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3 years ago

Identify the Lewis acid and Lewis base in each reaction:

Chemistry

2 answers:

nikdorinn [45]3 years ago

8 0

Answer:

The answers to the question are

(a)

Fe₃⁺ = Lewis acid

H₂O = Lewis base

(b)

H₂O = Lewis acid

H⁻ = Lewis base

(c)

CO⁺ = Lewis acid

Ni = Lewis base

Explanation:

Lewis theory of acids states that

An acid is a substance that accepts a pair of electrons an i doing so forms a covalent bond with the group that supplies the electrons

A Lewis base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared

(a) Here Fe₃⁺ is the Lewis acid as it accepts an electron pair from H₂O which is amphoteric an so can act either as a Lewis base or acid

Fe₃⁺ = Lewis acid

H₂O = Lewis base

(b) H₂O + H⁻ → H₂ + OH⁻

Here the water accepts a pair of electrons from H⁻ anion hence it is the Lewis acid while H⁻ anion donates a pair of electrons acting as the Lewis base

H₂O = Lewis acid

H⁻ = Lewis base

CO⁺ = Lewis acid

Ni = Lewis base

dybincka [34]3 years ago

5 0

Answer:

Explanation:

A Lewis acid is defined as a chemical species that contains an empty orbital which accepts an electron pair from a Lewis base to.

A Lewis base is defined as any species that has a filled orbital containing an electron pair which is not involved in bonding but forms a dative bond with a Lewis acid.

A. Fe3+ + 2H2O --> FeOH2+ + H30+

Fe3+ - Lewis acid

H2O - Lewis base

B. H2O + H- --> H2 + OH-

H- - Lewis acid

H2O - Lewis base

C. 4CO + Ni --> Ni(CO)4

CO - Lewis base

Ni - Lewis acid

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Answer:

2.29 g of N2

Explanation:

We have to start with the chemical reaction:

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The next step is to balance the reaction:

$2NaN_3~->~2Na~+~3N_2$

We can continue with the mol calculation using the molar mass of

$NaN_3$ (65 g/mol), so:

$3.55~g~NaN_3\frac{1~mol~NaN_3}{65~g~NaN_3}=0.054~mol~NaN_3$

Now, with the molar ratio between $NaN_3$ and $N_2$ we can calculate the moles of $N_2$ (2:3), so:

$0.054~mol~NaN_3\frac{3~mol~N_2}{2~mol~NaN_3}=0.0819~mol~N_2$

With the molar mass of $N_2$ we can calculate the grams:

$0.0819~mol~N_2=\frac{1~mol~N_2}{28~g~N_2}=2.29~g~N_2$

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Determine the primary structure of an octapeptide from the following data: acid-catalyzed hydrolysis gives 2 arg, leu, lys, met,

erma4kov [3.2K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The sequence of these amino acid would be

Lue Arg Lys Arg Met Phe Arg Ser

Explanation:

The acid-catalyzed hydrolysis from makes us to understand that the polypeptides contains 8 amino acids

From the question we are told that the Edman's reagent releases Leu it means that the N-terminal amino acid would be Lue(i.e Leucine)

Also from the question we are told that the Carboxypeptidase released Ser this mean the the C-terminal amino acid would be Ser ( i.e Serine)

The Amino acids would in the polypeptides would be arranged like this

Lue __ __ __ __ __ __ Ser

Now from the question we are told that treatment with cranogen bromide form two peptides.

Now generally cranogen bromide divides a peptide on the C- side(i.e the extreme left side ) of Met(Methionine)(This is an amino acid obtained by hydrolysis of most common peptides )

So this means that any peptides containing Met(Methionine) must have Methionine as a C- terminal amino acid(i.e at extreme left) and for peptides that does not contain Met must be C - terminal peptides

From the question we see that it is the second peptide that contain Met and it is a penta peptide(i.e it contains 5 amino acid)

Thus the fifth amino acid is Met

So the sequence of these amino acid would now be

Lue __ __ __ Met __ __ Ser

From the question we are told that the the Trypsin-catalyzed hydrolysis forms two amino acid and two peptides

Now generally Trypsin divides a peptide on the C- side(i.e the extreme left) of Arg(Arginine) and Lys (lysine) and any peptide that holds Arg or Lys must have them as their C- terminal amino acids

From the first peptide in the two peptide formed we see that Arg would be the Seventh amino acid of the octapeptide because commonly the trypsin that sticks to the C-side of Arg would for Ser

and Phe would be the sixth amino acid of the octapeptide

So the sequence of these amino acid would be

Lue __ __ __ Met Phe Arg Ser

Looking at the first amino acid formed from the Trypsin-catalyzed hydrolysis we see that Arg would be the fourth amino acid of octapeptide as Trypsin divides a peptide on the C- side(i.e the extreme left) of Arg(Arginine).

From the second peptide of the Trypsin-catalyzed hydrolysis we see that Lys would be the third amino acid of the octapeptide as trypsin divides on the C- side of Lys (lysine) and Tyr would be the second amino acid of the octapeptide

So the sequence of these amino acid would be

Lue Arg Lys Arg Met Phe Arg Ser

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