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Some calculations you may need
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PK
pH = b. Henderson (^) Hasselbranch (^) Equation [conjugate base]^ (forweakacid) pH = pKa+log (weakacid]
= 102 =
Amino (^) acid (^) D Ratio (D-]/[D] pK = 4.7 (^) pH= 4.
pH=pKa+log
antilog (base (^) (0)
=
==
In class assignment
units- name abbrev^ concentration millimolar mm 10-
100
=
= k, CoV2 -^ makes^4 bonds^ (12) 2 = 109K 2.81 = 1092
1284
nM=nanomolar (10-9) A + B (^) - AB
=
[R] Reactant Keg = - 9M
Easy way^ to^ solvethis^ problem: Think of^ what nM
nm= 10-9^ mol/L
positive ans would^ be 10"^ moll 1H-bond =^1 kcal/mole Given= each H bond contributes = 4.2K)Imole / key at^
k conc." x = 183
x=^1000 48 =
= 5.4 nM 3.6 amino (^) acid = 0.54 nM ->>
=^ amino^ acid^ can^ take^10 turns ----- around
3.6 x 10 =^36 2um ->^2000 nm
2 => 1000