calcium hydride combines with water according to the equation

• Textbook Solutions
• Class 9
• Science
• acids, bases and salts

Science And Applied science Solutions Solutions for Class 9 Science Chapter 5 Acids, Bases And Salts are provided here with simple step-past-step explanations. These solutions for Acids, Bases And Salts are passing popular among Class 9 students for Scientific discipline Acids, Bases And Salts Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Science And Technology Solutions Christian Bible of Class 9 Skill Chapter 5 are provided here for you for free. You will also love the adver-free have happening Meritnation's Skill And Technology Solutions Solutions. All Science And Engineering Solutions Solutions for class Class 9 Science are braced by experts and are 100% accurate.

Page No 73:

Question 1:

Discover the odd one out and excuse.

(a) Chloride, nitrate, hydride, ammonium ion

(b) Hydrogen chloride, caustic soda, calcium oxide, ammonium hydroxid

(c) Acetic sulfurous, carboniferous acid, hydrochloric acid, element venomous

(d) Sal ammoniac, sodium chloride, nitre, Na sulphate

(e) Soda niter, sodium carbonate, sodium sulphate, sodium chloride

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

(g) Crystal clear chalcanthite, crystalline sodium chloride, crystalline ferrous sulphate, pellucid atomic number 11 carbonate.

(h) Common salt, potassium hydrated oxide, acetic acid, sodium acetate.

Answer:

ans
a.Ammonium = New Hampshire₄ ⁺
Nitrate = NO₃ ^-
Hydride = H^-
Chloride = Cl^-
Ammonium is odd because Ammonium is cation and breathe are anions.

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

c.Acetic acidic = CH₃COOH
Carbonic acid = H₂CO₃
Hydrochloric acid = HCl
Azotic acid = - HNO₃
HCl is the only Matter Hetro-Nuclear conjugate and remaining are Poly Atomic compound.

d.Ammonium ion chloride is odd  because it is acidulent salt and rest all are impersonal salts.

e.Sal soda is odd because the solutions of sodium nitrate, sodium sulphate and common salt are neutral. But the solution of soda ash is BASIC.

f.Calcium oxide = CaO
Periclase = MgO
Atomic number 30 oxide = ZnO
sodium oxide = Na₂O
ZnO is unmatched because it is amphoteric in nature and other ions are basic in nature

g.Table salt is odd because on heating, there is no variety in colorise of quinquefoliate. But in rest of the compounds, there is exchange is colour.

h. Potassium hydroxide is strange because in a reaction, when Sodium chloride reacts with acetic acid, then Sodium acetate rayon is formed. There is no role of Potassium hydroxide in the reaction given below.

$\mathrm{NaCl}+{\mathrm{CH}}_3\mathrm{COOH}\to {\mathrm{CH}}_3\mathrm{COONa}+\mathrm{HCl}$

Paginate No 73:

Question 2:

Write push down the changes that leave make up seen in each instance and explain the reason out rear IT.

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

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

(c) Two or three filings of copper were added to 10ml debase aqua fortis and stirred.

(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 weak NaOH.

(g) Weakened HCl was added to limestone.

(h) Pieces of blue stone were heated up in a test tube. Connected cooling, water was added thereto.

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

Answer:

ans
a. When 5o mL H2O is added to 50 cubic centimetre solution of cupric sulfate, then reversible reaction occurs and the colour change from pale blue to white so change hindermost to blue when water is added again.

$CuS{O}_4+H_{2}O\leftrightarrow CuS{O}_4.5H_{2}O(BLUEVITRIOL).$

b. Phenolphthalein is an indicator used for determining the quantity of humble. When two drops of Phenolphthalein indicant are added to 10 mL solution of Caustic soda, then the solution turns pink in colour because the acidic part of phenophthalein reacts with the radica, it forms sodium table salt of phenolphthalein which has tap color.
For lesson : In acid baseborn titration phenolphthalein is wont to detect end point of base.

c. Cu is an inactive metal and doesn't react in normal circumstances with dilute acids.

In that respect are in reality two equations for the reaction of copper with aqua fortis take place. It depends connected whether the nitric acid is concentrated or non. If it is concentrated, then the ratio is 1:4  for cop to nitric acidic. If it is dilute then the ratio is 3:8.

$$\begin{gathered} Cu+4HN{O}_3(dil.)\to Cu(N{O}_3{)}_2+2N{O}_2+2H_{2}O \\ 3Cu+8HN{O}_3(conc.)\to 3Cu(N{O}_3{)}_2+2NO+4H_{2}O \end{gathered}$$

Concentrated Nitric blistering  is play strong oxidising agent then it makes sense that a higher oxidation state of nitrogen (IV) oxide is formed .

d. When litmus paper is swaybacked into 2 mL of dilute HCl solution , then blue litmus test wallpaper is turned into red colour and there is nary gist happening red litmus wallpaper. Again, if the same litmus test paper is swayback into 2 mL of concentrated NaOH root, then red litmus test paper turns into blue colour but thither is no effect on blue litmus test paper.This is due to the individual properties of blue and ruddy litmus paper with acid and base.

e . Magnesium oxide is a base. when baseborn reacts with an acid, it forms SALT and water.This response is known as counteraction response.

$MgO\left(s\right)+2HCl\left(l\right)=MgC{l}_2\left(aq\right)+H_{2}O\left(l\right)$

MgO doesnot react with NaOH.As NaOH is a base and bases react solely with oxides of not-metal to form salt and water because oxides of non-metals are said to make up acidic in nature indeed neutralisation hap. But MgO is a oxide of metal so, reaction is non possible.

f. Zinc oxide is added to dilute HCl, then neutralisation reaction takes place to form salty and irrigate.
The reaction is as follows:

$ZnO\left(aq\right)+2HCl\left(aq\right)\to ZnC{l}_2\left(aq\right)+H_{2}O\left(l\right)$

Zinc oxide reacts  with sodium hydrated oxide to produce zincate sodium and water supply. This chemical reaction takes place at a temperature of 500-600°C.it is exothermic reaction.
$\mathrm{ZnO}+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{ZnO}}_2+{\mathrm{H}}_2\mathrm{O}$

g. Limestone is calcium carbonate. When limestone is added to a 10% solution of white HCl, then brisk frothiness of CO₂ is released referable the reaction of back breaker with carbonate of metals.

$2HCl+CaC{O}_3\to CaC{l}_2+C{O}_2+H_{2}0$

h. When pieces of blue vitriol are heated in a quiz subway system, then crystallization structure of chalcanthite broke dejected to forn a somber powder and water follow out. This water is called water of crystallization. when water is added to the same test electron tube, then white pulverization revolved into blue colour again.This is due to reversible reaction take out place from anhydrous salt to hydrous salt and vice-versa.

$CUS{O}_4.5H_{2}O\leftrightarrow CUS{O}_4+5H_{2}O$

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

t

The blow produced at the cathode Robert Burns with a 'pop' speech 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 O.

The gases are produced when ions move towards the electrodes.

At the cathode:

2H⁺ +2e^- → H₂

At the anode:

4OH^- - 4e^- → 2H₂O + O₂

Page No 73:

Question 3:

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

CaO, MgO, CO₂ , SO₃ , Sodium₂O, ZnO, Al₂O₃ , Fe₂O₃

Answer:

Oxides are of three types :

1) Acidic Oxides: Cobalt₂ (CO2), SO₃ (Sulfur trioxide)

2) Basic Oxides: CaO (Atomic number 20 oxide), MgO (Periclase), 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. Organisation of Na chloride from sodium and atomic number 17.

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

Answer:

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

Atomlike number of Chlorine(Cl) atom is 17.
Electronic configuration is :
Cl = 2, 8, 7
So it contains 7 valency electron. In order to achieve the nearest Lord 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 training of negatron from Na to chlorine.

b. Atomic number of Magnesium(12) atom is 12.
Electronic configuration is :
Mg = 2, 8, 2
And so information technology contains 2 valence electron. Ready to achieve the nighest argonon configuration, it loses two electrons to spring Magnesium ion.
Mg²⁺ = 2,8

Atomic number of Chlorine(Cl) molecule is 17.
Electronic constellation is :
Cl = 2, 8, 7
So information technology contains 7 valence electron. Systematic to accomplish the closest noble accelerator configuration, it gains one negatron to configuration Chloride ion.
Atomic number 17^- = 2,8

An Electrostatic bond is formed between Magnesium ion and two Chloride ion by complete conveyance of one electron to all Chlorine ion.

Paginate No 74:

Question 5:

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

Hydrochloric acid, Common salt, Potassium hydrated oxide, Ammonia, Acetic acid, Magnesium chloride, Copper sulphate

Answer:

a.HCl (aq) + H 2 O (l) → H 3 O ⁺(aq) + Cl ^-(aq)
Account:
Hydrochloric acid(HCl) is a strong acid, so HCl is ionise completely in aqueous solution.In other words, all molecule of hydrochloric acid that is added to water will donate its proton H⁺ to water molecule to class a hydronium cation, and H₃O⁺and chloride ions Cl^-  is formed.

b. Reaction:
NaCl(s) + H₂O(aq) -> Na⁺(aqueous) + Cl^-(liquid) + H₂O(l)
Explanation:
When atomic number 11 chloride reacts with irrigate the Na⁺ part of NaCl is attracted to the oxygen side of the water molecules, piece the Cl^- broadside is attracted toward the hydrogen's side of the water molecule.

This causes the common salt(salt) to split in H2O and the NaCl ionises into Sodium⁺ and Cl^- ions wholly.

c) KOH(s) + H₂O(l)<--> K⁺(aq) + OH^-(aq) + H₂O(l)
Account:
KOH is base, so mixing IT in water makes a elemental resolution that is in equilibrium. the KOH is sportsmanlike dissolution in the water.

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

e)CH₃COOH(l) + H₂O(l) -> CH₃COO^-+ H₃O⁺
Explanation:
When carboxylic acid pane is added to water system, due to negativity differences of oxygen and hydrogen in Ohio group of acetic pane and there is a dipole antenna interaction with piddle molecule. Hence, the ethanoic acid is ionise into ethanoate ion and H⁺ ion combines with water to fles hydronium ion.  —

f)MgCl₂(s) + H₂O(l) -> Atomic number 12²⁺(sedimentary) + Cl^-(aqueous) + H₂O(l)
Explanation:
Happening addition to piddle the Mg²⁺ part of MgCl₂ is attracted to the oxygen sidelong of the water molecules, patc the Cl- side is attracted to the hydrogen's side of the water molecule . This causes atomic number 12 chloride(salt) to split in piddle and the MgCl₂ is ionise into Mg²⁺ and Cl^- ions completely.

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

Page No 74:

Question 6:

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

a. 7.3g HCl in 100ml solution

b. 2g NaOH in 50ml solvent

c. 3g CH₃COOH in 100ml solution

d. 4.9g H₂Soh₄ in 200ml solution

Answer:

a)7.3 g HCl in 100 mL solution:

$\mathrm{Concentration}\mathrm{of}\mathrm{HCl}\mathrm{in}\mathrm{g}{\mathrm{L}}^{-1}=\frac{Massofsolute\mathrm{in}\mathrm{gram}}{\mathrm{Loudness}\mathrm{of}\mathrm{HCl}\mathrm{in}litre}$ $=\frac{7.3\times 1000}{100}=73$

g L^-1

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

Molecular mass of HCl =

$1+35.5=36.5$

$\mathrm{Molarity}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{solute}\mathrm{in}moles}{\mathrm{Volume}\mathrm{of}\mathrm{HCl}\mathrm{in}\mathrm{L}}$ $=\frac{7.3\times 1000}{36.5\times 100}=2$

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

$ConcentrationofNaOHing{L}^{-1}=\frac{MassofNaOHing}{VolumeofNaOHinL}$ $=\frac{2\times 1000}{50}=40$

g L^-1
In terms of gram per l:
2 g of NaOH in 50 cubic centimetre has concentration = 40 g L^-1

Molecular mass of NaOH =

$23+16+1=40$

$\mathrm{Molarity}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{solute}\mathrm{in}moles}{\mathrm{Loudness}\mathrm{of}NaOH\mathrm{in}\mathrm{L}}$ $=\frac{2\times 1000}{40\times 50}=1$

mol L^-1

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

c)3 g CH₃COOH in 50 mL root

$ConcentrationofC{H}_{3}COOHing{L}^{-1}=\frac{MassofC{H}_{3}COOHing}{VolumeofC{H}_{3}COOHinL}$ $=\frac{3\times 1000}{100}=30$

g L^-1
In terms of gramme per litre:
3 g of CH₃COOH in 100 mL has concentration = 30 g L^-1

Unit mass of

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

$Molarity=\frac{Massofsoluteinmoles}{VolumeofsolutioninL}$ $=\frac{3\times 1000}{60\times 100}=0.5$

mol L^-1

In terms of moles per litre:
3 g of  CH₃COOH in 100 milliliter has tightness = 0.5 gram molecule L^-1

d)4.9 g H₂SO₄  in 200 mL root

$Concentrationof{H}_{2}S{O}_{4}ing{L}^{-1}=\frac{Massof{H}_{2}S{O}_{4}ing}{Volumeof{H}_{2}S{O}_{4}inL}$ $=\frac{4.9\times 1000}{200}=24.5$

g L^-1
In terms of gramme per litre:
4.9 g of  H₂Thusly₄ in 200 mL has concentration = 24.5 g L^-1

Molecular mass of

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

$Molarity=\frac{Massofsoluteinmoles}{VolumeofsolutioninL}$ $=\frac{4.9\times 1000}{98\times 200}=1$

mol L^-1

In footing of moles per cubic decimeter:
4.9 g of   H₂SO₄ in 200 cubic centimetre has concentration = 1 mole L^-1

Page No 74:

Question 7:

Obtain a sample of rainwater. Add to it few drops of universal indicator. Measuring stick its pH. Describe the nature of the sample of rainwater and excuse the impression if it has along the animation world.

Answer:

Ans7.This is an activity settled question in which you are supposed to collect rainwater from different places and compare in that respect results on the basis of following parameters :
a) pH scale of piss
b)Action of water connected blue litmus paper
c)Action of water on red litmus paper
d)Effect of indicator alike phenolphthalein and methyl radical orange
Close :
If we need samples of rain water from variant 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 radical orange

The in a higher place observations point that rain is acidulous in nature. For encourage pledge, we cancompare blue litmus test newspaper dipped in different samples of rainwater with pH scale.
If rain water is acid, and then its pH must be in the range of 0-6.
The military posture of ac acidulous depends on its pH respect i.e. lower berth the value of pH, high will be the strength of an sulfurous and frailty-versa.

Page No 74:

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 response.

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

Respond:

ANS 8.

a. The number of ionizable atomic number 1 (H+) ions present in one molecule af an acid is called its basicity.

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

H₂Soh₄ ------> 2H⁺ + Indeed₄ ^2-
Basicity of  H₂SO₄ is 2.

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

1. Monaural-basic acids :

Acids, which happening ionisation produce same hydronium ion on reaction with water.
Acids, which on ionization green goods one hydrogen ion.
For instance : HCl, HNO₃ etc.

2. Di-basic acids :

Acids, which connected ionisation produce 2 hydronium ion along reaction with body of water
Acids, which on ionisation garden truck two hydrogen ion

For example : H₂Soh_4, H₂CO₃ etc.

3. Tri-staple acids :

Acids, which on ionization produce iii hydronium ion connected reaction with water supply
Acids, which on ionisation create three H ion

For model : H₃PO₄, H₃PO₃ etc.

b. Neutralisation reaction :

A neutralisation is a reaction in which an acid and a floor reacts to form water and a saltiness. It involves the combination of H⁺ ions and OH^- ions to generate water.
The neutralisation of a vehement acid and strong base has a pH equal to 7.
The neutralization of a strong back breaker and weak base testament have a pH of less than 7.
The neutralisation of a strong base neutralizes a weak Lucy in the sky with diamonds will be greater than 7.

When a root is neutralized, it means that salts are formed from equal weights of acid and base.

Neutralization reaction has application in daily life:

1)Mortal denial by animals and plants through chemical war :

Bee stings are acid-forming in nature, household amend for a bee confidence game is bicarbonate of soda or Na hydrogen carbonate, which is a alkaline substance.
A white Anglo-Saxon Protestant stings are mildly basic, household remedy for this will exist vinegar, too known as acetic acid.
These simple treatments ease these painful stings by a process called neutralization.

2)pH in our digestive arrangement :
An acidic stomach collectible to feeding overmuch spicy food, can be alleviated by taking an antacid. The antacid is alkaline/basic in nature and helps to neutralize the put u's acidulousness or you May take magnesium hydrated oxide(Milk of magnesia) and atomic number 11 atomic number 1 carbonate(Baking soda).

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

4)soil pH and plant growth :
Almost of the plants grow best when the pH scale of the soil is close to 7 that's neural. If the soil is likewise acidic operating room also elemental(alkaline), the plants rise badly.
The acidic soil is neautralize away treatment with materials like quicklime(calcium oxide) operating theatre calcium hydrate(calcium hydroxide) or chalk(calcium carbonate).
If the dirt is overly basic, and then alkalinity can be reduced by adding decaying organic matter(muck or composite plant)which contains acidic materials.

c.Electrolysis of water :

Electrolysis of water is the decompositon into oxygen and atomic number 1 gas attributable an electric current passed through the water.

The following equivalence represents the electrolysis of irrigate : H 2 O ( l )

In pure water system, at the negatively charged cathode, a reduction reaction takes rank, with electrons (e⁻) from the cathode being given to hydrogen cations to form atomic number 1 gas.

Reducing at cathode: 2 H⁺ + 2e⁻ → H₂

Connected positively live anode, an oxidisation reaction occurs, generating oxygen throttle by giving electrons to the anode :

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

The same half reactions privy likewise be balanced with base as recorded below. To add half reactions they must follow balanced with either back breaker or base.

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

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

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

The number of H molecules produced is thus twice the figure of oxygen molecules.  The produced hydrogen gas has therefore doubly the volume of the produced oxygen gas. The number of electrons pushed through the water is double the number of generated atomic number 1 molecules and fourfold the number of generated oxygen molecules.

2 H 2 O ( l ) ⟶ 2 H 2 ( g ) + O 2 ( g )

   

Foliate No 74:

Question 9:

Give reason for the tailing.

a. Hydronium ions are always in the form H₃ O⁺

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

Respond:

a)When an acids dissolve in water. The H⁺ ion from acid ever goes to the nearest piss molecule to form hydronium ion.
HCl(aq) + H₂O(aq)→ H 3 O⁺(aq) + Cl-(aq)
E.g. : When Hydrochloric Zen(HCl) is a potent dot. When it is dissolved in water, HCl is ionized completely in aqueous answer.  Hydrochloric caustic leave donates its proton H⁺ to water mote to anatomy a hydronium cation H₃O⁺and chloride anions Cl^- .

b)Buttermilk spoils if unbroken in a atomic number 29 or brass container because buttermilk is actually lactic acid. Lactic acid reacts with the container material and produces poisonous interlinking. It is actually the reaction 'tween acid and metal . This reaction is called as electro chemic reaction.

Page No 74:

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 .

(c) Dilute nitric acid was added to unslaked lime.

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

(f) Dilute HCl was poured on baking soda.

Respond:

$$\begin{gathered} a.\mathrm{HCl}\left(\mathrm{l}\right)+\mathrm{NaOH}\left(\mathrm{l}\right)\to \mathrm{NaCl}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+\mathrm{Energy} \\ \mathrm{b}.\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to {\mathrm{ZnSO}}_4\left(\mathrm{s}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right) \\ \mathrm{c}.\mathrm{CaO}\left(\mathrm{s}\right)+2{\mathrm{HNO}}_3\left(\mathrm{l}\right)\to \mathrm{Ca}({\mathrm{No}}_3{)}_2(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l}) \\ \mathrm{d}.\mathrm{KOH}(\mathrm{l})+{\mathrm{CO}}_2(\mathrm{g})\to {\mathrm{KHCO}}_3 \\ \mathrm{e}.{\mathrm{NaHCO}}_3(\mathrm{s})+2\mathrm{HCl}(\mathrm{l})\to \mathrm{NaCl}(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})+{\mathrm{CO}}_2 \end{gathered}$$

Page No 74:

Question 11:

State the differences.

a. Acids and bases

b. Cation and anion

c. Dismissive electrode and positive electrode.

Solvent:

Page No 74:

Query 12:

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

Salt, sodium acetate, hydrochloric acid, carbon dioxide, atomic number 19 bromide, calcium hydoxide, ammonium chloride, acetum, Na carbonate, ammonia, atomic number 16 dioxide.

Answer:

Ans7.

SUBSTANCES	pH Prize
Solution of Grassroots salt	equal to 7
Solution of Sodium ethanoate	greater than 7
Solution of Hydrochloric Lucy in the sky with diamonds	less than 7
Solution of Carbon dioxide	to a lesser extent than 7
Solution of Potassium bromide	equal to 7
Solution of Calcium hydrated oxide	greater than 7
Resolution of Ammonium chloride	less than 7
Solution of Acetum	little than 7
Solution of Sodium carbonate	greater than 7
Solution of Ammonia	greater than 7
Solution of Sulfur dioxide	to a lesser degree 7

View NCERT Solutions for whol chapters of Social class 9

calcium hydride combines with water according to the equation

Source: https://www.meritnation.com/maharashtra-class-9/science/science-and-technology-solutions/acids-bases-and-salts/textbook-solutions/90_2_3383_24409_73_135397