Chemistry-
General
Easy

Question

IUPAC name of ethers is

  1. alkoxy alkane    
  2. alkanol    
  3. alkanal    
  4. alkyl alkanoate    

The correct answer is: alkoxy alkane

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The functional group present in acylchlorides is

The functional group present in acylchlorides is

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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
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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
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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

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

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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

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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

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

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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

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

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Chemistry-General
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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

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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

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

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A mixture of benzene and toluene can be separated by

A mixture of benzene and toluene can be separated by

Chemistry-General
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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
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