Answer:
Cd + Cu2+ -----> Cu + Cd2+
Explanation:
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:
I think we would need to see the animation.
This is an incomplete question, here is a complete question.
Calculate the volume in milliliters of a 1.29 mol/L iron(II) bromide solution that contains 275 mmol of iron(II) bromide . Round your answer to significant 3 digits.
Answer : The volume of iron(II) bromide solution is, 
Explanation : Given,
Concentration of iron(II) bromide = 1.29 mo/L
Moles of iron(II) bromide = 275 mmol = 0.275 mol
conversion used : 1 mmol = 0.001 mol
Now we have to calculate the volume of iron(II) bromide.
Now put all the given values in this formula, we get:
Thus, the volume of iron(II) bromide solution is,
