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From 3d sub-shell

Answer is p sub shell

Both Lithium and magnesium have small size and high charge density. The electronegativities of Li is 1.0 and Mg is 1.2. They are low and almost same. Their ionic radii are similar. Hence they show similarities which is known as diagonal relationship between first element of a group with the second element in the next higher group.

(b) 2/3 

The area under the Pdf curve must be unity. All three regions are equi -probable, thus area under each region must be 1/3

Area of region 1 = 2a *1/4

2a/4 = 1/3 = 2/3

Answer (d) [NiCl₄]^2-

Answer (c) Ni

Answer (a) Both the statements are true and Statement It is the correct explanation of statement I

Answer (B) 0

Answer (b) 0

The oxidation number of Ni = 0

The oxidation number of Nickel in Ni(CO)₄ is 0

 Cu²⁺ is paramagnetic

An alloy is a homogenous mixture of two or more metals or a metal and a non-metal. Two properties alloy are : 

(i) Electrical conductivity 

(ii) Melting point of an alloy is less than that of pure metal.

Correct option is (B) 2Al₂O₃ + 3C → 4Al + 3CO₂

Hall Heroult process is the major industrial process for extraction of aluminium.

The compound in which the distrce between tow adjacent carbon atoms is the largest is Ethone

The order of dehydration of alcohol is 1⁰ < 2⁰ < 3⁰

1. No.

2. 

• Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
• There is no force of attraction between the molecules of a gas.

3. If assumption (i) is correct, the p vs V graph of experimental data (real gas) and that theoritically calculated from Boyle’s law (ideal gas) should coincide. But this never happens. If assumption (ii) is correct, the gas will never liquify. But gases do liquify when cooled and compressed.

1. All gases are made up of very large numbers of minute particles called molecules.

2. Intermolecular forces of attraction (or) repulsion are negligible.

3. The pressure exerted by a gas is due to the collisions made by the gas molecules on the walls of the container.

4. The average kinetic energy of the molecules is directly proportional to the absolute temperature.

5. The molecules are involved in rapid, random movement. During their motion, they collide with each other and also against the walls of the container.

Correct option  (b) I is exothermic, II is endothermic 

Methane is the limiting reagent because the other reactant is oxygen of air which is always present in excess. Thus the amounts of carbon dioxide & water formed will depend upon the amount of methane burnt.

Molarity of water means the number of moles of water in 1 litre of water
1L of water = 1000 cm³ = 1000 g (1000 kg/m³ = 1g/cm³)
1000 g of water = 1000/18 = 55.56 moles
Molarity = 55.56 M

1 molar aqueous solution has higher concentration than 1 molal solution. A molar solution contains one mole of solute in one litre of solution while one molal solution contains one mole of solute in 1000 g of solvent. If density of water is 1, then one mole of solute is present in 1000 mL of water in 1 molal solution while one mole of solute is present in less than 1000mL of water in 1 molar solution (1000 mL sol = amount of solute + amount of solvent). Thus 1 molar solution is more concentrated.

\(\lim\limits_{n\to\infty} \frac{n^{98}}{n^x -(n - 1)^x} = \frac1{99}\)

⇒ \(\lim\limits_{n \to \infty} \cfrac{n^{98}}{n^x - \left(n^x -x.n^{x-1}+ \frac{x(x -1)}{2}n^{x -2}...... + (-1)^x\right)} = \frac1{99}\)

⇒ \(\lim\limits_{n \to \infty}\) \(\cfrac{n^{98}}{n^{x -1}\left(x -\frac{x(x-1)}{2n}.......+ \frac{(-1)^x}{n^{x-1}}\right)}= \frac1{99}\)

Limit exists if x -1 = 98

⇒ x = 99

\(\lim\limits_{x\to \infty}\frac{x^n}{e^n}=\lim\limits_{x\to \infty}(\frac xe)^n=\infty^n=\infty\) if n is fixed

not defined

• Four  conditions are necessary for satisfactory cross-pollination
• Pollinizer and main variety bloom periods must overlap.
• The pollinizer variety must have viable diploid pollen.
• The pollinizer variety must be located near the producing tree.
• Bees and other insects must be present in the orchard and be active at bloom.

\(\lim\limits_{x\to \infty}{\frac{x}{\sqrt{4x^2+1-1}}}\) = \(\lim\limits_{x\to \infty}\cfrac1{\sqrt{4+\frac1{x^2}-\frac1x}}\)

 = \(\cfrac1{\sqrt{4+\frac1{\infty}}-\frac1{\infty}}\) = \(\frac1{\sqrt4}\) = \(\frac12\) (\(\because\frac1{\infty}=0\))

\(\lim\limits_{x\to \infty}\) x(2x + (x³ + x² + 1)^1/3 + (x³ - x² + 1)^1/3)

= \(\lim\limits_{x\to \infty}\) x²(2 + (1 + 1/x + 1/x³)^1/3 + (1 - 1/x + 1/x²)^1/3)

 = \(\infty^2\)(2 + 1 + 1) \((\because 1/\infty=0)\)

 = \(\infty\)

 = not defined

Calculation:

We have to find the value of \(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)

\(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)       [Form \(\frac{∞}{∞}\)]

This limit is of the form \(\frac{∞}{∞}\), Here, We can cancel a factor going to ∞  out of the numerator and denominator.

\(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)

= \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x}{x\sqrt{\frac{1}{x^2}+2}}\)

Factor x becomes ∞ at x tends to ∞, So we need to cancel this factor from numerator and denominator.

= \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{1}{\sqrt{\frac{1}{x^2}+2}}\)

= \(\frac{1}{\sqrt{\frac{1}{\infty^2}+2}}=\frac{1}{\sqrt{0+2}}=\frac{1}{\sqrt 2}\)

\(\lim_\limits{x\to\infty}\frac{5x^3-6}{\sqrt{9+4x^6}}\) = \(\lim\limits_{x\to \infty}\cfrac{5-\frac6{x^3}}{\sqrt{\frac9{x^6}+4}}\)

 = \(\frac{5-0}{\sqrt{0+4}}=\frac52\)

\(\lim\limits_{x\to 0}\frac{1-cosx\sqrt{cos2x}}{x^2}\) (0/0 case)

\(=\lim\limits_{x\to 0}\frac{-cosx\times\frac1{2\sqrt{cos2x}}\times-2sin2x+sin x\sqrt{cos x}}{2x}\) (By using D.L.H. Rule)

\(=\lim\limits_{x\to 0}\frac{cos x}{\sqrt{cos2x}}\times\frac{sin 2x}{2x}+\lim\limits_{x\to 0}\frac12\frac{sin x}x\sqrt{cos2x}\) 

\(=\lim\limits_{x\to 0}\frac{cos x}{\sqrt{cos2x}}\) + \(\frac12\lim\limits_{x\to 0}\sqrt{cos2x}\) (\(\because\lim\limits_{x\to 0}\frac{sin x}x=1\))

 = \(\frac{cos 0}{\sqrt{cos 0}}+\frac12\sqrt{cos 0}\) 

 = 1 + 1/2 = 3/2

Hence, \(\lim\limits_{x\to 0}\frac{1-cosx\sqrt{cos2x}}{x^2}=\frac32\)

\(\lim\limits_{x \to 0}\frac{(1-x)^n-1}{x}\) (\(\frac{0}{0}\,case\))

= \(\lim\limits_{x \to 0}\frac{-n(1-x)^{n-1}}{1}\)

= -n

\(\lim\limits_{x \to π/4}\frac{sinx - cosx}{x - \frac{\pi}4}\) (0/0 type)

\(=\lim\limits_{x \to \pi/4}\frac{cosx + sinx}1\) 

\(=cos\frac{\pi}4+sin\frac{\pi}4\) 

\(=\frac{1}{\sqrt2}+\frac1{\sqrt2}=\frac2{\sqrt2}\) = \(\sqrt2\) 

\(\lim\limits_{x\to0}\frac{1-cos3x}{x^2}\) 

\(=\lim\limits_{x \to 0}\cfrac{1-(1-\frac{(3x)^2}{2!}+\frac{(3x)^4}{4!}-....)}{x^2}\)

\(=\lim\limits_{x \to 0}\cfrac{\frac{9x^2}2-\frac{81x^4}{24}+...}{x^2}\)

\(=\lim\limits_{x \to 0}\frac92-\frac{81}{24}x^2+....\)

\(=\frac92\) (By taking limit)

Ice at 0°C has a total latent heat of  Q=m*L = 540*80 = 43200 cal.  

Now for water to come at 0°C and it's state to remain as water, we need to remove heat from it which is given by,  

Q= m*Cp*(∆T) = 540* 1* 80 = 43200 cal.  

So, we can see that  

Heat released by water to reach from 80°C to 0°C = heat required by ice to melt into water at 0°C.  

So, the ice will melt to water and it's temperature will be 0°C.  

Hence, the final state would be water at 0°C.