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An ideal gas undergoes isothermal expansion at constant pressure.  During the process :
Option: 1  enthalpy increases but entropy decreases.
Option: 2  enthalpy remains constant but entropy increases.
Option: 3  enthalpy decreases but entropy increases.
Option: 4  Both enthalpy and entropy remain constant.  

\Delta H = nC_{p}\Delta T \\ \Delta S = nR ln (V_{f}/V_{i})\geqslant 0

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Posted by

vishal kumar

 


The major product obtained in the following reaction is : 
Option: 1


Option: 2

Option: 3 

Option: 4 

The major product is represented by the option (D). DIBAL−H reduces esters to aldehydes. It also reduces carboxylic acids to aldehydes. 

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Posted by

vishal kumar

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The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight (in kg) which a 75 kg person would gain if all ^{1}H atoms are replaced by ^{2}H atoms is :  
Option: 1 7.5
Option: 2 10
Option: 3 15
Option: 4 37.5
 

Given that

Mass of the person = 75 kg

Mass of 1H1 present in person = 10% of 75 kg = 7.5 kg

Since Mass of 1H2 is double the Mass of 1H1

So, Mass of 1H2 will be in person = 2 X 7.5 kg =15 kg

Thus, increase in weight = 15 - 7.5 = 7.5 kg

Therefore, Option (1) is correct

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Posted by

vishal kumar

A plane is inclined at an angle \alpha =30^{0} with respect to the horizontal. A particle is projected with a speed u=2ms^{-1}from the base of the plane , making an angle\theta =15^{0} With respect to the plane as shown in the  figure . The distance from the base ,at which the particle hits the plane is close to:\left ( Take \, \, g=10ms^{-2} \right )
Option: 1 20cm
Option: 2 18cm
Option: 3 26cm
Option: 4 14cm

As we know on inclined plane the range is given by

R=\frac{2u^{2}}{g}\frac{\sin \beta \cos \left ( \beta +\alpha \right )}{\cos ^{2}\alpha }

\beta =15^{0}\, \, ;\alpha =30^{0}\, \, ;u=2m/s

R=\frac{2\times 2^{2}\sin 15^{0}\cos \left ( 45 \right )}{10\times \cos \left ( 30^{0} \right )}

     = 20 cm \, \, approx

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Posted by

Ritika Jonwal

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The correct order of the atomic radii of C, Cs, Al and S is :
Option: 1 S<C<Cs<Al
Option: 2 C<S<Al<Cs
Option: 3 C<S<Cs<Al
Option: 4 S<C<Al<Cs
 

 

Periodicity of atomic radius and ionic radius in period -

In a period from left to right the effective nuclear charge increases because the next electron fills in the same shell. So the atomic size decrease.

- wherein

Li>Be>B>C>N>O>F

 

 

 

Electronegativity and atomic radius -

The attraction between the outer electrons and the nucleus increases as the atomic radius decreases in a period.

- wherein

Electronegativity\propto\frac{1}{atomic\:radius}

 

 

 

Size of atom and ion in a group -

In a group moving from top to the bottom the number of shell increases.So the atomic size increases.

- wherein

Li<Na<K<Rb<Cs

As we know that

From Left to right in a period size decreases and when going down the group size increases

C< S< Al< Cs

Therefore, Option(2) is correct

  

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Posted by

Ritika Jonwal

The IUPAC symbol for the element with atomic number 119 would be:
Option: 1 uue
Option: 2une
Option: 3 unh
Option: 4 uun

 

Nomenclature of elements with atomic number >100 -

The name is derived directly from the atomic number of the element using the following numerical roots:

0 = nil

1 = un

2 = bi

3 = tri

4 = quad

5 = pent

6 = hex

7 = sept

8 = oct

9 = enn

Eg:

 

Atomic number

Name

Symbol

101

Mendelevium (Unnilunium)

Md (Unu)

102

Nobelium (Unnilbium)

No (Unb)



 

-

uue

1  1  9

Un Un ennium

Therefore, Option(1) is correct.

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Posted by

Ritika Jonwal

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The dimension of stopping potential V_0 in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :   
Option: 1 h^{-2/3}c^{-1/3}G^{4/3}A^{-1}
 
Option: 2 [h]^0[G]^{-1}[c]^5[A]^{-1}

Option: 3 h^{2}G^{3/2}c^{1/3}A^{-1}  

Option: 4 h^{1/3}G^{2/3}c^{1/3}A^{-1}
 

 

 

 Let V_0=[h]^p[G]^q[c]^r[A]^s

Now, K_{max}=eV_0\Rightarrow [K_{max}]=[eV_0]\Rightarrow [V_0]=\frac{[K_{max}]}{[e]}\Rightarrow [V_0]=\frac{[ML^2T^{-2}]}{[AT]}\therefore [V_0]=[ML^2T^{-3}A^{-1}]

E=hf\Rightarrow [h]=\frac{[E]}{[f]}\Rightarrow \frac{[ML^2T^{-2}]}{[T^{-1}]}\Rightarrow [h]=[ML^2T^{-1}]

 

F=\frac{Gm^2}{r^2}\Rightarrow [G]=\frac{[F][r^2]}{[m^2]}=\frac{[MLT^{-2}][L^2]}{[M^2]}=[M^{-1}L^3T^{-2}]

So, [ML^2T^{-3}A^{-1}]=[ML^2T^{-1}]^p[M^{-1}L^3T^{-2}]^q[LT^{-1}]^r[A]^s

From here we will get: p-q=1---------(1)

2p+3q+r=2----------(2)

-p-2q-r=-3----------(3)

s=-1-----------(4)

From equation 1, 2,3 and 4 we will get: p=0,q=-1, r=5 and s=-1

So, [V_0]=[h]^0[G]^{-1}[c]^5[A]^{-1}

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Posted by

vishal kumar

The ammonia (NH_{3}) released on quantitative reaction of 0.6g, urea (NH_{2}CONH_{2}) with sodium hydroxide (NaOH) can be neutralised by:
Option: 1 200 ml of 0.2 N HCl
Option: 2200 ml of 0.4 N HCl
Option: 3100 ml of 0.1N HCl
Option: 4100 ml of 0.2N HCl
 

2 × mole of Urea = mole of NH_{3} ........(1)
mole of NH_{3} = mole of HCl ........(2)
\therefore mole of HCl = 2 × mole of Urea

mole of HCl =2\times \frac{0.6}{60}=0.02mol ...(i)

[We know , mole = M X V = N X n X V]

 \mathrm{100 \ ml\times 0.2N\times 1=0.02\ mol} ...as (i).

Therefore, Option(4) is correct.

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Posted by

Ritika Jonwal

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The average molar mass of chlorine is 35.5\: g\: mol^{-1}. The ratio of ^{35}Cl\; to\; ^{37}Cl in naturally occurring chlorine is close to :  
Option: 1 4:1
Option: 2 3:1
Option: 3 2:1
Option: 4 1:1

Given,

The average molar mass of chlorine is 35.5\: g\: mol^{-1}.

 

Let , the ratio of ^{35}Cl\; to\; ^{37}Cl in naturally occurring chlorine is close to x : y

 

Now, we know

\text { Av. molar mass }=\frac{\mathrm{n}_{1} \mathrm{M}_{1}+\mathrm{n}_{2} \mathrm{M}_{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)}

So, 

35.5=\frac{x \times 35+y \times 37}{x+y}

1.5 y=-0.5 x

\frac{x}{y}=\frac{3}{1}

Therefore, the correct option is (2).

 

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Posted by

Kuldeep Maurya

The minimum number of moles of O2 required for complete combustion of 1 mole of propane and 2 moles of butane is ______
 

Combustion reaction of 1 mole of propane and 2 moles of butane-

 

So, Total required mol of O2 = 5 + 13 = 18.

Ans = 18

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Posted by

Kuldeep Maurya

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