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Electric field a vector.

Linear density of charge is charge per unit length. Its SI unit is Cm^-1.

Electric field due to a charge at a point in space is defined as the force experienced by a unit positive charge placed at that point.  

Principle of superposition  

1C is the charge that when placed at a distance of 1m from another charge of the same magnitude, in vacuum, experiences an electrical force of repulsion of magnitude 9 x 10⁹ N.

The energy of electron is determined by the value of n in hydrogen atom and by n + l in multielectron atom. So for a given principal quantum number electrons of s, p, d and f orbitals have different energy.

Gold has highest malleability and ductility. Extremely very thin foil can be prepared by using Gold, which is not possible with other metals. So, we will not get same type of results using other metals.

• Bone china is made by adding some ash of animal bones in the mixture of china clay, feldspar and fine silica while making porcelain.
• This ceramic is harder than porcelain.

• Baking powder consist of 30% sodium bicarbonate, 5-12% monocalcium phosphate and 21-26% sodium aluminium sulphate. 
• Baking powder is a leavening agent and is used for increasing the volume and lightening the texture of baked goods.

1. Porcelain is a hard, translucent and white coloured ceramic. It is made by using the white clay called kaolin, found in China. Glass, granite and the mineral feldspar is mixed with kaolin and kneaded with water.

2. The resulting mixture is shaped and fired in a kiln at a temperature of 1200 to 1450 °C. On firing again after glazing, beautiful articles of porcelain are obtained.

Correct option is  D) 4

Correct option is  B) 2.5

It is the ability of an atom to attract a shared pairs of electrons towards itself. If the atoms forming a covalent bond have different electronegativities, the atom with higher electronegativity pulls the shared pair of electrons towards itself. Thus, the atom with the higher electronegativity develops a partial negative charge and the atom with the lower electronegative develops a partial positive charge. This bond with some polarity is called polar covalent bond.

No colour change is observed. It gives pink colour in basic solution

Gold is one of the most malleable metal we have and we could extract thin
foils from it but other metals are not that malleable. So, if we use a metal other than gold in the alpha-particle scattering experiment we would not being able to get a thin foil. A comparatively thick foil means that the number of atoms per unit volume would be large and hence the number of alpha particles passing through the foil without being deflected would decrease and those getting deflected would increase in comparison with the gold foil.

E = hv: Electromagnetic energy can have only certain discrete energy values which is given by the equation E = hv where h = Planck’s constant v = frequency of radiation

P₁ = X₁ x P_(total)

Both the reactions are in accordance with the principle of stoichiometry.

In the first reaction,

Both the reactants are completely consumed to form product according to reaction stoichiometry. 1 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride.

In the second reaction, 

Chlorine is the limiting reagent and it is completely consumed to form hydrogen chloride. 

Excess hydrogen remains unreacted at the end of the reaction. 

This reaction also follows principle of stoichiometry.

2 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride + 1 mol hydrogen

According to Charle's law, the volume of a given mass of a gas becomes zero at -273°C. Below this temperature the volume becomes negative, which has no significance. Therefore -273°C is the lowest possible temperature.

i. In the +2 oxidation state, Eu donates two electrons from its 6s-orbitals. In Eu⁺², 4f-orbitals are half-filled i.e., 4f⁷ which is a more stable state.

ii. Similarly in the +2 oxidation state, Yb donates two electrons from its 6s-orbitals. In Yb⁺², 4f-orbitals are completely filled i.e., 4f¹⁴ which is more stable.

Therefore, Eu and Yb show +2 oxidation states.

Option : d. All the above statements are correct.

{pV²T²}/{n} = {bar x (dm³)² x K²}/{mol}

= bar dm⁶ K² mol^-1

For ideal gas Z = 1.

a. Exactly 1.5 atm to pascals : 

1 atm = 101325 Pa 

∴ 1.5 atm = 1.5 × 101325 

= 151987.5 Pa

b. 89 kPa to newton per square metre (N m^-2) : 

1 Pa = 1 N m^-2 and 1 Pa = 10^-3 kPa 

∴ 10^-3 kPa = 1 N m^-2

∴ 89 kPa = \(\frac{1\times 89}{10^{-3}}Nm^{-2}\) 

c. 101.325 kPa to bar :

1 bar = 1.0 × 10⁵Pa

= 1.0 × 10² k Pa

∴ 100 kPa = 1 bar

∴ 101.325 kPa = \(\frac{1\times 101.325}{100}\) 

= 1.01325 bar

d. -100 °C to Kelvin :

T(K) = t°C + 273.15 

∴ T(K) = (- 100°C) + 273.15 

= 173.15 K

e. 0.124 torr to standard atmosphere :

1 atm = 760 torr

∴ 1 torr = \(\frac{1}{760}\) atm

∴ 0.124 torr = 0.124 × \(\frac{1}{760}\)

= 1.632 × 10^-4 atm

a. 10 atmosphere : 

1 atm = 101325 Pa 

∴ 10 atm = 1013250 Pa 

= 1.01325 × 10⁶ Pa

b. 1 kPa : 

1 kPa = 1000 Pa

c. 107000 Nm^-2 :

1 N m^-2 = 1 Pa 

∴ 107000 Nm^-2 = 107000 Pa 

= 1.07 × 10⁵ Pa

d. 1 atmosphere : 

1 atm = 101325 Pa 

= 1.01325 × 10⁵ Pa

1 atm = 101325 Pa or Nm^-2 , 1 bar = 10⁵ Pa. 

Correct option: (C) Zinc

i. All lanthanoids exhibit a common stable oxidation state of +3.

ii. In addition, some lanthanoids show +2 and +4 oxidation states also. 

iii. These are shown by those elements which by doing so attain the stable f⁰ , f⁷ or f¹⁴ configurations i.e., empty, half-filled and completely filled 4f sub-shells.

eg. Ce and Tb exhibit +4 oxidation states, Eu and Yb exhibit +2 oxidation states, while La, Gd and Lu exhibit only +3 oxidation states.

Note: Oxidation states of Lanthanoids are as shown in following table:

Because relative position of ligands attached to central atom are same with respect to one another.

i. The oxidation states of an element depends on its electronic configuration and the number of unpaired electrons.

ii. The electronic configuration of Fe, Co and Ni are as follows:

iii. Hence, the expected oxidation states of Fe, Co and Ni shall be +8, +9 and +10 respectively.

iv. Fe by the loss of two electrons from 4s subshell forms Fe²⁺ and further loss of one more electron from 3d-orbital forms Fe³⁺, which is very stable due to half filled orbital, 3d⁵ . Hence, there is no further loss of electrons from the 3d subshell and thus Fe does not show +8 oxidation.

v. Co forms Co²⁺ due to the loss of two electrons from 4s subshell and further loss of one more electron from 3d-orbital forms Co³⁺. In cobalt, +2 and +3 oxidation states are more stable and it does not lose electrons further. Hence, Co does not show +9 oxidation state.

vi. Ni forms Ni²⁺ by the loss of two electrons from 4s-orbital and by the loss of one more electron from 3dorbital, it forms Ni³⁺. The amount of energy required for the ionization of Ni to Ni²⁺ is less when compared to that required for the ionization of Ni to Ni⁴⁺. Hence, Ni²⁺ oxidation state is comparatively more stable. There is no further loss of electrons and therefore, Ni does not show +10 oxidation state.

(i) Normal urine has a surface tension of about 66 dynes/cm but if bile is present (a test for jaundice), it drops to about 55 dynes/cm. In this test, sulphur is sprinkled on the urine surface. If it floats, the test is negative but if it sinks, it is positive. 

(ii) Hot water is more effective because surface tension is lowered and it becomes a better wetting agent. 

(iii) Waxing changes the surface properties of the car. Water tends to adhere weakly to the wax and strongly to itself. 

(iv) In case of water taken on the glass surface the adhesive force between water and glass are greater than cohesive forces among water molecules but in case of mercury taken in glass tube, it is the reverse.

It 22400 ml.

Below Boyle Temperature.

At absolute zero (O K) temperature. 

Gram molar mass of both these gases is the same (44 g). According to Graham’s law of diffusion, the rates of diffusion of the gases are inversely proportional to the square root of their molar masses under similar conditions of temperature and pressure. Therefore, both these gases diffuse at the same rate.

P = ρ{RT}/{M}

If M is the molar mass of the gaseous oxide, we can write

2 bar = ρ{RT}/{M}

and for nitrogen

5 bar = ρ{RT}/{28 g/mol}

From these equation, we can write

{5 bar}/{2 bar} = {M}/{28 g.mol}

or, M = {5 bar}/{2 bar} x 28 g mol^-1

= 70 g mol^-1

(ii) Complete disorder of molecules
(iii) Random motion of molecules

PV = 1/3 mnc²/V

(i), (iv)

(i) 16 grams of gas

(iv) 11.2 litres of gas

8.3124 J K^-1 mol^-1

PV = 1/3 mnc²

The volume occupied by one mole of an ideal gas at STP is 22.414 L. This volume is known as molar volume.

a. P ∝ \(\frac{1}{V}\) 

b. The graph represents Boyle’s law as it gives relation between pressure and volume at constant temperature.

c. Statement of Boyle’s law : For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.

OR 

At constant temperature, 

The pressure of fixed amount (number of moles) of a gas varies inversely with its volume.

Ideal gas equation PV= nRT

or n/V = P/RT

or m/MV = P/RT (as nm = m/M)

or d/M = P/RT

or P = dRT/M

Relation between density of a gas and its molar mass:

According to Avogadro’s law, V ∝ n

Now, n = \(\frac{m}{M}\) (where, m is the mass of the gas and M is the molar mass of the gas)

∴ V ∝ \(\frac{m}{M}\)

M ∝ \(\frac{m}{V}\)

But, \(\frac{m}{V}\) = d (where, d is the density of the gas) 

∴ d ∝ M

Thus, density of a gas is directly proportional to its molar mass.

PV = nRT for n moles.

The ‘gas’ above the surface of liquid water is described as water vapour.

According to Boyle’s law,

V ∝ \(\frac{1}{P}\) (at constant T and n) ……….(1)

According to Charles’ law,

V ∝ T (at constant P and n) ……(2)

According to Avogadro’s law,

V ∝ n (at constant P and T) ……(3)

Combining relations (1), (2) and (3), we get

V ∝ \(\frac{nT}{P}\) 

Converting this proportionality into an equation by introducing a constant of proportionality (‘R’ known as gas constant), we get

∴ V = \(\frac{nRT}{P}\) 

On rearranging the above equation, we get 

PV = nRT

Where,

P = Pressure of gas, 

V = Volume of gas, 

n = number of moles of gas, 

R = Gas constant, 

T = Absolute temperature of gas. 

This is the ideal gas equation or equation of state.

[Note : In the ideal gas equation, R is called gas constant or universal gas constant, whose value is same for all the gases. In this equation, if three variables are known, fourth can be calculated. The equation describes the state of an ideal gas. Hence, it is also called as equation of state.]

The ideal gas equation is written as PV = nRT …(1)

On rearranging equation (1), we get,

∴ \(\frac{PV}{T}\) = nR = constant

∴  \(\frac{P_1V_1}{T_1}\) = \(\frac{P_2V_2}{T_2}\) …(2)

The ideal gas equation used in this form is called combined gas law.

Relation between density, molar mass and pressure: 

According to ideal gas equation,

PV = nRT …..(1) 

On rearranging equation (1), we get

\(\frac{n}{V}\) = \(\frac{P}{RT}\) ……….(2)

Now, n = \(\frac{m}M\)

On substituting the value of n, equation (2) becomes

\(\frac{m}{MV}\) = \(\frac{P}{RT}\)

\(\frac{d}M\) = \(\frac{P}{RT}\) ……….(3)

where d = \(\frac{m}V\) = density of the gas On rearranging the equation, we get

M = \(\frac{dRT}P\) ……(4)

This equation can be used to calculate molar mass of a gas in terms of its density.

1. C 

2. A 

3. D