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Natali [406]
4 years ago
9

Rank the following from highest to lowest priority according to the Cahn-Ingold-Prelog system: -CH2CH3, -CHCH2, -CCH, -CH3.

Chemistry
1 answer:
mario62 [17]4 years ago
7 0

Answer:

The order will be:

CCH > CHCH₂ > CH₂CH₃> CH₃

Explanation:

According to Cahn-Ingold-Prelog system we rank the groups based on the atomic number of directly attached atom with the chiral carbon.

For example: between C and H, we rank Carbon first.

If the same atoms are attached for different groups then we prioritized based on the second element with highest atomic number.

For example:

Among CH₃ and C₂H₅, the priority will be given to C₂H₅.

If an atom is double or triple bonded to the directly attached atom then each pi bond is considered to be a new atom.

Hence CH=CH₂ means, that there are two carbons attached to CH carbon.

So the order based on above selection rules will be:

CCH > CHCH₂ > CH₂CH₃> CH₃

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The dichloromethane (DCM) has less density than water and also the polarity of water is much more than DCM. So the mixture of water and dichloromethane will always be a heterogeneous mixture. In the mixture dichloromethane will be always up of the water layer. The volume of the separatory funnel which contains the mixture of DCM and water must have to be more than the total volume of the liquids thus the volume of the funnel will be more than (50+50) = 100mL.

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A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

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