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.
Answer:
Amount of HCL = 0.00318 L of 3.18 ml
Explanation:
Given:
HCL = 2.5 M
NaOH = 0.53 M
Amount of NaOH = 15 ml = 0.015 L
Find:
Amount of HCL
Computation:
HCL react with NaOH
HCl + NaOH ⇒ NaCl + H₂O
So,
Number of moles = Molarity × volume
Number of moles of NaOH = 0.53 × 0.015
Number of moles of NaOH = 0.00795 moles
So,
Number of moles of HCl needed = 0.00795 mol
es
So,
Volume = No. of moles / Molarity
Amount of HCL = 0.00795 / 2.5
Amount of HCL = 0.00318 L of 3.18 ml
Explanation:
I hope this helps Chemistry is so hard and I hate it
A. K+, OH-
B. C6H5CO+, OH-
C. NH4+, Cl-
D. Mg++, 2 NO3-
Everything has 1 except for the Nitrate ion in D, which has 2
Lurf shows helium. Jeptum shows oxygen. Noofnog shows neon. Phleegum shows sodium. Vorkel shows silicon. Dogzall shows chlorine. The metalloid is Vorkel.
Hope it helps