CHEMICAL REACTIONS AND EQUATIONS (CBSE 10TH CHEMISTRY QUESTIONS AND ANSWERS)

 

1.Why should a magnesium ribbon be cleaned before burning in air?

Answer:

• A magnesium ribbon should be cleaned before burning in air to remove the layer of magnesium oxide (MgO) that forms on its surface due to exposure to air.

• Magnesium reacts with oxygen in the air over time, forming a thin, dull coating of magnesium oxide.

• This oxide layer prevents or slows down the magnesium metal underneath from reacting efficiently when burned.

• Cleaning the ribbon (usually by scraping it with sandpaper) removes the oxide layer, exposing pure, shiny magnesium.

• The clean surface ensures that magnesium burns quickly and brightly, showing the characteristic intense white flame.

• Cleaning ensures a faster and more complete combustion.

2. Write the balanced equation for the following chemical reactions?

a.Hydrogen + Chloride----gives----Hydrogen chloride

b.Barium Chloride + Ammonium Sulphate----gives---Barium Sulphate + Ammonium Chloride

c.Sodium+Water----gives-----Sodium Hydroxide + Hydrogen

Answer

a.Hydrogen + Chlorine → Hydrogen chloride

Unbalanced:
H₂ + Cl₂ → HCl

Balanced:
H₂ + Cl₂ → 2HCl

b.Barium Chloride + Ammonium Sulphate → Barium Sulphate + Ammonium Chloride

Unbalanced:
BaCl₂ + (NH₄)₂SO₄ → BaSO₄ + NH₄Cl

Balanced:
BaCl₂ + (NH₄)₂SO₄ → BaSO₄ + 2NH₄Cl

c.Sodium + Water → Sodium Hydroxide + Hydrogen

Unbalanced:
Na + H₂O → NaOH + H₂

Balanced:
2Na + 2H₂O → 2NaOH + H₂

3. Write a balanced chemical equation with state symbols for the following reactions

i. Solution of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride

ii.Sodium hydroxide solution(in water) react with hydrochloric acid solution (in water) to produce sodium chloride solution and water

Answer:

i. Barium chloride solution reacts with sodium sulphate solution

Reaction:
Barium chloride (aq) + Sodium sulphate (aq) → Barium sulphate (s) + Sodium chloride (aq)

Balanced Equation with State Symbols:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

BaSO₄ precipitates as an insoluble solid.

ii. Sodium hydroxide solution reacts with hydrochloric acid solution

Reaction:
Sodium hydroxide (aq) + Hydrochloric acid (aq) → Sodium chloride (aq) + Water (l)

Balanced Equation with State Symbols:
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

This is a neutralization reaction.

4. A solution of X is used for white washing 

i. Name the substance X and write it's formula

ii. Write the reaction of the substance X named in (i) above with water

Answer:

i. Name the substance X and write its formula

• The substance X used for whitewashing is quicklime.

• Name: Calcium oxide

• Formula: CaO

ii. Reaction of calcium oxide (X) with water

When calcium oxide reacts with water, it forms calcium hydroxide (slaked lime) and releases heat. This is an exothermic reaction.

Balanced Chemical Equation:
CaO(s) + H₂O(l) → Ca(OH)₂(aq)

Calcium hydroxide forms a milky suspension in water and is used for whitewashing walls.

5.Why does copper sulphate solution change when an iron nail is dipped in it?

Answer:

Why the change occurs:

• Iron (Fe) is more reactive than copper (Cu).

• Therefore, iron displaces copper from copper sulphate (CuSO₄) solution.

• Chemical Reaction:

  Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)

What changes you observe:

The blue colour of the CuSO₄ solution fades because:

• Copper(II) sulphate (CuSO₄) is blue.
• Iron(II) sulphate (FeSO₄) is pale green.
• A reddish-brown layer of copper metal forms on the iron nail.

Type of reaction: This is a displacement reaction and also a redox reaction (iron gets oxidized, copper gets reduced).

6. Write one equation each for decomposition reactions where energy is supplied in the form of heat, light, or electricity?

Answer:

Here are three decomposition reactions, each involving a different form of energy:

A.Decomposition by Heat (Thermal Decomposition)

Example:
Calcium carbonate decomposes on heating to form calcium oxide and carbon dioxide.

Equation:
CaCO₃(s) → CaO(s) + CO₂(g)
(Heat is supplied)

B.Decomposition by Light (Photodecomposition)

Example:
Silver chloride decomposes in sunlight to form silver and chlorine gas.

Equation:
2AgCl(s) → 2Ag(s) + Cl₂(g)
(Light energy is supplied)

This reaction is used in photographic films.

C. Decomposition by Electricity (Electrolysis)

Example:
Electrolysis of water breaks it down into hydrogen and oxygen gases.

Equation:
2H₂O(l) → 2H₂(g) + O₂(g)
(Electric current is supplied)

7. What is the difference between displacement and double displacement reactions ? Write equations for these reactions.

Answer:

1. Displacement Reaction

Definition:
A displacement reaction is one in which a more reactive element displaces a less reactive element from its compound.

General form:
A + BC → AC + B

Example:
Zinc displaces copper from copper sulphate solution.

Equation:
Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

Single element (Zn) replaces another element (Cu) from its compound.

---

2. Double Displacement Reaction

Definition:
A double displacement reaction involves an exchange of ions between two compounds to form two new compounds.

General form:
AB + CD → AD + CB

Example:
Barium chloride reacts with sodium sulphate to form barium sulphate and sodium chloride.

Equation:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

Cations and anions are exchanged between the reactants.

Key Difference Summary: 

Feature                                            Displacement Reaction            Double Displacement Reaction

Number of exchanges               One element displaces another            Two compounds exchange ions

Type of reactants                       Element + Compound               Compound + Compound

Example reaction                  Zn + CuSO₄ → ZnSO₄ + Cu      BaCl₂ + Na₂SO₄ →BaSO₄ + 2NaCl 

   

DEFINE SARCOMERE AND LABEL ITS BANDS(CBSE 12TH BIOLOGY IMPORTANT QUESTION)

 

A sarcomere is the structural and functional unit of a myofibril in a striated muscle. It is the segment between two Z-lines and is responsible for muscle contraction through the sliding filament mechanism.

Bands of a Sarcomere (with Labels):

|←-- Sarcomere --→|

Z-line        I-band        A-band    H-zone   M-line H-zone  A-band   I-band   Z-line

  |              |            |         |          |        |          |            |            |

--Z--------I---------A-----H-----M----H------A--------I--------Z-- (repeats)

Description of Bands and Lines:

Z-line (Z-disc)-Ends of a sarcomere. Actin filaments are anchored here.

I-band (Isotropic band)-Light band with only actin (thin) filaments. Spans two sarcomeres.

A-band (Anisotropic band)-Dark band with overlapping actin and myosin. Remains constant during contraction

H-zone-Central part of A-band containing only myosin (thick filaments), disappears during contraction.

M-line-Middle of H-zone, holds myosin filaments together.

Important Points:

• The sarcomere shortens during muscle contraction.

• The A-band remains unchanged, while I-band and H-zone shorten.

• Z-lines move closer during contraction.

MUSCLE CONTRACTION- KEY NOTES(CLASS 12 CBSE BIOLOGY)

  

Mechanism of Muscle Contraction

   

1. Types of Muscles

• Skeletal muscles – Voluntary, striated, attached to bones.

• Smooth muscles – Involuntary, non-striated, found in internal organs.

• Cardiac muscles – Involuntary, striated, found only in the heart.

2. Structure of Skeletal Muscle

Muscle → Fascicles → Muscle fibers (cells) → Myofibrils → Sarcomeres

Sarcomere: Structural and functional unit (between two Z-lines)

Bands in Sarcomere:

• A-band: Dark band; both actin and myosin.

• I-band: Light band; only actin.

• H-zone: Only myosin; center of A-band.

• M-line: Middle line of H-zone.

• Z-line: Boundaries of sarcomere.

3. Types of Proteins

• Contractile proteins:

  Actin (thin filament): Two F-actins + tropomyosin + troponin.
• Myosin (thick filament): Heavy chains (tail) + light chains (head).
• Regulatory proteins: Tropomyosin and Troponin.
• Accessory proteins: Help in alignment and elasticity (e.g., titin).

3. Mechanism of Muscle Contraction

Sliding Filament Theory (by Huxley and Niedergerke):

"Actin filaments slide over myosin filaments, shortening the sarcomere."

Steps:

• Nerve impulse → reaches neuromuscular junction.

• Release of Acetylcholine (ACh) → depolarizes sarcolemma.

• Ca²⁺ released from sarcoplasmic reticulum.

• Ca²⁺ binds to troponin, exposing myosin-binding sites on actin.

• Myosin head binds to actin → cross-bridge formation.

• ATP hydrolysis → myosin head pulls actin (power stroke).

• New ATP binds → myosin head detaches → recocks → cycle repeats.

4. Role of ATP and Calcium

• ATP: Energy source for detachment of myosin head and resetting it.

• Calcium ions: Initiate contraction by binding to troponin.

5. Muscle Relaxation

• Calcium ions pumped back into SR.

• Troponin-tropomyosin complex blocks binding sites.

• Muscle returns to resting state.

6.Types of Muscle Contractions

• Isotonic: Muscle changes length (e.g., lifting weight).

• Isometric: Muscle length constant, tension changes (e.g., holding posture).

7.Disorders Related to Muscles (for extras)

• Myasthenia Gravis: Autoimmune disorder; affects neuromuscular junction.

• Muscular Dystrophy: Genetic disease; progressive muscle degeneration.

• Fatigue: Due to lactic acid buildup after anaerobic respiration.

WHAT WILL HAPPEN IF A PLANT IS KEPT IN THE DARK FOR 72 HOURS? EXPLAIN WITH REASON(CBSE 10TH BIOLOGY)

  

If a plant is kept in the dark for 72 hours, it will become destarched — meaning it will use up the stored starch and not produce any new starch.

• Photosynthesis requires sunlight to produce glucose, which is stored as starch.
• In the absence of light, photosynthesis stops, so the plant cannot make food.
• However, the plant continues to respire, using up the stored starch for energy.
• As a result, after 72 hours in the dark, no starch will be present in the leaves.
• This is why destarching is done before photosynthesis experiments  to ensure that any starch found afterward is due to new photosynthesis.

WHY IS SUNLIGHT CONSIDERED ESSENTIAL FOR PHOTOSYNTHESIS?(CBSE 10TH BIOLOGY)

 

Sunlight is considered essential for photosynthesis because it provides the energy needed to drive the process.

• During the light-dependent stage of photosynthesis, chlorophyll absorbs sunlight.

• This light energy is then used to:

  Split water molecules into hydrogen and oxygen (a process called photolysis).
• Convert ADP to ATP and NADP⁺ to NADPH, which are energy-rich molecules.
• These energy carriers are then used in the dark reaction to form glucose from carbon dioxide.
• Without sunlight, the plant cannot produce ATP or NADPH, so photosynthesis cannot occur.

WHY DO DESERT PLANTS TAKE UP CARBONDIOXIDE AT NIGHT?(CBSE 10TH BIOLOGY)

 

Desert plants take up carbon dioxide at night to reduce water loss.

• In deserts, temperatures are very high during the day, and opening stomata (tiny pores on leaves) would lead to excessive water loss due to evaporation.

• To avoid this, desert plants keep their stomata closed during the day and open them at night when it’s cooler and humidity is higher.

• At night, they take in carbon dioxide and store it as organic acids.

• During the day, with stomata closed, they use the stored CO₂ to carry out photosynthesis.

• This special adaptation is known as CAM (Crassulacean Acid Metabolism) photosynthesis.

WHAT IS PHOTOSYNTHESIS? WRITE THE BALANCED CHEMICAL EQUATION FOR PHOTOSYNTHESIS?(CBSE 10TH BIOLOGY)

 

Photosynthesis is the process by which green plants, algae, and some bacteria use sunlight, carbon dioxide, and water to produce glucose (food) and release oxygen.

It occurs in the chloroplasts of plant cells and is essential for the survival of most life on Earth.

Balanced Chemical Equation of Photosynthesis:

6CO2​+6H2​O ----(sunlight, chlorophyll) → C6​H12​O6​+6O2

(Carbon dioxide + Water → Glucose + Oxygen)​

TWO MAIN STAGES OF PHOTOSYNTHESIS (CBSE 10TH BIOLOGY SHORT ANSWER QUESTION)

 

The two main stages of photosynthesis are:

• Light-dependent stage (Light reaction):

  • Occurs in the thylakoid membranes of chloroplasts.

  • Light energy is absorbed by chlorophyll and converted into chemical energy (ATP and NADPH).

  • Oxygen is released from the splitting of water (photolysis).

• Light-independent stage (Dark reaction or Calvin cycle):

• Occurs in the stroma of chloroplasts.

• Uses ATP and NADPH (from the light reaction) to fix carbon dioxide (CO₂) and form glucose.

• Does not require light directly.

SOAPS AND DETERGENTS IMPORTANT NOTES CBSE 10TH IMPORTANT PORTION CHEMISTRY

   

1. What are Soaps?

Soap is a sodium or potassium salt of a long-chain fatty acid.

• Common example: Sodium stearate
  (C₁₇H₃₅COONa)

 Made by:

• Saponification: Reaction of fat/oil + NaOH/KOH → Soap + Glycerol

• Fat/Oil + Alkali (NaOH) → Soap + Glycerol

2. Structure of Soap Molecule

Two parts:

• Hydrophobic (Tail) – long hydrocarbon chain, repels water, attracts dirt/oil.

• Hydrophilic (Head) – ionic part (COO⁻Na⁺), attracts water.

   

STRUCTURE OF SOAP

 3. How Soaps Work

• Dirt/oil is greasy and does not dissolve in water.

• Soap molecules form a micelle:

• Tails trap the dirt/oil.

• Heads face outward, attracting water.

• These micelles lift dirt and rinse it away.

• Micelle: Cluster of soap molecules trapping dirt inside.

4. Limitations of Soaps

• Not effective in hard water (water with Ca²⁺ and Mg²⁺ ions).

• Forms insoluble scum with these ions and reduces cleaning power.

2. Detergents

What are Detergents?

• Detergents are synthetic cleansing agents.

• Made from petrochemicals (not natural fats/oils).

• Example: Sodium lauryl sulfate

Advantages of Detergents

• Work well in hard water

• More foam, more cleaning efficiency

• No scum formation

Types of Detergents

• Anionic – used in laundry (common type)

• Cationic – used in hair conditioners

• Non-ionic – used in dishwashing liquids

Properties of  Soaps and Detergents

   

Feature	Soap

Source

Natural oils/fats

In Hard Water

Forms scum, less effective

Cost

Usually cheaper

Environment

Biodegradable

                      

Feature	Detergent

                               

Source                             Petrochemicals (synthetic)

       

In Hard Water

Works well, no scum

Cost	Often costlier

Environment

Some may not be biodegradable

                             

                                                              

STRUCTURE OF DETERGENT

LAWS OF REFLECTION AND LAWS OF REFRACTION(CBSE 10TH PHYSICS)

 

1. Laws of Reflection of Light

• First Law:
  The incident ray, the reflected ray, and the normal to the reflecting surface at the point of incidence all lie in the same plane.
  Second Law (Law of Reflection):
  The angle of incidence is equal to the angle of reflection.

  • $$\angle i=\angle r$$

2. Laws of Refraction of Light (Snell’s Laws)

These apply when light passes from one medium to another (like from air to glass).
First Law:
The incident ray, the refracted ray, and the normal to the interface of two media at the point of incidence all lie in the same plane.
Second Law (Snell’s Law):
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant (for a given pair of media).

sini/sinr=constant=μ

• $$\frac{\mathrm{<br>}}{}$$
•   i= angle of incidence
• $r$ = angle of refraction
• $\mu$ = refractive index (e.g., glass with respect to air)
• When light travels from a rarer to a denser medium, it bends towards the normal.
• When light travels from a denser to a rarer medium, it bends away from the normal.

"MONOCOT STEM FEATURES FOR CLASS XI (BIOLOGY) - EASY NCERT NOTES + LABELLED DIAGRAM"

 Important Notes on Monocot Stem

General Characteristics

• Belongs to monocotyledonous plants (e.g., maize, sugarcane, bamboo, grasses).

• No secondary growth due to the absence of a vascular cambium.

• Vascular bundles are scattered throughout the ground tissue.

• The stem is usually herbaceous, but some may become woody (e.g., palms).

Anatomical Features (TS - Transverse Section)

1. Epidermis

   • Outermost layer; single-layered, made of compactly arranged cells.

   • Covered with a cuticle to prevent water loss.

   • May contain stomata and trichomes (in some species).

2. Hypodermis

   • Usually consists of sclerenchyma cells (mechanical support).

   • Lies just beneath the epidermis.

3. Ground Tissue

   • No clear differentiation into cortex, endodermis, pericycle, etc.

   • Entire area between hypodermis and vascular bundles is parenchymatous.

   • Stores food and may help in photosynthesis.

4. Vascular Bundles

   • Scattered in the ground tissue (atactostele).

   • Each bundle is conjoint, collateral, and closed (no cambium).

   • Surrounded by a sclerenchymatous bundle sheath for strength.

   • Xylem is oriented towards the center (endarch), phloem towards the outside.

5. Pith

   • Not distinct as in dicots; ground tissue is uniform.

Examples of Monocot Stems: Maize, Wheat, Sugarcane, Banana, Bamboo

                       

Key Differences from Dicot Stem

Feature	Monocot Stem	Dicot Stem
Vascular bundles	Scattered	Arranged in a ring
Cambium	Absent	Present
Secondary growth	Absent	Present
Ground tissue	Undifferentiated	Differentiated (cortex, pith)
Bundle type	Closed	Open

Group 1 vs Group 17 Elements – Class 10 CHEMISTRY | Easy Comparison Table

 

Property	Group 1: Alkali Metals	Group 17: Halogens
Valence Electrons	1	7
Valency	+1	–1
Reactivity	Highly reactive metals	Highly reactive non-metals
Nature	Metallic	Non-metallic
Tendency	Lose 1 electron easily (form cations)	Gain 1 electron easily (form anions)
State at Room Temp	Solid (except hydrogen, which is not a metal)	Gases (e.g., Cl₂, F₂), liquid (Br₂), solid (I₂)
Melting & Boiling Points	Low (but higher than halogens)	Low (decreases down the group)
Appearance	Shiny, soft metals	Colored and poisonous
Reaction with Water	React violently to form hydroxides and H₂ gas	Do not react with water but can displace oxygen
Examples	Lithium (Li), Sodium (Na), Potassium (K)	Fluorine (F), Chlorine (Cl),   Bromine (Br)

PROPERTIES OF ETHANOL AND ETHANOIC ACID (CBSE 10TH CHEMISTRY IMPORTANT QUESTIONS)

 

Ethanol (C₂H₅OH)

Physical Properties:

• It is a colorless liquid with a pleasant smell.

• Soluble in water in all proportions.

• Boiling point ≈ 78°C.

Chemical Properties:

a) Combustion (Burning):

Ethanol burns in air to produce carbon dioxide and water.

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O  

Exothermic reaction – releases heat and light.

b) Reaction with Sodium:

Ethanol reacts with sodium metal to form sodium ethoxide and hydrogen gas.

2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑

c) Dehydration (with Conc. H₂SO₄):

Ethanol can be dehydrated to form ethene in the presence of concentrated sulfuric acid.

C₂H₅OH  →  C₂H₄ + H₂O  

  (Conc. H₂SO₄, heat)

Ethanoic Acid (CH₃COOH)

Physical Properties:

• Also known as acetic acid.

• Has a sour taste and pungent smell.

• Freezes to form ice-like crystals below 16.7°C – called glacial acetic acid.

2. Chemical Properties:

a) Reaction with Metals (e.g., Zinc):

Forms a salt (zinc acetate) and hydrogen gas.

2CH₃COOH + Zn → (CH₃COO)₂Zn + H₂↑

b) Reaction with Carbonates/Bicarbonates:

Reacts to form a salt, water, and carbon dioxide gas.

CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑

c) Esterification Reaction (with Alcohols):

Reacts with ethanol to form an ester (ethyl ethanoate) and water (in presence of acid catalyst).

CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O  
             (Conc. H₂SO₄)

CARBON AND ITS COMPOUNDS (CBSE 10TH CHEMISTRY IMPORTANT QUESTIONS REPEATED)

 

1. What is the valency of carbon?

The valency of carbon is 4. Carbon has 4 electrons in its outermost shell and needs 4 more to complete its octet. Therefore, it forms four covalent bonds with other atoms.

2.Give the IUPAC name of CH₃CH₂OH?

The IUPAC name of CH₃CH₂OH is Ethanol.

It is an alcohol with a two-carbon chain (ethane) and an –OH functional group, hence the name ethanol.

3. What is catenation?

Catenation is the ability of an element to form bonds with atoms of the same element, forming chains or rings. Carbon shows a strong tendency for catenation, allowing it to form long chains, branched structures, and rings due to the strong C–C bonds.

4. Draw the electron dot structure of ethene (C₂H₄)?

Molecular formula: C₂H₄

Ethene contains:

• 2 carbon atoms connected by a double bond (C=C)

• 4 hydrogen atoms, each forming a single covalent bond with carbon

• Electron Dot Structure:

•   H       H

       \     /

     C = C

      /     \

   H       H

5. What are homologous series? Give one example.

A homologous series is a group of organic compounds having the same functional group, similar chemical properties, and a general formula, where each successive member differs by a –CH₂– group (i.e., 14 amu).

Example:
The alkane series:
Methane (CH₄), Ethane (C₂H₆), Propane (C₃H₈), Butane (C₄H₁₀), etc.

6. Why do carbon compounds have low melting and boiling points?

Carbon compounds (especially covalent compounds) have low melting and boiling points because:

• They are held together by weak intermolecular forces (van der Waals forces).

• They do not have strong ionic or metallic bonds that require more energy to break.