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ozzi
3 years ago
10

Which of the statements are true? Triacylglycerols (triglycerides) are carboxylic acids. Fats that contain more unsaturated fatt

y acid residues than saturated fatty acid residues are more likely to be liquid at room temperature. Saturated fats have lower melting points than unsaturated fats. Triacylglycerols (triglycerides) are composed of fatty acid residues and glycerol. Unsaturated fats are more likely than saturated fats to be liquid at room temperature.
Chemistry
1 answer:
astraxan [27]3 years ago
7 0
<h2>The correct answer is given below:</h2>

Explanation:

  • Fats or lipid can be defined as high energy containing organic macro-molecules.
  • The basic molecule which makes up fat or lipid is Triacylglycerol.
  • A Triacylglcerol molecule is made up of one molecule of glycerol attached to three fatty acids molecules.
  • Here, the carboxylic acid groups of three fatty acid molecules are attached with the help of ester bonds to the three hydroxyl groups of one glycerol molecule.
  • Fatty acids are of two types, Saturated and Unsaturated.
  • Saturated fatty acid molecules can be defined as those in which the carbon atoms forming the hydrocarbon chain of the fatty acid molecule are linked to each other by single covalent bonds. The structure of saturated fatty acid molecule is linear or straight.
  • Unsaturated fatty acid molecules can be defined as those in which there are presence of one or multiple double covalent bonds in between the carbon atoms forming the hydrocarbon chain of the fatty acid molecule.
  • These double bonds in unsaturated fatty acid molecules form kinks or bend in the structure of the fatty acid .
  • In lipid or fat, the fatty acid molecules associate with each other by hydrophobic or Van der Waals forces of interactions.
  • In saturated fat, due to the linear structure of the fatty acids, the fatty acid molecules associate close to each other thereby forming a compact fat molecule with high melting point. This causes them become solidified at room temperature.
  • In unsaturated fat, the presence of kinks or bends in the fatty acid molecules results in the generation of fat molecules whose structure is not compact due to the presence of spaces in between the fatty acid molecules. These fats have a lower melting point which causes them to be liquid at room temperature.
  • The first statement is FALSE as triacylglycerols are formed by the esterification of three molecules of fatty acids with one molecule of glycerol.
  • The second statement is TRUE as unsaturated fatty acids have lower melting point than saturated fatty acids.
  • The third statement is FALSE as saturated fats have higher melting point than unsaturated fats.
  • The fourth statement is TRUE as triacylglycerols are formed by the esterification of three molecules of fatty acids with one molecule of glycerol.
  • The fifth statement is TRUE as unsaturated fats have lower melting point and so they are liquid at room temperature.
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Answer:

Question 1: <u>1 s after the motion starts</u>

Question 2: <u>0 (just when the motion starts)</u>

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

<u>1. From the speedometer shown at the right.</u>

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

  • Free fall equation: Vf = Vo - gt

  • Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

  • t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

<u>2. Time when the speedometer displays a reading of 20 m/s</u>

First, calculate the launching speed:

  • Vf = Vo - gt

Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

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  • Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

  • 20 m/s = 30 m/s - 10 m/s² × t

  • t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

<u />

<u>3. Time when the speedometer displays a reading of 30 m/s</u>

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

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