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kobusy [5.1K]
3 years ago
7

Five structural isomers, or constitutional isomers, have the formula C6H14 . Draw the indicated isomers, grouped by number of ca

rbon atoms in the main chain.Draw two isomers with 5 carbon atoms in the main chain. Be sure to include all hydrogen atoms.

Chemistry
1 answer:
pantera1 [17]3 years ago
3 0

Answer:

See figure 1

Explanation:

For this question, we have to remember that definition of an <u>"isomer"</u> in an isomer we have the<u> same condensed formula</u> (in this case C_6H_14) and <u>different structures</u>. The first structure is a linear structure of 6 carbons (<u>hexane</u>). Then we can have a 5 carbon linear structure in which we have to add a methyl group. This methyl group can be attached to carbon 2 or carbon 3 (<u>2-methylpentane and 3-methylpentane</u>). Finally, we can have a 4 carbon linear structure in which we have to add 2 methyl groups. We can do this addition in carbon 2 (<u>2,2-dimethylbutane</u>) or we can do this addition in carbon 2 and carbon 3 (<u>2,3-dimethylbutane</u>).

See figure 1

I hope it  helps!

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What is the percent composition of N H S O in (NH4)2SO4
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Answer:

The percent composition is 21% N, 6% H, 24% S and 49% O.

Explanation:

1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.

In 1 mole of (NH4)2SO4, there are:

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2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:

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- 8 moles of H (1g/mol) = 8g

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3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:

\begin{gathered} \text{ Nitrogen:} \\ 132g-100\% \\ 28g-x=\frac{28g*100\%}{132g} \\ x=21\% \end{gathered}\begin{gathered} \text{ Hydrogen:} \\ 132g-100\operatorname{\%} \\ 8g-x=\frac{8g*100\operatorname{\%}}{132g} \\ x=6\% \end{gathered}\begin{gathered} \text{ Sulfur:} \\ 132g-100\operatorname{\%} \\ 32g-x=\frac{32g*100\operatorname{\%}}{132g} \\ x=24\% \end{gathered}\begin{gathered} \text{ Oxygen:} \\ 100\%-21\%-6\%-24\%=49\% \\  \\  \end{gathered}

In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.

So, the percent composition is 21% N, 6% H, 24% S and 49% O.

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