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Zepler [3.9K]
4 years ago
15

A protein has the amino acid sequence

Chemistry
1 answer:
kherson [118]4 years ago
6 0

Answer:

Number of peptide fragments resulting from cleaving with cyanogen bromide? A: Three peptide fragments

Number of peptide fragments resulting from cleaving with trypsin? A: Four peptide fragments

Which of these reagents gives the smallest single fragment (in number of amino acid residues)? A: CnBr, a dipeptide fragment consisting of AL (Alanine-Leucine)

Explanation:

Cyanogen bromide cleaves the methionine C-terminus, then we have a first fragment of 8 amino acids: DSRLSKTM, a second fragment of 15 aas YSIEAPAKLDWEQNM, and a last fragment of only 2 aas is produced, AL

Trypsin cuts the C-terminus of Arginine and Lysine, then we'll have a first fragment of 3 aas DSR, a second fragment consisting of also 3 aas LSK, a third fragment of 10 aas TMYSIEAPAK, and a last fragment of 9 aas LDWEQNMAL.  All produced in three cut sites.

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7 0
3 years ago
How many grams of AlCl3 can be prepared from 3.5 moles of HCl gas?
poizon [28]

Answer:

156 g of AlCl₃ will be produced from 3.5 moles of HCl.

Explanation:

Given data:

Number of moles of HCl = 3.5 mol

Grams of AlCl₃ produced = ?

Solution:

Balanced Chemical equation:

3HCl + Al(OH)₃   →  AlCl₃ + 3H₂O

Now we will compare the moles of AlCl₃  with HCl from balanced chemical equation.

                   HCl          :           AlCl₃

                     3            :              1

                   3.5           :           1/3×3.5 = 1.17 mol

Mass of  AlCl₃ produced:

Mass = number of moles ×molar mass

Mass = 1.17 mol ×   133.341 g/mol

Mass =  156 g

Thus 156 g of AlCl₃ will be produced from 3.5 moles of HCl.

4 0
3 years ago
Hydrogen ions _____.
andreev551 [17]

are hydrogen atoms which have lost protons

5 0
3 years ago
Read 2 more answers
What mass of silver chloride (m = 143.4) will dissolve in 1.00 l of water? the ksp of agcl is 1.8 × 10–10 ?
trapecia [35]
Given that solubility product of AgCl = 1.8 X 10^-10

Dissociation of AgCl can be represented as follows,

AgCl(s)             ↔      Ag+(ag)             +           Cl-(aq)

Let, [Ag+] = [Cl-] = S

∴Ksp = [Ag+][Cl-] = S^2

∴ S = √Ksp = √(1.8 X 10^-10) = 1.34 x 10^-5 mol/dm3

Now, Molarity of solution = \frac{\text{weight of solute}}{\text{Molecular weight X volume of solution (l)}}
              ∴  1.34 x 10^-5     = \frac{\text{weight of AgCl}}{143.4 X 1}
∴ Weight of AgCl present in solution = 1.92 X 10^-3 g

Thus, mass of AgCl that will dissolve in 1l water = 1.92 x 10^-3 g
7 0
3 years ago
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