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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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What type of reaction is this? A. double replacemment B. Single replacement

irina [24]

Answer:

its B for sure

Hope it helps

3 0

3 years ago

What is Saponification!?

Tom [10]

Answer:

<h3>Saponification is a process that involves conversion of fat, oil or lipid into soap and alcohol by the action of heat in the presence of aqueous alkali. Soaps are salts of fatty acids and fatty acids are monocarboxylic acids that have long carbon chains e.g. sodium palmitate.</h3>

8 0

3 years ago

A motor cyclist starts from rest and reaches a speed of 6m/s after travelling with uniform acceleration for 3s. What is his acce

choli [55]

Answer:

2m/s²

Explanation:

When an object starts or at its state of rest it has an Initial speed U = 0

Final speed = 6m/s

total time taken for the acceleration = 3s

Acceleration =?

Acceleration is the change in velocity (speed) with time

Time rate of change of velocity

Acceleration = Change in Speed(velocity)

Time taken

Hence,

Acceleration =  V - U 

a = 6 - 0

a = 6 

a = 2m/s²

4 0

3 years ago

Based upon the mass of baking soda (NaHCO3) and using an excess of HCl in this experiment, you will 1) determine the mass of CO2

weqwewe [10]

Equation of reaction

NaHCO3 + HCl ---------> NaCl + H2O + CO2(g)

1)The mass(actual yield) of CO2 can be gotten by isolating it from other products and getting its mass.

Assume 1.0g of CO2 was gotten as a product of the experiment

2) For the theoretical yield

The mass of NaHCO3 was not stated, but for the purpose of this solution, I would assume 2g of NaHCO3 was used for the experiment. You can substitute any parameter by following the steps I follow

Number of mole of NaHCO3 = Mass of NaHCO3/ Molar Mass of NaHCO3

Number of mole of NaHCO3 = 2/84.01 = 0.0238 moles

1 mole of NaHCO3 yielded 1 mole of CO2

0.0238 moles of NaHCO3 will yield 0.0238 moles of CO2

Mass of CO2 = Number of moles * Molar Mass

Mass of CO2 = 0.0238 * 44 = 1.0472g

Theoretical yield of CO2 = 1.0472 grams

3) Percentage yield of CO2 = Actual yield/Theoretical yield *100%

Percentage yield of CO2 = 1.0/1.0472 *100

Percentage yield of CO2 = 95.49%

8 0

3 years ago

Read 2 more answers

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

3 years ago