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2 years ago

Can you draw the peptide structure at pH 7.0 (including peptide bonds)

Chemistry

1 answer:

ELEN [110]2 years ago

7 0

Answer:

See image below

Explanation:

The image is labeled according to the sequence N'-Trp-Ser-Asg-Gly-Cys-His-COOH' which means that in the main chain of the peptide, the amino group of the Tryptophan and the carboxylic group of the Histidine are free and thus its charge depends on the pH; other groups that rely on the pH are the side groups of the Cysteine and the Histidine.

Overall, ionizable groups in this peptide are:

Amino Group of the Tryptophan (pKa = 9.39)

SH group of the Cysteine (pKa = 8.18)

Secondary amine of the Histidine (pKa = 6.00)

Carboxylic Group of the Histidine (pKa = 1.82)

Then, the amino group of Trp and SH group of Cys are protonated since the peptide is at a pH below the pKa. The secondary amine of the Histidine is deprotonated because the pH is greater than the pKa, as well with its carboxylic acid group.

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An atom has 2 protons in its nucleus . Which of the following must be true for this atom to have no net electrical charge

ehidna [41]

If an atom has 2 protons, then it must also have 2 electrons for there to be no charge. An extra electron will cause for you to have a negative charge, and if you have more protons than electrons, then you will have a positive charge. :)

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2 years ago

Homolysis, or homolytic bond dissociation, produces a very specific type of product under certain reaction conditions. In Part 1

vekshin1

Answer:

1) R₃CO , H, H₃C, a carbon free radical

2) high temperature, ultraviolet irradiation

Explanation:

1) Homolysis leads to the formation of free radicals (species having a free electron). Thus, answer is :

R₃CO

H₃C

a carbon free radical

2) Homolysis require high temperature, ultraviolet irradiation.

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3 years ago

The exoskeleton of a housefly is made up of

OverLord2011 [107]

Answer:

Chitin

Explanation:

A fly's main body is made of a thorax and an abdomen. These parts are covered by a hard exoskeleton made of a bony substance called chitin.

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3 years ago

Read 2 more answers

Can someone help?? This is really hard.

Sergeu [11.5K]

Answer:

See Explanation

Explanation:

Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.

2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4

2^1 = 2^1

The rate of reaction is first order with respect to Hbn

Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.

1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4

3^1 = 3^1

The reaction is also first order with respect to CO

b) The overall order of reaction is 1 + 1=2

c) The rate equation is;

Rate = k [CO] [Hbn]

d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4 /[5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4/6.7 * 10^-7

k = 4.7 * 10^2 mmol-1 L s-1

e) The reaction occurs in one step because;

1) The rate law agrees with the experimental data.

2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.

8 0

2 years ago

A 10.0 mL sample of HNO3 was exactly neutralized by 13.5 mL of 1.0 M KOH. What is the molarity of the HNO3? Use the titrations f

Kruka [31]

Answer: Thus molarity of $HNO_3$ is 1.35 M

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HNO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=1\\M_1=?M\\V_1=10.0mL\\n_2=1\\M_2=1.0M\\V_2=13.5mL$

Putting values in above equation, we get:

$1\times M_1\times 10.0=1\times 1.0\times 13.5\\\\M_1=1.35M$

Thus molarity of $HNO_3$ is 1.35 M

7 0

3 years ago