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pychu [463]
2 years ago
9

Your lab partner named this compound 3-methyl-4-n-propylhexane, but that is not correct.

Chemistry
1 answer:
loris [4]2 years ago
4 0

<u>Answer:</u> The correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

<u>Explanation:</u>

The IUPAC nomenclature of alkanes are given as follows:

  • Select the longest possible carbon chain.
  • For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
  • A suffix '-ane' is added at the end of the name.
  • If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.

We are given:

An alkane having chemical name as 3-methyl-4-n-propylhexane. This will not be the correct name of the alkane because the longest possible carbon chain has 7 Carbon atoms, not 6 carbon atoms

The image of the given alkane is shown in the image below.

Hence, the correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

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For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
Nikolay [14]

Explanation:

Molar mass of KClO_{3} = 39.1 + 35.5 + 3(16.0) = 122.6 g

Molar mass of KCl = 39.1 + 35.5 = 74.6 g

Molar mass of O_{2} = 32.0 g

According to the equation, 2 moles of KClO_{3} reacts to give 3 moles of oxygen.

Therefore, 2 (122.6) = 245.2 g of KClO_{3} will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of KClO_{3} gives 96.0 g of oxygen.

(a) Calculate the amount of oxygen given by 2.72 g of KClO_{3} as follows.

       \frac {96g}{245.2g} \times 2.72 g = 1.06 g of O_{2}


(b) Calculate the amount of oxygen given by 0.361 g of KClO_{3} as follows.

    \frac {96g}{245.2g} \times 0.361 g = 0.141 g of O_{2}


c) Calculate the amount of oxygen given by 83.6 kg KClO_{3} as follows.  

    \frac {96g}{245.2g} \times 83.6 g = 32.731 kg of O_{2}

Convert kg into grams as follows.

    \frac{32.731 kg}{1 kg} \times 1000 g = 32731 g of O_{2}


(d) Calculate the amount of oxygen given by 22.5 mg of KClO_{3} as follows.  

    \frac {96g}{245.2g} \times 22.5 mg = 8.79 mg

Convert mg into grams as follows.

  \frac{8.79 mg}{1 mg}\times 10^{-3} g = 8.79 \times 10^{-3} g of O_{2}

8 0
3 years ago
A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent
Degger [83]

Answer :  The % of (+) limonene isomer = 79%


                The % of (-) limonene isomer = 0%


                The % of enantiomeric excess = 58%


Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.


Given,


% of pure limonene enantiomer = The % of (+) limonene isomer = 79%


Therefore, The % of (-) limonene isomer = 0%


Formula used :  

\%(+)\text{ isomer}=\frac{ee}{2}+50\%


Where,         ee → enantiomeric excess


Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


     {ee}=\frac{\%(+)-50\%}{2}


            =\frac{79\%-50\%}{2}


              = 58%



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