If the value of E degree cell is 2.10 V for the reaction F_2(g) + 2Fe^2+(aq) rig
ID: 517520 • Letter: I
Question
If the value of E degree cell is 2.10 V for the reaction F_2(g) + 2Fe^2+(aq) rightarrow 2Fe^3+(aq) + 2F^-(aq) what is the value of E degree cell for F^-(aq) + Fe^3+(aq) rightarrow Fe^2+(aq) + 1/2F_2(aq) A) -4.20 V B) -2.10V C) -1.05 V D) 1.05 V E) 2.10 V The reaction, A + 2B rightarrow B_2 + A, proceeds by the following mechanism: (A is a catalyst) A + B rightarrow AB AB + B rightarrow B_2 A What is the rate law expression for this reaction? a. Rate = k[A] b. Rate = k[B] c. Rate = k[A][B] d. Rate = k[A][B]^2 e. Rate = k[A]^2[B] Excess Ag_2SO_4(s) is placed in water at 25 degree C. At equilibrium, the solution contains 0.029 M Ag^+(aq). What is the equilibrium constant for the reaction below? Ag_2SO_4(aq) 2Ag^+(aq) + SO^2-_4(aq) a. 1.8 times 10^-7 b. 6.1 times 10^-6 c. 1.2 times 10^-5 d. 2.4 times 10^-5 e. 8.4 times 10^-4 What is the pH of the final solution when 25 mL of 0.016 M HCl has been added to 35 mL of 0.041 M HCl at 25 degree C? a. 3.2 b 2.0 c. 1.5 d. 2.7 e 3.5.Explanation / Answer
F2(g) + 2Fe+2 (aq) ---------> 2Fe+3 + 2F- E0cell = 2.10V
1/2 F2(g) + Fe+2 (aq) ---------> Fe+3 + F- E0cell = 2.10V
Fe+3 + F- ---------> 1/2 F2(g) + Fe+2 (aq) E0cell = -2.10v
B. -2.10V
38. b. rate = k[B]
39. Ag2SO4 -----------> 2Ag+ (aq) + So42- (aq)
2s s
Ksp = [Ag+]2[SO42-]
= (2s)^2*s
= 4s^3
= 4*(0.029)^3
= 9.76*10^-5
40 M = M1V1 + M2V2/V1 + V2
= 0.016*25 + 0.041*35/25+35
= 1.835/60 = 0.0306M
PH = -log[H+]
= -log0.0306 = 1.5 >>>>answer c