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Igoryamba
2 years ago
11

With examples of amino acids, explain the type of isomerism that exists in amino acids.​

Chemistry
1 answer:
Maslowich2 years ago
3 0

Answer:

All amino acids are stereoisomers with the exception of glycine (because it has no chiral centers) and the two types are enantiomers and diastereomers

Explanation:

Not sure how in depth you need but the most fundamental categories are:

Enantiomers: non superimposable images which means that they cannot be placed on top of one another and look perfectly identical and instead are structurally the same but flipped in the opposite direction. An example being D-alanine and L-alanine.

Diastereomers: The molecules are superimposable which means they have an identical structure that will look the same placed on top of one another however, the compounds attached to the structure are placed in different orders an example being, L-isoleucine and D-allo-isoleucine (compounds in same place but isoleucine has two hydrogens positioned forward while allo-iso have one positioned forward and one positioned in the back)

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A student predicts that a solution of ethanol (C2H5OH) and water will have a lower density at room temperature than that of pure
ohaa [14]

Answer:

It would probably be, something that can take up moisture to test it.

Explanation:

(to see if it can evaporate)

4 0
2 years ago
Read 2 more answers
What is the name of the ionic compound AlBr3?
eduard
The ionic compound for AlBr3 is ↓

                                         Aluminum Bromide 
8 0
3 years ago
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Find the volume of 0.100M hydrochloric acid necessary to react completely with 1.51g Al(OH)3.
shtirl [24]
Reaction equation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
Moles of Al(OH)₃:
moles = mass/Mr
= 1.51 / (27 + 17 x 3)
= 0.019
Molar ratio Al(OH)₃ : HCl = 1 : 3
Moles of HCl required = 0.019 x 3
=0.057
concentration = moles/volume
volume = 0.057 / 0.1
= 0.57 dm³
= 570 ml
3 0
3 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
A package contains 1.33 lb of ground round. If it contains 29% fat, how many grams of fat are in the ground round? The book is s
Effectus [21]

Answer:

To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).

Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.

In this way, your reasoning is correct and it is probably a mistake in the book.

6 0
3 years ago
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