Question Solved1 Answer Consider the following half-reactions: Half-reaction E° (V) Cl2(g) + 2e 2Cl(aq) 1.360V Co2+(aq) + 2e Co(s) -0.280V Zn2+(aq) + 2e = Zn(s) -0.763V → (1) The weakest oxidizing agent is: enter formula (2) The strongest reducing agent is: (3) The strongest oxidizing agent is: (4) The weakest reducing agent is: (5) Will Cl(aq) reduce Zn2+ (aq) to Zn(s)? Co2+(aq)? () Which species can be oxidized by If none, leave box blank. Consider the following half-reactions: Half-reaction E° (V) Ag+ (aq) + 2Ag(s) 0.799V Pb2+(aq) + 2e → Pb(s) -0.126V Mn? 12+(aq) + 2e —— Mn(s) -1.180V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Agt(aq) oxidize Mn(s) to Mn2+(aq) yes no (6) Which species can be reduced by Pb(s)? If none, leave box blank.

Sublimation reaction of an acid hydrate is shown below. A.nH2O(K) → A(g)+ nH2O(g) For the reactions in which n=2 and 3 the AG and asº at 435 K are given in the following table. For these reactions compute the 4.Gº at 337 K. n T=435 K AS (kJ molilK) 0.539 0.655 A.G(ki mol ) 109.5 174.2 2 3

Problem 4.11 Part A Use the node-voltage method to find v1 in the circuit(Figure 1) if v 130 V, i 2.2 A Express your answer with the appropriate units. 四? Value Units Submit Request Answer Part B Use the node-voltage method to find 2 in the circuit. Express your answer with the appropriate units. Figure 1 of 1 oz= Value Units Submit Request Answer 4Ω 80Ω Problem 4.11 Part A Use the node-voltage method to find v1 in the circuit(Figure 1) if v 130 V, i 2.2 A Express your answer with the appropriate units. 四? Value Units Submit Request Answer Part B Use the node-voltage method to find 2 in the circuit. Express your answer with the appropriate units. Figure 1 of 1 oz= Value Units Submit Request Answer 4Ω 80Ω Provide Feedback t) 102 5Ω

5. Evaluate the following integrals: a. Sg(+ b)8(t – t)dt -00 b. $(2t? + 8)8(2 – t)dt c. S. cos (2x – 1)]8(3x – 2)dx - Hint: 8(x) is located at x = 0. For example, 8(1-t) is located at 1 – t=0, that is, at t = 1, and so on.

How much heat is released when 36,0 g of water at 20.0°C is cooled to form ice at 20.0°C? Specific heat capacity of H2O(s) - 2.09 J/g.°C Specific heat capacity of H2O(l) - 4.18 J/g.°C Specific heat capacity of H2O(g) - 1.84 J/g.°C Molar mass of H2O - 18.0 g/mol AHtus -6.02kJ/mol AHvap 40.7kJ/mol . . 14.3 kJ 16.6 kJ 18.8 kJ 21.1 kJ 23.3 kJ

Please assist me with the bottom part ONLY! 5Br (aq) + BrO3(aq) + 6H+(aq) + 3Br2(1) + 3H20(1) The above reaction is expected to obey the mechanism: BrO3(aq) + H+(aq) HBrO3(aq) Fast equilibrium HBrO3(aq) + H+ (aq) H2BrO3+ (aq) Fast equilibrium Slow H2BrO3+ (aq) + Br (aq) (Br-BrO2)(aq) + H20 (1) (Br-Bro2)(aq) + 4H+(aq) + 4Br"(aq) kes , products Fast Choose, from the list below, correct expressions for the overall rate law which are completely consistent with the above mechanism. Correct: - d[Br"]/dt = k[Br"][H+]2[BrO3-] Correct: -d[BrO3-]/dt = k[BrO3"][Br"][+]2 Incorrect: d[Br-Bro2]/dt = k[Br"][BrO3"][H+]2 Incorrect: k[BrO 3 ) = -d[Br"]/dt Correct: -d[H+]/dt = k[H+]2[BrO 3 ][Br"] Incorrect: -d[Br"]/dt = k[Br"][BrO 3 ][H+] You are correct. Previous Tries Your receipt no. is 151-3441 For each of the given rate expressions choose the correct expression for the rate constant (k) from the list below. A. K = 5x[(K1+ka+k3)/(k-1+k-2)] B. K = 5x(k1k2k3-k-1k-2) C. K = 6xk3 D. k = [(k1k2k3)/(k-1k-2)] E. K = 6x([(k1k2k3)/(k-1k-2)]) F. none of the above -d[BrO3']/dt -d[Br"]/dt -d[H+]/dt

Point charge q1 = -5.00 nC is on the x-axis at x = -0.400 m. Point P is on the x-axis at x=0.200m. Point charge q2 is at the origin. What are the sign and magnitude of q2 if the resultant electric field at point P is zero? Express your answer in nanocoulombs.