Extra Credit 42.

17.6.2 Part One

Step One: Recall how sacrificial anodes work.

Sacrificial anodes are metals that are very easily oxidized, compared to the metal they are protecting. This means that they have more negative standard reduction potentials compared to their counterpart.

Step Two: Apply to the problem.

Of all of the metals to chose from in this problem, the only one that has a more negative reduction potential compared to Iron is Zinc. So, Zinc will serve as the sacrificial anode for Iron; it will oxidize, or corrode, before the Iron will.

17.6.2 Part Two

Step One: Recall how Aluminum reacts with Oxygen.

When Aluminum reacts with the environment (or high levels of Oxygen), it will produce Aluminum Oxide Aluminum Oxide allows for passivation to occur, which means that it will serve as a coating for the aluminum metal, and will shield it from unwanted oxidation (corrosion).

Since Iron does not have these properties, it has a lower resistance to corrosion.

12.3.5

For increasing the pressure:

Step One: Pick values CO.

Let CO= 1

Step Two: Calculate the rate at .1 atm and at .3 atm. Determine how the rate changes.

For .1: rate = k(1)(.1) = .1k

For .3: rate = k(1)(.3) = .3k

The rate increased by a factor of 3.

For increasing the concentration:

Step One: Pick values for k and NO2.

Let NO2=1

Step Two: Calculate the rate at .02 M and .06 M and determine how the rate changes.

For .02: k(.02)(1)= .02k

For .06: k(.06)(1) = .06k

The rate increased by a factor of three.

12.5.14

Step One: Recall the Arrhenius equation.

ln(k2/k1) = -Ea/R (1/T2-1/T1)

Step Two: Plug in the given values.

ln(.1/.054)=-Ea/8.314 (1/298-1/293)

Ea= 1294 J.

21.4.9

Step One: Recall the rules for nuclear chemistry and the belt of stability.

Alpha decay occurs when the atomic number, or number of protons, is above 83.

Beta decay occurs when the atomic number is lower than 83 (above the belt of stability.

Positron emission occurs when the isotope is below the band of stability, and there are more protons than there are neutrons.

Step Two: Apply to the problem.

a. Since the number of protons is lower than 83, the nucleus should show beta decay.

b. Since the number of protons is above 83, the nucleus should show alpha decay.

c. Since there are more protons than there are neutrons, the nucleus should show positron emission.

d. Since the number of protons is less than 83, the nucleus should show beta decay.

e. Since the number of protons is above 83, the nucleus should show alpha decay.

20.2.13

Step One: Understand what constitutes the products and what constitutes the reactants in terms of the question.

Reactants: Zinc, Water

Products: Hydrogen Gas

Step Two: Write the reaction and determine the rest of the products.

Zn (s) + H2O (l) ---> ZnO (s) + H2 (g)

20.5.8

Step One: Recall how electrodes work.

Electrodes are metal surfaces where oxidation and reduction can occur in order to establish equilibrium between the electrode and the solution it is placed in. Oxidation occurs at the anode electrode and reduction occurs at the cathode electrode.

Step Two: Apply understanding to the problem.

A clinical sample of blood will serve as the solution that the electrode is placed in. As mentioned in the problem, the PCO2 electrode will have a semi-permeable membrane covering its tip. The carbon dioxide will react with water to produce H+ ions in such a way that is proportional to the partial pressure of carbon dioxide.

24.6.4

Step One: Recall the rules of strong and weak field ligands.

A general rule of thumb is that strong field ligands are "low spin" and weak field ligands are "high spin". There are many ways to determine if a ligand is strong or weak field. Strong field ligands absorb light at longer wavelengths (or lower energy). Weak field ligands absorb light at shorter wavelengths (or higher energy). Another method of determining if something is strong or weak field is by looking at the spectrochemical series.

Step Two: Apply the rules to the question.

If the complex has six d electrons, and it is low spin, it will completely fill up the bottom row of the CFT diagram for octahedral complexes. If the complex has six d electrons, and it is high spin, it will have 4 electrons on the bottom row of the CFT diagram, and two on the top. The low spin complex is diamagentic (no unpaired electrons) and the high spin complex is paramagnetic (unpaired electrons are present).

14.7.10

Step One: Determine what is being used up and what is being made in the reaction.

Used: Will have negative rate constants.

Made: Will have positive rate constants.

Step Two: Write the rate of disappearance for the enzyme-substrate complex.

Δ[ES]Δt=−(k2+k−1)[ES]+k1[E][S]+k−2[E][P]≈0.

Step Three: Determine the order.

Since everything cancels out, the order is zero.