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General
chemistry-

What will be the most probable product when compound ‘X’ is treated with two equivalents of NaOH,

What will be the most probable product when compound ‘X’ is treated with two equivalents of NaOH,

chemistry-General
General
chemistry-

Aniline reacts with mixed acid (conc. HNO3 and conc. H2SO4 ) at 288 K to give P (51%), Q (47%) and R (51%), Q(47%) and R (2%). The major product(s) of the following reaction sequence is/are

Aniline reacts with mixed acid (conc. HNO3 and conc. H2SO4 ) at 288 K to give P (51%), Q (47%) and R (51%), Q(47%) and R (2%). The major product(s) of the following reaction sequence is/are

chemistry-General
General
chemistry-

The major product of the following reaction is

The major product of the following reaction is

chemistry-General
General
chemistry-

In the following reaction sequences V and W are respectively :

In the following reaction sequences V and W are respectively :

chemistry-General
General
chemistry-

GivenC left parenthesis s right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis C O subscript 2 end subscript left parenthesis g right parenthesis plus 94.2Kcal H subscript 2 end subscript left parenthesis g right parenthesis plus 1 divided by 2 O subscript 2 end subscript left parenthesis g right parenthesis rightwards arrow stack H subscript 2 end subscript O with _ below left parenthesis l right parenthesis plus 68.3Kcal C H subscript 4 end subscript left parenthesis g right parenthesis plus 2 O subscript 2 end subscript left parenthesis g right parenthesis times C O subscript 2 end subscript left parenthesis g right parenthesis plus 2 stack stack H with _ below subscript 2 end subscript O with _ below left parenthesis l right parenthesis plus 210.8Kcal thheaoformatioomethane in Kcawilbe-

GivenC left parenthesis s right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis C O subscript 2 end subscript left parenthesis g right parenthesis plus 94.2Kcal H subscript 2 end subscript left parenthesis g right parenthesis plus 1 divided by 2 O subscript 2 end subscript left parenthesis g right parenthesis rightwards arrow stack H subscript 2 end subscript O with _ below left parenthesis l right parenthesis plus 68.3Kcal C H subscript 4 end subscript left parenthesis g right parenthesis plus 2 O subscript 2 end subscript left parenthesis g right parenthesis times C O subscript 2 end subscript left parenthesis g right parenthesis plus 2 stack stack H with _ below subscript 2 end subscript O with _ below left parenthesis l right parenthesis plus 210.8Kcal thheaoformatioomethane in Kcawilbe-

chemistry-General
General
maths-

A set of (2n + 1) elements is given. The no. of subsets of the set which contain atmost n elements -

No. of subsets
= 2n+1C0 + 2n+1C1 + 2n+1C2 +...........+ 2n+1Cn = N
(let) We have 2N = 2n+1C0 + 2n+1C1 .................2n+1C2n +1
2N = 22n + 1
N = 22n

A set of (2n + 1) elements is given. The no. of subsets of the set which contain atmost n elements -

maths-General
No. of subsets
= 2n+1C0 + 2n+1C1 + 2n+1C2 +...........+ 2n+1Cn = N
(let) We have 2N = 2n+1C0 + 2n+1C1 .................2n+1C2n +1
2N = 22n + 1
N = 22n
General
chemistry-

  under Hofmann conditions will give :

  under Hofmann conditions will give :

chemistry-General
General
chemistry-

 Which step is rate determining ste

 Which step is rate determining ste

chemistry-General
General
chemistry-

Which reagent (X) is used to convert I to II

Which reagent (X) is used to convert I to II

chemistry-General
General
chemistry-

An organic compound A upon reacting with NH3 gives B. On heating B gives C. C in presence of KOH reacts with Br2 to given CH3CH2NH2 . A is :

An organic compound A upon reacting with NH3 gives B. On heating B gives C. C in presence of KOH reacts with Br2 to given CH3CH2NH2 . A is :

chemistry-General
General
Maths-

If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r squared s to the power of 4 t squared, then the number of ordered pair (p, q) is-

Step by step solution :
It is given that L C M space left parenthesis p comma q right parenthesis space equals space r squared s to the power of 4 t squared
That is, at least one of p and q must have r squareds to the power of 4 and t squared in their prime factorizations.
Now, consider the cases for power of r as follows:
Case 1 : p contains  s to the power of 4 then q has  s to the power of K with K = (0,1,2,3)
That is, number of ways=4.
Case 2 : q contains  s to the power of 4 then q has  s to the power of K with K = (0,1,2,3)
That is, number of ways=4.
Case 3 : Both p and q contains s to the power of 4 
Then, number of ways=1.
Therefore, exponent of r may be chosen in 2+2+1=5 ways.
Similarly, exponent of s may be chosen in 4+4+1=9 ways and 
Exponent of s may be chosen in 2+2+1=5 ways
Thus, the total number of ways is:
 5 space cross times 9 space cross times 5 space equals space 225
Hence, the number of the ordered pair (pq) is 225.

If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r squared s to the power of 4 t squared, then the number of ordered pair (p, q) is-

Maths-General
Step by step solution :
It is given that L C M space left parenthesis p comma q right parenthesis space equals space r squared s to the power of 4 t squared
That is, at least one of p and q must have r squareds to the power of 4 and t squared in their prime factorizations.
Now, consider the cases for power of r as follows:
Case 1 : p contains  s to the power of 4 then q has  s to the power of K with K = (0,1,2,3)
That is, number of ways=4.
Case 2 : q contains  s to the power of 4 then q has  s to the power of K with K = (0,1,2,3)
That is, number of ways=4.
Case 3 : Both p and q contains s to the power of 4 
Then, number of ways=1.
Therefore, exponent of r may be chosen in 2+2+1=5 ways.
Similarly, exponent of s may be chosen in 4+4+1=9 ways and 
Exponent of s may be chosen in 2+2+1=5 ways
Thus, the total number of ways is:
 5 space cross times 9 space cross times 5 space equals space 225
Hence, the number of the ordered pair (pq) is 225.
General
Maths-

A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

We have, number of ways of selecting vertical sides  1+3+5+....+(2m1which is equal to m squared
Similarly, the number of ways of selecting horizontal sides  1+3+5+....+(2n1which is equal to n squared
So, the number of rectangles possible with odd side length =
 number of ways of selecting vertical sides × number of ways of selecting horizontal sides.
Therefore, we get the number of rectangles possible with odd side length = m squared cross timesn squared

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.

A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

Maths-General
We have, number of ways of selecting vertical sides  1+3+5+....+(2m1which is equal to m squared
Similarly, the number of ways of selecting horizontal sides  1+3+5+....+(2n1which is equal to n squared
So, the number of rectangles possible with odd side length =
 number of ways of selecting vertical sides × number of ways of selecting horizontal sides.
Therefore, we get the number of rectangles possible with odd side length = m squared cross timesn squared

Error converting from MathML to accessible text.
.
General
chemistry-

Observe the esterification mechanisms for primary and tertiary alcohols
Type-1:

Type-2: 
Mechanism: 

In the above reaction (P) and (Q) are respectively :

Observe the esterification mechanisms for primary and tertiary alcohols
Type-1:

Type-2: 
Mechanism: 

In the above reaction (P) and (Q) are respectively :

chemistry-General
General
physics-

Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically work done is

   

Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically work done is

physics-General
   
General
physics-

A thermodynamic system is taken from state A to B along ACB and is brought back to Along BDA as Shown in the PV diagram. The net work done during the complete cycle is given by the area.

   

A thermodynamic system is taken from state A to B along ACB and is brought back to Along BDA as Shown in the PV diagram. The net work done during the complete cycle is given by the area.

physics-General