Chemistry-
General
Easy

Question

In the given reaction sequence,

the final product [BJ will be:

  1.    
  2.    
  3.    
  4.    

The correct answer is:

Related Questions to study

General
Chemistry-

Arrange the following compounds in the decreasing order of reactivity for hydrolysis reaction:
1) C subscript 6 end subscript H subscript 5 end subscript C O C l
2)
3)
4)

Arrange the following compounds in the decreasing order of reactivity for hydrolysis reaction:
1) C subscript 6 end subscript H subscript 5 end subscript C O C l
2)
3)
4)

Chemistry-General
General
Chemistry-

 The product obtained in above reaction will be :

 The product obtained in above reaction will be :

Chemistry-General
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The percentage of nitrogen in urea is:

The percentage of nitrogen in urea is:

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In a packet of 40 pens, 12 are red. So, what % age are red pens?

In a packet of 40 pens, 12 are red. So, what % age are red pens?

Maths-General
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Chemistry-

Consider the following compounds, which of these will release CO2 with 5% NaHCO3 ?
i)
ii)
iii) 

Consider the following compounds, which of these will release CO2 with 5% NaHCO3 ?
i)
ii)
iii) 

Chemistry-General
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Chemistry-

Identify the product' Y' in the following reaction sequence:

Identify the product' Y' in the following reaction sequence:

Chemistry-General
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General
Chemistry-

In homogeneous catalytic reactions, there are three alternative paths A,B and C (shown in the figure) which one of the following indicates the relative ease with which the reaction can take place?

In homogeneous catalytic reactions, there are three alternative paths A,B and C (shown in the figure) which one of the following indicates the relative ease with which the reaction can take place?

Chemistry-General
General
Chemistry-

In a spontaneous adsorption process

In a spontaneous adsorption process

Chemistry-General
General
Chemistry-

Graph between log open parentheses fraction numerator x over denominator m end fraction close parentheses and log P is a straight line at angle 0 45 with intercept OA as shown Hence ,blank open parentheses fraction numerator x over denominator m end fraction close parentheses at a pressure of 2 atm is

Graph between log open parentheses fraction numerator x over denominator m end fraction close parentheses and log P is a straight line at angle 0 45 with intercept OA as shown Hence ,blank open parentheses fraction numerator x over denominator m end fraction close parentheses at a pressure of 2 atm is

Chemistry-General
parallel
General
Maths-

If 0 less than alpha less than fraction numerator pi over denominator 2 end fraction and  alpha to the power of pi left parenthesis cosec invisible function application 2 alpha right parenthesis times cos invisible function application 2 alpha plus pi left parenthesis cosec invisible function application 2 alpha sin invisible function application 2 alpha right parenthesis end exponent then

If 0 less than alpha less than fraction numerator pi over denominator 2 end fraction and  alpha to the power of pi left parenthesis cosec invisible function application 2 alpha right parenthesis times cos invisible function application 2 alpha plus pi left parenthesis cosec invisible function application 2 alpha sin invisible function application 2 alpha right parenthesis end exponent then

Maths-General
General
Maths-

Assertion : For a equals fraction numerator 1 over denominator square root of 3 end fraction the volume of the parallel piped formed by vectors i with ˆ on top plus a j with ˆ on top comma a i with ˆ on top plus j with ˆ on top plus k with ˆ on top and j with ˆ on top plus a k with ˆ on top is maximum (The vectors form a right-handed system)
Reason: The volume of the parallel piped having three coterminous edges stack a with ‾ on top comma stack b with ‾ on top and stack c with ‾ on top

For such questions, we should know the formula of a parallelepiped. We should also know how to take a scalar product.

Assertion : For a equals fraction numerator 1 over denominator square root of 3 end fraction the volume of the parallel piped formed by vectors i with ˆ on top plus a j with ˆ on top comma a i with ˆ on top plus j with ˆ on top plus k with ˆ on top and j with ˆ on top plus a k with ˆ on top is maximum (The vectors form a right-handed system)
Reason: The volume of the parallel piped having three coterminous edges stack a with ‾ on top comma stack b with ‾ on top and stack c with ‾ on top

Maths-General

For such questions, we should know the formula of a parallelepiped. We should also know how to take a scalar product.

General
Maths-

Assertion (A): Let a with rightwards arrow on top equals 3 i with ˆ on top minus j with ˆ on top comma b with rightwards arrow on top equals 2 i with ˆ on top plus j with ˆ on top minus 3 k with ˆ on top. If stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript such that stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top is possible, then b with rightwards arrow on top subscript 2 equals i with ˆ on top plus 3 j with ˆ on top minus 3 k with ˆ on top.
Reason (R): If stack a with rightwards arrow on top and stack b with rightwards arrow on top are non-zero, non-collinear vectors, then stack b with rightwards arrow on top can be expressed as stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript where stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top

Assertion (A): Let a with rightwards arrow on top equals 3 i with ˆ on top minus j with ˆ on top comma b with rightwards arrow on top equals 2 i with ˆ on top plus j with ˆ on top minus 3 k with ˆ on top. If stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript such that stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top is possible, then b with rightwards arrow on top subscript 2 equals i with ˆ on top plus 3 j with ˆ on top minus 3 k with ˆ on top.
Reason (R): If stack a with rightwards arrow on top and stack b with rightwards arrow on top are non-zero, non-collinear vectors, then stack b with rightwards arrow on top can be expressed as stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript where stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top

Maths-General
parallel
General
Maths-

Statement negative 1 colon If a comma b comma c are distinct non-negative numbers and the vectors â plus a j with ˆ on top plus c k with ˆ on top comma i with ˆ on top plus k with ˆ on top and c i with ˆ on top plus c j with ˆ on top plus b k with ˆ on top are coplanar then c is arithmetic mean of a and b.
Statement -2: Parallel vectors have proportional direction ratios.

Statement negative 1 colon If a comma b comma c are distinct non-negative numbers and the vectors â plus a j with ˆ on top plus c k with ˆ on top comma i with ˆ on top plus k with ˆ on top and c i with ˆ on top plus c j with ˆ on top plus b k with ˆ on top are coplanar then c is arithmetic mean of a and b.
Statement -2: Parallel vectors have proportional direction ratios.

Maths-General
General
Maths-

If a with ‾ on top equals i plus j minus k comma b with ‾ on top equals 2 i plus j minus 3 k and stack r with ‾ on top is a vector satisfying 2 stack r with ‾ on top plus stack r with ‾ on top cross times stack a with ‾ on top equals stack b with ‾ on top.
Assertion left parenthesis A right parenthesis colon stack r with ‾ on top can be expressed in terms of stack a with ‾ on top comma stack b with ‾ on top and stack a with ‾ on top cross times stack b with ‾ on top.
Reason left parenthesis R right parenthesis colon r with ‾ on top equals 1 over 7 left parenthesis 7 i plus 5 j minus 9 k plus a with ‾ on top cross times b with ‾ on top right parenthesis

If a with ‾ on top equals i plus j minus k comma b with ‾ on top equals 2 i plus j minus 3 k and stack r with ‾ on top is a vector satisfying 2 stack r with ‾ on top plus stack r with ‾ on top cross times stack a with ‾ on top equals stack b with ‾ on top.
Assertion left parenthesis A right parenthesis colon stack r with ‾ on top can be expressed in terms of stack a with ‾ on top comma stack b with ‾ on top and stack a with ‾ on top cross times stack b with ‾ on top.
Reason left parenthesis R right parenthesis colon r with ‾ on top equals 1 over 7 left parenthesis 7 i plus 5 j minus 9 k plus a with ‾ on top cross times b with ‾ on top right parenthesis

Maths-General
General
Maths-

If stack a with minus on top comma stack b with minus on top are non-zero vectors such that vertical line stack a with minus on top plus stack b with minus on top vertical line equals vertical line stack a with minus on top minus 2 stack b with minus on top vertical line then
Assertion left parenthesis A right parenthesis : Least value of stack a with ‾ on top times stack b with ‾ on top plus fraction numerator 4 over denominator vertical line stack b with ‾ on top vertical line to the power of 2 end exponent plus 2 end fraction is 2 square root of 2 minus 1
Reason (R): The expression stack a with minus on top times stack b with minus on top plus fraction numerator 4 over denominator vertical line stack b with minus on top vertical line to the power of 2 end exponent plus 2 end fraction is least when magnitude of stack b with minus on top is square root of 2 t a n invisible function application open parentheses fraction numerator pi over denominator 8 end fraction close parentheses end root

If stack a with minus on top comma stack b with minus on top are non-zero vectors such that vertical line stack a with minus on top plus stack b with minus on top vertical line equals vertical line stack a with minus on top minus 2 stack b with minus on top vertical line then
Assertion left parenthesis A right parenthesis : Least value of stack a with ‾ on top times stack b with ‾ on top plus fraction numerator 4 over denominator vertical line stack b with ‾ on top vertical line to the power of 2 end exponent plus 2 end fraction is 2 square root of 2 minus 1
Reason (R): The expression stack a with minus on top times stack b with minus on top plus fraction numerator 4 over denominator vertical line stack b with minus on top vertical line to the power of 2 end exponent plus 2 end fraction is least when magnitude of stack b with minus on top is square root of 2 t a n invisible function application open parentheses fraction numerator pi over denominator 8 end fraction close parentheses end root

Maths-General
parallel

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