Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Chemistry-

General

Easy

Question

The molecules
are:

enantiomers
diastereomers
structural isomers
none of these

The correct answer is: enantiomers

Related Questions to study

(A) The boiling point 6f cis-l,2-dichloroethene is higher than corresponding trans-isomer.
(R) The dipole moment of cis-l,2-dichloroethene is higher than trans-isomers.

(A) The boiling point 6f cis-l,2-dichloroethene is higher than corresponding trans-isomer.
(R) The dipole moment of cis-l,2-dichloroethene is higher than trans-isomers.

Chemistry-General

(A) Metamers can also be chain or position isomers :
(R) The term tautomerism was introduced to explain the reactivity of a substance . according to two possible structures.

(A) Metamers can also be chain or position isomers :
(R) The term tautomerism was introduced to explain the reactivity of a substance . according to two possible structures.

Chemistry-General

(A) Following amide exist in two Structural forms:
I)
II)
(R) Rotation about carbon nitrogen bond is restricted due to resonance.·

(A) Following amide exist in two Structural forms:
I)
II)
(R) Rotation about carbon nitrogen bond is restricted due to resonance.·

Chemistry-General

parallel

If $square root of 2$ cosA = cosB + cos³B, $square root of 2$ sinA = sinB - sin³B. Then |sin(A-B)| =

If $square root of 2$ cosA = cosB + cos³B, $square root of 2$ sinA = sinB - sin³B. Then |sin(A-B)| =

Which of the following is chiral?

Which of the following is chiral?

Chemistry-General

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?

parallel

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

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

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

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

parallel

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 $open parentheses c o t invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o t invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses c o t invisible function application alpha subscript n end subscript close parentheses equals 1$

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

parallel

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

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

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.