Activity Based questions for Class 10 Chemical reaction and equation QueryExpress

Activity Based questions for Class 10 Chemical reaction and equation- Here, we will share Activity Based questions for Class 10 Chemical reaction and equation chapter 1 chemistry. if you are searching for case based or Activity Based questions for Class 10 Chemical reaction and equation, then you are in the right place. queryexpress provides the most important case based and Activity Based questions for Class 10 Chemical reaction and equation for you.

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Activity Based questions for Class 10 Chemical reaction and equation

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Class 10 Activity Based questions on Chemical reaction and equation

Case- A student observed that a piece of iron kept in a humid environment showed signs of rusting after a few days.

Question: Explain the chemical reaction involved in the rusting of iron. What are the conditions responsible for rusting?

Answer: Rusting of iron occurs due to the reaction of iron with oxygen and water in the presence of moisture and air. The chemical equation for rusting is: Iron(s)+Oxygen(g)+Water(moisture)→Iron Oxide (Rust)

Fe(solid) + O_2(gas) + H₂O → Fe₂O₃

The conditions necessary for rusting to occur are – Presence of oxygen (from air), Presence of water (moisture), and Presence of electrolytes (such as salts)

Case- During a laboratory experiment, when magnesium ribbon was burnt in air, a bright white light was observed.

Question: Explain the type of chemical reaction that occurs when magnesium reacts with oxygen. Write a balanced chemical equation for the reaction.

Answer: The burning of magnesium in air is a combustion reaction where magnesium reacts with oxygen to form magnesium oxide. The balanced chemical equation is: 2Mg + O₂ → 2MgO

Case based question and answers class 10 chemistry Chemical reaction and equation

Activity- Take 2 g of ferrous sulphate crystals in a dry boiling tube and then Heat the boiling tube. Now Observe the colour of the crystals before and after heating.

Question: Describe the observations made during the activity with ferrous sulphate crystals before and after heating.

Answer: Before heating, the ferrous sulphate crystals are typically green in color. Upon heating, the crystals lose their water of crystallization and turn white or pale yellow. This change in color indicates the dehydration of ferrous sulphate crystals.

Question: Explain the chemical change that occurs when ferrous sulphate crystals are heated.

Answer: When ferrous sulphate crystals (FeSO₄·7H₂O) are heated, they undergo thermal decomposition, losing water molecules. The chemical equation for this reaction is: FeSO 4 . 7H₂O→ FeSO₄ + 7H₂O

Note-Here, ferrous sulphate heptahydrate decomposes to form anhydrous ferrous sulphate and water vapor.

Activity- Take 2 gram of lead nitrate powder in a boiling tube. Now heat the boiling tube over a flame.

Question: Describe the observations made during the heating of lead nitrate powder in the boiling tube.

Answer: Upon heating lead nitrate (Pb(NO₃)₂) powder in a boiling tube, it undergoes thermal decomposition. Initially, the powder may melt and then decompose with formation of bubbles due to the release of brown nitrogen dioxide gas(NO₂) and oxygen gas(O₂). The boiling tube may also become coated with a yellow residue of lead oxide (PbO).

Question: Explain the chemical change that occurs when lead nitrate powder is heated.

Answer: The thermal decomposition of lead nitrate can be represented by the chemical equation:

2Pb(NO₃)₂ →2PbO+4NO₂ +O₂

Question: What safety precautions should be taken while performing this activity?

Answer: Safety precautions include wearing safety goggles and conducting the experiment in a well-ventilated area or under a fume hood to avoid inhalation of nitrogen dioxide gas (N₂) , which is toxic. The boiling tube should be handled with tongs to prevent burns from the heat.

Activity based questions on chemical reactions and equations worksheet

Activity- Take 2gram of silver chloride in a china dish. obseve its colour. Now Place this china dish in sunlight for some time. Now again Observe the colour of the silver chloride after some time.

Question: What is the initial color of silver chloride when taken in a china dish?

Answer: Silver chloride (AgCl) is initially white or pale yellow in color.

Question: Describe the observations made after placing the china dish with silver chloride in sunlight for some time.

Answer: After exposure to sunlight, silver chloride gradually darkens in color due to the formation of silver metal. This change occurs as silver chloride decomposes into silver metal and chlorine gas (Cl2) under the influence of sunlight.

Question: Explain the chemical change that occurs when silver chloride is exposed to sunlight.

Answer: The exposure of silver chloride to sunlight leads to the following photochemical reaction:

2AgCl → 2Ag + Cl2

In this reaction, silver chloride decomposes into silver metal and chlorine gas. The silver metal deposited on the surface of silver chloride darkens its color over time.

Question: Why does silver chloride darken upon exposure to sunlight?

Answer: Silver chloride darkens due to the reduction of silver ions (Ag⁺) to silver metal (Ag) by the energy from sunlight. This reduction process leads to the visible formation of metallic silver particles within the silver chloride crystal lattice.

Activity based questions on chemical reactions and equations class 10

Activity- Take three iron nails and clean them. Now Take two test tubes A and B. Now take 10 mL copper sulphate solution In each test tube. Now put two iron nails carefully in the copper sulphate solution in test tube B for about 20 minutes. Keep one iron nail aside for comparison. After 20 minutes, take out the iron nails from the copper sulphate solution.

Question: Why does the iron nail become brownish in color after being immersed in copper sulfate solution?

Answer: The brownish color observed on the iron nail after immersion in copper sulfate solution is due to the deposition of metallic copper (Cu) on the surface of the iron. This occurs because iron (Fe) is more reactive than copper (Cu) and displaces copper from copper sulfate according to the following reaction:

Fe (𝑠)+CuSO₄(𝑎𝑞)→FeSO₄(𝑎𝑞)+Cu(𝑠)

The metallic copper deposited on the iron nail gives it a brownish appearance.

Question: Why does the blue color of copper sulfate solution fade after immersing iron nails in it?

Answer: The blue color of copper sulfate solution fades because iron nails displace copper ions (Cu²⁺) from copper sulfate solution. The displaced copper ions react with the iron to form metallic copper, which precipitates out of the solution. As a result, the concentration of copper ions decreases in the solution, leading to a decrease in the intensity of the blue color.

Question: Compare the intensity of the blue color in test tubes (A) and (B) after the activity. Explain the differences observed.

Answer: Test tube (A) contains copper sulfate solution without any iron nails, so its blue color remains unaffected. In test tube (B), where iron nails were immersed in copper sulfate solution, the blue color fades due to the displacement of copper ions by iron. This observation demonstrates the chemical reaction where iron displaces copper from copper sulfate solution, resulting in the fading of its blue color.

Question: Compare the color of the iron nails dipped in copper sulfate solution with the one kept aside. Explain the differences observed.

Answer: The iron nail kept aside remains unaffected and retains its metallic grey color. However, the iron nails dipped in copper sulfate solution turn brownish due to the deposition of metallic copper on their surface. This difference illustrates the chemical change where iron reacts with copper sulfate solution, leading to the formation of a new substance (metallic copper) on the surface of the iron nails.

activity based questions class 10 chemistry chapter 1

Activity-Take about 3 mL of sodium sulfate solution in a test tube. take about 3 mL of barium chloride solution In another test tube. Now Mix the two solutions.

Question-Explain the chemical reaction that occurs when sodium sulfate and barium chloride are mixed.

Answer-The reaction between sodium sulfate and barium chloride results in the formation of barium sulfate (BaSO₄), which is insoluble in water and precipitates out of the solution. According to the solubility rules, most sulfate salts are soluble in water, but barium sulfate is an exception and is insoluble.

Question-Write the balanced chemical equation for this reaction and identify the precipitate formed.

Answer-Na2SO₄(𝑎𝑞)+BaCl₂(𝑎𝑞)→2NaCl(𝑎𝑞)+BaSO₄(𝑠)

Question-Identify the type of chemical reaction taking place when these two solutions are mixed.

Answer-This reaction is a double displacement reaction, specifically a precipitation reaction, where barium sulfate (BaSO₄) precipitates out of the solution as a white solid.

Activity-When lead(II) nitrate and potassium iodide solutions are mixed, the following observations and reactions occur:

Question-What was the colour of the precipitate formed? Write the name of the compound

Answer-The precipitate formed is yellow in colour. The compound precipitated is lead(II) iodide (PbI₂).

(ii) Write the balanced chemical equation for this reaction.

Answer-The balanced chemical equation for the reaction between lead(II) nitrate and potassium iodide is: Pb(NO3)₂(𝑎𝑞)+2KI(𝑎𝑞)→PbI₂(𝑠)+2KNO₃(𝑎𝑞)

(iii) Write down the type of Reaction

This reaction is called displacement reaction or a precipitation reaction. In this type of reaction, the cations and anions of the two reactants exchange partners, resulting in the formation of a new insoluble compound (precipitate). Here, lead(II) nitrate and potassium iodide exchange their ions to form lead(II) iodide and potassium nitrate.