Chemistry-
General
Easy

Question

The functional group present in acylchlorides is

  1. -Cl    
  2.    
  3.    
  4.    

The correct answer is:

Related Questions to study

General
Maths-

If Z subscript 1 end subscript comma Z subscript 2 end subscript comma Z subscript 3 end subscript are complex numbers such that A open parentheses Z subscript 1 end subscript close parentheses comma B open parentheses Z subscript 2 end subscript close parentheses comma C open parentheses Z subscript 3 end subscript close parentheses are vertices of a triangle ABC,
angle A equals theta comma fraction numerator A C over denominator A B end fraction equals lambda and open parentheses 1 plus lambda to the power of 2 end exponent minus 2 lambda c o s invisible function application theta close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 3 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses open parentheses lambda to the power of 2 end exponent minus lambda c o s invisible function application theta close parentheses Z subscript 2 end subscript equals left parenthesis 1 minus lambda c o s invisible function application theta right parenthesis Z subscript 3 end subscript close parentheses plus 2 lambda c o s invisible function application theta Z subscript 2 end subscript Z subscript 3 end subscript then
If open parentheses 1 plus lambda to the power of 2 end exponent close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 2 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses lambda to the power of 2 end exponent Z subscript 2 end subscript plus Z subscript 3 end subscript close parentheses then the triangle is

If Z subscript 1 end subscript comma Z subscript 2 end subscript comma Z subscript 3 end subscript are complex numbers such that A open parentheses Z subscript 1 end subscript close parentheses comma B open parentheses Z subscript 2 end subscript close parentheses comma C open parentheses Z subscript 3 end subscript close parentheses are vertices of a triangle ABC,
angle A equals theta comma fraction numerator A C over denominator A B end fraction equals lambda and open parentheses 1 plus lambda to the power of 2 end exponent minus 2 lambda c o s invisible function application theta close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 3 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses open parentheses lambda to the power of 2 end exponent minus lambda c o s invisible function application theta close parentheses Z subscript 2 end subscript equals left parenthesis 1 minus lambda c o s invisible function application theta right parenthesis Z subscript 3 end subscript close parentheses plus 2 lambda c o s invisible function application theta Z subscript 2 end subscript Z subscript 3 end subscript then
If open parentheses 1 plus lambda to the power of 2 end exponent close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 2 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses lambda to the power of 2 end exponent Z subscript 2 end subscript plus Z subscript 3 end subscript close parentheses then the triangle is

Maths-General
General
Maths-

The general procedure for solving inequation containing modulus functions is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach.
For example:
i) The inequation f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater than 0 has no solution.
ii) Solving the inequaiton f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater than 0 is equivalent to solving the equations 0 less than f left parenthesis x right parenthesis less than infinity
iii) Solving the inequation vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater than 0is equivalent to solving the equations negative infinity less than f left parenthesis x right parenthesis less than 0
iv) Solving the inequation f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0is equivalent to solving the equations 0 less or equal than f left parenthesis x right parenthesis less than infinity
v) The inequation f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0is true for all x element of domain of f left parenthesis x right parenthesis
vi) The inequation vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater or equal than 0is true for all x element of domain of f left parenthesis x right parenthesis
Set of all real values of x for which fraction numerator 1 over denominator square root of left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis ∣ negative left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis end root end fraction is defined, is

The general procedure for solving inequation containing modulus functions is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach.
For example:
i) The inequation f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater than 0 has no solution.
ii) Solving the inequaiton f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater than 0 is equivalent to solving the equations 0 less than f left parenthesis x right parenthesis less than infinity
iii) Solving the inequation vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater than 0is equivalent to solving the equations negative infinity less than f left parenthesis x right parenthesis less than 0
iv) Solving the inequation f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0is equivalent to solving the equations 0 less or equal than f left parenthesis x right parenthesis less than infinity
v) The inequation f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0is true for all x element of domain of f left parenthesis x right parenthesis
vi) The inequation vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater or equal than 0is true for all x element of domain of f left parenthesis x right parenthesis
Set of all real values of x for which fraction numerator 1 over denominator square root of left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis ∣ negative left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis end root end fraction is defined, is

Maths-General
General
Maths-

Ten different letters of English alphabet are given. Words with five letters are formed from these given letters. Then, the number of words which have at least one letter repeated is

Ten different letters of English alphabet are given. Words with five letters are formed from these given letters. Then, the number of words which have at least one letter repeated is

Maths-General
parallel
General
Maths-

If two real valued functions f(x) & g(x) are given, for fog(x), g(x) is substituted in place of x in f(x) taking care of the domain of fog(x) which is subset of Domain of g(x), only those values of x such that g(x) lies in the Domain of f. Now, If f left parenthesis x right parenthesis equals s i n to the power of 2 end exponent invisible function application x plus s i n to the power of 2 end exponent invisible function application open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses plus c o s invisible function application x times c o s invisible function application open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses text  and  end text g open parentheses fraction numerator 5 over denominator 4 end fraction close parentheses equals 1 then g of (x) is

If two real valued functions f(x) & g(x) are given, for fog(x), g(x) is substituted in place of x in f(x) taking care of the domain of fog(x) which is subset of Domain of g(x), only those values of x such that g(x) lies in the Domain of f. Now, If f left parenthesis x right parenthesis equals s i n to the power of 2 end exponent invisible function application x plus s i n to the power of 2 end exponent invisible function application open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses plus c o s invisible function application x times c o s invisible function application open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses text  and  end text g open parentheses fraction numerator 5 over denominator 4 end fraction close parentheses equals 1 then g of (x) is

Maths-General
General
Maths-

where x, y, t R If t (5, k], then greatest value of k for which x is one-one function of t is

where x, y, t R If t (5, k], then greatest value of k for which x is one-one function of t is

Maths-General
General
Maths-

STATEMENT-I : If [4x] = [x], then the largest set of values of x is. open square brackets negative fraction numerator 1 over denominator 4 end fraction comma fraction numerator 1 over denominator 4 end fraction close parentheses and
STATEMENT-II : If [nx] = [x], n element of I to the power of plus end exponent then the largest set of values of x is open square brackets negative fraction numerator 1 over denominator n end fraction comma fraction numerator 1 over denominator n end fraction close parentheses

STATEMENT-I : If [4x] = [x], then the largest set of values of x is. open square brackets negative fraction numerator 1 over denominator 4 end fraction comma fraction numerator 1 over denominator 4 end fraction close parentheses and
STATEMENT-II : If [nx] = [x], n element of I to the power of plus end exponent then the largest set of values of x is open square brackets negative fraction numerator 1 over denominator n end fraction comma fraction numerator 1 over denominator n end fraction close parentheses

Maths-General
parallel
General
Maths-

x to the power of 4 end exponent minus 4 x minus 1 equals 0 text end texthas

x to the power of 4 end exponent minus 4 x minus 1 equals 0 text end texthas

Maths-General
General
Chemistry-

ClCH2COOH is heated with fuming HNO3 in the presence of AgNO3 in carius tube After filtration and washing a white precipitate is obtained The precipitate is of

ClCH2COOH is heated with fuming HNO3 in the presence of AgNO3 in carius tube After filtration and washing a white precipitate is obtained The precipitate is of

Chemistry-General
General
Chemistry-

Sodium extract gives blood red colour when treated with FeCl3 Formation of blood red colour confirms the presence of

Sodium extract gives blood red colour when treated with FeCl3 Formation of blood red colour confirms the presence of

Chemistry-General
parallel
General
Maths-

STATEMENT-I : The graph ofblank y equals left square bracket square root of 1 minus c o s invisible function application 2 x end root right square bracket consists of only two line segments. Where [.] represent greatest integer function and
STATEMENT-II : The graph ofblank y equals square root of 1 minus c o s invisible function application 2 x end root is

STATEMENT-I : The graph ofblank y equals left square bracket square root of 1 minus c o s invisible function application 2 x end root right square bracket consists of only two line segments. Where [.] represent greatest integer function and
STATEMENT-II : The graph ofblank y equals square root of 1 minus c o s invisible function application 2 x end root is

Maths-General
General
Maths-

Statement 1: Letblank f colon left parenthesis a comma b right parenthesis rightwards arrow left square bracket c comma d right square bracket be a continuous and onto function, then it cannot be one-one. And
Statement 2: If for a continuous and differentiable function f,blank f to the power of ´ end exponent left parenthesis x right parenthesis greater than 0 then f(x) is one-one

Statement 1: Letblank f colon left parenthesis a comma b right parenthesis rightwards arrow left square bracket c comma d right square bracket be a continuous and onto function, then it cannot be one-one. And
Statement 2: If for a continuous and differentiable function f,blank f to the power of ´ end exponent left parenthesis x right parenthesis greater than 0 then f(x) is one-one

Maths-General
General
Maths-

Number of solutions of the equation |3x 2| = 3[x] + 2{x}, where [.] and {.} denote the integral and fractional parts of x respectively, is

Number of solutions of the equation |3x 2| = 3[x] + 2{x}, where [.] and {.} denote the integral and fractional parts of x respectively, is

Maths-General
parallel
General
Chemistry-

A mixture of benzene and toluene can be separated by

A mixture of benzene and toluene can be separated by

Chemistry-General
General
Maths-

The solution set of the inequality log1/5 (2x + 5) + log5 (16 x 2) 1 is

The solution set of the inequality log1/5 (2x + 5) + log5 (16 x 2) 1 is

Maths-General
General
Maths-

Let f be an injective map with domainblank left curly bracket x comma y comma z right curly bracket and rangeblank left curly bracket 1 , 2 comma 3 right curly bracket such that exactly one of the following statements is correct and the remaining are falseblank f left parenthesis x right parenthesis equals 1 comma f left parenthesis y right parenthesis not equal to 1 comma f left parenthesis z right parenthesis not equal to 2 The value ofblank f to the power of negative 1 end exponent left parenthesis 1 right parenthesis blankis

For such questions, we should know concept of injective function.

Let f be an injective map with domainblank left curly bracket x comma y comma z right curly bracket and rangeblank left curly bracket 1 , 2 comma 3 right curly bracket such that exactly one of the following statements is correct and the remaining are falseblank f left parenthesis x right parenthesis equals 1 comma f left parenthesis y right parenthesis not equal to 1 comma f left parenthesis z right parenthesis not equal to 2 The value ofblank f to the power of negative 1 end exponent left parenthesis 1 right parenthesis blankis

Maths-General

For such questions, we should know concept of injective function.

parallel

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