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3 years ago

How does the structure of amino acids allow them to form a polypeptide

1 answer:

7 0

Amino acids which are known to be linked by peptide bonds they form polypeptide chains.

Proteins are linear polymers are formed by way of linking an a-carboxy group of one amino acid to a-amino of different amino acids which have peptide bond. The formation which results from two amino acids which result in a loss of a water molecule. The best process of the reaction is hydrolysis.

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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0 mL of 1 M H2SO4. A 25.00 mL aliquot is anal

Olenka [21]

Answer:

The weight percent in the sample is 17,16%

Explanation:

The dissolution of the Ce(IV) salt provides free Ce⁴⁺ that reacts, thus:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03247M Na₂S₂O₃ = 4,228x10⁻⁴ moles of S₂O₃²⁻.

As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,228x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^{2-}}$ = 2,114x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,114x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = 4,228x10⁻⁴ moles of Ce(IV).

These moles are:

4,228x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = 0,05924 g of Ce(IV)

As was taken an aliquot of 25,00mL from the solution of 250,0mL:

0,05924 g of Ce(IV)× $\frac{250,0mL}{25,00mL}$ =0,5924g of Ce(IV) in the sample

As the sample has 3,452g, the weight percent is:

0,5924g of Ce(IV) / 3,452g × 100 = 17,16 wt%

I hope it helps!

3 0

3 years ago

each of the following reactions has been reported in the chemical literature. predict the product in each case, showing stereoch

poizon [28]

(a): This is an elimination reaction. The reaction proceeds in 3 steps; Step 1: Protonation of alcohol, step 2: Loss of leaving group, and step 3: Deprotonation to form an alkene. Elimination (E1) reaction dominates than substitution (SN1) reaction because the formed HSO4- anion is a poor nucleophile and tends not to add to the carbocation intermediate formed. Hence the product formed is I, is trans.

(b) This is a dihydroxylation of the alkenes reaction. A catalytic amount of OsO4 is used along with an oxidizing agent, which oxidizes the reduced osmium(VI) into osmium(VIII) to regenerate the catalyst. The reaction proceeds in 2 steps; Step 1: Cis addition of OsO4 to double bond and form 4 member cyclic osmate ester. Step 2: Hydrolysis of the intermediate to 1.2-diol. Hence the product formed is II, is cis.

(c) This is hydroboration - oxidation reaction. The reaction proceeds through the addition of H- on the more substituted carbon and BH2 on the less substituted carbon of the double bond. The reaction is a syn addition. Hence the product formed is III.

(d) This a reduction reaction where carboxylic acid (-COOH) is reduced to 1o alcohol (-CH2OH). Hence the formed product is IV.

Stereochemistry, a subdiscipline of chemistry, involves the take a look at the relative spatial association of atoms that form the shape of molecules and their manipulation.

The observation of stereochemistry specializes in stereoisomers, which by means of definition have an equal molecular method and series of bonded atoms (constitution), however, differ within the three-dimensional orientations of their atoms in space. for that reason, it's also referred to as 3-d chemistry three-dimensionality.

Learn more about Stereochemistry here:-brainly.com/question/13266152

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Disclaimer:- your question is incomplete, please see below for the complete question.

Each of the following reactions has been reported in the chemical literature. Predict the product in each case, showing stereochemistry where appropriate.

5 0

10 months ago

The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38

adell [148]

Answer:

$1.21x10^{-3}$ M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x $10^{-3}$

H = $\frac{0.21}{1.38x10^{-3} }$

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = $1.21x10^{-3}$ M

7 0

3 years ago

The surface of the earth is divided into more than 50 large plates

Romashka-Z-Leto [24]

Major Plates

Africa Plate
Antarctic Plate
Indo-Australian Plate
Australian Plate
Eurasian Plate
North American Plate
South American Plate
Pacific Plate
Minor Plates
There are dozens of smaller plates, the seven largest of which are:
Arabian Plate
Caribbean Plate
Juan de Fuca Plate
Cocos Plate
Nazca Plate
Philippine Sea Plate
Scotia Plate

5 0

2 years ago

Methyl violet is an indicator that changes color over a range from pH=0.0 to pH=1.6. What is Ka of methyl violet?

Tju [1.3M]

it's basic in nature

7 0

3 years ago