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charle [14.2K]
3 years ago
12

Modify the given fatty acid so that it represents the 18‑carbon fatty acid designated 18:2(Δ9,12). Draw any double bonds in the

cis configuration. Utlizise Groups when appropriate.

Chemistry
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

See explanation

Explanation:

In this case, we have to remember the meaning of the nomenclature "18:2Δ9,12".  Where 18 is the <u>number of carbon atom</u>s, 2 is the <u>number of double bonds,</u> and the numbers successive to Δ "delta" the position of the double bonds <u>starting</u> to count from the carboxylic -COOH end of the molecule.

In other words, the main functional group is a <u>carboxylic acid</u>. We have a total of 18 carbons. Additionally, we have 2 double bonds. On carbons 9 and 12.

Lets see figure 1

I hope it helps!

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Missing question: what is the density of 53.4 wt% aqueous NaOH if 16.7 mL of the solution diluted to 2.00L gives 0.169 M NaOH?

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c₁(NaOH) = ?; molarity of concentrated sodium hydroxide.

V₁(NaOH) = 16.7 mL; volume of concentrated sodium hydroxide.

c₂(NaOH) = 0.169 M; molarity of diluted sodium hydroxide.

V₂(NaOH) = 2.00 L · 1000 mL/L = 2000 mL; volume of diluted sodium hydroxide.  

Use equation: c₁V₁  = c₂V₂.

c₁ = c₂V₂ / V₁.

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The mass fraction is the ratio of one substance (in this example sodium hydroxide) with mass to the mass of the total mixture (solution).

Make proportion: m(NaOH) : m(solution) = 53.4 g : 100 g.

m(solution) = 1516 g in one liter of solution.

d(solution) = 1516 g/L = 1.52 g/mL.

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