2008 Exam 2 key

2008 Exam 2 key - Biology 131 Examination 2 February 27,...

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Biology 131 Examination 2 February 27, 2008 You may wish to remove pages 1 and 2 for easy reference. If you do, be careful to leave the other pages stapled together. J = -D Δ C/ Δ x; J = -P Δ C ln [] o x i z RT z FZ x V x = Vm V X X mV X X x o i o i == 25 58 ln log 21 2 1 1 2 ( ) ( ), at equilibrium, = 0. PP ss ΔΨ=Ψ −Ψ = Ψ −Ψ − Ψ −Ψ ΔΨ mi quations must be divided 2 2 2 2 23 -6 -9 3 ,for a monovalent cation, X. For a diavalent ion, the above e by 2 (z=2), thus VV V =− V = -58 mV log (Cl /Cl ). F = 10 coulomb/mol (RT/F) = 25 mV V Ca = 29 mV log (Ca o /Ca i ). For a monovalent anion such as Cl , z = -1, so Cl o i 5 - o 3 2.5 10 2.5 s jn n nn Joules L MPa R TC Ψ=− C C mole mol ∑∑ () Ps P Ψ=Ψ +Ψ = −Π where C n refers to the concentration of the n different solutes in the solution. in 3, 2 and 1 dimensions, respectively. LD tD t D t 42 ,, = 6 Thus, t increases with L , so t = L /6D, L /4D, L /2D. The pressure generated by the meniscus of water: P = -2T/r c , where T is the surface tension of water (7 x 10 -8 MPa m) and r c is the radius of curvature of the meniscus. 1 Pa = 1 joule/m 3 ; 1 atm = 0.1 MPa = 760 mmHg
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2008 Exam 2 key - Biology 131 Examination 2 February 27,...

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