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Mars2501 [29]
2 years ago
12

If the last carbon atom of an unsaturated fatty acid is bonded to two hydrogen atoms and one carbon atom, what type of bond must

exist between the last carbon and the other carbon atom
Chemistry
2 answers:
marissa [1.9K]2 years ago
9 0

Answer:

Double bond

Explanation:

its in the book

Debora [2.8K]2 years ago
4 0

Answer:

Double bond

Explanation:

Carbon is a tetravalent atom. This implies that the carbon atom always forms four covalent bonds.

These four covalent  bonds may be single, double or triple bonds. Compounds that contain double and triple bonds are said to be "unsaturated".

If the last carbon atom of an unsaturated fatty acid is bonded to two hydrogen atoms and one carbon atom, it means that the bond between the last carbon and the other carbon atom must be a double bond because carbon must be tetravalent and we have already been told that the fatty acid is unsaturated.

The last carbon atom of the unsaturated fatty acid must form two covalent bonds to the other carbon atom in order to respect the tetravalency of carbon.

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Percentage Weight-in-volume is defined as the <em><u>number of grams of a solute in a 100 ml (milliliters) solution.</u></em>

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<u>Percentage Weight-in-volume</u> can tell us about the <em>degree of concentration of a given solution.</em>

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The solute can be <em>crystalline or non-crystalline in nature.</em>

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The <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.

  • This question is based on a Percentage Weight-in-volume. The formula states that:

a% of a glucose solution =<u> a grams of glucose in a 100 mL solution</u>

Hence, 5% glucose solution = 5 grams of glucose / 100 mL solution

Therefore, the <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.

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brainly.com/question/8482854

6 0
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Explanation:

Glad to help :)

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Determine if each of the following statements about chiral molecules is True or False and select the best answer from the dropdo
oee [108]

Answer:

See explanation

Explanation:

-)<u> True or False: All R stereocenters are dextrorotatory. </u>

The absolute configuration is based is in specific rules that are not related to the ability to deflect polarized light. <u>FALSE</u>

-)<u> True or False: Chiral centers in organic molecules will have 4 different groups attached to a carbon. </u>

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A chiral carbon by definition is a carbon with 4 groups. <u>TRUE</u>

-)<u> True or False: A racemic mixture has an optical activity of 0. </u>

In a racemic mixture, we have equal amounts of enantiomers, and this cancels out the optical activity. <u>TRUE</u>

-)<u> True or False: Normal linear amines can be chiral centers. </u>

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In primary amines, we have 2 hydrogens. Therefore all the groups can not be different. So, is <u>TRUE</u>

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-)<u>True or False: Compound C has an optical activity of 0. </u>

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We need to know the structure of the compound

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-)<u>True or False: If the R isomer of a molecule has an optical rotation of + 25.7 degree then the S isomer of the molecule will have an optical rotation of -25.7 degree. </u>

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If we have the exact opposite (as we have in this case) the magnitud of the optical activity value will remain and the sign will change. <u>TRUE</u>

-)<u>True or False: A CN is a higher priority than a CH2OH. </u>

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In this case, a carbon is directly bonded to the chiral carbon. The carbon on CN is bonded to a nitrogen atom and the carbon on CH2OH is bonded to an oxygen. So, the CH2OH will have more priority because O has a higher atomic number. <u>FALSE</u>

<u />

-)<u>True or False: All molecules with chiral centers are optically active. </u>

<u />

We can have for example mesocompounds in which the optical activity is canceled out due to symmetry planes. <u>FALSE</u>

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-)<u>True or False: To have an enantiomer a molecule must have at least two chiral centers. </u>

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A pair of enantiomers is made between at least 1 chiral carbon. Enantiomer R and enantiomer S. <u>FALSE</u>

<u />

-)<u>True or False: Chiral molecules are always optically active. </u>

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We can have racemic mixtures or mesocompounds with chiral carbons but without optical activity. <u>FALSE</u>

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-)<u>True or False: A CH2CH2Br is higher priority than a CH2F. </u>

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-)<u>True or False: Meso molecules with two stereocenters have a R,S configuration. </u>

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On the compounds with R and S configuration at the same time can have symmetry planes so, we will not have optical activity. <u>TRUE</u>

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<u>True or False: Diastereomers have the same physical properties except in a chiral environment. </u>

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All diastereomers have the same physical properties. <u>TRUE</u>

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<u>True or False: Compound H has an optical activity of 0. </u>

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We have to have the structure of the compound.

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<u>True or False: A C=C double bond is higher priority than a -CH(CH3)2.</u>

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In the case of C=C, we can say that is equivalent to two carbon bonds without hydrogens. Therefore C=C has higher priority. <u>TRUE</u>

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