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Chemistry-

General

Easy

Question

Which of the following is chiral?

${C l C H}_{2} {C H}_{2} {C H}_{2} {C H}_{2} {C H}_{3}$

$C l C H = {C H C H}_{2} {C H}_{2} {C H}_{3}$

The correct answer is:

Related Questions to study

statement I: Minimum value of $left parenthesis 3 s i n invisible function application theta minus 4 c o s invisible function application theta right parenthesis left parenthesis 3 c o s invisible function application theta plus 4 s i n invisible function application theta right parenthesis$ is $negative fraction numerator 25 over denominator 2 end fraction$
statement II: Minimum value of $9 t a n to the power of 2 end exponent invisible function application x plus 16 c o t to the power of 2 end exponent invisible function application x$ is 24 Which of the above statements is correct?

statement I: Minimum value of $left parenthesis 3 s i n invisible function application theta minus 4 c o s invisible function application theta right parenthesis left parenthesis 3 c o s invisible function application theta plus 4 s i n invisible function application theta right parenthesis$ is $negative fraction numerator 25 over denominator 2 end fraction$
statement II: Minimum value of $9 t a n to the power of 2 end exponent invisible function application x plus 16 c o t to the power of 2 end exponent invisible function application x$ is 24 Which of the above statements is correct?

$open curly brackets open x element of R divided by l o g open square brackets open parentheses 1.6 close parentheses to the power of 1 minus x 2 end exponent minus open parentheses 0.625 close parentheses to the power of 6 open parentheses 1 plus x close parentheses end exponent close square brackets element of R close curly brackets equals close$

$open curly brackets open x element of R divided by l o g open square brackets open parentheses 1.6 close parentheses to the power of 1 minus x 2 end exponent minus open parentheses 0.625 close parentheses to the power of 6 open parentheses 1 plus x close parentheses end exponent close square brackets element of R close curly brackets equals close$

The extreme values of $4 c o s invisible function application open parentheses x to the power of 2 end exponent close parentheses c o s invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus x to the power of 2 end exponent close parentheses c o s invisible function application open parentheses fraction numerator pi over denominator 3 end fraction minus x to the power of 2 end exponent close parentheses$ over $R$ are

The extreme values of $4 c o s invisible function application open parentheses x to the power of 2 end exponent close parentheses c o s invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus x to the power of 2 end exponent close parentheses c o s invisible function application open parentheses fraction numerator pi over denominator 3 end fraction minus x to the power of 2 end exponent close parentheses$ over $R$ are

parallel

The period of $open parentheses t a n invisible function application theta minus fraction numerator 1 over denominator 3 end fraction t a n to the power of 3 end exponent invisible function application theta close parentheses open parentheses fraction numerator 1 over denominator 3 end fraction minus t a n to the power of 2 end exponent invisible function application theta close parentheses to the power of negative 1 end exponent$ , where $t a n to the power of 2 end exponent invisible function application theta not equal to fraction numerator 1 over denominator 3 end fraction$ is

The period of $open parentheses t a n invisible function application theta minus fraction numerator 1 over denominator 3 end fraction t a n to the power of 3 end exponent invisible function application theta close parentheses open parentheses fraction numerator 1 over denominator 3 end fraction minus t a n to the power of 2 end exponent invisible function application theta close parentheses to the power of negative 1 end exponent$ , where $t a n to the power of 2 end exponent invisible function application theta not equal to fraction numerator 1 over denominator 3 end fraction$ is

The minimum value of $27 t a n to the power of 2 end exponent invisible function application theta plus 3 c o t to the power of 2 end exponent invisible function application theta$ is

The minimum value of $27 t a n to the power of 2 end exponent invisible function application theta plus 3 c o t to the power of 2 end exponent invisible function application theta$ is

The maximum value of $open parentheses c o s invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o s invisible function application alpha subscript 2 end subscript close parentheses$ $open parentheses c o s invisible function application alpha subscript n end subscript close parentheses$ , under the restrictions $0 less or equal than alpha subscript 1 end subscript comma alpha subscript 2 end subscript comma horizontal ellipsis. alpha subscript n end subscript less or equal than fraction numerator pi over denominator 2 end fraction$ and $open parentheses c o t invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o t invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses c o t invisible function application alpha subscript n end subscript close parentheses equals 1$

The maximum value of $open parentheses c o s invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o s invisible function application alpha subscript 2 end subscript close parentheses$ $open parentheses c o s invisible function application alpha subscript n end subscript close parentheses$ , under the restrictions $0 less or equal than alpha subscript 1 end subscript comma alpha subscript 2 end subscript comma horizontal ellipsis. alpha subscript n end subscript less or equal than fraction numerator pi over denominator 2 end fraction$ and $open parentheses c o t invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o t invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses c o t invisible function application alpha subscript n end subscript close parentheses equals 1$

parallel

The maximum value of $s i n invisible function application left parenthesis c o s invisible function application x right parenthesis$ is equal to

The maximum value of $s i n invisible function application left parenthesis c o s invisible function application x right parenthesis$ is equal to

Range of $c o s to the power of 4 end exponent invisible function application x minus s i n to the power of 4 end exponent invisible function application x$ is

Range of $c o s to the power of 4 end exponent invisible function application x minus s i n to the power of 4 end exponent invisible function application x$ is

The maximum value of $16 c o s to the power of 5 end exponent invisible function application theta minus 20 c o s to the power of 3 end exponent invisible function application theta plus 5 c o s invisible function application theta$ is

The maximum value of $16 c o s to the power of 5 end exponent invisible function application theta minus 20 c o s to the power of 3 end exponent invisible function application theta plus 5 c o s invisible function application theta$ is

parallel

The range of $c o s invisible function application x plus 4 square root of 2 s i n invisible function application open parentheses x minus fraction numerator pi over denominator 4 end fraction close parentheses plus 6$ is

The range of $c o s invisible function application x plus 4 square root of 2 s i n invisible function application open parentheses x minus fraction numerator pi over denominator 4 end fraction close parentheses plus 6$ is

Period of $s i n invisible function application left parenthesis s i n invisible function application x right parenthesis plus s i n invisible function application left parenthesis c o s invisible function application x right parenthesis$ is

Period of $s i n invisible function application left parenthesis s i n invisible function application x right parenthesis plus s i n invisible function application left parenthesis c o s invisible function application x right parenthesis$ is

Period of $t a n invisible function application open parentheses x plus 8 x plus 27 x plus midline horizontal ellipsis plus n to the power of 3 end exponent x close parentheses$ is

Period of $t a n invisible function application open parentheses x plus 8 x plus 27 x plus midline horizontal ellipsis plus n to the power of 3 end exponent x close parentheses$ is

parallel

$A equals s i n to the power of 8 end exponent invisible function application theta plus c o s to the power of 14 end exponent invisible function application theta$ then for all values of $theta$

$A equals s i n to the power of 8 end exponent invisible function application theta plus c o s to the power of 14 end exponent invisible function application theta$ then for all values of $theta$

Period of $s i n invisible function application left parenthesis x plus 2 x plus 3 x plus midline horizontal ellipsis plus n x right parenthesis$ is

Period of $s i n invisible function application left parenthesis x plus 2 x plus 3 x plus midline horizontal ellipsis plus n x right parenthesis$ is

The function $f colon R rightwards arrow R g colon R rightwards arrow R$ are defined as follows $f left parenthesis x right parenthesis equals 0 left parenthesis x$ is rational $right parenthesis comma g left parenthesis x right parenthesis equals negative 1 left parenthesis x$ is rational $right parenthesis equals$ $1 left parenthesis x$ is irrational $equals 0$ ( $x$ is irrational $right parenthesis$ then fog $left parenthesis pi right parenthesis plus g o f invisible function application left parenthesis e right parenthesis$

The function $f colon R rightwards arrow R g colon R rightwards arrow R$ are defined as follows $f left parenthesis x right parenthesis equals 0 left parenthesis x$ is rational $right parenthesis comma g left parenthesis x right parenthesis equals negative 1 left parenthesis x$ is rational $right parenthesis equals$ $1 left parenthesis x$ is irrational $equals 0$ ( $x$ is irrational $right parenthesis$ then fog $left parenthesis pi right parenthesis plus g o f invisible function application left parenthesis e right parenthesis$

parallel

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