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Chemistry-

General

Easy

Question

In the reaction,

the compound (C) is:

${C H}_{3} - {C H}_{2} {C H}_{2} I$

${C H}_{3} - C H = C H I$

The correct answer is: $C H subscript 3 end subscript minus C H subscript 2 end subscript C H subscript 2 end subscript I$

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If $open vertical bar table row a b 0 row 0 a b row b 0 a end table close vertical bar equals 0$ , then

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$not stretchy integral e to the power of sin invisible function application x end exponent. fraction numerator x cos to the power of 3 end exponent invisible function application x minus sin invisible function application x over denominator cos to the power of 2 end exponent invisible function application x end fraction d x$ is equal to

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