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Paladinen [302]
3 years ago
15

A team of researchers is investigating a new carbon-rich molecule important to life. They find evidence for the presence of both

cis and trans isomers of the molecule in a mixture of near-pure molecules. Which of the following must be contained in the new molecule?
A. A double bond
B. An aliphatic hydrocarbon
C. A hydrocarbon structure
Chemistry
1 answer:
ser-zykov [4K]3 years ago
6 0

Answer:

A. A double bond

Explanation:

Trans and cis Isomerism is a form of isomerism that specifically refers to alkenes, which is to say, aliphatic chains containing a <em>double bond</em> between two carbon atoms. Technically speaking, options B. and C. are both correct, because an alkene contains aliphatic hydrocarbons. However, neither of those options necessarily relate to cis and trans isomers.

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Explanation:

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The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C
Kay [80]

<u>Answer:</u> The percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 77.5% and 22.5% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{17}^{35}\textrm{Cl} isotope be 'x'. So, fractional abundance of _{17}^{37}\textrm{Cl} isotope will be '1 - x'

  • <u>For _{17}^{35}\textrm{Cl} isotope:</u>

Mass of _{17}^{35}\textrm{Cl} isotope = 35 amu

Fractional abundance of _{17}^{35}\textrm{Cl} isotope = x

  • <u>For _{17}^{37}\textrm{Cl} isotope:</u>

Mass of _{17}^{37}\textrm{Cl} isotope = 37 amu

Fractional abundance of _{17}^{37}\textrm{Cl} isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775

Percentage abundance of _{17}^{35}\textrm{Cl} isotope = 0.775\times 100=77.5\%

Percentage abundance of _{17}^{37}\textrm{Cl} isotope = (1-0.775)=0.225\times 100=22.5\%

Hence, the percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 77.5% and 22.5% respectively.

6 0
3 years ago
A parachute on a racing dragster opens and changes the speed of the car from 85 m/s to 45 m/s in a period of 4.5 seconds. What i
LiRa [457]

Answer:

- 8.89 m/s^{2}

Explanation:

\frac{(Vf-Vo)}{time} =  \frac{(45-85 m/s)}{4.5s} = - 8.89 m/s^{2}

7 0
3 years ago
Read 2 more answers
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