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2 years ago

Which atoms in the amino acids become the h2o molecule produced by their action in model?

1 answer:

4 0

Atoms in the amino acids become the h₂O molecule produced by their action in the model and come off from the central carbon and nitrogen but not from the carboxyl, R side chain, or amine.

An amino acid is a group of organic molecules that consist of a basic acidic carboxyl group (―COOH), amino group (―NH2), and an organic R group (or side chain) that is different from each amino acid. Amino acid, the term is a short form of α-amino [alpha-amino] carboxylic acid.

Whereas, the peptide bond is the chemical bond which is a chemical bond formed between two molecules when the carboxyl group of a particular molecule reacts with the amino group of the other molecule, leading to releasing a molecule of water (H2O).

Each molecule consists of a central carbon atom referred to as the α-carbon, to which both a carboxyl group and amino are attached. The remaining two bonds of the α-carbon atom are generally occupied by the R group and a hydrogen (H) atom .

To know more about amino acids refer to the link brainly.com/question/14583479?referrer=searchResults.

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Which of the following is true for a reliable scientific source?

ipn [44]

It cites valid data because a reliable scientific source needs data to back up what they’re proving.

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3 years ago

Read 2 more answers

Important of thermometer

Mama L [17]

Answer:

to see if something or someone has either a low or high temperature

Explanation:

because you wanna know if some has a fever so you dont get the coronna virus

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3 years ago

Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

D. It will decrease by a factor of 4

Explanation:

According to the question , the equation follows :

$A+B\rightarrow C+D$

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

$Rate\ \alpha [A]^{a}[B]^{b}$

$r=[A]^{a}[B]^{b}$ .................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

$r'=[A']^{a}[B']^{b}$

Put the value of [A'] , [B'] and r' in the above equation:

$2r=[2A]^{a}[B]^{b}$ ...........(2)

Divide equation (1) by (2) we , get

$\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}$

$2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}$

Here A and A cancel each other

B and B cancel each other

We get,

$2= 2^{a}\times 1^{b}$

1^b = 1 ( power of 1 = 1)

$2= 2^{a}$

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1 = 2

b = 2

Hence the rate law becomes :

$r=[A]^{a}[B]^{b}$

$r=[A]^{1}[B]^{2}$ .............(3)

Look in the question now, it is asked to calculate the concentration of [B],if cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

$r'=[A]^{1}[B']^{2}$

$r'=[A]^{1}(\frac{1}{2}[B])^{2}$

$r'=\frac{1}{4}[A]^{1}[B]^{2}$ ............(a)

But

$r=[A]^{a}[B]^{b}$ ..............(b)

Compare equation (a) and (b) , we get

new rate r' =

r' = 1/4 r

7 0

3 years ago

The equation for another reaction used in industry isCO(g) + H₂O(g) <img src="https://tex.z-dn.net/?f=%5Crightleftharpoons" id="

Sloan [31]

Answer:

(i) CO = 0.4 mol; H₂O = 1.6 mol; Kc = 4

(ii) CO = 0.67 mol; H₂O = 0.67 mol; CO₂ = 1.33 mol; H₂ = 1.33 mol

Explanation:

(i) For the equation given let's make a table of the concentrations for equilibrium (the volume is constant, so, we can do it with moles number)

CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)

2.0 mol 3.2 mol 0 0 Initial

-x -x +x +x Reacts (stoichiometry is 1: 1: 1: 1)

2.0-x 3.2-x x x Equilibrium

In the equilibrum, the moles number of hydrogen and carbon dioxide are 1.6 mol, so x = 1.6 mol

The amounts of CO and H₂O are:

CO = 2.0 - 1.6 = 0.4 mol

H₂O = 3.2 - 1.6 = 1.6 mol

The constant of the equilibrium is the multiplications of the concentrations of products divided by the multiplication of the concentration of the reactants (all the concentrations elevated to the coefficient). So:

Kc = (1.6x1.6)/(0.4x1.6)

Kc = 1.6/0.4

Kc = 4

(ii) Kc must remais constant (it only changes with the temperature), so let's construct a new table of equilibrium:

CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)

2.0 mol 2.0 mol 0 0 Initial

-x -x +x +x Reacts (stoichiometry is 1: 1: 1: 1)

2.0-x 2.0-x x x Equilibrium

Kc = (x*x)/((2.0-x)*(2.0-x))

4 = x²/(4 - 4x + x²)

16 - 16x + 4x² = x²

3x² - 16x + 16 = 0

Using Baskhara's equation:

Δ =(-16)² - 4x3x16

Δ = 256 - 192

Δ = 64

x = (-(-16) +/- √64)/(2*3)

x' = (16 + 8)/6 = 4

x'' = (16 - 8)/6 = 1.33

x must be small than 2.0, so x = 1.33 mol, which is the amount of hydrogen and carbon dioxide at equilibrium. The both reactants has 2.0 - 1.33 = 0.67 mol at equilibrium.

5 0

3 years ago

Anyone tryna help mee

devlian [24]

Answer:

It is a solid

4 0

3 years ago