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snow_lady [41]
3 years ago
7

A sample of the amino acid tyrosine is barely soluble in water. Would a polypeptide containing only Tyr residues, poly(Tyr), be

more or less soluble, assuming the total number of Tyr residues is the same in both cases? Explain.
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
4 0

Answer:

Explanation:

The first thing we need to know is that the amino acid of tyrosine has an aromatic ring within its structure.

Also, recall that aromatics hydrocarbons are considered to be unsaturated and non-polar.

So, since like dissolve like. Then, The non-polar character of tyrosine is enhanced by its aromatic ring. As a result, Tyrosine amino acid is much more soluble in less polar solvents like methanol and ethanol than in moderate highly polar water.

From the given information:

Since water is a very polar solvent, and tyrosine amino acid is not particularly polar, it is just barely soluble in it.

In addition to the aforementioned, peptides are strongly polar in nature. However, as the amount of tyrosine amino acids in a peptide increases, the region of the non-polar portion of the peptide increases (owing to the rise of the aromatic ring), and thus the non-polar value of the peptide will also rise. And this non-polar nature of peptide will make them more soluble in non-polar solvent and decrease its solubility in water (because water is polar in nature). And since peptides are non-polar, they would be more soluble in non-polar solvents and have less solubility in water.

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Ksivusya [100]

Answer:

1 x 10^28

Explanation:

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2 years ago
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Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?
german

Answer : The correct option is, (A) AlCl_3

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) AlCl_3

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of AlCl_3 will be,

AlCl_3\rightarrow Al^{3+}+3Cl^{-}

So, Van't Hoff factor = Number of solute particles = Al^{3+}+3Cl^{-} = 1 + 3 = 4

(B) KI

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of KI will be,

KI\rightarrow K^{+}+I^{-}

So, Van't Hoff factor = Number of solute particles = K^{+}+I^{-} = 1 + 1 = 2

(C) CaCl_2

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of CaCl_2 will be,

CaCl_2\rightarrow Ca^{2+}+2Cl^{-}

So, Van't Hoff factor = Number of solute particles = Ca^{2+}+2Cl^{-} = 1 + 2 = 3

(D) MgSO_4

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of MgSO_4 will be,

MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}

So, Van't Hoff factor = Number of solute particles = Mg^{2+}+SO_4^{2-} = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, AlCl_3

7 0
3 years ago
The human body on average contains 6 liters of blood. If 20 drops are equal to 1milliliter, how many drops of blood are in the a
garri49 [273]

Answer:

120000drops

Explanation:

Average blood in human blood = 6L - 6*1000 = 6000ml

1 ml of blood is equal to 20 drops

6000ml of blood makes 20*6000 = 120000 drops

5 0
2 years ago
Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.
Svetradugi [14.3K]

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%

\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%

\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

3 0
3 years ago
9.8 x 10​^-6 regular notation
topjm [15]

Answer:

0.0000098 should be the answer

Explanation:

3 0
3 years ago
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