abcdnotes

Friday, October 13, 2023

Acids, Bases and salts

(a) Chloride, nitrate, hydride, ammonium

(b) Hydrogen chloride, sodium hydroxide, calcium oxide, ammonia

(d) Ammonium chloride, sodium chloride, potassium nitrate, sodium sulphate

(e) Sodium nitrate, sodium carbonate, sodium sulphate, sodium chloride

(f) Calcium oxide, magnesium oxide, zinc oxide, sodium oxide.

(g) Crystalline blue vitriol, crystalline common salt, crystalline ferrous sulphate, crystalline sodium carbonate.

(h) Sodium chloride, potassium hydroxide, acetic acid, sodium acetate.

ANSWER:

ans
a.Ammonium = NH₄⁺
Nitrate = NO₃^-
Hydride = H^-
Chloride = Cl^-
Ammonium is odd because Ammonium is cation and rest are anions.

b.Hydrogen chloride is odd because Hydrogen chloride is acid and rest are base.

c.Acetic acid = CH₃COOH
   Carbonic acid = H₂CO₃
   Hydrochloric acid = HCl
   Nitric acid = - HNO₃
   HCl is the only Diatomic Hetro-Nuclear compound and remaining are Poly Atomic compound.

d.Ammonium chloride is odd because it is acidic salt and rest all are neutral salts.

e.Sodium carbonate is odd because the solutions of sodium nitrate, sodium sulphate and sodium chloride are neutral. But the solution of sodium carbonate is BASIC.

f.Calcium oxide = CaO
   Magnesium oxide = MgO
   Zinc oxide = ZnO
   sodium oxide = Na₂O
   ZnO is odd because it is amphoteric in nature and other ions are basic in nature

g.Common salt is odd because on heating, there is no change in color of compound. But in rest of the compounds, there is change is colour.

h. Potassium hydroxide is odd because in a reaction, when Sodium chloride reacts with acetic acid, then Sodium acetate is formed. There is no role of Potassium hydroxide in the reaction given below.
$NaCl + {CH}_{3} COOH \to {CH}_{3} COONa + HCl$

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Question 2:

Write down the changes that will be seen in each instance and explain the reason behind it.

(a) 50ml water is added to 50ml solution of copper sulphate.

(b) Two drops of the indicator phenlphthalein were added to 10ml solution of sodium hydroxide.

(d) A litmus paper was dropped into 2ml dilute HCl. Then 2ml concentrated NaOH was added to it and stirred.

(e) Magnesium oxide was added to dilute HCl and magnesium oxide was a added to dilute NaOH.

(f) Zinc oxide was added to dilute HCl and zinc oxide was added to dilute NaOH.

(g) Dilute HCl was added to limestone.

(h) Pieces of blue vitriol were heated in a test tube. On cooling, water was added to it.

(i) Dilute H 2 SO 4 was taken in an electrolytic cell and electric current was passed through it.

ANSWER:

ans
a. When 5o mL water is added to 50 mL solution of copper sulphate, then reversible reaction occurs and the colour change from pale blue to white and then change back to blue when water is added again.

$C u S O_{4} + H_{2} O \leftrightarrow C u S O_{4} . 5 H_{2} O (B L U E V I T R I O L) .$

b. Phenolphthalein is an indicator used for determining the quantity of base. When two drops of Phenolphthalein indicator are added to 10 mL solution of Sodium hydroxide, then the solution turns pink in colour because the acidic part of phenophthalein reacts with the base, it forms sodium salt of phenolphthalein which has pink color.
For example : In acid base titration phenolphthalein is used to detect end point of base.

c. Copper is an unreactive metal and doesn’t react in normal circumstances with dilute acids.

There are actually two equations for the reaction of copper with nitric acid take place. It depends on whether the nitric acid is concentrated or not. If it is concentrated, then the ratio is 1:4 for copper to nitric acid. If it is dilute then the ratio is 3:8.

C u + 4 H N O_{3} (d i l .) \to C u (N O_{3})_{2} + 2 N O_{2} + 2 H_{2} O 3 C u + 8 H N O_{3} (c o n c .) \to 3 C u (N O_{3})_{2} + 2 N O + 4 H_{2} O

Concentrated Nitric acid is act as strong oxidising agent so it makes sense that a higher oxidation state of nitrogen (IV) oxide is formed .

d. When litmus paper is dipped into 2 mL of dilute HCl solution , then blue litmus paper is turned into red colour and there is no effect on red litmus paper. Again, if the same litmus paper is dipped into 2 mL of concentrated NaOH solution, then red litmus paper turns into blue colour but there is no effect on blue litmus paper.This is due to the respective properties of blue and red litmus paper with acid and base.

e. Magnesium oxide is a base. when base reacts with an acid, it forms salt and water.This reaction is known as neutralisation reaction.

$M g O (s) + 2 H C l (l) = M g C l_{2} (a q) + H_{2} O (l)$

MgO doesnot react with NaOH.As NaOH is a base and bases react only with oxides of non-metal to form salt and water because oxides of non-metals are said to be acidic in nature so neutralization reaction take place. But MgO is a oxide of metal so, reaction is not possible.

f. Zinc oxide is added to dilute HCl, then neutralization reaction takes place to form salt and water.
The reaction is as follows:

$Z n O (a q) + 2 H C l (a q) \to Z n C l_{2} (a q) + H_{2} O (l)$

Zinc oxide reacts with sodium hydroxide to produce zincate sodium and water. This reaction takes place at a temperature of 500-600°C.it is exothermic reaction.
$ZnO + 2 NaOH \to {Na}_{2} {ZnO}_{2} + H_{2} O$

g. Limestone is calcium carbonate. When limestone is added to a 10% solution of dilute HCl, then brisk effervescence of CO₂ is released due to the reaction of acid with carbonate of metals.

$2 H C l + C a C O_{3} \to C a C l_{2} + C O_{2} + H_{2} 0$

h. When pieces of blue vitriol are heated in a test tube, then crystal structure of blue vitriol broke down to forn a colourless powder and water come out. This water is called water of crystallization. when water is added to the same test tube, then white powder turned into blue colour again.This is due to reversible reaction take place from anhydrous salt to hydrous salt and vice-versa.
$C U S O_{4} . 5 H_{2} O \leftrightarrow C U S O_{4} + 5 H_{2} O$

i. When a dilute solution of sulfuric acid is electrolysed, gases are produced at both the anode and the cathode electrode.

Hydrogen gas forms at the cathode, oxygen gas forms at the anode.

The gas produced at the cathode burns with a 'pop' sound, when a sample is lit with a lighted splint. This shows that the gas is hydrogen.

The gas produced at the anode relights a glowing splint dipped into a sample of the gas. This shows that the gas is oxygen.

The gases are produced when ions move towards the electrodes.

At the cathode:

2H⁺ +2e^- → H₂

At the anode:

4OH^- - 4e^- → 2H₂O + O₂

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Question 3:

Classify the following oxides into three types and name the types.

CaO, MgO, CO₂ , SO₃, Na₂O, ZnO, Al₂O₃ , Fe₂O₃

ANSWER:

Oxides are of three types :

1) Acidic Oxides: CO₂ (Carbon dioxide), SO₃ (Sulfur trioxide)

2) Basic Oxides: CaO (Calcium oxide), MgO (Magnesium oxide), Na₂O (Sodium oxide)

3) Amphoteric Oxides: ZnO (Zinc oxide), Al₂O₃(Aluminium oxide), Fe₂O₃(Ferric oxide)

Page No 73:

Question 4:

Explain by drawing a figure of the electronic configuration.

a. Formation of sodium chloride from sodium and chlorine.

b. Formation of a magnesium chloride from magnesium and chlorine.

ANSWER:

a. Atomic number of Sodium(Na) atom is 11.
Electronic configuration is :
Na = 2, 8, 1
So it contains 1 valence electron. In order to achieve the nearest noble gas configuration, it loses one electron to form Sodium ion.
Na⁺= 2,8

Atomic number of Chlorine(Cl) atom is 17.
Electronic configuration is :
Cl = 2, 8, 7
So it contains 7 valence electron. In order to achieve the nearest noble gas configuration, it gains one electron to form Chloride ion.
Cl^-= 2,8
An Ionic bond is formed between sodium ion and chloride ion by complete transfer of electron from sodium to chlorine.

b. Atomic number of Magnesium(12) atom is 12.
Electronic configuration is :
Mg = 2, 8, 2
So it contains 2 valence electron. In order to achieve the nearest noble gas configuration, it loses two electrons to form Magnesium ion.
Mg²⁺= 2,8

Atomic number of Chlorine(Cl) atom is 17.
Electronic configuration is :
Cl = 2, 8, 7
So it contains 7 valence electron. In order to achieve the nearest noble gas configuration, it gains one electron to form Chloride ion.
Cl^-= 2,8
An Ionic bond is formed between Magnesium ion and two Chloride ion by complete transfer of one electron to each Chlorine ion.

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Question 5:

Show the dissociation of the following compounds on dissolving in water, with the help of chemical equation and write whether the proportion of dissociation is small or large .

Hydrochloric acid, Sodium chloride, Potassium hydroxide, Ammonia, Acetic acid, Magnesium chloride, Copper sulphate

ANSWER:

$a.HCl$ (aq) $+ H_{2} O$ (l) $\to H_{3} O$ ⁺(aq) $+ {Cl}$ ^-(aq)
Explanation:
Hydrochloric acid(HCl) is a strong acid, so HCl is ionize completely in aqueous solution.In other words, every molecule of hydrochloric acid that is added to water will donate its proton H⁺ to water molecule to form a hydronium cation, and H₃O⁺and chloride ions Cl^- is formed.

b.Reaction:
NaCl(s) + H₂O(aq) -> Na⁺(aqueous) + Cl^-(aqueous) + H₂O(l)
Explanation:
When sodium chloride reacts with water the Na⁺ part of NaCl is attracted to the oxygen side of the water molecules, while the Cl^- side is attracted toward the hydrogen's side of the water molecule.

This causes the sodium chloride(salt) to split in water and the NaCl ionises into Na⁺ and Cl^-ions completely.

c)KOH(s) + H₂O(l)<--> K⁺(aq) + OH^-(aq) + H₂O(l)
Explanation:
KOH is base, so mixing it in water makes a basic solution that is in equilibrium. the KOH is just dissolving in the water.

d)NH₃(base) + H₂O(acid)<--> NH₄⁺(conjugate acid) + OH^-(conjugate base)
Explanation:
when ammonia dissolves in water. In an aqueous solution, ammonia acts as a base, acquiring hydrogen ions from H₂O to yield ammonium cation and hydroxide ions.

e)CH₃COOH(l) + H₂O(l) -> CH₃COO^-+ H₃O⁺
Explanation:
When acetic acid is added to water, due to electronegativity differences of oxygen and hydrogen in OH group of acetic acid and there is a dipole interaction with water molecule. Hence, the acetic acid is ionise into acetate ion and H⁺ ion combines with water to form hydronium ion. —

f)MgCl₂(s) + H₂O(l) -> Mg²⁺(aqueous) + Cl^-(aqueous) + H₂O(l)
Explanation:
On addition to water the Mg²⁺ part of MgCl₂ is attracted to the oxygen side of the water molecules, while the Cl- side is attracted to the hydrogen's side of the water molecule . This causes magnesium chloride(salt) to split in water and the MgCl₂is ionise into Mg²⁺ and Cl^-ions completely.

g)CuSO₄ (s) + H₂O (l) --> Cu⁺² (aq) + SO₄^-2 (aq) + H₂O(l)
Explanation:
when a compound dissolves in water, it dissociates to form ions.The reaction between anhydrous copper(II) sulphate(white) and water turns blue in the presence of water.

Page No 74:

Question 6:

Write down the concentration of each of the following solutions in g/L and mol/L.

a. 7.3g HCl in 100ml solution

b. 2g NaOH in 50ml solution

c. 3g CH₃COOH in 100ml solution

d. 4.9g H₂SO₄in 200ml solution

ANSWER:

a)7.3 g HCl in 100 mL solution:

$Concentration of HCl in g L^{- 1} = \frac{M a s s o f s o l u t e in gram}{Volume of HCl in l i t r e}$

                                       $= \frac{7.3 \times 1000}{100} = 73$ g L^-1

In terms of gram per litre:
7.3 g of HCl in 100 mL has concentration = 73 g L^{-1

Molecular mass of HCl = $1 + 35.5 = 36.5$}

$Molarity = \frac{Mass of solute in m o l e s}{Volume of HCl in L}$

             $= \frac{7.3 \times 1000}{36.5 \times 100} = 2$ mol L^-1

In terms of moles per litre:
7.3 g of HCl in 100 mL has concentration = 2 mol L^-1

b)2g NaOH in 50 mL solution

$C o n c e n t r a t i o n o f N a O H i n g L^{- 1} = \frac{M a s s o f N a O H i n g}{V o l u m e o f N a O H i n L}$

                                       $= \frac{2 \times 1000}{50} = 40$ g L^-1
In terms of gram per litre:
2 g of NaOH in 50 mL has concentration = 40 g L^{-1

Molecular mass of NaOH = $23 + 16 + 1 = 40$}

$Molarity = \frac{Mass of solute in m o l e s}{Volume of N a O H in L}$

             $= \frac{2 \times 1000}{40 \times 50} = 1$ mol L^-1

In terms of moles per litre:
2 g of NaOH in 50 mL has concentration = 1 mol L^-1

c)3 g CH₃COOH in 50 mL solution

$C o n c e n t r a t i o n o f C H_{3} C O O H i n g L^{- 1} = \frac{M a s s o f C H_{3} C O O H i n g}{V o l u m e o f C H_{3} C O O H i n L}$

                                       $= \frac{3 \times 1000}{100} = 30$ g L^-1
In terms of gram per litre:
3 g of CH₃COOH in 100 mL has concentration = 30 g L^{-1

Molecular mass of} CH₃COOH^{= $2 \times 12 + 2 \times 16 + 4 \times 1 = 60$}

$M o l a r i t y = \frac{M a s s o f s o l u t e i n m o l e s}{V o l u m e o f s o l u t i o n i n L}$

             $= \frac{3 \times 1000}{60 \times 100} = 0.5$ mol L^-1

In terms of moles per litre:
3 g of CH₃COOH in 100 mL has concentration = 0.5 mol L^-1

d)4.9 g H₂SO₄ in 200 mL solution

$C o n c e n t r a t i o n o f H_{2} S O_{4} i n g L^{- 1} = \frac{M a s s o f H_{2} S O_{4} i n g}{V o l u m e o f H_{2} S O_{4} i n L}$

                                                 $= \frac{4.9 \times 1000}{200} = 24.5$ g L^-1
In terms of gram per litre:
4.9 g of H₂SO₄ in 200 mL has concentration = 24.5 g L^{-1

Molecular mass of} H₂SO₄^{= $2 \times 12 + 2 \times 16 + 4 \times 1 = 60$}

$M o l a r i t y = \frac{M a s s o f s o l u t e i n m o l e s}{V o l u m e o f s o l u t i o n i n L}$

             $= \frac{4.9 \times 1000}{98 \times 200} = 1$ mol L^-1

In terms of moles per litre:
4.9 g of   H₂SO₄ in 200 mL has concentration = 1 mol L^-1

Page No 74:

Question 7:

Obtain a sample of rainwater. Add to it a few drops of universal indicator. Measure its pH. Describe the nature of the sample of rainwater and explain the effect if it has on the living world.

ANSWER:

Ans7.This is an activity based question in which you are supposed to collect rainwater from different places and compare there results on the basis of following parameters :
a) pH of water
b)Action of water on blue litmus paper
c)Action of water on red litmus paper
d)Effect of indicator like phenolphthalein and methyl orange
Conclusion :
If we take samples of rain water from different places, we observe the following results :

pH of water is in between 1-6
Blue litmus paper turns red
No change in colour of red litmus paper
No change in colour of rain water on adding phenolphthalein
rain water turns red in colour on adding methyl orange

The above observations show that rainwater is acidic in nature. For further assurance, we cancompare blue litmus paper dipped in different samples of rainwater with pH scale.
If rain water is acidic, then its pH must be in the range of 0-6.
The strength of ac acid depends on its pH value i.e. lower the value of pH, higher will be the strength of an acid and vice-versa.

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Question 8:

Answer the following questions.

a. Classify the acids according to their basicity and give one examplen of each type.

b. What is meant by neutralization? Give two examples from everyday life of the neutralization reaction.

c. Explain what is meant by electrolysis of water. Write the electrode reactions and explain them.

ANSWER:

Ans 8.

a. The number of ionizable hydrogen (H+) ions present in one molecule af an acid is called its basicity.

For example :
HCl ---------> H⁺+ Cl^-
Basicity of HCl is 1.

H₂SO₄ ------> 2H⁺ + SO₄^2-
Basicity of H₂SO₄ is 2.

Based on Basicity acids were classified into different types :
1. Mono-basic acids
2. Di-basic acids
3. Tri-basic acids

1. Mono-basic acids :

Acids, which on ionisation produce one hydronium ion on reaction with water.
Acids, which on ionisation produce one hydrogen ion.
For example : HCl, HNO₃ etc.

2. Di-basic acids :

Acids, which on ionisation produce two hydronium ion on reaction with water
Acids, which on ionisation produce two hydrogen ion

For example : H₂SO_4,H₂CO₃etc.

3. Tri-basic acids :

Acids, which on ionisation produce three hydronium ion on reaction with water
Acids, which on ionisation produce three hydrogen ion

For example : H₃PO₄, H₃PO₃ etc.

b. Neutralization reaction :

A neutralization reaction is a reaction in which an acid and a base reacts to form water and a salt. It involves the combination of H⁺ ions and OH^- ions to generate water.
The neutralization of a strong acid and strong base has a pH equal to 7.
The neutralization of a strong acid and weak base will have a pH of less than 7.
The neutralization of a strong base neutralizes a weak acid will be greater than 7.
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When a solution is neutralized, it means that salts are formed from equal weights of acid and base.

Neutralization reaction has application in daily life:

1)Self defence by animals and plants through chemical warfare :

Bee stings are acidic in nature, household remedy for a bee sting is baking soda or sodium bicarbonate, which is a basic substance.
A wasp stings are mildly basic, household remedy for this will be vinegar, also known as acetic acid.
These simple treatments ease these painful stings by a process called neutralization.

2)pH in our digestive system :
An acidic stomach due to eating too much spicy food, can be relieved by taking an antacid. The antacid is alkaline/basic in nature and helps to neutralize the stomach's acidity or you may take magnesium hydroxide(Milk of magnesia) and sodium hydrogen carbonate(Baking soda).

3) pH change as the cause of tooth decay :
When we eat food containing sugar, then bacteria present in our mouth break down the sugar to form acids(such as lactic acid). Thus acid is formed in the mouth after digestion. This will lead to the cause of tooth decay. The best way to prevent tooth decay is to clean the mouth after eation food with toothpaste, which is basic in nature. This will result in neutralization of acid by base.

4)soil pH and plant growth :
Most of the plants grow best when the pH of the soil is close to 7 that's neural. If the soil is too acidic or too basic(alkaline), the plants grow badly.
The acidic soil is neautralize by treatment with materials like quicklime(calcium oxide) or slaked lime(calcium hydroxide) or chalk(calcium carbonate).
If the soil is too basic, then alkalinity can be reduced by adding decaying organic matter(manure or composite)which contains acidic materials.

c.Electrolysis of water :

Electrolysis of water is the decompositon into oxygen and hydrogen gas due to an electric current passed through the water.

The following equation represents the electrolysis of water : $H 2 O (l)$ In pure water, at the negatively charged cathode, a reduction reaction takes place, with electrons (e⁻) from the cathode being given to hydrogen cations to form hydrogen gas.

Reduction at cathode: 2 H⁺ + 2e⁻ → H₂

On positively charged anode, an oxidation reaction occurs, generating oxygen gas by giving electrons to the anode :

Oxidation at anode: 2 H₂O → O₂(g) + 4 H⁺(aq) + 4e⁻

The same half reactions can also be balanced with base as listed below. To add half reactions they must be balanced with either acid or base.

Cathode (reduction):	2 H₂O(l) + 2e⁻	→	H₂(g) + 2 OH⁻(aq)
Anode (oxidation):	4 OH⁻(aq)	→	O₂(g) + 2 H₂O(l) + 4 e⁻

Combining either half reaction pair yields the same overall decomposition of water into oxygen and hydrogen:

Overall reaction: 2 H₂O(l) → 2 H₂(g) + O₂(g)

The number of hydrogen molecules produced is thus twice the number of oxygen molecules. The produced hydrogen gas has therefore twice the volume of the produced oxygen gas. The number of electrons pushed through the water is twice the number of generated hydrogen molecules and four times the number of generated oxygen molecules.

2 H 2 O (l) âŸ¶ 2 H 2 (g) + O 2 (g)

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Question 9:

Give reason for the following.

a. Hydronium ions are always in the form H₃O⁺

b. Buttermilk spoils if kept in a copper or brass container.

ANSWER:

a)When an acids dissolve in water. The H⁺ ion from acid always goes to the nearest water molecule to form hydronium ion.

HCl(aq)

+ H2O(aq)\to H_{3} {O+(aq)} + {Cl-(aq)}

For example : When Hydrochloric acid(HCl) is a strong acid. When it is dissolved in water, HCl is ionized completely in aqueous solution. Hydrochloric acid will donates its proton H⁺ to water molecule to form a hydronium cation H₃O⁺and chloride anions Cl^- .b)Buttermilk spoils if kept in a copper or brass container because buttermilk is actually lactic acid. Lactic acid reacts with the container material and produces poisonous complex. It is actually the reaction between acid and metal . This reaction is called as electro chemical reaction.

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Question 10:

Write the chemical equations for the following activities.

(a) NaOH solution was added to HCl solution.

(b) Zinc dust was added to dilute H₂ SO₄ .

(e) Carbon dioxide gas was passed through KOH solution.

(f) Dilute HCl was poured on baking soda.

ANSWER:

a . HCl (l) + NaOH (l) \to NaCl (s) + H_{2} O (l) + Energy b . Zn (s) + H_{2} {SO}_{4} (l) \to {ZnSO}_{4} (s) + H_{2} (g) c . CaO (s) + 2 {HNO}_{3} (l) \to Ca ({NO}_{3})_{2} (s) + H_{2} O (l) d . KOH (l) + {CO}_{2} (g) \to {KHCO}_{3} e . {NaHCO}_{3} (s) + 2 HCl (l) \to NaCl (s) + H_{2} O (l) + {CO}_{2}

Page No 74:

Question 11:

State the differences.

a. Acids and bases

b. Cation and anion

c. Negative electrode and positive electrode.

ANSWER:

Page No 74:

Question 12:

Classify aqueous solutions of the following substances according to their pH into three groups : 7, more than 7, less than 7.

Common salt, sodium acetate, hydrochloric acid, carbon dioxide, potassium bromide, calcium hydoxide, ammonium chloride, vinegar, sodium carbonate, ammonia, sulphur dioxide.

ANSWER:

Ans7.

SUBSTANCES	pH Value
Solution of Common salt	equal to 7
Solution of Sodium acetate	greater than 7
Solution of Hydrochloric acid	less than 7
Solution of Carbon dioxide	less than 7
Solution of Potassium bromide	equal to 7
Solution of Calcium hydroxide	greater than 7
Solution of Ammonium chloride	less than 7
Solution of Vinegar	less than 7
Solution of Sodium carbonate	greater than 7
Solution of Ammonia	greater than 7
Solution of Sulphur dioxide	less than 7

Tuesday, October 10, 2023

metals and non metals

Question 1:

Complete the table

Property of metal	Use in every day life
i. Ductility
ii. Malleability
iii. Conduction of heat
iv. Conduction of electricity
v. Sonority

ANSWER:

Property of metal	Use in every day life
i. Ductility	In electrical wires, cable wires etc.
ii.Malleability	Aluminium foil
iii. Conduction of heat	Cooking wares, microwave, electric press, straightening machine, electric belts
iv. Conduction of electricity	Bulb, tubelight, lamp, refrigerator, television
v. Sonority	Cymbals, doorbells

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Question 2:

Identify the odd term
A. Gold, silver, iron, diamond
B. Ductility, brittelness, sonority, malleability
C. Carbon, bromic, sulphur, phosphorus
D. Brass, bronze, iron, steel

ANSWER:

a.Iron is odd one out because iron is non-lustrous in nature and others are lustrous.
b.Brittleness is odd one out because it is a property of non-metals and rest are the properties of metals.
c.Bromine is odd one out because it is liquid non-metal and others are solid non-metals.
d.Iron is odd one out because it is not an alloy and others are alloys.

Page No 53:

Question 3:

Write scientific reasons.
A. The stainless steel vessels in Kitchen have copper coating on the bottom.
B. Copper and brass vessels are cleaned with lemon.
C. Sodium metal is kept in kerosene.

ANSWER:

a.The stainless steel vessels in Kitchen have copper coating on the bottom because copper is good conductor of heat and electricity, by using copper cookware, any meal can be prepared in a perfect and gentle way. They are the best pots and pans for cooking and roasting. This is particularly due to the fact that copper has excellent material properties and is is cheap metal as compare to other metals.

b.Copper vessels get tarnished due to corrosion of copper metal. It forms a layer of copper oxide, which is basic in nature. When lemon juice which contains citric acid is added to it, neutralisation reaction takes place thus the vessel gets cleaned.

c. Sodium metal is kept in kerosene because sodium is very reactive metal. It is kept in kerosene to prevent it from coming in contact with oxygen and moisture. If this happens, it will react with the moisture present in air and form Sodium hydroxide. This is a strongly exothermic reaction, and lot of heat is generated. It will result out in the form of fire.

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Question 4:

Answer the following
A. What is done to prevent corrosion of metals?
B. What are the metals that make the alloys brass and bronze ?
C. What are the adverse effects of corrosion ?
D. What are use of Noble metals. ?

ANSWER:

a.We can prevent corrosion by selecting the following things:

metal type
protective coating
environmental measures
sacrificial coatings
design modification

METAL TYPE :
ONE SIMPLE WAY TO PREVENT CORROSION IS TO USE A CORROSION RESISTANT METAL SUCH AS ALUMINIUM OR STAINLESS STEEL. DEPENDING ON THE APPLICATION, THESE METALS CAN BE USED TO REDUCE THE NEED FOR ADDITIONAL CORROSION PROTECTION.

PROTECTIVE COATINGS :
THE APPLICATION OF A PAINT COATING IS A COST-EFFECTIVE WAY OF PREVENTING CORROSION. PAINT COATINGS ACT AS A BARRIER BETWEEN METAL AND ATMOSPHERIC MOISTURE TO AVOID ITS CONTACT. FOR EXAMPLE : BLACK PAINT.
ANOTHER POSSIBILITY IS APPLYING A POWDER COATING. IN THIS PROCESS, A DRY POWDER IS APPLIED TO THE CLEAN METAL SURFACE TO AVOID ITS CONTACT WITH SURROUNDING OXYGEN. FOR EXAMPLE : ACRYLIC, POLYESTER, EPOXY, NYLON AND URETHANE.

ENVIRONMENTAL MEASURES :
CORROSION IS CAUSED BY A CHEMICAL REACTION BETWEEN THE METAL AND GASES IN THE SURROUNDING ENVIRONMENT. BY TAKING MEASURES TO CONTROL THE ENVIRONMENT, THESE UNWANTED REACTIONS CAN BE MINIMIZED. FOR EXAMPLE : THIS WOULD BE DECREASE BY TREATING THE WATER IN WATER BOILERS WITH SOFTENERS TO ADJUST HARDNESS, ALKALINITY OR OXYGEN CONTENT.

SACRIFICIAL COATINGS :
SACRIFICIAL COATING INVOLVES COATING THE METAL WITH AN ADDITIONAL METAL TYPE THAT IS MORE LIKELY TO OXIDIZE.
THERE ARE TWO MAIN TECHNIQUES FOR SACRIFICIAL COATING:
CATHODIC PROTECTION : THE MOST COMMON EXAMPLE OF CATHODIC PROTECTION IS THE COATING OF IRON ALLOY STEEL WITH ZINC, THIS PROCESS IS KNOWN AS GALVANIZING. ZINC IS A MORE ACTIVE METAL THAN STEEL AND WHEN IT STARTS TO CORRODE IT OXIDES WHICH INHIBITS THE CORROSION OF THE STEEL. THIS METHOD IS KNOWN AS CATHODIC PROTECTION
ANODIC PROTECTION : ANODIC PROTECTION INVOLVES COATING THE IRON ALLOY STEEL WITH A LESS ACTIVE METAL, SUCH AS TIN. TIN WILL NOT CORRODE, SO THE STEEL WILL BE PROTECTED AS LONG AS THE TIN COATING IS IN PLACE. THIS METHOD IS KNOWN AS ANODIC PROTECTION

Design Modification :
Design modifications can help reduce corrosion and improve the durability of any existing protective anti-corrosive coatings. Ideally, designs should avoid trapping dust and water, encourage movement of air, and avoid open crevices. Ensuring the metal is accessible for regular maintenance will also increase longevity.

b.Brass: It is an alloy, that contains copper and zinc as major constituents.
Bronze: It is an alloy, that contains copper and tin as major constituents.

c.The adverse effects of corrosion are as follows:

Damage to commercial airplanes that could result in possible in-flight problems
Damage to oil pipelines that could cause a costly and dangerous rupture that creates significant environmental damage.
Damage to bridge supports that could cause a bridge failure
Release of harmful pollutants from iron corrosion that contaminates the air
Costs of repairing or replacing household equipment that fails
Lose of efficiency
Contamination of product
Damage of metallic equipments
Inability to use metallic materials
Lose of valuable materials such as blockage of pipes, mechanical damage of underground water pipes
Accidents due to mechanical lose of metallic cars, aircrafts etc.
Causes pollution due to escaping products from corrosion
Depletion of natural resource

d.Uses of Noble metals are:

Uses of Silver : It is a shiny, heavy metal, and the best conductor of electricity.

It is used for making silver ornaments and expensive utensils such as glasses, mugs, etc.
It is used for making coins.
Salts of silver like silver chloride are used for making photographic films.
Silver foils are used for decorating sweets.
Silver is also used for making mirrors using a process called sputtering.

Uses of gold : Gold is bright yellow and a highly malleable and ductile metal.

Gold is used as the index of wealth. The countries which have more gold reserve are considered to be wealthy.
It is used for making ornaments.
It is used for making high-value coins and medals.
It is used for covering the mainframe of artificial satellites.

Uses of platinum : Platinum is silvery white, a highly malleable and ductile metal.

It is used for making ornaments and watches.
It is used as a catalyst in the manufacture of sulphuric acid and nitric acid.
It is used in platinum catalytic converters.
It is also used in chemical laboratories.

Page No 53:

Question 5:

Three experiments to study the process of rusting are given below. Observe the three test tubes and answer the following questions.

A. Why the nail in the test tube 2 is not rusted ?
B. Why is the nail in the test tube 1 is rusted highly ?
C. Would the nail in the test tube 3 get rusted ?

ANSWER:

a.Essential requirement for corrosion are:

Supply of oxygen
Presence of water
Material itself

a.The nail in the test tube 2 is not rusted because in test tube 2, we take boiling water. So, all the dissolved oxygen is removed from water and if the iron nail donot get supply of oxygen then corrosion is not carried out.

b.The nail in the test tube 1 is rusted highly because in the test tube 1, iron nail meets with all the requirements which is essential for the process of corrosion. That's Supply of oxygen in dissolved form, presence of water and material itself.

c.The nail in the test tube 3 is not rusted because calcium chloride is one of the best absorbent. So, it absorbs all the dissoved oxygen present in water. Hence corrosion process is not take place.

Monday, July 31, 2023

Std.7 Math (PRACTICE PAPER)

PRACTICE PAPER

Q.I) Choose the correct alternative with the alphabet against the sub-question number: 04 marks

1.The LCM of 4 and 5 is ________

A) 20 B) 15 C) 5

2.If the sum of the measures of two angles is 90° then they are known as ........... angles.

A) supplementary B) complementary C) zero

3) The quotient of two negative integers is a ........... integer.

A) positive B) negative C) infinite

4. Two numbers which have only 1 as a common factor are said to be ______ number.

A) composite number B) co-prime number C) small number

Q) II Attempt the following (Any 3) 06 marks

1. Divide: A) (- 98) ÷ (- 14) B) (-10) ÷ (-10)

2.Find the LCM of 15 and 30

3.The product of two digit numbers is 765 and the GCD = 3. What is their LCM ?

4. Name the pairs of opposite rays in the figure alongside.

Q.III) Solve (Any 2) 06 marks

1.Write the measures of the supplements of the angles given below

a) 30° b) 60° c) 40°

2.a) Which number is neither a prime nor a composite number.?

b) Which are the even prime numbers?

c) Say true or false.

2 is the HCF of any two consecutive even numbers.

3.Observe the figure and complete table.

Q.IV) Solve (Any 1) 04 marks

1.Write four divisions of integers such that the fractional form of each will be 24/5.

2. A)Draw a number line and show the following numbers on number line

-3 , 2

B) Find HCF by the division method and reduce to the simplest form.

161/69

Sunday, July 30, 2023

Std.7 Practice Paper (Science)

PRACTICE PAPER

_________________________________________________________________________________________

Q.1 A Choose the correct alternative and write it along with its allotted alphabet. [2 marks]

1) Plant food is produced by a special process known as __________

(a) photosynthesis (b) evaporation (c) condensation

2) The part that grows from inside the seed towards the soil is called the __________

(a) radical (b) herbs (c) plumule

B] Answer the following: [2 marks]

1) True or False:

Androecium is a male reproductive organ.

2) Match the following:

Q.2 A. Give scientific reasons (Any 1) [2marks]

1) Why do living organisms need nutrition ?

2) Corn is an example of monocot seed.

3) Nowadays, food is served buffet style during large gatherings.

Q.2 B] Answer the following (Any 2) [4 marks]

1) Draw a neat and labelled diagram of parts of leaf.

2) Name the following:(Who am I ?)

i) I transport minerals and water from the root to all aerial parts of the plant.

ii) I transport the food (glucose, etc.) from the leaves to other parts of the plant where it is either consumed or stored..

3) Complete the table:

Q.3) Answer the following (Any 2) [6 marks]

1) Choose the appropriate word from the bracket and rewrite the paragraph.

(host ,chlorophyll ,Scavengers, photosynthesis, parasitic, decomposers, saprophytic, insects.)

Due to the absence of ........., the cuscuta is completely dependent on the ........ plant. Hence, it is said to be a completely ........ plant.

Plants which obtain the food from dead and decaying bodies of other organisms are called ...... plants.

........ obtain their food from dead bodies of animals, for example, vulture,

crow, hyena, etc. and. ....... are some microbes which obtain their food by decomposing the dead bodies of organisms or other materials.

2) Write a note on Pasteurisation.

3) Describe any three parts of flower.

4. Classify according to food-type.

cow, lion, sparrow, frog, cockroach, tick.

Q.4) Answer the following (Any 1) [4 marks]

1.Write the practical tests to find out if food has been adulterated or not.

For Milk,Rava,Turmeric powder and Red chilly powder.

2. Identify the parts of stem and write any two functions of stem.

*****************************************************************************************

Sunday, July 9, 2023

Poster making competition

Dear Students,

SAGH is glad to announce Poster making and slogan writing competition on the occasion of swastha bharat abhiyan.

As you are aware about harmful effects of intoxicant material, highlighting the health and other risks associated with tobacco use.

On this occasion of great significance, we are glad to announce this competition to explore the creative instinct of students from class7th to 8th and encourage them to portray their ideas on “health and other risks associated with tobacco use”.

PARTICIPATION GUIDELINES

The creative participation is open for students of classes as per following two categories;(Only SAGH, Chembur)

1. 7th

2. 8th

LANGUAGE

Creative entries must be written in English,Marathi or Hindi only.

IMPORTANT DATES

• Start Date: 12th July 2023

• Last Date: 14th July, 2023

MODE OF SUBMISSION:

• A3 or A4 sheet may be used for Poster.

• Write-up text (slogan) of the Poster should be submitted in readable font of English/Hindi/Marathi and it should be written on the poster itself.

Thank you

Image credit:Internet

Friday, July 7, 2023

Std. 8. Force and Pressure

Question 1:

Write the word in the blank space.
A. The SI unit of force is_________ (Dyne, Newton, Joule)
B. The air pressure on our body is equal to __________ pressure (Atmospheric, sea bottom, space)
C. For a given object, the buoyant force in liquids of different ________ is__________ (the same, density, different, area)
D. The SI unit of pressure is_______ (N/m³, N/m², kg/m², pa/m²)

ANSWER:

A. The SI unit of force is Newton.
B. The air pressure on our body is equal to atmospheric pressure.
C. For a given object, the buoyant force in liquids of different density is different.
D. The SI unit of pressure is N/m².

Page No 22:

Question 2:

Make a match

A group	B group
1. Fluid	a. Higher pressure
2. Blunt knife	b. Atmospheric pressure
3. sharp needle	c. Specific gravity
4.Relative density	d. Lower pressure
5. Hecto pascal	e. Same pressure in all directions

ANSWER:

A group	B group
1. Fluid	e. Same pressure in all directions
2. Blunt knife	d. Lower pressure
3. sharp needle	a. Higher pressure
4. Relative density	c. Specific gravity
5. Hecto pascal	b. Atmospheric pressure

Page No 22:

Question 3:

Answer the following questions in brief.
A. A plastic cube is released in water. Will it sink of come to the surface of watet?
B. Why do the load carrying heavy vehicles have large number of wheels?
C. How much pressure do we carry on our heads? why don't we feel it?

ANSWER:

(a) The plastic cube is going to float on the surface of water as its density is less than that of water.

(b) We know, $Pressure = \frac{Force}{Area}$
So, greater the area of contact between two surfaces, lesser will be the pressure. So, the load carrying heavy vehicles have large number of wheels so that the pressure on the road is reduced due to larger contact area. Also, using large number of wheels ensures that the force due to the load is shared among the tyres and no single tyre is under stress.

(c) We carry atmospheric pressure of about 10⁵ Pa on our heads. We don't feel it because this atmospheric pressure is balanced by the pressure created by the air and blood inside our body.

Page No 22:

Question 4:

Why does it happen?
A. A ship dips to a larger depth in fresh water as compared to marine water.
B. Fruits can easily be cut with a sharp knife.
C. The wall of a dam is broad at its base.
D. If a stationary bus suddenly speeds up, passengers are thrown in the backward direction.

ANSWER:

A. The density of marine water is more than the fresh water due to which the buoyant force on the ship in marine water is more than the fresh water. Hence, the ship dips to a larger depth in fresh water as compared to marine water.

B. Since the sharp edged knife makes lesser contact with the fruits to be cut, thus the pressure exerted by it on the fruit is very large. Because of this large pressure, fruits are easily cut with a sharp knife.

C. We know pressure of liquid increases with depth. So, the wall of a dam is made broader at its base so that it is able to withstand the heavy pressure exerted by the river water.

D. If a stationary bus suddenly speeds up, passengers are thrown in the backward direction. This is because initially the whole body of a passenger inside the bus was in the state of rest. But, when the bus suddenly starts or speeds up, the lower half of the passenger's body comes in motion in the forward direction but the upper half still remains at rest due to inertia of rest. Hence, the passengers are thrown backward when a stationary bus suddenly speeds up.

Page No 22:

Question 5:

Complete the following tables.

Mass(Kg)	Volume (m³)	Density(kg/m³)
350	175	-
-	190	4

Density of Metal (kg/m³)	Density of water (kg/m³)	Relative Density
	10³	5
8.5 $\times$ 10³	10³	-

weight(N)	Area (m²)	Pressure(Nm⁻²)
-	0.04	20,000
1500	500	-

ANSWER:

We know, $Density = \frac{Mass}{Volume}$
So, using above formula, we can find one quantity if other two quantities are given.

Mass (kg)	Volume (m³)	Density (kg/m³)
350	175	2
760	190	4

We know,

Relative density of substance = \frac{Density of substance}{Density of water}

So, using above formula, we can find one quantity if other two quantities are given.

Density of Metal (kg/m³)	Density of water (kg/m³)	Relative Density
5 $\times$ 10³	10³	5
8.5 $\times$ 10³	10³	8.5

We know,

Pressure = \frac{Force}{Area}

So, using above formula, we can find one quantity if other two quantities are given.

Weight (N)	Area (m²)	Pressure (N m⁻²)
800	0.04	20,000
1500	500	3

Page No 22:

Question 6:

The density of a metal is $10.8 \times 10^{3} kg / m^{3}$ . Find the relative density of the metal.

ANSWER:

Given:
Density of metal = $10.8 \times 10^{3} kg / m^{3}$
We know, density of water = 1000 kg/m³
$Relative density of substance = \frac{Density of substance}{Density of water}$
$Relative density of substance = \frac{10.8 \times 10^{3}}{1000} = 10.8$

Page No 22:

Question 7:

Volume of an object is 20 cm³ and the mass in 50 g. Density of water is 1 g cm^{$-$ 3}. Will the object float on water or sink in water?

ANSWER:

Given:
Volume of object = 20 cm³
Mass of object = 50 g
Density of object = $\frac{Mass of object}{Volume of object} = \frac{50}{20} = 2.5 g / {cm}^{3}$
Now, we know density of water = 1 g/cm³
Since, density of object > density of water, therefore the object is going to sink in water.

Page No 22:

Question 8:

The volume of a plastic covered sealed box is 350 cm³ and the box has a mass 500 g. Will the box float on water or sink in water? what will be the mass of water displaced by the box?

ANSWER:

Given:
Volume of box = 350 cm³
Mass of box = 500 g
Density of object = $\frac{Mass of object}{Volume of object} = \frac{500}{350} = 1.43 g / {cm}^{3}$
Now, we know density of water = 1 g/cm³
Since, density of box > density of water, therefore the object is going to sink in water.
Now, volume of liquid displaced = Volume of the object = 350 cm^{3
$\Rightarrow \frac{Mass of liquid displaced}{Density of liquid displaced} = 350 {cm}^{3} \Rightarrow Mass of liquid = 350 \times 1 = 350 g$}

Parameter	Acid	Base
Arrhenius Definition	substance which when dissolved in water gives hydrogen ion	substance which when dissolved in water can accept hydrogen ions
Bronstead Lowry Definition	substance which donates a proton	substance which accepts a proton
Strength	depends on the concentration of the hydronium ions	depends on the concentration of the hydroxide ions
Characteristics (Physical)	depends upon the temperature as it can be solid, liquid or in the form of gas have a sour taste	can be solid in nature except ammonia which is gaseous have a bitter taste can be slippery in touch
Dissociation	releases hydrogen ions (H⁺) when mixed with water	releases hydroxide ions(OH^-) when mixed with water
pH value	less than 7.0	greater than 7.0
Litmus paper	blue litmus paper turns red	red litmus paper turns blue
Chemical Formula	has a chemical formula with H at the beginning of it. For example, HCl (Hydrochloric Acid). There is one exception to this rule, CH₃COOH = Acetic Acid (vinegar)	has a chemical formula with OH at the end of it. For example, NaOH (Sodium Hydroxide)

Cations	Anions
positively charged particles	negatively charged particles
formed by loss of electrons from metals	formed by gain of electrons from non-metals
during electrolysis, it moves towards cathode	during electrolysis, it moves towards anode
size of cation is smaller than its parent atom	size of anion is same or larger than its parent atom
for examples : Na⁺, Mg²⁺,Ca²⁺ etc.	for examples :Cl^-, Br^-,S^2- etc.

Negative Electrode	Positive Electrode
refers to a piece of electrochemical cell that is the negative pole	refers to a piece of electrochemical cell that is the positive pole
connects to negative terminal of a battery by means of wire	connects to positive terminal of a battery by means of wire
also called as cathode	also called as anode
positive charged cations moves toward it	negative charged anions moves toward it
accepts electrons to deposite	donates electron

Friday, October 13, 2023

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Tuesday, October 10, 2023

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Question 4:

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METAL TYPE :ONE SIMPLE WAY TO PREVENT CORROSION IS TO USE A CORROSION RESISTANT METAL SUCH AS ALUMINIUM OR STAINLESS STEEL. DEPENDING ON THE APPLICATION, THESE METALS CAN BE USED TO REDUCE THE NEED FOR ADDITIONAL CORROSION PROTECTION.

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Question 5:

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Monday, July 31, 2023

Sunday, July 30, 2023

Sunday, July 9, 2023

Friday, July 7, 2023

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METAL TYPE :
ONE SIMPLE WAY TO PREVENT CORROSION IS TO USE A CORROSION RESISTANT METAL SUCH AS ALUMINIUM OR STAINLESS STEEL. DEPENDING ON THE APPLICATION, THESE METALS CAN BE USED TO REDUCE THE NEED FOR ADDITIONAL CORROSION PROTECTION.