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Lina20 [59]
3 years ago
14

This dipeptide is designated as: there is a structure for h3nchcnhchco with ch2oh group attached to the first (from left to righ

t) carbon, two oxygen atoms attached to the second and fourth carbons by double bonds, and ch2sh group attached to the third carbon. nitrogen has a charge of 1 plus. oxygen atom, which is attached to the fourth carbon by a single bond, has a charge of 1 minus.

Chemistry
1 answer:
oksian1 [2.3K]3 years ago
4 0

A dipeptide is formed when 2 amino acids are joined together by a peptide linkage.

Each amino acid molecule contains an amino group ( -NH₂) and a carboxyl ( -COOH) group. During peptide formation, a carboxyl group of one amino acid reacts with amino group of the other , resulting in the formation of a linkage known as "peptide linkage". A water molecule is also eliminated in this reaction.

A general peptide formation reaction can be written as

NH_{2}-CH_{2}-COOH + H_{2}N-CH_{2}-COOH \rightarrow NH_{2}-CH_{2}-CONH-CH_{2}-COOH +H_{2}O

In the above reaction, we can see a peptide linkage ( -CONH-) getting formed between 2 amino acid molecules.

In the given molecule the peptide linkage is shown by drawing a box around it. Please refer to the attached image.

If we want to know the amino acid molecules from which this peptide is formed, we will simply break the CO-NH bond.

We will add -OH group to CO part and -H to NH part to get 2 amino acid molecules as shown in the picture.

The two amino acids thus formed are Serine & Cysteine.

Serine is designated as "Ser" and Cysteine is designated as "Cys"

Therefore the given dipeptide is designated as "Ser-Cys"

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Answer:

8.37g

Explanation:

Step 1 :

The balanced equation for the reaction. This is given below:

N2 + 2O2 —> 2NO2

Step 2:

Data obtained from the question.

Volume (V) of N2 = 2L

Pressure (P) = 840mmHg

Temperature (T) = 24°C

Number of mole (n) of N2 =?

Step 3:

Conversion to appropriate unit.

For pressure:

760mmHg = 1atm

840mmHg = 840/760 = 1.11 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 24°C

T(K) = 24°C + 273

T(K) = 297K

Step 4:

Determination of the number of mole N2.

The number of mole of N2 can be obtained by using the ideal gas equation as follow:

Volume (V) of N2 = 2L

Pressure (P) = 1.11 atm

Temperature (T) = 297K

Number of mole (n) of N2 =?

Gas constant (R) = 0.082atm.L/Kmol

PV = nRT

Divide both side by RT

n = PV / RT

n = 1.11 x 2 / 0.082 x 297

n = 0.091 mole

Therefore, the number of mole of N2 that reacted is 0.091 mole

Step 5:

Determination of the mass of NO2 produced from the reaction. This is illustrated below:

N2 + 2O2 —> 2NO2

From the balanced equation above,

1 mole of N2 produced 2 moles of NO2.

Therefore, 0.091 mole of N2 will produce = 0.091 x 2 = 0.182 mole of NO2.

Finally, we will convert 0.182 mole of NO2 to gram as shown below:

Number of mole NO2 = 0.182 mole

Molar mass of NO2 = 14 + (16x2) = 46g/mol

Mass = number of mole x molar mass

Mass of NO2 = 0.182 x 46

Mass of NO2 = 8.37g

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