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(d) H₂(g) + I₂(g) ⇌ 2HI(g)

In the reaction H₂(g) + I₂(g) ⇌ 2HI(g) the volumes are equal on both sides and so pressure has no effect.

(c) N₂(g) + O₂(g) ⇌ 2NO(g) 

K_P = K_C.RT^Δn_g 

In equation N₂(g) + O₂(g) ⇌ 2NO(g) 

Δn_g = 0 

K_P = K_C.RT° 

K_P = K_C

A catalyst will increase the rate of a chemical reaction by lowering the activation energy

(a) Both (A) and (R) are correct and (R) is the correct explanation of (A). 

(b) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate

Oxalate ions of oxalic acid form complex with ferric ions thus decrease its concentration thus, concentration of red complex in product decreases.

In a closed system, A(s) ⇌ 3B(g) + 3C(g) If partial pressure of C is doubled, then partial pressure of B will be 1 / 2√2 times the original value.

(d) PCI₅ ⇌ (g) PCl_3(g) + Cl_2(g)

For reactions given in options (a), (b) and (c) ∆n_g = 0 

For option (d) ∆n_g = 2 – 1 = 1 

∴ K_p = K_c (RT)

(d) Both A and R are false

(c) All the physical processes stop at equilibrium

Correct statement – Physical processes occurs at the same rate at equilibrium.

(b) (i)-3, (ii)-1, (iii)-4, (iv)-2

When sugar is added to water at a particular temperature. it dissolves to form sugar solution. When more sugar is added to that solution, a particular stage sugar remains as solid and results in the formation of saturated solution. Here a dynamic equilibrium is established between the solute molecules ii the solid phase and in the solution phase.

Sugar_(solid) ⇌ Sugar_(Solution)

(c) 2

Δn_g = n_p – n_r = 2 – 0 = 2

(i) Sodium (Na): 

They are found in our blood plasma and the interstitial fluids around the cells. They help in

(a) Transmission of nerve signals. 

(b) They regulate the flow of water across the membranes of the neighboring cells. 

(c) Transport sugars and amino acids from and to cells.

(ii) Potassium (K): 

They are found mostly in the cell fluids in greater quantities. They help in 

(a) Activating enzymes. 

(b) Oxidising glucose to form ATP. 

(c) Transmitting nerve signals.

(iii) Magnesium (Mg) and calcium (Ca): 

They are also called as macro-minerals named so because of their abundance in our body. Mg helps in 

(a) Relaxing nerves and muscles. 

(b) Building and strengthening bones. 

(c) Maintaining blood circulation in our body. 

Ca helps in 

(a) coagulation of blood 

(b) Maintaining homeostasis

(b) x + 0.5

Total no. of moles at equilibrium = 0.5 – x + x + x 

= 0.5 + x

1. Equilibrium will be shifted in the forward direction. 

2. Equilibrium will be shifted in the backward direction. 

3. Equilibrium will be shifted in the backward direction. 

4. Equilibrium will be shifted in the forward direction.

(c) All the physical processes stop at equilibrium 

(i) When the total number of moles of gaseous reactants and gaseous products are equal, the change in pressure has no effect on system at equilibrium. 

H₂(g) + I₂(g) ⇌ 2HI(g)

Here the number of moles of reactants and products are equal. So the pressure has no effect on such equilibrium with ∆n_g = O.

1. In a pregnant women, the oxygen supply for the fetus is provided by the maternal blood in the placenta where the blood vessels of both mother and fetus arc in close proximity. 

Both fetal and maternal hemoglobin binds to oxygen reversibly as follows.

Hb_(mother) + O₂ ⇌ HbO_2(mother)

Hb_(fetus) + O₂ ⇌ HbO_2(fetus)

2. In the above two equilibrium, the equilibrium constant value for the oxygenation of fetal hemoglobin is higher which is due to its higher affinity for oxygen compared to adult hemoglobin. Hence in placenta, the oxygen from the mother’s blood s effectively transferred to fetal hemoglobin.

Due to its smallest size, Li⁺ can polarize water molecules easily than the other alkali metal ions.

LiF has a greater ionic character than LiCl which disturbs the balance between hydration energy and lattice energy. This balance is crucial for the solvability of ions in solution. Due to greater covalent character and lower lattice energy, dissolution of LiCl is more exothermic in nature than that of LiF.

(d) increasing the total pressure container

It favours backward reaction i.e., formation of H₂ molecule.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 

In the above equilibrium, if the pressure is increased, the volume wiLl decreases. The system responds to this effect by reducing the number of gas molecules. 

i.e., it favours the formation of ammonia. If the pressure is reduced, the volume will increases. It favours the decomposition of ammonia.

In this equilibrium, 

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ∆H = – 92.2 kJ.

Forward reaction is exothermic while the reverse reaction is endothermic. If the temperature of the system increased, the system responds by decomposing some of ammonia molecules to nitrogen and hydrogen by absorbing the supplied heat energy.

Similarly, the system responds to a drop in the temperature by forming more ammonia molecule from nitrogen and hydrogen which release heat energy.

(a) 2NO_(g) + O_2(g) ⇌ 2NO_2(g)

All the other three are physical equilibrium. Only (a) is chemical equilibrium.

Equilibrium constant changes with Both temperature and pressure.

(b) 9

\(Q=\frac{[HI]^2}{[H_2]\,[I_2]}=\frac{0.6\times0.6}{0.2\times0.2}=9\)

(c) H₂O_(l) ⇌ H₂O_(g) 

Physical states arc in equilibrium 

i.e., liquid – vapour equilibrium.

(c) PCI₅ ⇌ PCI₃ + Cl₂ 

0.02

Forward reaction equilibrium constant K₁ = 50

Reverse reaction equilibrium constant K₂ = ? 

K₂ = \(\frac{1}{K_1}\) = \(\frac{1}{50}\) = 0.02

For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium constant K is \(\frac{[HI]^2}{[H_2][I_2]}\).

(i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass.

Mol. wt. of mixture.of NO₂ and N₂O₄ = 38.3 x 2 = 76.6 

Let a mole of NO₂ be present in 100 mole mixture 

wt. of NO₂ + wt. of N₂ O₄ = wt. of mixture, 

a x 46 + (100 - a ) x 92= 100 x 76.6 

a = 33.48 mole 

a. Na⁺ – Sodium ion, K⁺ – Potassium ion

b. Na⁺ – Sodium ion, K⁺ – Potassium ion, Ag⁺ – Silver ion

c. \(SO_4^{2-}, NH_4^{+}\)

(a) Iron (Ferrum) 

(i) Fe²⁺ – Ferrous [Iron – II] 

(ii) Fe³⁺ – Ferric [Iron – III]

(b) Copper (Cuprum)

(i) Cu⁺ – Cuprous [Copper -1] 

(ii) Cu²⁺ – Cupric [Copper – II]

(c) Mercury (Hydragyrum) 

(i) Hg⁺ – Mercurous [Mercury -1] 

(ii) Hg²⁺ – Mercuric [Mercury – II]

e. O^2- – Oxide, S^2- – Sulphide,\(CO_3^{2-}\) – Carbonate

f. Al³⁺ – Aluminium, Cr³⁺ – Chromium, Fe³⁺ – Ferric.

Correct answer is 

0.1 N

Neutral

Correct option is (d) 0.3

Correct option is (d) 0.36 N

1. M.P. 

2. Rajasthan 

3. Jharkhand.

The minerals which do not have any metalic constituents are called non – metalic minerals. 

Example: limestone, dolomite, mica, etc.

1. Ferrous minerals 

2. Non – Ferrous minerals.

The minerals which have metal constituents are called metalic minerals. Example: iron ore, copper, gold and manganese.

1. It is mainly used in making electrical apparatus. 

2. It is mixed with gold to give strength to ornaments.

Those metalic minerals which have traces of iron metal are called ferrous metalic minerals.

Examples: iron – ore, manganese, tungsten and iron etc.

To Japan country, India’s iron-ore is mostly exported

Mormugao port.

In India the types of iron – ore are – magnetite, hematite, limonite and siderite.

Dharwar and Cuduppah rock groups.

(a) Charle’s law. 

(b) Isobar. 

(c) - 273.15°C, absolute zero is a lowest hypothetical temperature at which gases are supposed to occupy zero volume.

(d) p₁ < p₂ < p₃ < p₄ 

Reason: For a particular value of temperature: If volume is more it means pressure is less hence p₁ is lesser than p₂ and so on.

(c) (l₁T₂-l₂ T₁)/(T₂-T₁)

Explanation: 

If the original length of the wire be l and the area of the cross-section a, then  

l₁-l = T₁*l/Ya 

And l₂-l = T₂*l/Ya 

Dividing we get, 

(l₁ -l)*T₂ = (l₂-l)*T₁ 

→l₁T₂ -lT₂= l₂ T₁-T₁l 

→l(T₂-T₁) = l₁T₂-l₂T₁

 →l = (l₁T₂-l₂T₁)/(T₂-T₁) 

Hence the option (c).