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Due to the ease with which it liberates atoms of nascent oxygen, it acts as a powerful oxidising agent. O3→O₂+O (nascent oxygen).

Third estate. Because they paid taxes and had low social status.

Faraday's first law states that the mass of a substance that is deposited or liberated at an electrode is directly proportional to the amount of charge passed.

W=ZQ

Where W is the mass of the substances and Q is the charge and Z is the electrochemical equivalent.

Solution: Isotopes of Mg. 

Atomic mass = Mg²⁴ = 23.99 × 78.99/100 = 18.95 

Atomic mass = Mg²⁶ = 24.99 × 10/100 = 2.499 

Atomic mass = Mg²⁵ = 25.98 × 11.01/100 = 2,860 

Average Atomic mass = 24.309 

Average atomic mass of Mg = 24.309

Correct option (B) Jose De San Martin 

1. The attack on, the marble of buildings by acid rain is called stone leprosy. 

2. Acid rain causes extensive damage to buildings made up of marble.

CaCO + H₂SO₄ → CaSO₄ + H₂O + CO₂ ↑

O₃ being endothermic compound readily decomposes on heating to give dioxygen or nascent oxygen.

O₃ → O₂ +O (nascent oxygen)

since nascent oxygen is very reactive therefore O act as a powerful oxidising agent.

1. The electron in an atom can be characterized by a set of four quantum numbers, namely – principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s). 

2. When Schrodinger equation is solved for a wave function φ, the solution contains the first three quantum numbers n, 1 and m. 

3. The fourth quantum number arises due to the spinning of the electron about its own axis.

State General 

The First Estate  –  285 

The Second Estate  –  308 

The Third Estate  –  621 

One vote for 1^st and second estate. In the third a vote for each member 

Vote for first estate – 285 

Vote for second estate – 308 

Vote for third estate – 621 

If one vote is given for each member of each estate the third estate will have the majority of one vote is given for each estate, when they stay united the first and second can easily beat the third.

The bond enthalpy is defined as the minimum amount of energy required to break one mole of a particular bond in molecules in their gaseous state. Larger the bond enthalpy stronger will be the bond.

We will use the ideal gas equation for this calculation as below:

P = \(\frac{nRT}{V}\) = image 7 = 9.39 atm.

1. If the solute is dissolved in the solvent water, the resultant solution is called as an aqueous solution, e.g., salt in water. 

2. If the solute is dissolved in the solvent other than water such as benzene, ether, CCl₄ etc,

the resultant solution is called a non aqueous solution, e.g., Br₂ in CCl₄.

(I) The first law of thermodynamics states that “the total energy of an isolated system . remains constant though it may change from one form to another” (or) Energy can neither be created nor destroyed, but may be converted from one form to another.

(II) The complete ethylene combustion reaction can be written as,

C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(I) 

ΔU = -1406 kJ 

Δn = n_p(g) – n_r(g) 

Δn = 2 – 4 – 2 

ΔH = ΔU + RTΔn_g 

ΔH = -1406 + (8.314 × 10^-3 × 300 × (-2)) ΔH 

= -1410.9 kJ

1. Higher lattice energy shows greater electrostatic attraction and therefore a stronger bond in the solid. 

2. The lattice enthalpy is greater for ions of higher charge and smaller radii.

(I) 1. The value of equilibrium constant K_C tells us the extent of the reaction i.e., it indicate how far the reaction has proceeded towards product formation at a given temperature.

2. A large value of K_C indicates that the reaction reaches equilibrium with high product yield on the other hand, lower value of K_C indicates that the reaction reaches equilibrium with low product yield.

3. If K > 10^-3 , the reaction proceeds nearly to completion. 

4. If K_C < 10^-3 the reaction rarely proceeds. 

5. It the K_C is in the range 10^-3 to 10³, significant amount of both reactants and products are present at equilibrium.

(II) Addition of catalyst does not affect the state of equilibrium. The catalyst increases the rate of both the forward and reverse reactions to the same extent. Hence it does not change the equilibrium composition of the reaction mixture.

 All the bonds that are present in PCl₅ no similar. It has three equatorial and two axial bond, the equatorial bonds are stronger than axial one, therefore when PCl₅ is heated strongly.

It decomposes to form:

 PCl₅ rightwards arrow with heat on top PCl₃ + Cl₂

IUPAC name of the complex K₃[Cr(C₂O₄)₃] is Potassium trioxalatochromate(III). Here, oxalate is a bidentate ligand thus oxidation state of Cr will be 3.

A solution is said to be ideal if each of its components obeys Raoult's law for the entire range of concentration (composition). In the preparation of ideal solution, no thermal change is observed. i.e.

△H_mix=0

The force of interaction between A−A, B−B and A−B are of same order. The volume of mixing (△V_mix) is also zero. i.e. The volume of the solution will be equal to sum of the volumes of the two components.

For example, the solution of benzene in CCl₄ shows same attractive forces as are present in benzene and CCl₄ separately. Another examples are benzene and toluene, bromoethane and chloroethane.

Solutions are non-ideal if they do not obey Raoult's law. i.e. 

△H_mix≠0  and  △V_mix≠0

Therefore these solution deviate from ideality and depending on type of deviation from ideal behaviour, non-ideal solutions may be classified as showing negative deviation or positive deviation.

The order of reaction can be defined as the power dependence of rate on the concentration of all reactants. For example, the rate of a first-order reaction is dependent solely on the concentration of one species in the reaction. Some characteristics of the reaction order for a chemical reaction are listed below.

• Reaction order represents the number of species whose concentration directly affects the rate of reaction.
• It can be obtained by adding all the exponents of the concentration terms in the rate expression.
• The order of reaction does not depend on the stoichiometric coefficients corresponding to each species in the balanced reaction.
• The reaction order of a chemical reaction is always defined with the help of reactant concentrations and not with product concentrations.
• The value of the order of reaction can be in the form of an integer or a fraction. It can even have a value of zero.

There are several different methods which can be followed in order to determine the reaction order. Some of these methods are described in this subsection.

• Initial Rates Method
• Integral Method
• Differential Method
• Different Values of Reaction Order

First-Order Reactions

The rates of these reactions depend on the concentration of only one reactant, i.e. the order of reaction is 1.

In these reactions, there may be multiple reactants present, but only one reactant will be of first-order concentration while the rest of the reactants would be of zero-order concentration.

Example of a first-order reaction: 2H₂O2 → 2H₂O + O₂

The mixture of cryolite and aluminium oxide has a lower melting point than pure aluminium oxide. This means a lower amount of energy is required to establish effective conditions for electrolysis and thus makes it more cost effective.

Molten cryolite serves as a solvent for the molten aluminium oxide and increases the conductivity of the solution.

PH₃ has a lower boiling point than ammonia due to its inability to assorted via intermolecular hydrogn bonding. This is because phosphorus is bigger have bigger size as compare to nitrogen thus have weak bonding with hydrogen.

1. Repeater: Also called a regenerator, it is an electronic device that operates only at physical layer. It receives the signal in the network before it becomes weak, regenerates the original bit pattern and puts the refreshed copy back in to the link.

2. Bridges: These operate both in the physical and data link layers of LANs of same type. They divide a larger network in to smaller segments. They contain logic that allow them to keep the traffic for each segment separate and thus are repeaters that relay a frame only the side of the segment containing the intended recipent and control congestion

3. Routers: They relay packets among multiple interconnected networks (i.e. LANs of different type). They operate in the physical, data link and network layers. They contain software that enable them to determine which of the several possible paths is the best for a particular transmission.

4. Gateways: They relay packets among networks that have different protocols (e.g. between a LAN and a WAN). They accept a packet formatted for one protocol and convert it to a packet formatted for another protocol before forwarding it. They operate in all seven layers of the OSI model.

The main reasons are:

(i) Six large chloride ions cannot be accommodated around Si⁴⁺ due to limitation of its size.

(ii) Interaction between lone pair of chloride ion and Si⁴⁺ is not very strong.

The data unit in the LLC level is called the protocol data unit (PDU). The PDU contains of four fields a destination service access point (DSAP), a source service access point (SSAP), a control field and an information field. DSAP, SSAP are addresses used by the LLC to identify the protocol stacks on the receiving and sending machines that are generating and using the data. The control field specifies whether the PDU frame is a information frame (I - frame) or a supervisory frame (S - frame) or a unnumbered frame (U - frame).

It is a project started by IEEE to set standards to enable intercommunication between equipment from a variety of manufacturers. It is a way for specifying functions of the physical layer, the data link layer and to some extent the network layer to allow for interconnectivity of major LAN protocols.

It consists of the following:

1. 802.1 is an internetworking standard for compatibility of different LANs and MANs across protocols.

2. 802.2 Logical link control (LLC) is the upper sublayer of the data link layer which is non-architecture-specific, that is remains the same for all IEEE-defined LANs. 

3. Media access control (MAC) is the lower sublayer of the data link layer that contains some distinct modules each carrying proprietary information specific to the LAN product being used. The modules are Ethernet LAN (802.3), Token ring LAN (802.4), Token bus LAN (802.5). 

4. 802.6 is distributed queue dual bus (DQDB) designed to be used in MANs.

Threshold Energy-

The minimum energy which the colliding molecules must have in order that the collision between them may be effective is called threshold energy. Threshold Energy = Activation Energy = Average kinetic energy of the reactants.

Electro Chemical Equivalent-

If Q = 1 coulomb, I = 1 ampere, t = 1 second them W = Z

Electrochmical equivalent of a substance may be defined as the mass of the substance deposited when a current of one ampere is passed for one second i.e, a quantity of electricity equal to one coulomb is passed.

Faraday's first law of electrolysis-

It states that- The amount of chemical reaction and hence the mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte (solution of melt). Thus, it W gram of the substance is deposited on passing Q coulombs of electricity then W ∝ R ⇒ W = ZQ

When z is a constant of proportionally and is called electrochemical equivalent of the substance deposited.

If a current of 1 amperes is passed for t seconds, then Q = I x t

So W = Z x Q = z x I x t

In forest ecosystem decrease Primary consumers.

According to evolutional view, we are more similar to Chimpanese.

In human being inhaling and exhaling phenomenon is called as Breathing.

Male and female gamete combine to form Zygote.

Concept:

Binomial expansion for \(\rm\left ( x+a \right )^{n}-\left ( x-a \right )^{n}\)  is given by

\(\rm\left ( x+a \right )^{n}-\left ( x-a \right )^{n}\) = 2{nC1xn-1a1 + nC3xn-3a3 + nC5xn-5a5 +....}

\(\rm\left ( x+a \right )^{n}-\left ( x-a \right )^{n}\) = 2{sum of terms at even places}

Calculation:

For this expression \(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\)  assume \(\rm \sqrt{3} =a\)

\(\rm(a+1)^5-(a-1)^5\) = 2{5C1 a5-1 11 + 5C3 a5-3 13 + 5C5 a5-5 15}

\(\rm(a+1)^5-(a-1)^5\) = 2{5C1a4 + 5C₃a²+ 5C5}

\(\rm(a+1)^5-(a-1)^5\) = 2{5a4 + 10a2+ 1}

Put  \(\rm a =\sqrt{3}\)

\(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\) = 2{5(3²) + 10(3)+ 1}

\(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\) = 2{5.3² + 10.3 + 1}

\(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\) = 2{45 + 30 +1}

\(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\) = 152

Concept:

Binomial expansion for \(\rm\left ( x+a \right )^{n}+\left ( x-a \right )^{n}\)  is given by

\(\rm\left ( x+a \right )^{n}+\left ( x-a \right )^{n}\) = 2{ⁿC₀ x^n-0 a⁰ + nC₂ xn-2 a² + nC4 xn-4 a⁴ +....}

\(\rm\left ( x+a \right )^{n}+\left ( x-a \right )^{n}\) = 2{sum of terms at odd places}

Calculation:

For this \(\rm\left ( x+\sqrt{x} \right )^{6}+\left ( x-\sqrt{x} \right )^{6}\) assume  \(\rm a=\sqrt{x}\)

\(\rm\left ( x+a \right )^{6}+\left ( x-a \right )^{6}\) = 2{6C0 x6-0 a0 + 6C2 x6-2 a2 + 6C4 x6-4 a4 + 6C6 x6-6 a6}

\(\rm\left ( x+a \right )^{6}+\left ( x-a \right )^{6}\) = 2{x6 + 15x⁴a2 + 15x2a4 + a⁶}

Put \(\rm a=\sqrt{x}\)

\(\rm\left ( x+\sqrt{x} \right )^{6}+\left ( x-\sqrt{x} \right )^{6}\) = 2{x6 + 15x⁴.x + 15x².x²+ x3}

\(\rm\left ( x+\sqrt{x} \right )^{6}+\left ( x-\sqrt{x} \right )^{6}\) = 2{x6 + 15x5 + 15x4+ x3}

\(\rm\left ( x+\sqrt{x} \right )^{6}+\left ( x-\sqrt{x} \right )^{6}\) = 2x6 + 30x5 + 30x4 +2x3

Blood vessels which carry pure or oxygenated blood from heart to different body parts are called as Arteries.

Hypothesis testing is an act in statistics whereby an analyst tests an assumption regarding a population parameter. The methodology employed by the analyst depends on the nature of the data used and the reason for the analysis.

Hypothesis testing is used to assess the plausibility of a hypothesis by using sample data. Such data may come from a larger population, or from a data-generating process.

Answer: d. All of the above

Area of the rectangle = 225 cm²

Breadth = 25 cm

Length = ?

Area of rectangle = Length × Breadth

225 cm² = Length x 25 cm

\(\frac{225\, cm^2}{25\, cm}\) = Length

9 cm = Length

Length = 9 cm

Correct answer is b) Microsoft Windows 

Sixth term = T₆ = -1/32

ninth term = T₉ = 1/256

then , common ratio = R = ?

11^th term = T₁₁ = ?
R^p-q = Tp/Tq

R^9-6 = (1/256)/-1/32

R³ = 1/256 × -32

R³ = 1/-8

R³ = -1/8

R³ = (-1/2)³

R = -1/2 .

T₆ = a(R)^n-1

-1/32 = aR^6-1

-1/32 = aR⁵

-1/32 = a × (-1/2)⁵

-1/32 = a × -1/32

(-1/32)/(-1/32) = a

-1/32 × -32 = a

-1×-1 = a

1 = a

a = 1

T₁₁ = aR^n-1

T₁₁ = (1)(-1/2)^11-1

T₁₁ = (-1/2)¹⁰

T₁₁ = 1/1024

tan 45 = tan ( 32 + 13)

= (tan 32 + tan 13)/( 1- tan 32 tan 13)

Therefore tan 32 + tan 13

= tan(45)( 1 - tan32 tan 13)

= 1 - tan32 tan13 ..................... since tan 45 = 1

and tan 32 + tan 13 + tan 32 tan 13 = 1

Similarly tan 23 + tan 22 + tan 23 tan 22 = 1

Now (1+Tan32°)(1+tan13°)(1+tan23°)(1+tan22°)

= ( 1+Tan32° + tan13° + tan32 tan 12)(1+tan23° + tan 22 + tan 23 tan 22)

= ( 1 + 1 )(1 + 1)

= 4

b) A drawback/feature for this approach is that the learning is static. 

d) In learning based approach, the relationship or patterns in data are defined by the developer.