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3 years ago

Determine the amino acid sequence of a polypeptide from the following data:

Chemistry

1 answer:

svlad2 [7]3 years ago

7 0

Solution :

1. It is given that when Edman's reagent is used, it releases PTH-Glycin. The degradation of Edmans is used for the amino acids to sequencing of the peptide/protein and is also removes N terminal amino acid and thus gives us the PTH amino acid product.

Therefore, the first amino acid is Glycine.

2. The Carboxypeptidase A releases the C terminal amino acid, and therefore the last amino acid is Phe.

3. The Cyanogen Bromide helps to chop after the Med residue, therefore the sequence is 1 - 2 - 3 or 2 - 1 - 3.

4. The trysin chops after the basic amino acid - Arg and the Lys unless followed by Proline and so the sequence of the peptide is 1 - 2 - 3 4.

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The following reactions are used in batteries:Reaction I is used in fuel cells, II in the automobile lead-acid battery, and III

elixir [45]

The highest ratio is -0.613kj/g which is for automobile lead-acid battery .

Given ,

Reaction-1 used in fuel cells

Reaction-2 used in automobile lead-acid battery

Reacttion-3 used in experimental high-temperature battery for powering electric vehicles .

1) Fuel cells :

2H2(g) + O2(g) ==> 2H20 (l) , E^0cell = 1.23V

We know ,

Free energy is the maximum work done .

delta G = Wmax

=> Wmax = delta G = -4.75*10^5j/mol e^- = -4.75*10^2 kj/ mol e^-

Molecular mass of 1 mole of H2 = 2 .016 g / mol e^-

Molecular mass of 2 mole of H2 = 4.032 g/mol e^-

Molecular mass of O2 = 32 g/mol e^-

Total mass of the reactants = (4.032+32)g/mol e^- = 36.032 g/mol e^-

Ratio of Wmax to the total mass of the reactants

=Wmax/ total mass of the reactants

=( -4.75*10^2 kj/mol e^-)/36.032 g/mol e^-

=-13.2kj/g

Therefore , the ratio of Wmax to total mass of the reactants is -13.2kj/g .

2) Automobile lead-acid battery :

Pb(s) + PbO2(s) + 2H2SO4(aq) ==> 2PbSO4 (s) + 2H2O (l) , E^0cell = 2.04V

We know ,

Wmax = delta G = -3.94*10^5j/mol e^- = -3.94*10^2kj/mol e^-

molecular mass of Pb = 207.2 g /mol e^-

molecular mass of Pbo2 = 239.2g/mol e^-

molecular mass of H2SO4 = 98.008g/mol e^-

molecular mass of 2 mole of H2SO4 = 196.016g/mol e^-

total mass of the reactants = (207.2+239.2+196.016)g/mol e^- = 642.416g/mol e^-

Ratio of Wmax to total mass of the reactants

= Wmax/total mass of reactants

=(-3.94*10^2kj/mol e^-)/642.416g/mol e^-

=-0.613kj/g

Hence ,the ratio of Wmax to the total mass of the reactants of automobile lead-acid battery is -0.613kj/g .

3) Experimental high temperature battery :

2Na(l) +FeCl2(s) ==> 2NaCl (s) + Fe(s) , E^0 cell = 2.35V

We know ,

Wmax = delta G = -4.53*10^5j/mol e^- = -4.53*10^2kj/mol e^-

molecular mass of Na = 22.99 g/mol e^-

molecular mass of 2 moles of Na = 45.98g/mol e^-

molecular mass of FeCl2 = 126.85 g/mol e^-

total mass of the reactants =( 45.98+ 126.85)g/mol e^- = 172.83 g/mol e^-

Ratio of Wmax to the total mass of the reactants

= Wmax / total mass of reactants

=(-4.53*10^2kj/mol e^-)/ 172.83g/mol e^-

=-2.62kj/g

Therefore , the ratio of Wmax to the total mass of the reactants in an experimental high temperature battery is -2.62kj/g .

Hence , the highest ratio is -0.613kj/g which is of lead -acid battery .

Learn more about fuel cell here :

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4 0

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How many moles of H2O are needed to produce 5.6 mol of NaOH?<br> Na2O + H2O --> 2NaOH

navik [9.2K]

Answer: 2.8 moles

Explanation:

The balanced equation below shows that 1 mole of sodium oxide reacts with 1 mole of water to form 2 moles of sodium hydroxide respectively.

Na2O + H2O --> 2NaOH

1 mole of H2O = 2 moles of NaOH

Let Z moles of H2O = 5.6 mole of NaOH

To get the value of Z, cross multiply

5.6 moles x 1 mole= Z x 2 moles

5.6 = 2Z

Divide both sides by 2

5.6/2 = 2Z/2

2.8 = Z

Thus, 2.8moles of H2O are needed to produce 5.6 mol of NaOH

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valentina_108 [34]

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