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Triss [41]
3 years ago
11

Reaction of (r)-2-chloro-4-methylhexane with excess nai in acetone gives racemic 2-iodo-4-methylhexane. what is the explanation

that best describes this transformation?
Chemistry
1 answer:
saveliy_v [14]3 years ago
4 0

I believe this question has the following five choices to choose from:

 

>an SN2 reaction has occurred with inversion of configuration

>racemization followed by an S N 2 attack

>an SN1 reaction has taken over resulting in inversion of configuration

>an SN1 reaction has occurred due to carbocation formation

>an SN1 reaction followed by an S N 2 “backside” attack

 

 

The correct answer is:

an SN1 reaction has occurred due to carbocation formation 

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Aqueous concentrated nitric acid is 69% hno3 by weight and has a density of 1.42 g/ml.
OleMash [197]

Answer: -

15.55 M

35.325 molal

Explanation: -

Let the volume of the solution be 1000 mL.

Density of nitric acid = 1.42 g/ mL

Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid

= 1000 mL x 1.42 g/ mL

= 1420 g.

Percentage of HNO₃ = 69%

Amount of HNO₃ = \frac{69} {100} x 1420 g

= 979.8 g

Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol

Number of moles of HNO₃ = \frac{979.8 g}{63 g/ mol}

= 15.55 mol

Molarity is defined as number of moles per 1000 mL

We had taken 1000 mL as volume and found it to contain 15.55 moles.

Molarity of HNO₃ = 15.55 M

Mass of water = Total mass of nitric acid solution - mass of nitric acid

= 1420 - 979.8

= 440.2 g

So we see that 440.2 g of water contains 15.55 moles of HNO₃

Molality is defined as number of moles of HNO₃ present per 1000 g of water.

Molality of HNO₃ = \frac{15.55 x 1000}{440.2}

= 35.325 molal

3 0
3 years ago
The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichl
nevsk [136]

Complete Question

A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?

Answer:

The mass of  caffeine extracted is  P =  39.8 \ mg

Explanation:

From the question above  we are told that

    The  K value is  K =  4.6

     The  mass of the caffeine is  m  = 40 mg

      The  volume of water is  V  = 2 mL

      The volume  of caffeine is  v_c =  2 mL

     The number of times the extraction was done is  n =  3

Generally the mass of  caffeine that will be  extracted is  

           P =  m  *  [\frac{V}{K *  v_c + V} ]^3

substituting values  

           P =  40   *  [\frac{2}{4.6 *  2 + 2} ]^3

           P =  39.8 \ mg

6 0
3 years ago
Based on the balanced equation 2C2H2 + 5O2 → 4CO2 + 2H2O calculate the number of excess reagent units remaining when 52 C2H2 mol
erica [24]

Answer:

20 molecules of oxygen gas remains after the reaction.

Explanation:

2C_2H_2 + 5O_2\rightarrow 4CO_2 + 2H_2O

Molecules of ethyne = 52

Molecules of oxygen gas = 150

According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.

Then 52 molecules of ethyne will react with:

\frac{5}{2}\times 52=130 molecules of oxygen gas.

As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.

Remaining molecules of recessive reagent = 150 - 130 = 20

20 molecules of oxygen gas remains after the reaction.

5 0
3 years ago
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