapter 3

Class 10 Science Chapter 3 Metals and Non-Metals InText Questions and Answers

In-text Questions (Page 40)

Question 1.
Give an example of a metal which
(i) is a liquid at room temperature.
(ii) can be easily cut with a knife.
(iii) is the best conductor of heat.
(iv) is a poor conductor of heat.
Answer:
(i) Mercury,
(ii) Sodium,
(iii) Silver,
(iv) Lead.

Question 2.
Explain the meanings of malleable and ductile.
Answer:
Metals are malleable i.e., they can be hammered into sheets because layer of atoms slide over each other. Gold and silver and most malleable while copper and aluminium are also highly malleable. Metals such as aluminium can be rolled into thin sheets, because the regular layers of atoms can slide over one another.

Metals have ductile nature i.e., they can be drawn into wires. The abiltiy of metals to be drawn into thin wires is called ductility. Gold and silver are the most malleable metals, A wire of about 2 km length can be drawn from one gram of gold.

In-text Questions (Page 46)

Question 1.
Why is sodium kept in kerosene oil?
Answer:
Sodium is highly reactive metal and it combines with oxygen of air and water (moisture) to form oxides and hydroxide vigorously.

In order to store these metals these are kept in kerosene oil.

Question 2.
Write equations for the reactions of
(i) Iron with steam
(ii) Calcium and potassium with water
Answer:

Question 3.
Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows:

Use the table above to answer the following questions about metals A. B, C and D.
(i) Which is the most reactive metal?
(ii) What would you observe if B is added to a solution of copper (II) sulphate?
(iii) Arrange the metals A, B, C and D in the order of decreasing reactivity?
Answer:
(i) ‘B’ is the most reactive metal.
(ii) ‘B’ displaces copper from copper sulphate.
(iii) Order of decreasing reactivity of metal A, B, C and D.
B > A > C > D

Question 4.
Which gas is produced when dilute hydrochloric acid is added to a reactive metal?
Write the chemical reaction when iron reacts dilute H₂SO₄.
Answer:
Hydrogen gas is produced when dilute hydrochloric acid is added to a metal.

Question 5.
What would you observe when zinc is added to a solution o iron (II) sulphate ? Write the chemical reaction that takes place.
Answer:
FeSO₄(aq) + Zn(s) → ZnSO₄(aq) + Fe(s)
Iron (Il) sulphate solutIon has greenish colour. A more reactive metal can displace a less reactive metal from its salt solution. Zinc is more reactive than iron so it displaces iron from iron sulphate solution and the greenish colour of solution changes into colourless because zinc sulphate solution is colourless.

In-text Questions (Page 49)

Question 1.
(i) Write the electron dot structures for sodium oxygen and magnesium.
(ii) Show the formation of Na₂O and MgO by the transfer of electrons.
(iii) What are the ions present in these compounds ?
Answer:

Question 2.
Why do ionic compounds have high melting point ?
Answer:
Ionic compounds have high melting point because ions are tightly held together by strong electrostatic forces of attraction and to break this strong interionic attraction a large amount of energy is needed.

In-text Questions (Page 53)

Question 1.
Define the terms :
(i) mineral (ii) ore and (iii) gangue
Answer:
(i) Mineral: In combined state metals are found in the earth crust as oxides, carbonates, sulphides, silicates, phosphates etc. These compounds are known as minerals. All the ores are minerals but all the minerals are not ores.
(ii) Ore : The mineral from which the metal can be economically and conveniently extracted is called ore. For example, bauxite is the ore of aluminium.
(iii) Gangue : The eartly, rocky and silicious impurities associated with the minerals are collectively known as gangue or matrix.

Question 2.
Name two metals which are found in nature in the free state.
Answer:
Gold, Silver.

Question 3.
What chemical process is used for obtaining a metal from its oxide ?
Answer:
Reduction Process.

In-text Questions (Page 55)

Question 1.
Metallic oxides of zinc, magnesium and copper were heated with the following metals:

In which cases will you find displacement reactions taking place?
Answer:
Magnesium is more reactive than zine and copper so it can displace zinc from zinc oxide and copper from copper oxide.

Zinc is more reactive than copper so it can displace copper from copper oxide.
\(\mathrm{CuO}+\mathrm{Zn} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{ZnO}+\mathrm{Cu}\)

So when ZnO and CuO react with Mg and CuO reacts with ‘Zn’ they gives displacement reactions.

Question 2.
Which metals do not corrode easily?
Answer:
Least reactive metals such Au (Gold), PL (Platinum), Ag (Silver).

Question 3.
What are alloys?
Answer:
An alloy is a homogenous mixture of two as more metals or a metal with a non-metal. E.g. Brass, Bronze etc.

Class 10 Science Chapter 3 Metals and Non-Metals Textbook Questions and Answers

Page No. 56, 57

Question 1.
Which of the following pairs will give displacement reactions:
(a) NaCl solution and copper metal
(b) MgCl₂ solution and aluminium metal
(c) FeSO₄ solution and silver metal
(d) AgNO₃ solution and copper metal
Answer:
(d) AgNO₃ solution and copper metal.

Question 2.
Which of the following methods is suitable for preventing an iron frying pan from rusting.
(a) applying grease
(b) applying paint
(c) applying a coating of zinc
(d) all of the above
Answer:
(d) All of the above

Question 3.
An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be:
(a) calcium
(b) carbon
(c) silicon
(d) iron
Answer:
(d) Calcium.

Question 4.
Food cans are coated with tin and not with zinc because:
(a) zinc is costlier than tin
(b) zinc has a higher melting point than tin
(c) zinc is more reactive than tin
(d) zinc is less reactive than tin
Answer:
(a) Zinc is costlier than tin

Question 5.
You are given a hammer, a battery, a bulb, wires and a switch.
(a) How could you use them to distinguish between metals and non metals?
(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.
Answer:
(a) A hammer can be used to check the malleability of metals. Metals are malleable i.e., they can be hammered into sheets because layer of atoms slide over each other. But a non-metal cannot be hammered into sheets. However non-metals are brittle in nature. When a metal is connected with a battery and a bulb through wires, bulb does not glow.

(b) Metals are malleable but non-metals are brittle. Malleability is an important property of metals. Metals can be distinguished by using hammer.

Metals are good conductors of electricity but non-metals are bad conductors of electricity. So conductivity method can also be used to distinguish between metals and non-metals. Except graphite all non-metals are bad conductors of electricity.

Question 6.
What are amphoteric oxides? Give two examples of amphoteric oxides.
Answer:
Amphoteric oxide: A metal oxide which shows both acidic as well as basic character i.e., such metal oxides react with both acids as well as bases to produce salts and water is known as amphoteric oxides.

For examples Aluminum oxide (Al₂O₃) and Zinc oxide (ZnO).

Question 7.
Name two metals which will displace hydrogen from dilute acids, and two metals which will not.
Answer:
Metals which will displace hydrogen from dilute acids : Zn, Mg.
Metals which will not displace hydrogen from dilute acids: Cu, Ag.

Question 8.
In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte.
Answer:
In the electrolyte refining of a metal ‘M’, the crude metal is made anode whereas the thin sheet of pure metal ‘M’ is made cathode. Electrolyte is the solution of some salt of the metal ‘M’.

Question 9.
Pratyush took Sulphur power on a spatula and heated it. He collected a gas evolved by inverting a test tube over it as shown in figure:

(a) What will be the action of gas on
(i) dry litmus paper?
(ii) moist litmus paper?
(b) Write a balanced chemical equation for the reaction taking place.
Answer:
(a) (i) No effect on dry litmus paper.
(ii) It turns the colour of moist blue litmus paper into red because the evolved gas has acidic nature.

Sulphurous acid turns blue litmus to red.

Question 10.
State two ways to prevent the rusting of iron.
Answer:
The rusting of iron can be prevented by painting, oiling, greasing, galvanishing, chromeplating, anodising or making alloys.
1. Barrier Protection : In barrier protection, a protective film is introduced between iron and atmospheric oxygen and water. Barrier protection can be achieved by any of the following methods:

• By painting the surface
• By coating the surface with a thin film of oil or grease.
• By electroplating iron with some non-conosive metal such as nickel, copper etc.

2. Galvanisation: Zinc metal is generally used for protecting iron and the process is called galvanisation. Galvanised iron sheets maintain their shine due to the formation of a protective layer of basic zinc carbonate due to the reaction between zinc, oxygen, carbon dioxide and moisture in air.

Question 11.
Give reasons:
(a) Platinum, gold and silver are used to make jewellery.
(b) Sodium, potassium and lithium are stored under oil.
(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking
(d) Carbonate and sulphides ores are usually converted into oxides during the process of extraction.
Answer:
(a) Platinum, gold and silver have high metallic lustre and are less reactive metals. So these metals are not corroded when exposed to air and moisture. That is why these metals are used to make jewellery.

(b) Sodium, potassium and lithium are highly reactive metals. When these metals are exposed to atmosphere, they react with oxygen, moisture and carbon dioxide present in the air forming Oxides, hydroxides and carbonates. In order to prevent the reaction, the alkali metals i.e., sodium, potassium and lithium are stored under oil.

(c) Aluminium is a highly reactive metal. It readily reacts with oxygen of air to form oxide and once a thin layer of oxides formed on surface then it protect the metal from further action and simultaneously it a good conductor of heat.

Due to these reasons aluminium is used to make utensils for cooking

(d) The metals in the middle of the activity series such as iron, zinc, lead, copper etc., are moderately reactive. These metals are usually present as Sulphides or carbonates in nature. It is easier to obtain a metal from its oxide, as compared to its sulphides and carbonates. So, prior to reduction, the metal sulphides and carbonates must be converted into metal oxides.

Question 12.
You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.
Answer:
Copper reacts with moist carbondioxide in the air and slowly loses its shiny brown surface and gains a green coat. This green substance is copper carbonate (CuCO₃). Sour substances like lemon or tamarind juice contains acids such as Citric acid or methanoic and copper carbonate is basic in nature so it reacts with acids to form water soluble salts and thus copper vessels can be cleaned.

Question 13.
Differentiate between metal and non-metal on the basis of their chemical properties.
Answer:

Question 14.
A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments, an unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?
Answer:
Gold is a least reactive metal. It does react with oxygen, Carbondioxide, water, even cone. Acids. But it is dissolved in aqua regia. Aqua regia (Latin name for “royal water”) is a freshly prepared mixture of concentrated hydrocloric acid and concentrated nitric acid in the ratio of 3: 1. It is one of the few reagents that is able to dissolve gold and platinum. So the solution used by the man is aqua regia.

Question 15.
Give the reason why copper is used to make hot water tanks but steel (an alloy of iron) is not.
Answer:
Metals at the top of the reactivity series are most reactive and are not available in nature in free state and all metals placed above hydrogen can displace hydrogen from their compounds such as water and acids.

According to reactivity series iron is more reactive than copper when iron is exposed to hot water it corroded vigorously but copper is not. That is why copper is used to make hot water tanks but steel (an alloy of iron) is not.

Class 10 Science Chapter 3 Metals and Non-Metals Textbook Activities

Activity 3.1 (Page 37)

• Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample.
• Clean the surface of each sample by rubbing them with sand paper and note their appearance again.

Observation: Metals, in their pure state, have a shining surface. This property is called metallic lustre,

Activity 3.2 (Page 37)

• Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations.
• Hold a piece of sodium metal with a pair of tongs.

CAUTION: Always handle sodium metal with care. Dry it by pressing between the folds of a filter paper.

• Put it on a watch-glass and try to cut it with a knife.

Question 1.
What do you observe?
Answer:
Observation: Metals are generally hard, but sodium is a soft metal. We observe that sodium metal cuts with the help of knife.

Activity 3.3 (Page 38)

• Take pieces of iron, zinc, lead and copper.
• Place any one metal on a block of iron and strike it four or five times with a hammer. What do you observe?
• Repeat with other metals.
• Record the change in the shape of these metals.

Observations: We find that some metals (Iron, Zinc, Copper) can be beaten into thin sheets. This property of metals is called malleability. Gold and silver are the most malleable metals?

Activity 3.4 (Page 38)

• Consider some metals such as iron, copper, aluminium, lead etc.
• Which is the above metals are also available in the form of wires ?

Observation : Iron, Copper and Aluminium metals are available in the form of wires. The ability of metals to be drawn into thin wires is called ductility. Gold is the most ductile metal.

Activity 3.5 (Page 38)

• Take an aluminium or copper wire. Clamp this wire on a stand, as shown in Fig.
• Fix a pin to the free end of the wire using wax.
• Heat the wire with a spirit lamp, candle or a burner near the place where it is clamped.
• What do you observe after some time?

• Note your observations. Does the metal wire melt?

Observation : Metal wire does not melt but the wax is melted and the pin fall down on the earth. This activity shows that metals are good conductors of heat and have high melting point.

Activity 3.6 (Page 39)

• Set up an electric circuit as shown in Fig.
  
• Place the metal to.be tested in the circuit between terminals A and B as shown.
• Does the bulb glow? What does this indicate?

Observation : Yest the bulb glows. This indicates that metals are good conductors of electricity.

Activity 3.7 (Page 39)

• Collect samples of carbon (coal or graphite), sulphur and iodine.
• Carry out the Activities 3.1 to 3.6 with these non-metals and record your observations.

Compile your observations regarding metals and non-metals in Table 3.1.

Activity 3.8 (Page 40)

• Take a magnesium ribbon and some sulphur powder.
• Bum the magnesium ribbon. Collect the ashes formed and dissolve them in water.
• Test the resultant solution with both red and blue litmus paper.
• Is the product formed on burning magnesium acidic or basic?
• Now burn sulphur powder. Place a test tube over the burning sulphur to collect the fumes produced.
• Add some water to the above test tube and shake.
• Test this solution with blue and red litmus paper.
• Is the product formed on burning sulphur acidic or basic?
• Can you write equations for these reactions?

Observation: The product formed on burning of magnesium is basic in nature because it turns red litmus to blue.
The produt formed on burning of sulphur is acidic in nature because it turns blue litmus to red

Magnesium hydroxide is basis in nature.

Sulphurous acid is acidie in nature.

Activity 3.9 (Page 41)

CAUTION : The following activity needs the teacher’s assistance. It would be better if students wear eye protection.

• Hold any of the samples taken above with a pair of tongs and try burning over a; flame. Repeat with the other metal samples.
• Collect the product if formed.
• Let the products and the metal surface cool down.
• Which metals burn easily?
• What flame colour did you observe when the metal burnt?
• How does the metal surface appear after burning?
• Arrange the metals in the decreasing order of their reactivity towards oxygen.
• Are the products soluble in water?

Observation: Let us take sodium, magnesium calcium and iron. All metals combine with oxygen to form metal oxides. Oxide of sodium is soluble in water but the oxides of magnesium and aluminium is insoluble in water.
Order of Reactivity
Na > Ca > Mg > Fe

Activity 3.10 (Page 42)

CAUTION: This Activity needs the teacher’s assistance.

• Collect the samples of the same metals as in Activity 3.9.
• Put small pieces of the samples separately ¡n beakers half-filled with cold water.
• Which metals reacted with cold water? Arrange them in the increasing order of their reactivity with cold water.
• Did any metal produce fire on water?
• Does any metal start floating after some time?
• Put the metals that did not react with cold water in beakers half-filled with hot water.
• For the metals that did not react with hot water, arrange the apparatus as shown in Fig. and observe their reaction with steam.

• Which metals did not react even with steam?
• Arrange the metals in the decreasing order of reactivity with water.

Observation: Sodium metal reacts with cold water

Sodium metal produces fire on water because the reaction between sodium and water is so voilent and exothermic that the evolved hydrogen immediately catches fire.

Calcium starts floating, because the bubbles of hydrogen gas formed stick to the surface of the metal. Magnesium reacts with hot water and forms magnesium hydroxide and hydrogen.

Metals like iron and aluminium do not react either with cold or hot water. But they react with steam to form the metal oxide and hydrogen.
3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)
Decreasing order of reactivity of different metals with water.
Na > Ca > Mg > Fe

Activity 3.11 (Page 44)

• Collect all the metal samples except sodium and potassium again. If the samples are tarnished, rub them clean with sand paper.
  CAUTION: Do not take sodium and potassium as they react vigorously even with cold water.
• Put the samples separately in test tubes containing dilute hydrochloric acid
• Suspend thermometers in the test tubes, so that their bulbs are dipped in the acid.
• Observe the rate of formation of bubbles carefully.
• Which metals reacted vigorously with dilute hydrochloric acid?
• With which metal did you record the highest temperature?
• Arrange the metals in the decreasing order of reactivity with dilute acids.

Observation: The rate of formation of bubbles is fastest in the case of calcium and reaction is also most exothermic in this case.
This order of reactivity of metals with dilute HCl is:
Ca > Mg > Fe

Activity 3.12 (Page 44)

• Take a clean wire of copper and an iron nail.
• Put the copper wire in a solution of iron sulphate and the iron nail in a solution of copper sulphate taken in test tubes (Fig.).
• Record your observations after 20 minutes.
• In which test rube did you find that a reaction has occurred?
• On what basis can you say that a reaction has actually taken place?
• Can you correlate your observations for the Activities 3.9, 3.10 and 3.11?
• Write a balanced chemical equation for the reaction that has taken place.
• Name the type of reaction.

Observation : Reactive metals can displace less reactive metals from their compounds in solution or molten state.

Iron displaces copper from copper sulphate solution because iron is more reactive than copper cannot displace iron from, iron sulphate solution because copper is less reactive than iron.
Iron is more reactive than copper.
FeSO₄(aq) + Cu(s) → No reaction
CuSO₄(aq) + Fe(s) → FeSO₄ + Cu(s)
This is a displacement reaction.

Sodium can displace calcium from calcium sulphate, calcium can displace magnesium from magnesium sulphate and magnesium can displace iron from iron sulphate solution.

Activity 3.13 (Page 48)

• Take samples of sodium chloride, potassium iodide, barium chloride or any other salt from the science laboratory.
• What is the physical state of these salts?
• Take a small amount of a sample on a metal spatula and heat directly on the flame (Fig.). Repeat with other samples.
• What did you observe? Did the samples impart any colour to the flame? Do these compounds melt?
• Try to dissolve the samples in water, petrol and kerosene. Are they soluble?
• Make a circuit as shown in Fig. and insert the electrodes into a solution of one salt. What did you observe? Test the other salt samples too in this manner.
• What is your inference about the nature of these compounds?

Observation : All there salts are solid and some what hard due to the strong forces of attraction between the positive and negative ions.

NaCl crystals on heating break up with a crackling noise. The same noise is observed in all the cases and salts are dried. No colour is imparted by the sample to the flame. These salts are not melted.

These salts are fairly soluble in water but insoluble in petrol and kerosene.
Blub glow in all the salt solutions in water. It means ionic compounds conduct electricity and they are made of ions.

Activity 3.14 (Page 53)

• Take three test tubes and place clean iron nails in each of them
• Label these test tubes A, 13 and C. Pour some water in test tube A and cork it.
• Pour boiled distilled water in test tube B, add about 1 mL of oil and cork it. The oil will float on water and prevent the aIr from dissolving in the water.
• Put some anhydrous calcium chloride in test tube C and cork it. Anhydrous calcium chloride will absorb the moisture, if any, from the air. Leave these test tubes for a few days and then observe (Fig.).

Fig. : investigatIng the conditions under which Iron rusts. In tube A, both air and water present. In tube B, there is no air dissolved In the water. In tube C, the air is dry

Observation : We observe that iron nails rust in test tube A, but they do not rust in test tubes B and C. In the test tube A, the nails are exposed to both air and water. In the test tube B, the nails are exposed to only water, and the nails in test tube C are exposed to dry air. It suggested that air and water are necessary for corrosion of iron or a metal.

Class 10 Science Chapter 3 Metals and Non-Metals Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Give the name of a metal, which is in liquid state?
Answer:
Mercury.

Question 2.
Give the name of a non-metal, which is in liquid state?
Answer:
Bromine.

Question 3.
Give the name of a non-metal, which is in solid state?
Answer:
Iodine.

Question 4.
What is malleability?
Answer:
Metals can be hammered into sheets because layer of atoms slide over each other. This property of metals is called malleability.

Question 5.
Give two examples of soft metals.
Answer:
Sodium and Potassium.

Question 6.
What is ductility?
Answer:
The property of a metal in which they can be drawn into wires is called ductility.

Question 7.
Which is the most ductile metal?
Answer:
Gold.

Question 8.
Which are the most malleable metals?
Answer:
Gold and Silver

Question 9.
What is sonorocity?
Answer:
Metals produce a metallic sound when struck this property of metals is called sonorocity.

Question 10.
What is tensile strength?
Answer:
Metals can be stretched to some degree without breaking this property is called as tensile strength.

Short Answer Type Questions

Question 1.
What happens when:
(a) Sodium reacts with oxygen
(b) Magnesium reacts with oxygen
(c) Cu heated with oxygen
(d) Iron and Aluminium react with oxygen.
Give chemical reactions also.
Answer:

Question 2.
How ‘Al₂O₃‘ is an amphoteric oxide ?
Answer:
Some metal oxides which show acidic as well as basic character are known as amphoteric oxides, amphoteric metal oxides react with both adds as well as bases to produce salt and water. For example Al₂O₃ and ZnO.

Aluminium oxide is an amphoteric oxide because it reacts with acids as well as bases and form salts and water.

Question 3.
Are all metal oxides in soluble in water? If not, then given two examples of water soluble oxides, with chemical equations.
Answer:
No, Most metal oxides are insoluble in water but some of these dissolve to form alkalis. For example sodium oxide and potassium oxide dissolve in water to produce alkalis as follows :

Long Answer Type Questions

Question 1.
Name the chief ore of the following metals in the earth’s crust (i) aluminium (ii) calcium (iii) sodium (iv) mercury
How electrolytic reduction is carried out ?
Answer:
(i) Aluminium : Bauxite
(ii) Calcium : Lime stone
(iii) Sodium : Rock salt
(iv) Mercury : Cinnabar

Electrolytic reduction : Certain highly electropositive metals such as alkali metals (Na, K etc), alkaline earth metals and aluminium etc. are commonly extracted by the electrolysis of their fused salts. The process of extraction of metals by electrolysis process is called electrometallurgy. For example, sodiums extracted from fused sodium chloride by electrolysis as:
\(\mathrm{NaCl} \stackrel{\text { ionisation }}{\rightarrow} \mathrm{Na}^{+}+\mathrm{Cl}^{-}\)
At cathode : Na⁺ + e → Na
At anode : Cl^– → Cl + e
Cl + Cl → Cl₂
Thus sodium is liberated at cathode and Cl₂ gas liberated at Anode.

Multiple Choice Questions

Question 1.
The wires that carry current in your homes have a coating of:
(a) PVC
(b) BUNa-S
(c) Polybutadienes
(d) Bakelite
Answer:
(a) PVC

Question 2.
Choose non-metal from the following:
(a) S
(b) K
(c) Na
(d) Fe
Answer:
(a) S

Question 3.
Zinc can displace:
(a) Cu from CuSO₄
(b) Ag from AgNO₃
(c) Mg from MgSO₄
(d) a and b
Answer:
(d) a and b

Question 4.
The metal used in tinning is:
(a) Sn
(b) Zn
(c) Cu
(d) Pt
Answer:
(a) Sn

Question 5.
A noble metal is:
(a) Pt
(b) Cu
(c) Zn
(d) Pd
Answer:
(a) Pt

These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.1

Question 1.
Fill in the blanks by using the correct word given in brackets.
(i) All circles are …………….. (congruent/similar)
(ii) All squares are ……………. (similar/congruent)
(iii) All …………….. triangles are similar. (isosceles/equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are …………… and (b) their corresponding sides are …………… (equal/proportional)
Solution:
(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are equal and
(b) their corresponding sides are proportional.

Question 2.
Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.
Solution:
(i) similar figures:

Both circles have equal radii so they are similar.

Both squares have equal sides so they are similar.

(ii) non-similar figures

They are not similar.

Question 3.
State whether the following quadrilaterals are similar or not.

Solution:
They are not similar because both figures have different sides.

In this Page, you will learn how to solve questions of Ex 1.1 Class 10 Maths NCERT Solutions recommended in CBSE 10 Board Maths Exam syllabus.

These NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

Question 1.
Euclid’s division algoritgm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 225
Solution:
(i) 135 and 225
Since 225 > 135, we apply the division leema to 225 and 135, to get
225 = 135 x 1 + 90
Since the remainder 90 ≠ 0, we can apply the division lemma to 135 and 90, to get 135 = 90 x 1 + 45
We consider the new divisor 90 and new remainder 45, and apply division lemmat to get 90 = 45 x 2 + 0
The remainder has now become zero, so the HCF of 135 and 225 is 45.

(ii) 196 and 38220
Since 38220 > 196, we apply the division leema to 38220 and 196, to get
38220 = 196 x 195 + 0
The remainder has now become zero, so the HCF of 38220 and 196 is 196

(iii) 867 and 255
Since 867 > 255, we apply the division lemma to 867 and 225, to get
867 = 255 x 3 + 102
Since the remainder 102 = 0, we apply the division lemma to 255 and 102, to get 255 = 102 x 2 + 51
We consider the new divisor 102 and new remainder 51, and apply division lemma to get 102 = 51 x 2 + 0
The remainder has now become zero, so the HCF of 867 and 255 is 51.

Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let a be any positive odd integer. We apply the division algorithm with a and b = 6.
Since 0 < r < 6, the positive remainders are 0,1, 2, 3, 4, 5.
Now putting the values r = 0,1, 5 we get
⇒ a = bq + r
⇒ a = 6q + 0 ⇒ a = 6q [r = 0]
⇒ a = 6q + 1 ⇒ a = 6q +1 [r = 1]
Similarly integers of the form 6q, 6q + 1,6q + 2, 6q + 3, 6q + 4, 6q + 5.

Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
The maximum number of columns is HCF of (616,32).
Now, let us apply Euclid’s division algorithm to find their HCF
616 = 32 x 16 + 8
Since the remainder 8 ≠ 0, we apply the division lemma, to get
32 = 8 x 4 + 0
So, the HCF of 616 and 32 is 8.
Therefore, the maximum number of columns is 8.

Question 4.
Use Euclid’s division lemma to show, that the square of any positive integer is either of the form 3m or 3m +1 for some integer m.
Solution:
[Hint : Let x be any positive integer then it is of the form 3q, 3q +1, or 3q + 2. Now square each of these and show that they can be written in the form 3m or 3m+ 1]

In this Page, you will learn how to solve questions of Ex 5.1 Class 10 Maths NCERT Solutions recommended in CBSE 10th Board Maths Exam syllabus.

These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac { 1 }{ 4 }\) of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.
Solution:
(i) Yes, 15, 23, 31, …. forms an AP as each succeeding term is obtained by adding 8 in its preceeding term.
(ii) No, volumes are V, \(\frac { 3V }{ 4 }\), (\(\frac { 3V }{ 4 }\))², L.
(iii) Yes, 150,200,250,…. form an A.P.
(iv) No, Amounts are

10000(1 + \(\frac { 8 }{ 100 }\)), 10000(1 + \(\frac { 8 }{ 100 }\))², 10000(1 + \(\frac { 8 }{ 100 }\))², L

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
(n) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = \(\frac { 1 }{ 2 }\)
(v) a = -1.25, d = -0.25
Solution:
(i) Given: a = 10, d = 10
a₁ = 10,
a₂ = 10 + 10 = 20
a₃ = 20 + 10 = 30
a₄ = 30 + 10 = 40
Thus, the first four terms of the AP are 10, 20, 30, 40.

(ii) Given: a = – 2, d = 0
The first four terms of the AP are -2, -2, -2, -2.

(iii) a₁ = 4, d = -3
a₂ = a₁ + d = 4 – 3 = 1
a₃ = a₂ + d = 1 – 3 = -2
a₄ = a₃ + d = -2 – 3 = -5
Thus, the first four terms of the AP are 4, 1, -2, … -5.

(iv)

(v) a₁ = -1.25, d = -0.25
a₂ = a₁ + d = -1.25 – 0.25 = -1.50
a₃ = a₂ + d = -1.50 – 0.25 = -1.75
a₄ = a₃ + d = -1.75 – 0.25 = -2.00
Thus, the first four terms of the AP are -1.25, -1.50, -1.75, -2.

Question 3.
For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3, ……
(ii) -5, -1, 3, 7, ……
(iii) \(\frac { 1 }{ 3 }\) , \(\frac { 5 }{ 3 }\) , \(\frac { 9 }{ 3 }\), \(\frac { 13 }{ 3 }\) , ……..
(iv) 0.6, 1.7, 2.8, 3.9, …….
Solution:
(i) a₁ = 3, a₂ = 1
d = a₂ – a₁ = 1 – 3 = -2
where, a₁ = first term and d = common difference
a₁ = 3, d = -2

(ii) a₁ = -5, a₂ = -1
d = a₂ – a₁ = -1 – (-5) = -1 + 5 = 4
So, first term a₁ = -5 and common difference d = 4

(iii) a₁ = \(\frac { 1 }{ 3 }\), a₂ = \(\frac { 5 }{ 3 }\)
d = \(\frac { 5 }{ 3 }\) – \(\frac { 1 }{ 3 }\) = \(\frac { 4 }{ 3 }\)
So, first term a₁ = \(\frac { 1 }{ 3 }\) and common difference d = \(\frac { 4 }{ 3 }\)
a₁= 0.6, a₂ = 1.7
d = a₂ – a₁ = 1.7 – 0.6 = 1.1
So, first term a₁ = 0.6 and common difference d = 1.1

Question 4.
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, …….
(ii) 2, \(\frac { 5 }{ 2 }\) , 3, \(\frac { 7 }{ 2 }\) , …….
(iii) -1.2, -3.2, -5.2, -7.2, ……
(iv) -10, -6, -2,2, …..
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …..
(vi) 0.2, 0.22, 0.222, 0.2222, ……
(vii) 0, -4, -8, -12, …..
(viii) \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , …….
(ix) 1, 3, 9, 27, …….
(x) a, 2a, 3a, 4a, …….
(xi) a, a2, a3, a4, …….
(xii) √2, √8, √18, √32, …..
(xiii) √3, √6, √9, √12, …..
(xiv) 1², 3², 5², 7², ……
(xv) 1², 5², 7², 7², ……
Solution:
(i) 2, 4, 8, 16, ……
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
i.e., d = a_n+1 – a_n is not same every time, so the given series do not form of an A.P.

(ii) We have given the series,
i.e., d = a_n+1 – a_n an is same everything, so the given list of numbers form an AP.
So, common difference (d) = \(\frac { 1 }{ 2 }\)
Next three terms after the last given term are
\(\frac { 7 }{ 2 }\) + \(\frac { 1 }{ 2 }\) = 4
4 + \(\frac { 1 }{ 2 }\) = \(\frac { 9 }{ 2 }\) and \(\frac { 9 }{ 2 }\) + \(\frac { 1 }{ 2 }\) = 5
Thus, we obtain three terms as 4, \(\frac { 9 }{ 2 }\), 5.

(iii) We have given the series,
– 1.2, – 3.2, – 5.2, – 7.2, …
∴ a₂ – a₁ = – 3.2 – (1.2) = – 2
a₃ – a₂ = – 5.2 – (- 3.2) = – 2
i.e., d = a_n+1 – a<sub<n is same every time, so the given list of numbers form an AP.
So, common difference (d) = – 2
Next three terms after the last term are,
– 7.2 + (- 2) = – 9.2
– 9.2 + (- 2) = – 11.2 and – 11.2 + (- 2), – 13.2
Thus, we obtain three terms as – 9.2, 11.2, – 13.2.

(iv) We have gives series,
10, – 6, -2, 2
∴ a₂ – a₁ = – 6 – (- 10) = 4
a₃ – a₂ = – 2 -( – 6) = 4
i.e., d = a_n+1 is same every time, so the given list of numbers form an AP.
So, common difference (d) = 4
Next three terms after after last given term
are,
2 + 4= 6
6 + 4 = 10 and 10 + 4 = 14 Thus, we obtain three terms as 6,10,14.

(vi) We have given the series,
0.2, 0.22, 0.222, 0.2222,…
∴ a₂ – a₁ = 0.2 – 0.20 = 0.02
a₃ – a₂ = 0.222 – 0.22 = 0.002
i.e., d = a_n+1 is not same everytime, so the given series do not form an AP.

(vii) We have given the series,
0, – 4, – 8, – 12, …
= a₂ – a₁ = – 4 – 0 = – 4
a₃ – a₂ = – 8 – (- 4) = – 4
i. e., d = a_n+1 – a_n is same everytime, so the given series is in the form of an AP.
So, common difference (d) = – 4
Next three terms after the last given terms
are,
– 12 + (- 4)= – 16
– 16 + (- 4) = – 20 and – 20 + (- 4) = – 24
Thus, we obtain three as – 16, – 20, – 24.

(viii) We have given the series,

i.e., d = a_n+1 – a_n is same everytime, i.e., 0.
so the given series do not form an AP.
Next three terms after the last given terms are
\(\frac { -1 }{ 2 }\) + 0 = \(\frac { -1 }{ 2 }\), \(\frac { -1 }{ 2 }\) + 0 = \(\frac { -1 }{ 2 }\) and \(\frac { -1 }{ 2 }\) + 0 = \(\frac { -1 }{ 2 }\)

(ix) We have given the series, 1, 3, 9, 27,…
∴ a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
i.e., d = a_n+1 – a_n is not always same.
So, the given series are, do not form an AP.

(x) We have given the series, a, 2a, 3a, 4a, …
∴ a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
i.e., d = a_n+1 – a_n is same everytime, so the given series is in the form of an A.P.
So, common difference (d) = a
Next three terms after the last given term are, 4a + a = 5a
5a + a = 6a and 6a + a = 7a Thus, we obtain three terms as 5a, 6a, 7a.

(xi) We have given the series,
a₁, a², a³, a⁴, ….
∴ a²₂ – a₁ = a (a – 1)
a³ – a² = a² (a – 1)
i.e., d = a_n+1 – a_n is not always same.
So the given series are do not an A.P.

(xii) We have given the series,

i.e., d = a_n+1 – a_n is same everytime so, the given series is in the form of an AP.
So, common difference (d) = \(\sqrt{2}\)
Next three terms after the last terms ate,

Thus, we obtain three terms as
\(\sqrt{50}\), \(\sqrt{72}\), \(\sqrt{98}\).

(xiii) We have gives the series

i.e., d = a_n+1 – a_n is not same evervtime, so the given sereis are do not form an AP.

(xiv) We have given the series, 1², 3², 5², 7²
∴ a₂ – a₁ = 3² – 1² = 9 – 1 = 8
a₃ – a₂ = 5² – 32 = 25 – 9 = 16
i.e., d = a_n+1 – a_n is not same everytime, so the given series do not form an AP.

(xv) We have given the series, 1², 5², 7², 7², …
∴ a₂ – a₁ = 5² – 1² = 25 – 1 = 24
a₃ – a₂ = 7² – 5² = 49 – 25 = 24
a₄ – a₃ = 7³ – 7² = 73 – 49 = 24
i.e., d = a_n+1 – a_n is same everytime, so the given series are in the form of an AP.
So, common difference (d) = 24
Next three terms after the last term are,
73 + 24 = 97
97 + 24 = 121 and 121 + 24 = 145
Thus we obtain three terms as 97, 121, 145.