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#### 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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#### 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 $\dpi{100} ^{1}H$ atoms are replaced by $\dpi{100} ^{2}H$ 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

#### A plane is inclined at an angle with respect to the horizontal. A particle is projected with a speed from the base of the plane , making an angle 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:Option: 1Option: 2Option: 3Option: 4

As we know on inclined plane the range is given by

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

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

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

- wherein

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

As we know that

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

Therefore, Option(2) is correct

#### 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

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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#### The dimension of stopping potential  in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :    Option: 1   Option: 2  Option: 3    Option: 4

Let

Now,

So,

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,

#### 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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#### 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 4:1Option: 2 3:1Option: 3 2:1Option: 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).

#### 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