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leva [86]
2 years ago
14

How many cis trans isomers does gammalinoleic acid has

Chemistry
1 answer:
Tresset [83]2 years ago
6 0

There must be two cis and two trans isomers possible of gammalinoleic acid.

<h3>What are cis and trans isomers?</h3>

The cis-trans isomers are the property shown by the clarity of the carbon which is asymmetric in structure.

The acid has 4 branches attached with different functional groups if two are on the same side then cis is formed and if the other is on a different side then trans is formed.

Therefore, only 2 isomers of cis and 2 isomers of trans are possible for the acid.

Learn more about cis and trans isomers, here:

brainly.com/question/16289453

#SPJ1

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lina2011 [118]

Answer:   " 0.69 g / mL "

____________________________________________

Explanation:  

______________________________________________

The "pre-lab question" given is:

_____________________________________________

"The volume of an unknown liquid is 15 ml, and the mass of the liquid and the graduated cylinder it is in is 55.2g.   If the mass of the graduated cylinder is 44.8g , what is the density of the unknown liquid? " .

____________________________________________

Note:  The volume of the unknown liquid is 15 mL ;  regardless of whether or not the "unknown liquid" is in the graduated cylinder.

The density of the unknown liquid is measured in:  "g / mL " ;  

  that is, "grams per mL" .

____________________________________________

Note:  "D = m / V " ;   that is;  "Density  = mass/ volume" ;

                                  that is:   ["Density = mass per 'unit volume' ]" .

____________________________________________

So;  to find the density, "D" , of the "unknown liquid" ;  we would have to find the "mass" of the "unknown liquid" by "subtracting" :

       the "known mass of liquid when the liquid is not in the cylinder";  that is:  " 44.8g" ;  From:

       the "known combined value of the: 'mass of the liquid PLUS the mass of the cylinder" ;  that is:  "55.2g" ;

→     " 55.2g - 44.8g = 10.4g " .

___________________________________________

So:   " D =  m / V " ; (55.2g - 44.8 g) / 15 mL  " ;

              =  (55.2g - 44.8 g) / 15 mL  " ;

              =  (10.4g) / 15mL  ;

____________________________________________

Note that the "Density" =  mass per "unit volume" ;

→ So:  D = m / V ;  in units of:  " g / mL " (grams per millileter) ;

            = (10.4g) / 15mL  ;

             = [ (10.4) / 15 }  g/ mL ;  

            =  0.693333333333333.... g / mL  ;

       →  We round to "2 (two) significant figures" ;

       → since we have:  "10.4 g / 15 mL " ;

             and 15 mL is considered a measured/experimental value;

             and:  10.4g is considered a measured/experimental value;

→  so, the least precise value;  15 mL (has only 2 (two) significant figures ;

      compared to other value:  10.4g (which has 3 (three) significant figures;

→  so we shall round off to 2 (two) significant figures:

             =  0.69 g / mL

____________________________________________

              →  which is our answer:  " 0.69 g / mL " .

____________________________________________

3 0
3 years ago
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