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Nesterboy [21]
4 years ago
8

Which of the following is the correct ranking of the three bonds and interactions in order from highest to lowest in terms of th

eir bond strength between two side chains of a protein in their tertiary structure?
I. Disulfide bond between two cystines
II. Hydrophobic interactions between two leucines
III. H-bonding in water
Chemistry
1 answer:
Nesterboy [21]4 years ago
6 0

Answer:

I > III > II

Explanation:

I) A disulfide bond between two cystines is created when a sulfur atom from one cystine forms a strong, single covalent bond with a sulfur atom from a second cystine. When a disulfide bond is created, each cystine loses one hydrogen atom. The atom count is 11 for a cystine in mid-chain, but changes to 10 if the cystine joins with another in a disulfide bond. This lead to a much more stable intermolecular interaction.

III) Hydrogen Bonding in water

These hydrogen bonds are at best an interaction, inducing slight positive and negative charges in the Hydrogen and Oxygen/Nitrogen atoms.

The Hydrophilic amino acids have O & N atoms, which form hydrogen bonds with water. These atoms have an uneven distribution of electrons, creating a polar molecule that can interact and form hydrogen bonds with water.

The hydrogen bonds aren't as strong as the covalent bonds in disulfides.

II) Hydrophobic interactions between two leucines

A hydrophobic interaction is formed between two nonpolar molecules.

It describes the preference of nonpolar molecular surfaces to interact with other nonpolar molecular surfaces, thereby displacing water molecules from the interacting surfaces.

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Read 2 more answers
The analysis of a hydrocarbon revealed that it was 85.7% C and 14.3% H by mass. When 1.77 g of the gas was stored in a 1.500-L f
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Answer:

The formula of hydrocarbon = C_3H_6

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 85.7

Molar mass of C = 12.0107 g/mol

% moles of C = 85.7 / 12.0107 = 7.14

% of H = 14.3

Molar mass of H = 1.00784 g/mol

% moles of H = 14.3 / 1.00784 = 14.19

Taking the simplest ratio for C and H as:

7.14 : 14.19 = 1 : 2

The empirical formula is = CH_2

Also, Given that:

Pressure = 508 Torr

Temperature = 17 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (17 + 273.15) K = 290.15 K  

Volume = 1.500 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637 L.torr/K.mol

Applying the equation as:

508 Torr × 1.500 L = n × 62.3637 L.torr/K.mol × 290.15 K  

⇒n = 0.0421 moles

Given that :  

Amount  = 1.77 g

Molar mass = ?

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0421\ moles= \frac{1.77\ g}{Molar\ mass}

Molar mass of the hydrocarbon = 42.04 g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12 + 2×1= 14 g/mol

Molar mass = 42.04 g/mol

So,  

Molecular mass = n × Empirical mass

42.04 = n × 14

⇒ n = 3

<u>The formula of hydrocarbon = C_3H_6</u>

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