Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Pb2+ + 2e- → Pb -0.13 V

How would the following changes alter the potential of the cell?

Adding Ag+ ions to the silver half reaction (assume no volume change)

Adding Pb2+ ions to the lead half reaction (assuming no volume change).

Removing Pb2+ ions from solution by precipitating them out of the lead half reaction (assume no volume change). Adding equal amounts of water to both half reactions.

increase, decrease or no change in potential

A) An aqueous solution of methyl alcohol is made by transfering 2.39 mL of liquid methyl alcohol to a 100. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the volume/volume percentage of methyl alcohol in the solution?

B) An aqueous solution of acetone is made by transfering 7.17 mL of liquid acetone to a 100. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the volume/volume percentage of acetone in the solution?

C) How many mL of propyl alcohol are required to make a 5.57% (v/v) aqueous solution in a 100. mL volumetric flask?

A student measures the potential of a cell constructed by immersing a copper strip into a l M CuSO4 solution and a silver wire immersed in a l M AgNO3 solution. The two solutions are connected by a salt bridge. She measures a potential of 0.45 V, with the Cu electrode being the negative electrode. 7. The student adds 6 M NH3 to be CuSO4 solution until the Cu2+ ion is essentially all converted to Cu(NH3)4 2+ ion. The potential of the cell, Ecell, goes up to 0.92 V and the Cu electrode is still negative. What is the concentration of the Cu2+ ion under these conditions?

A 0.1002 g sample containing only CCl4 and CHCl3 was dissolved in methanol and electrolyzed at the surface of a mercury electrode at –1.0 V, which required 50.20 s at a constant current of 0.200 A. The potential of the cathode was then adjusted and held constant at –1.80 V. Completion of the titration at this potential required 382.35 s at a constant current of 0.200 A. Calculate the respective percentages of CCl4 and CHCl3 in the mixture. The electrolysis reactions are given below. 2CCl4 + 2H+ + 2e- + 2Hg(l) → 2CHCl3 + Hg2Cl2(s) [occurs at –1.0V] 2CHCl3 + 6H+ + 6e- + 6Hg(l) → 2CH4 + 3Hg2Cl2(s) [occurs at –1.80V]

The relationship between the speed of a car and its stopping distance can be modelled by the function

D= 0.0575v2, where D is the stopping distance, in metres, and v is the speed, in kilometres per hour.

a) Express the speed, v, as a function of D, in the form of y = af[k(x-d)] + c

b) Explain the meaning of the inverse function.

c) Graph each relation on separate axes. Label axes and title each graph.

d) State the domain and range for both functions. Give reasons for your selections.

e) Give evidence to confirm that the inverse function you found in part (a) is correct

1. Consider the three vectors, u and v are 10 degrees above and below the x-axis respectively, and ||u|| = 1, ||v|| = 2, and ||w|| = 3. Arrange the dot products taken among these vectors from least to greatest

2. Let A be a 4 x 6 matrix. Find the elimination matrix E that corresponds with the row operation "switch rows 1 and 3, and scale row 4 by a factor of 6.

3. Find the formula for the entry in the ith row and j column of the product AB in terms of the entries of A and B. Assume A is m x n and B is n x p.

Calculate the final concentration of each of the following diluted solutions:

A) 0.50 L of a 4.0 M HNO3 solution is added to water so that the final volume is 8.0 L

B) Water is added to 0.55 L of a 6.0 M KOH solution to make 2.0 L of a diluted KOH solution.

C) A 35.0 mL sample of 8.0 % (m/v) NaOH is diluted with water so that the final volume is 200.0 mL.

D) A 8.0 mL sample of 50.0 % (m/v) acetic acid (HC2H3O2) solution is added to water to give a final volume of 25 mL .

Elastic unequal mass collision: We have a similar situation as in the previous question but now truck A has mass mA and initial velocity uA. Truck B has mass mB and is initially at rest. The collision is elastic. (a) Show the subsequent velocities vA, vB in the laboratory frame are: vA = (mA − mB)uA/(mA + mB), vB = 2mAuA/(mA + mB). On the basis of these results, find the velocities in the two limiting cases: (i) mA >> mB; (ii) mA << mB. (b) Find the subsequent initial and final velocities u′ A, u′ B and v′ A, v′ B in the CM Frame