Extra Credit 44

Q17.6.3

Question: If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon.

S17.6.3

In order to figure out this phenomenon, one must first understand that there is an anode in this case, and that an anode oxidizes. In the current case, one must look at the voltages as looking at the voltages can show that zinc has the higher negative potential compared to iron, meaning that it is the anode and that it corrodes. On the other hand, iron has the higher negative potential compared to that of copper, which is why it is the anode in that case meaning that it corrodes.

Q12.3.7

Question: Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:

\(^{32}_{15}P\rightarrow^{32}_{16}S+e^{-}\)

\(Rate ={4.85\times10^{-2}}day^{-1}\) \([^{32}P]\)

What is the instantaneous rate or production of electrons in a sample with a phosphorus concentration of 0.0033M?

S12.3.7

The student will first notice that this questions states that it is in \(days^{-1}\) meaning that it is a first order reaction. Thus one will notice that you can plug in the numbers accordingly to the rate law of \(instantaneous\)\(rate = [k][A]\) therefore :

\(4.85\times10^{-2}day^{-1}\times(0.0033)\) \(=\) \(1.601\times10^-4\)

Therefore the instantaneous rate when phosphorus is \(0.0033M\) is \(1.601\times10^-4\)

*fixed one of the equations, but no other corrections needed

Q12.5.16

Question: Use the PhET Reactions & Rates interactive simulation to simulate a system. On the "Single collision" tab of the simulation applet, enable the "Energy view" by clicking the "+" icon. Select the first A + BC ----> AB + C reaction (A is yellow, B is purple, and C is navy blue). Using the "straight shot" default option,try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?

S.12.5.16

When the Total Energy line at launch is below the transition state of the Potential Energy line, the total energy is unable to go over the activation energy bump. This is because there is not enough force to provide energy to push the A(ball). Without enough force, it doesn't move a lot and thus does not pass the activation energy bump. When the Total Energy line at launch is above the transition state of the Potential Energy line, the total energy is able to go over the activation energy bump. This is because there is enough force to provide energy to push the A(ball) and with enough force it moves a lot and passes the activation energy bump.

*reworded to make it easier to understand but no other corrections needed

Q21.4.11

Question: Write a nuclear reaction for each step in the formation of \(^{218}_{84}Po\) from \(^{238}_{98}U\) , which proceeds by a series of decay reactions involving the step-wise emission of α, β, β, α, α, α particles, in that order.

S.21.4.11

1. (Alpha particle emitted) \(^{238}_{92}U\rightarrow^{234}_{91}Th+^{4}_{2}He\)

2. (Beta particle emitted) \(^{234}_{90}Th\rightarrow^{234}_{91}Pa+^{0}_{-1}e\)

3. (Beta particle emitted) \(^{234}_{91}Pa\rightarrow^{234}_{92}U+^{0}_{-1}e\)

4. (Alpha particle emitted) \(^{234}_{92}U\rightarrow^{230}_{90}Th+^{4}_{2}He\)

5. (Alpha particle emitted)\(^{230}_{90}Th\rightarrow^{226}_{88}Ra+^{4}_{2}He\) \(^{226}_{88}Ra\rightarrow^{222}_{88}Rn+^{4}_{2}He\)

6. (Alpha particle emitted) \(^{222}_{86}Rn\rightarrow^{218}_{84}Po+^{4}_{2}He\)

*Fixed formatting but no other corrections needed

Q20.2.15

Question: Classify each reaction as an acid-base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:

a. \(Pt^{2+}(aq)\) + \(Ag(s)\rightarrow\)

b. \(HCN(aq) + NaOH(aq)\rightarrow\)

c. \(Fe(NO_3)_3(aq)\) + \(NaOH(aq)\rightarrow\)

d. \(CH_4(g)\) + \(O_2(g)\rightarrow\)

S.20.5.15

a. \(Pt^{2+}(aq)\) + \(Ag(s)\rightarrow\) \(Pt(aq) + 2Ag(s)\)

This is a precipitation reaction. One can tell that this is a precipitation reaction because a solid is formed.

b. \(HCN(aq) + NaOH(aq)\rightarrow\) \(H_20(l) + NaCN(aq)\)

This is an acid base reaction. One can tell that this is an acid base reaction because NaOH is a strong base and

HCN is a strong acid.

c. \(Fe(NO_3)_3(aq)\) + \(3NaOH(aq)\rightarrow\) \(Fe(OH)_3(s) + (3NaNO_3)(aq)\)

This is an acid base reaction. One can tell that it is an acid base reaction because NaOH is a strong base and \(NO_3\) is a strong acid.

d. \(CH_4(g)\) + \(2O_2(g)\rightarrow\) \(CO_2(g) + 2H_2O (g)\) This is a redox reaction since the oxidation numbers change. The Oxygen is being reduced while Methane is being oxidized into carbon dioxide.

*No corrections needed

Q20.5.10

Question: Describe how an electrochemical cell can be used to measure the solubility of a sparingly soluble salt.

S.20.5.10

The Nernst Equation, \(E^{o}_{cell}=^{RT}_{nF}lnK\) can be used in this case where a cell at equilibrium containing a sparingly soluble salt as a cathode and anode can be used to find the \(E^{o}_{cell}\) , this can then be converted to \(K_{sp}\).

*No corrections needed

Q.24.6.6

Question: Do strong-field ligands favor a tetrahedral or a square planar structure? Why?

S.24.6.6

Strong-field ligands favor a square planar structure because unlike weak-field ligands, strong field ligands fill up the rows starting from the bottom going up. In a tetrahedral complex, it is split into only two rows while in a square planar complex it is split into 4 rows with one containing two columns. Thus more rows can be filled up in a square planar complex leaving less lone pairs. This is favorable for a strong field ligand since a strong field ligand must fill up the rows starting from the bottom up unlike that of a weak ligand which just needs to fill up the rows evenly amongst the structure leaving an unknown number of lone pairs.

*Reworded for better understanding but no other corrections needed

Q.14.7.11

Question: A particular reaction was found to proceed via the following mechanism:

What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates.

- A + B -> C + D

- 2C -> E

- E+ A -> 3B + F

S.14.7.11

The overall reaction in this mechanism is 2A + C -> D + F. This reaction is catalytic and the catalyst is B in this reaction while the intermediate is E and C. One can figure this out because a catalyst is something that is a reaction in the beginning and results as a product in the end. An intermediate, on the other hand, is something that does not appear in the first line; however, it comes in the middle of the reaction and appears at the end product which is the overall reaction.

*No corrections needed