Answer:
[N2] = [O2] = 0.841M
And [NO] = 0.00172M
Explanation:
The equilibrium constant of this reaction, Kc, is:
Kc = 2400 = [N2] [O2] / [NO]²
<em>Where [] are the equilibrium concentration of each specie.</em>
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The initial concentration of [N2] = [O2] = 0.850M. The equilibrium will shift to the left in order to produce NO. The equilibrium concentrations are:
[N2] = [O2] = 0.850M - X
And [NO] = 2X
Replacing:
2400 = [0.850-X]² / [2X]²
9600X² = 0.7225 - 1.7 X + X²
0 = 0.7225 - 1.7 X - 9599X²
Solving for X:
X = -0.0088M. False solution, there is no negative concentrations.
X = 0.00859M. Right solution.
Replacing:
[N2] = [O2] = 0.850M - 0.00859M
And [NO] = 2*0.00859M
[N2] = [O2] = 0.841M
And [NO] = 0.00172M
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:
Sound waves transfer energy by the motion of particles
Explanation: guessing lol