Explore topic-wise InterviewSolutions in .

1. No interest on capital is payable to any partner.

2. No interest is charged on drawing made by the partner. 

3. Profit should be distributed equally. 

4. No remuneration is payable to any partner. 

5. Interest on loan is payable at 6% per annum.

(a) Equal ratio

(b) Interest on partners’ capital is allowed at 7% per annum

1. (i) White

2. (i) TRUE

3. (ii) Double displacement

4. (iii) Both (i)  and (ii)

5. (i) TRUE

(d) A colourless and odourless gas is

(c) to avoid action by sunlight

(i) Brown fumes. white residue.

(ii) Decomposition reaction 

2Pb(NO₃)₂ →2PbO+4NO₂+O₂

The characteristic test for 

(a) 

• Carbon dioxide (CO₂) gas turns lime water milky when passed through it due to the formation of insoluble calcium carbonate. 
• Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
• Lime water Carbon Calcium dioxide carbonate 

(b) 

• Sulphur dioxide (SO₂) gas when passed through acidic potassium permanganate solution (purple in colour) turns
• it colourless because SO₂ is a strong reducing agent 
• 2KMnO₄ + 2H₂O + 5SO₂→ K₂SO₄ + 2MnSO₄ + 2H₂SO₄
• Potasssium Sulphur Potassium Manganese permanganate dioxide sulphate sulphate (Purple) (Colourless) (Colourless).
• Sulphur dioxide gas when passed through acidic dichromate solution (orange in colour) turns it to green because sulphur dioxide is a strong reducing agent. 

(c) 

• The evolution of oxygen (O₂) gas during a reaction can be confirmed by bringing a burning candle near the mouth of the test tube containing the reaction mixture.
• The intensity of the flame increases because oxygen supports burning.

 (d) 

• Hydrogen (H₂) gas burns with a pop sound when a burning candle is brought near it.

C) Silver bromide

(i) 2AgBr(s) + sunlight  → 2Ag(s)+Br₂(g)

(ii) 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

Correct answer is (d) Ag₂S

(a) Balanced chemical equation 

\(2Cu(NO_3)_2(s)\overset{Heat}\rightarrow 2CuO(s) +O_2(g)+4NO_2(g)\)

(b) The brown gas X evolved is nitrogen dioxide (NO₂ ) 

(c) This is a decomposition reaction 

(d) Nitrogen dioxide dissolves in water to form acidic solution because it is an oxide of non-metal. Therefore, pH of this solution is less than 7.

i. Silver reacts with hydrogen sulphide gas present in air and forms a compound Ag2S. This compound is black in colour. Hence, silver articles turn black when kept in the open for a few days. This phenomenon is called corrosion.
ii. The black substance formed issilver sulphide. Its chemical formula is Ag2S.

(a) Metals such as silver when attacked by substances around it such as moisture, acids, gases etc, are said to corrode and this phenomenon is called corrosion. 

(b) The black substance is formed because silver (Ag) reacts with H₂S present in air. It forms thin black coating of silver sulphide (Ag₂S)

(i) Silver article reacts with sulphur compounds such as H₂S present in the air. The phenomenon is called corrosion. For silver particularly, it is called tarnishing of silver.

(ii) The black substance is silver sulphide (Ag₂S)

Correct option is D) Non essential nutrients

(a) Copper sulphate. 

(b) Blue colour. 

(c) Copper sulphide. 

(d) CuSO₄ (aq) + H₂S (g)—–>CuS (s) + H₂SO₄ (aq) 

(e) Double displacement reaction. 

C) stainless steel

A) Oxidation

C) Chemical displacement

B) To prevent the spoilage by oxidation

B) oxidation

True 

Keeping food in airtight containers helps to slow down oxidation. Airtight containers prevent the food materials from getting oxidized. 

Hence, the given statement is true.

Mark A, B, C on the test tubes. Put red litmus in each test tube. If there is no change in test tube it is acid, if litmus changes to blue it is an acid, if there is little change in litmus it is distilled water.

(A) When solutions are kept in copper container

(a) Dilute HCl

Copper does not react with dilute HCl. Therefore, it can be kept.

(b) Dilute HNO₃

Nitric acid acts as a strong oxidising agent and reacts with copper vessel, therefore cannot be kept.

(c) ZnCl₂

Zinc is more reactive than copper (Cu) therefore, no displacement reaction occurs and hence can be kept.

(B) When solutions are kept in aluminium containers

(a) Dilute HCl

Aluminium reacts with dilute HCl to form its salt and hydrogen
is evolved. Therefore, cannot be kept.

2Al + 6HCl → 2 AlCl₃ + 3H₂

(b) Dilute HNO₃

Aluminium gets oxidised by dilute HNO₃ to form a layer of Al₂O₃ and can be kept.

(c) ZnCl₂

Aluminium being more reactive than zinc can displace zinc
ion from the solution. Therefore, the solution cannot be kept.

2Al + 3ZnCl₂ → 2 AlCl₃ + 3Zn

(d) H₂O

Aluminium does not react with cold or hot water. Therefore,
water can be kept

Aluminium is attacked by steam to form aluminium oxide
and hydrogen

2Al (s) + 3H₂O (g) → Al₂O₃ (s) + 3H₂ (g)

Put blue litmus paper in all the three test tubes. In one test tube, it will turn red that is acid. 

Now Put red Litmus paper in other two test tubes. The one which turns blue is base and the third one is distilled water, as it turned paper neither red nor blue.

During free fall:

Acceleration of the boy = Acceleration of mass M = g

Acceleration of mass M w.r.t. boy, a = 0

So, the force exerted by the box on the boy’s head = M × a = 0

Yes, the force greatly increases during the period he balances himself after striking the ground because of the weight of the box.

Air applies a velocity-dependent force on the parachute in upward direction when the parachute opens. This force opposes the gravitational force acting on the spy. Hence, the net force in the downward direction decreases and the spy decelerates.

In the game of tug, till both the teams pull the rope with the same force, the net force on the system remains zero. As the force applied by one team increases than the other one, the whole system starts moving in the direction of net force. Thus one team wins and the other one loses.

The forces on the rope must be equal and opposite, according to Newton’s third law. But not all the forces acting on each team are equal. The friction between one team and the ground does not depend on the other team and can be larger on one side than on the other. In addition, the grips on the rope need not be equal and opposite. Thus, the net force acting on each team from all sources need not be equal.

b) Equal Force is being applied in opposite direction.

The tension in the rope if a condition of equilibrium exists 1000 kg wt.

Indicators : Substances which are used to test acids or bases are called acid base indicators.

Turmeric, hibiscus, litmus, red cabbage, are some Natural indicators.

(i) Acidic strength: HOCl < HClO₂ < HCIO₃ < HCIO₄

(ii) Acidity: Ga₂O₃ < GeO₂ < AsO₃ < CIO₂

(iii) Bond angle: SbH₃ < AsH₃ < PH₃ < NH₃

(iv) Acidic strength: HF < HCl < HBr < HI

(v) Ionic character: MI< MBr < MCl < MF

Option : (b) HClO₄

Correct option: 3. HClO₃ + HClO₄   

Explanation:

HClO₃ + HClO₄ → Cl₂O₆ + H₂O

(i) CCl₄ is a covalent compound while H₂O is a polar compound. Therefore, it is insoluble in water. Alternatively, CCl₄ is insoluble in water because carbon does not have orbitals to accommodate the electrons donated by oxygen atom of water molecules. As a result, there is no interaction between CCl₄ and water molecules and hence CCl₄ is insoluble in water. On the other hand, SiCl₄ has d-orbitals to accommodate the lone pair of electrons donated by oxygen atom of water molecules. As a result, there is a strong interaction between SiCl₄ and water molecules. Consequently, SiCl₄ undergoes hydrolysis by water to form silicic acid.

(b)  The bond dissociation energy decreases rapidly as the atomic size increases. Since the atomic size of carbon is much smaller (77 pm) as compared to that of silicon (118 pm), therefore, carbon-carbon bond dissociation energy is much higher (348 kJ mol^-1) than that of silicon-silicon bond (297 kJ mol^-1). Hence, because C - C bonds are much stronger as compared to Si-Si bonds, carbon has a much higher tendency for catenation than silicon.

BCl₃ and AlCl₃ being electron deficient due to incomplete octet of central metal atom behave as Lewis acids.

(i) TlCl is more stable. It is an ionic compound. Inert pair effect is maximum in thallium and therefore, +1 oxidation state is more stable than +3 state. 

(ii) AlCl₃ is more stable. Anhydrous state is covalent while aqueous state is ionic in nature. +3 oxidation state is more stable than +1 oxidation state. Inert pair effect is negligible. 

(iii) lnCl₃ is more stable. Anhydrous state is more covalent less ionic in nature. Inert pair effect is considerable but less than in thallium. 

BF₃ exists as a monomer due to pπ−pπback bonding. Fluorine transfers two electrons to vacant 2p-orbital of boron. The delocalisation reduces the deficiency of electrons on boron thereby increasing the stability of BF₃ molecule. Due to absence of lone pair of electrons on H, the back bonding does not occur in BH₃. In other words, electron deficiency of boron remains and BH₃ does not exist. To reduce electron deficiency BH₃ dimerises to form B₂H₆. Hence, boron hydride exists in dimeric form and known as diborane. Structure of diborane is already given in exercise 8.19(NCERT). 

NH₃ forms hydrogen bonds with water and hence is soluble in water. But PH₃ cannot form hydrogen bonds with water and hence is not soluble in water. It escapes as gas.

Option : (d) ClF₃

Interhalogen compounds are more reactive than halogen compounds. But in case of fluorine due to the small size of fluorine, it has high electronegativity and low bond energy so it is more reactive than ClF₃. Therefore ClF₃ is more reactive than CI₂.

Since H-F bond is strongest with higher bond dissociation energy than HCl, hence it is weakest acid among all the halogen acids.

Due to small size, the electron-electron repulsions in the relatively compact 2p-subshell of F are quite strong and hence the incoming electron is not accepted with the same ease as in case of bigger Cl atom where repulsions are comparatively weak. Thus, electron gain enthalpy of chlorine is more negative as compared to that of fluorine.

Bond enthalpy of F-F is smaller due to greater repulsive interactions between the lone pair of one F atom with those of other. The repulsive interaction arise due to greater concentration of electron density on each F atom because of its extremely small size.

Property of catenation depends upon the strength of element-element bond which, in turn, depends upon the size of the element. Since the atomic size of carbon is much smaller than that of lead, therefore, carbon-carbon bond strength is much higher than that of leadlead bond. Due to stronger C-C than Pb-Pb bonds, carbon has a much higher tendency for catenation than lead.

(i) H -I < H - Br < H - Cl < H-F

(ii) H₂O < H₂S < H₂Se < H₂Te

Bond enthalpy of F-F is smaller due to greater repulsive interactions between the lone pair of one F atom with those of other. The repulsive interaction arise due to greater concentration of electron density on each F atom because of its extremely small size.