Saturated fatty acids exhibit a linear structure while unsaturated fatty acids bend, or kink, due to double bonds within the chemical foundation.
Answer:
(1) Bromination, (2) E2 elimination and (3) epoxidation
Explanation:
- In the first step, -OH group in cyclopentanol is replaced by more facile leaving group Br by treating cyclopentanol with

- In the second step, E2 elimination in presence of strong base e.g. NaOEt/EtOH produce cyclopentene
- In the third step, treatment of cyclopentene with mCPBA produces 1,2-epoxycyclopentane
- Full reaction scheme has been shown below
Agar is used to assist establish an anaerobic environment that promotes nitrate reduction.
Nitrate Reduction test:
- The nitrate in the broth is converted to nitrite by organisms that can produce the nitrate reductase enzyme, which can then be further converted to nitric oxide, nitrous oxide, or nitrogen.
- Anaerobic respiration and denitrification are two processes that can convert nitrate to a variety of compounds.
- While denitrification only reduces nitrate to molecular nitrogen, anaerobic respiration employs nitrate as the bacterium's final electron acceptor, reducing it to a range of chemicals.
- The nitrate reduction test is based on the detection of nitrite and its capacity to produce a red precipitate (prontosil), which is a water-soluble azo dye, when it combines with sulfanilic acid to create a complex (nitrite-sulfanilic acid).
Learn more about the Nitrate reduction test with the help of the given link:
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The food darkens because the color is changing in the food.
Answer:
0.6410g of HgS (mercury (II) sulfide) are formed
Explanation:
First you should write the balanced chemical equation, so we have:

Where:
is the formula for the sodium sulfide
is the formula for the mercury (II) nitrate
is the formula for the mercury (II) sulfide
and
is the formula for the sodium nitrate
Then you should calculate the amount of each substance in each solution, so:
- For the
:


moles of 
- For the
:


moles of 
As the quantity of
is smaller than the quantity of
. The
is the limiting reagent and you should work with this quantity, so we have:
moles of HgS
And as the molar mass of the HgS is
you can calculate the mass of HgS that is produced:
HgS